3.5.35 \(\int \frac {x^{5/2} (c+d x^2)^3}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=374 \[ \frac {d x^{3/2} \left (5 a^2 d^2-11 a b c d+7 b^2 c^2\right )}{2 b^4}+\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {3 (b c-5 a d) (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {3 (b c-5 a d) (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {3 d^2 x^{7/2} (11 b c-5 a d)}{14 b^3}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {15 d^3 x^{11/2}}{22 b^2} \]

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Rubi [A]  time = 0.41, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {466, 467, 570, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {d x^{3/2} \left (5 a^2 d^2-11 a b c d+7 b^2 c^2\right )}{2 b^4}+\frac {3 d^2 x^{7/2} (11 b c-5 a d)}{14 b^3}+\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {3 (b c-5 a d) (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {3 (b c-5 a d) (b c-a d)^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {15 d^3 x^{11/2}}{22 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(c + d*x^2)^3)/(a + b*x^2)^2,x]

[Out]

(d*(7*b^2*c^2 - 11*a*b*c*d + 5*a^2*d^2)*x^(3/2))/(2*b^4) + (3*d^2*(11*b*c - 5*a*d)*x^(7/2))/(14*b^3) + (15*d^3
*x^(11/2))/(22*b^2) - (x^(3/2)*(c + d*x^2)^3)/(2*b*(a + b*x^2)) - (3*(b*c - 5*a*d)*(b*c - a*d)^2*ArcTan[1 - (S
qrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(19/4)) + (3*(b*c - 5*a*d)*(b*c - a*d)^2*ArcTan[1 + (Sq
rt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(1/4)*b^(19/4)) + (3*(b*c - 5*a*d)*(b*c - a*d)^2*Log[Sqrt[a] - S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(19/4)) - (3*(b*c - 5*a*d)*(b*c - a*d)^2*Log
[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(1/4)*b^(19/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 467

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^{5/2} \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^6 \left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (c+d x^4\right )^2 \left (3 c+15 d x^4\right )}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b}\\ &=-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\operatorname {Subst}\left (\int \left (\frac {3 d \left (7 b^2 c^2-11 a b c d+5 a^2 d^2\right ) x^2}{b^3}+\frac {3 d^2 (11 b c-5 a d) x^6}{b^2}+\frac {15 d^3 x^{10}}{b}+\frac {3 \left (b^3 c^3-7 a b^2 c^2 d+11 a^2 b c d^2-5 a^3 d^3\right ) x^2}{b^3 \left (a+b x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{2 b}\\ &=\frac {d \left (7 b^2 c^2-11 a b c d+5 a^2 d^2\right ) x^{3/2}}{2 b^4}+\frac {3 d^2 (11 b c-5 a d) x^{7/2}}{14 b^3}+\frac {15 d^3 x^{11/2}}{22 b^2}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b^4}\\ &=\frac {d \left (7 b^2 c^2-11 a b c d+5 a^2 d^2\right ) x^{3/2}}{2 b^4}+\frac {3 d^2 (11 b c-5 a d) x^{7/2}}{14 b^3}+\frac {15 d^3 x^{11/2}}{22 b^2}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}-\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^{9/2}}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 b^{9/2}}\\ &=\frac {d \left (7 b^2 c^2-11 a b c d+5 a^2 d^2\right ) x^{3/2}}{2 b^4}+\frac {3 d^2 (11 b c-5 a d) x^{7/2}}{14 b^3}+\frac {15 d^3 x^{11/2}}{22 b^2}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b^5}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 b^5}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}\\ &=\frac {d \left (7 b^2 c^2-11 a b c d+5 a^2 d^2\right ) x^{3/2}}{2 b^4}+\frac {3 d^2 (11 b c-5 a d) x^{7/2}}{14 b^3}+\frac {15 d^3 x^{11/2}}{22 b^2}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}+\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {\left (3 (b c-5 a d) (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}\\ &=\frac {d \left (7 b^2 c^2-11 a b c d+5 a^2 d^2\right ) x^{3/2}}{2 b^4}+\frac {3 d^2 (11 b c-5 a d) x^{7/2}}{14 b^3}+\frac {15 d^3 x^{11/2}}{22 b^2}-\frac {x^{3/2} \left (c+d x^2\right )^3}{2 b \left (a+b x^2\right )}-\frac {3 (b c-5 a d) (b c-a d)^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {3 (b c-5 a d) (b c-a d)^2 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}-\frac {3 (b c-5 a d) (b c-a d)^2 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} \sqrt [4]{a} b^{19/4}}\\ \end {align*}

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Mathematica [C]  time = 2.22, size = 377, normalized size = 1.01 \begin {gather*} \frac {-385 a^3 \left (130321 c^3+390963 c^2 d x^2+390963 c d^2 x^4+124561 d^3 x^6\right )-330 a^2 b x^2 \left (112027 c^3+336081 c^2 d x^2+350865 c d^2 x^4+114907 d^3 x^6\right )+385 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\frac {b x^2}{a}\right ) \left (a^3 \left (130321 c^3+390963 c^2 d x^2+390963 c d^2 x^4+124561 d^3 x^6\right )+9 a^2 b x^2 \left (16875 c^3+50625 c^2 d x^2+52033 c d^2 x^4+16875 d^3 x^6\right )+3 a b^2 x^4 \left (14641 c^3+41235 c^2 d x^2+43923 c d^2 x^4+14641 d^3 x^6\right )+b^3 x^6 \left (3553 c^3+7203 c^2 d x^2+7203 c d^2 x^4+2401 d^3 x^6\right )\right )-45 a b^2 x^4 \left (122993 c^3+299987 c^2 d x^2+322515 c d^2 x^4+109553 d^3 x^6\right )-32768 b^3 x^6 \left (c+d x^2\right )^3}{887040 a b^4 x^{9/2}}-\frac {128 b x^{11/2} \left (c+d x^2\right )^3 \, _5F_4\left (2,2,2,2,\frac {11}{4};1,1,1,\frac {27}{4};-\frac {b x^2}{a}\right )}{72105 a^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^(5/2)*(c + d*x^2)^3)/(a + b*x^2)^2,x]

[Out]

(-32768*b^3*x^6*(c + d*x^2)^3 - 45*a*b^2*x^4*(122993*c^3 + 299987*c^2*d*x^2 + 322515*c*d^2*x^4 + 109553*d^3*x^
6) - 330*a^2*b*x^2*(112027*c^3 + 336081*c^2*d*x^2 + 350865*c*d^2*x^4 + 114907*d^3*x^6) - 385*a^3*(130321*c^3 +
390963*c^2*d*x^2 + 390963*c*d^2*x^4 + 124561*d^3*x^6) + 385*(b^3*x^6*(3553*c^3 + 7203*c^2*d*x^2 + 7203*c*d^2*
x^4 + 2401*d^3*x^6) + 3*a*b^2*x^4*(14641*c^3 + 41235*c^2*d*x^2 + 43923*c*d^2*x^4 + 14641*d^3*x^6) + 9*a^2*b*x^
2*(16875*c^3 + 50625*c^2*d*x^2 + 52033*c*d^2*x^4 + 16875*d^3*x^6) + a^3*(130321*c^3 + 390963*c^2*d*x^2 + 39096
3*c*d^2*x^4 + 124561*d^3*x^6))*Hypergeometric2F1[3/4, 1, 7/4, -((b*x^2)/a)])/(887040*a*b^4*x^(9/2)) - (128*b*x
^(11/2)*(c + d*x^2)^3*HypergeometricPFQ[{2, 2, 2, 2, 11/4}, {1, 1, 1, 27/4}, -((b*x^2)/a)])/(72105*a^3)

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IntegrateAlgebraic [A]  time = 0.56, size = 280, normalized size = 0.75 \begin {gather*} \frac {x^{3/2} \left (385 a^3 d^3-847 a^2 b c d^2+220 a^2 b d^3 x^2+539 a b^2 c^2 d-484 a b^2 c d^2 x^2-60 a b^2 d^3 x^4-77 b^3 c^3+308 b^3 c^2 d x^2+132 b^3 c d^2 x^4+28 b^3 d^3 x^6\right )}{154 b^4 \left (a+b x^2\right )}+\frac {3 (5 a d-b c) (a d-b c)^2 \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}}+\frac {3 (5 a d-b c) (a d-b c)^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{19/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(5/2)*(c + d*x^2)^3)/(a + b*x^2)^2,x]

[Out]

(x^(3/2)*(-77*b^3*c^3 + 539*a*b^2*c^2*d - 847*a^2*b*c*d^2 + 385*a^3*d^3 + 308*b^3*c^2*d*x^2 - 484*a*b^2*c*d^2*
x^2 + 220*a^2*b*d^3*x^2 + 132*b^3*c*d^2*x^4 - 60*a*b^2*d^3*x^4 + 28*b^3*d^3*x^6))/(154*b^4*(a + b*x^2)) + (3*(
-(b*c) + a*d)^2*(-(b*c) + 5*a*d)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(4*Sqrt[2]*a
^(1/4)*b^(19/4)) + (3*(-(b*c) + a*d)^2*(-(b*c) + 5*a*d)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + S
qrt[b]*x)])/(4*Sqrt[2]*a^(1/4)*b^(19/4))

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fricas [B]  time = 1.73, size = 2542, normalized size = 6.80

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(d*x^2+c)^3/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

1/616*(924*(b^5*x^2 + a*b^4)*(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^3*b^9*c^9*d^3 +
10015*a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7 + 76111*a^8*b^4*
c^4*d^8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12)/(a*b^19))^(1/4
)*arctan((sqrt((b^18*c^18 - 42*a*b^17*c^17*d + 801*a^2*b^16*c^16*d^2 - 9200*a^3*b^15*c^15*d^3 + 71220*a^4*b^14
*c^14*d^4 - 394392*a^5*b^13*c^13*d^5 + 1619684*a^6*b^12*c^12*d^6 - 5050512*a^7*b^11*c^11*d^7 + 12147630*a^8*b^
10*c^10*d^8 - 22765820*a^9*b^9*c^9*d^9 + 33419166*a^10*b^8*c^8*d^10 - 38446992*a^11*b^7*c^7*d^11 + 34503236*a^
12*b^6*c^6*d^12 - 23888280*a^13*b^5*c^5*d^13 + 12508500*a^14*b^4*c^4*d^14 - 4790000*a^15*b^3*c^3*d^15 + 126562
5*a^16*b^2*c^2*d^16 - 206250*a^17*b*c*d^17 + 15625*a^18*d^18)*x - (a*b^21*c^12 - 28*a^2*b^20*c^11*d + 338*a^3*
b^19*c^10*d^2 - 2316*a^4*b^18*c^9*d^3 + 10015*a^5*b^17*c^8*d^4 - 28856*a^6*b^16*c^7*d^5 + 57148*a^7*b^15*c^6*d
^6 - 78968*a^8*b^14*c^5*d^7 + 76111*a^9*b^13*c^4*d^8 - 50220*a^10*b^12*c^3*d^9 + 21650*a^11*b^11*c^2*d^10 - 55
00*a^12*b^10*c*d^11 + 625*a^13*b^9*d^12)*sqrt(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^
3*b^9*c^9*d^3 + 10015*a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7
+ 76111*a^8*b^4*c^4*d^8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12
)/(a*b^19)))*b^5*(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^3*b^9*c^9*d^3 + 10015*a^4*b^
8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7 + 76111*a^8*b^4*c^4*d^8 - 50
220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12)/(a*b^19))^(1/4) + (b^14*c^
9 - 21*a*b^13*c^8*d + 180*a^2*b^12*c^7*d^2 - 820*a^3*b^11*c^6*d^3 + 2190*a^4*b^10*c^5*d^4 - 3606*a^5*b^9*c^4*d
^5 + 3716*a^6*b^8*c^3*d^6 - 2340*a^7*b^7*c^2*d^7 + 825*a^8*b^6*c*d^8 - 125*a^9*b^5*d^9)*sqrt(x)*(-(b^12*c^12 -
28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^3*b^9*c^9*d^3 + 10015*a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d
^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7 + 76111*a^8*b^4*c^4*d^8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^1
0*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12)/(a*b^19))^(1/4))/(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*
b^10*c^10*d^2 - 2316*a^3*b^9*c^9*d^3 + 10015*a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 -
78968*a^7*b^5*c^5*d^7 + 76111*a^8*b^4*c^4*d^8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b
*c*d^11 + 625*a^12*d^12)) - 231*(b^5*x^2 + a*b^4)*(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 23
16*a^3*b^9*c^9*d^3 + 10015*a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5
*d^7 + 76111*a^8*b^4*c^4*d^8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12
*d^12)/(a*b^19))^(1/4)*log(27*a*b^14*(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^3*b^9*c^
9*d^3 + 10015*a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7 + 76111*
a^8*b^4*c^4*d^8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12)/(a*b^1
9))^(3/4) - 27*(b^9*c^9 - 21*a*b^8*c^8*d + 180*a^2*b^7*c^7*d^2 - 820*a^3*b^6*c^6*d^3 + 2190*a^4*b^5*c^5*d^4 -
3606*a^5*b^4*c^4*d^5 + 3716*a^6*b^3*c^3*d^6 - 2340*a^7*b^2*c^2*d^7 + 825*a^8*b*c*d^8 - 125*a^9*d^9)*sqrt(x)) +
231*(b^5*x^2 + a*b^4)*(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^3*b^9*c^9*d^3 + 10015*
a^4*b^8*c^8*d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7 + 76111*a^8*b^4*c^4*d^
8 - 50220*a^9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12)/(a*b^19))^(1/4)*log(
-27*a*b^14*(-(b^12*c^12 - 28*a*b^11*c^11*d + 338*a^2*b^10*c^10*d^2 - 2316*a^3*b^9*c^9*d^3 + 10015*a^4*b^8*c^8*
d^4 - 28856*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 78968*a^7*b^5*c^5*d^7 + 76111*a^8*b^4*c^4*d^8 - 50220*a^
9*b^3*c^3*d^9 + 21650*a^10*b^2*c^2*d^10 - 5500*a^11*b*c*d^11 + 625*a^12*d^12)/(a*b^19))^(3/4) - 27*(b^9*c^9 -
21*a*b^8*c^8*d + 180*a^2*b^7*c^7*d^2 - 820*a^3*b^6*c^6*d^3 + 2190*a^4*b^5*c^5*d^4 - 3606*a^5*b^4*c^4*d^5 + 371
6*a^6*b^3*c^3*d^6 - 2340*a^7*b^2*c^2*d^7 + 825*a^8*b*c*d^8 - 125*a^9*d^9)*sqrt(x)) + 4*(28*b^3*d^3*x^7 + 12*(1
1*b^3*c*d^2 - 5*a*b^2*d^3)*x^5 + 44*(7*b^3*c^2*d - 11*a*b^2*c*d^2 + 5*a^2*b*d^3)*x^3 - 77*(b^3*c^3 - 7*a*b^2*c
^2*d + 11*a^2*b*c*d^2 - 5*a^3*d^3)*x)*sqrt(x))/(b^5*x^2 + a*b^4)

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giac [A]  time = 0.60, size = 552, normalized size = 1.48 \begin {gather*} -\frac {b^{3} c^{3} x^{\frac {3}{2}} - 3 \, a b^{2} c^{2} d x^{\frac {3}{2}} + 3 \, a^{2} b c d^{2} x^{\frac {3}{2}} - a^{3} d^{3} x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 11 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{7}} + \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 11 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{7}} - \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 11 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{7}} + \frac {3 \, \sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 7 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 11 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{7}} + \frac {2 \, {\left (7 \, b^{20} d^{3} x^{\frac {11}{2}} + 33 \, b^{20} c d^{2} x^{\frac {7}{2}} - 22 \, a b^{19} d^{3} x^{\frac {7}{2}} + 77 \, b^{20} c^{2} d x^{\frac {3}{2}} - 154 \, a b^{19} c d^{2} x^{\frac {3}{2}} + 77 \, a^{2} b^{18} d^{3} x^{\frac {3}{2}}\right )}}{77 \, b^{22}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(d*x^2+c)^3/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(b^3*c^3*x^(3/2) - 3*a*b^2*c^2*d*x^(3/2) + 3*a^2*b*c*d^2*x^(3/2) - a^3*d^3*x^(3/2))/((b*x^2 + a)*b^4) + 3
/8*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 7*(a*b^3)^(3/4)*a*b^2*c^2*d + 11*(a*b^3)^(3/4)*a^2*b*c*d^2 - 5*(a*b^3)^(3/
4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a*b^7) + 3/8*sqrt(2)*((a*b^3)^(
3/4)*b^3*c^3 - 7*(a*b^3)^(3/4)*a*b^2*c^2*d + 11*(a*b^3)^(3/4)*a^2*b*c*d^2 - 5*(a*b^3)^(3/4)*a^3*d^3)*arctan(-1
/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a*b^7) - 3/16*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 7*(a
*b^3)^(3/4)*a*b^2*c^2*d + 11*(a*b^3)^(3/4)*a^2*b*c*d^2 - 5*(a*b^3)^(3/4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1
/4) + x + sqrt(a/b))/(a*b^7) + 3/16*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 7*(a*b^3)^(3/4)*a*b^2*c^2*d + 11*(a*b^3)^
(3/4)*a^2*b*c*d^2 - 5*(a*b^3)^(3/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^7) + 2/77*
(7*b^20*d^3*x^(11/2) + 33*b^20*c*d^2*x^(7/2) - 22*a*b^19*d^3*x^(7/2) + 77*b^20*c^2*d*x^(3/2) - 154*a*b^19*c*d^
2*x^(3/2) + 77*a^2*b^18*d^3*x^(3/2))/b^22

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maple [B]  time = 0.02, size = 748, normalized size = 2.00 \begin {gather*} \frac {2 d^{3} x^{\frac {11}{2}}}{11 b^{2}}-\frac {4 a \,d^{3} x^{\frac {7}{2}}}{7 b^{3}}+\frac {6 c \,d^{2} x^{\frac {7}{2}}}{7 b^{2}}+\frac {a^{3} d^{3} x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b^{4}}-\frac {3 a^{2} c \,d^{2} x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {3 a \,c^{2} d \,x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {c^{3} x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b}+\frac {2 a^{2} d^{3} x^{\frac {3}{2}}}{b^{4}}-\frac {4 a c \,d^{2} x^{\frac {3}{2}}}{b^{3}}+\frac {2 c^{2} d \,x^{\frac {3}{2}}}{b^{2}}-\frac {15 \sqrt {2}\, a^{3} d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{5}}-\frac {15 \sqrt {2}\, a^{3} d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{5}}-\frac {15 \sqrt {2}\, a^{3} d^{3} \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{5}}+\frac {33 \sqrt {2}\, a^{2} c \,d^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{4}}+\frac {33 \sqrt {2}\, a^{2} c \,d^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{4}}+\frac {33 \sqrt {2}\, a^{2} c \,d^{2} \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{4}}-\frac {21 \sqrt {2}\, a \,c^{2} d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}-\frac {21 \sqrt {2}\, a \,c^{2} d \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}-\frac {21 \sqrt {2}\, a \,c^{2} d \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{3}}+\frac {3 \sqrt {2}\, c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}+\frac {3 \sqrt {2}\, c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}}+\frac {3 \sqrt {2}\, c^{3} \ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 \left (\frac {a}{b}\right )^{\frac {1}{4}} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(d*x^2+c)^3/(b*x^2+a)^2,x)

[Out]

2/11*d^3*x^(11/2)/b^2-4/7*d^3/b^3*x^(7/2)*a+6/7*d^2/b^2*x^(7/2)*c+2*d^3/b^4*x^(3/2)*a^2-4*d^2/b^3*x^(3/2)*a*c+
2*d/b^2*x^(3/2)*c^2+1/2/b^4*x^(3/2)/(b*x^2+a)*a^3*d^3-3/2/b^3*x^(3/2)/(b*x^2+a)*a^2*c*d^2+3/2/b^2*x^(3/2)/(b*x
^2+a)*a*c^2*d-1/2/b*x^(3/2)/(b*x^2+a)*c^3-15/16/b^5/(a/b)^(1/4)*2^(1/2)*a^3*d^3*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1
/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))-15/8/b^5/(a/b)^(1/4)*2^(1/2)*a^3*d^3*arctan(2^(1
/2)/(a/b)^(1/4)*x^(1/2)+1)-15/8/b^5/(a/b)^(1/4)*2^(1/2)*a^3*d^3*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+33/16/b^
4/(a/b)^(1/4)*2^(1/2)*a^2*c*d^2*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+
(a/b)^(1/2)))+33/8/b^4/(a/b)^(1/4)*2^(1/2)*a^2*c*d^2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+33/8/b^4/(a/b)^(1/4
)*2^(1/2)*a^2*c*d^2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)-21/16/b^3/(a/b)^(1/4)*2^(1/2)*a*c^2*d*ln((x-(a/b)^(1
/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))-21/8/b^3/(a/b)^(1/4)*2^(1/2)*a*c
^2*d*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-21/8/b^3/(a/b)^(1/4)*2^(1/2)*a*c^2*d*arctan(2^(1/2)/(a/b)^(1/4)*x^(
1/2)-1)+3/16/b^2/(a/b)^(1/4)*2^(1/2)*c^3*ln((x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*2^(1/2)
*x^(1/2)+(a/b)^(1/2)))+3/8/b^2/(a/b)^(1/4)*2^(1/2)*c^3*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+3/8/b^2/(a/b)^(1/
4)*2^(1/2)*c^3*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 2.36, size = 337, normalized size = 0.90 \begin {gather*} -\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{\frac {3}{2}}}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {3 \, {\left (b^{3} c^{3} - 7 \, a b^{2} c^{2} d + 11 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, b^{4}} + \frac {2 \, {\left (7 \, b^{2} d^{3} x^{\frac {11}{2}} + 11 \, {\left (3 \, b^{2} c d^{2} - 2 \, a b d^{3}\right )} x^{\frac {7}{2}} + 77 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{\frac {3}{2}}\right )}}{77 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(d*x^2+c)^3/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

-1/2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*x^(3/2)/(b^5*x^2 + a*b^4) + 3/16*(b^3*c^3 - 7*a*b^2*c
^2*d + 11*a^2*b*c*d^2 - 5*a^3*d^3)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))
/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/
4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b
^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(
b)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/b^4 + 2/77*(7*b^2*d^3*x^(11/2) + 11*(3*b^2*c*d^2 - 2*a*b*d^3)*x^(7/2) + 77*
(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x^(3/2))/b^4

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mupad [B]  time = 0.45, size = 681, normalized size = 1.82 \begin {gather*} x^{3/2}\,\left (\frac {2\,c^2\,d}{b^2}+\frac {2\,a\,\left (\frac {4\,a\,d^3}{b^3}-\frac {6\,c\,d^2}{b^2}\right )}{3\,b}-\frac {2\,a^2\,d^3}{3\,b^4}\right )-x^{7/2}\,\left (\frac {4\,a\,d^3}{7\,b^3}-\frac {6\,c\,d^2}{7\,b^2}\right )+\frac {2\,d^3\,x^{11/2}}{11\,b^2}+\frac {x^{3/2}\,\left (\frac {a^3\,d^3}{2}-\frac {3\,a^2\,b\,c\,d^2}{2}+\frac {3\,a\,b^2\,c^2\,d}{2}-\frac {b^3\,c^3}{2}\right )}{b^5\,x^2+a\,b^4}-\frac {3\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (5\,a\,d-b\,c\right )\,\left (25\,a^7\,d^6-110\,a^6\,b\,c\,d^5+191\,a^5\,b^2\,c^2\,d^4-164\,a^4\,b^3\,c^3\,d^3+71\,a^3\,b^4\,c^4\,d^2-14\,a^2\,b^5\,c^5\,d+a\,b^6\,c^6\right )}{{\left (-a\right )}^{1/4}\,\left (125\,a^{10}\,d^9-825\,a^9\,b\,c\,d^8+2340\,a^8\,b^2\,c^2\,d^7-3716\,a^7\,b^3\,c^3\,d^6+3606\,a^6\,b^4\,c^4\,d^5-2190\,a^5\,b^5\,c^5\,d^4+820\,a^4\,b^6\,c^6\,d^3-180\,a^3\,b^7\,c^7\,d^2+21\,a^2\,b^8\,c^8\,d-a\,b^9\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (5\,a\,d-b\,c\right )}{4\,{\left (-a\right )}^{1/4}\,b^{19/4}}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (5\,a\,d-b\,c\right )\,\left (25\,a^7\,d^6-110\,a^6\,b\,c\,d^5+191\,a^5\,b^2\,c^2\,d^4-164\,a^4\,b^3\,c^3\,d^3+71\,a^3\,b^4\,c^4\,d^2-14\,a^2\,b^5\,c^5\,d+a\,b^6\,c^6\right )\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}\,\left (125\,a^{10}\,d^9-825\,a^9\,b\,c\,d^8+2340\,a^8\,b^2\,c^2\,d^7-3716\,a^7\,b^3\,c^3\,d^6+3606\,a^6\,b^4\,c^4\,d^5-2190\,a^5\,b^5\,c^5\,d^4+820\,a^4\,b^6\,c^6\,d^3-180\,a^3\,b^7\,c^7\,d^2+21\,a^2\,b^8\,c^8\,d-a\,b^9\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (5\,a\,d-b\,c\right )\,3{}\mathrm {i}}{4\,{\left (-a\right )}^{1/4}\,b^{19/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(5/2)*(c + d*x^2)^3)/(a + b*x^2)^2,x)

[Out]

x^(3/2)*((2*c^2*d)/b^2 + (2*a*((4*a*d^3)/b^3 - (6*c*d^2)/b^2))/(3*b) - (2*a^2*d^3)/(3*b^4)) - x^(7/2)*((4*a*d^
3)/(7*b^3) - (6*c*d^2)/(7*b^2)) + (2*d^3*x^(11/2))/(11*b^2) + (x^(3/2)*((a^3*d^3)/2 - (b^3*c^3)/2 + (3*a*b^2*c
^2*d)/2 - (3*a^2*b*c*d^2)/2))/(a*b^4 + b^5*x^2) - (3*atan((b^(1/4)*x^(1/2)*(a*d - b*c)^2*(5*a*d - b*c)*(25*a^7
*d^6 + a*b^6*c^6 - 14*a^2*b^5*c^5*d + 71*a^3*b^4*c^4*d^2 - 164*a^4*b^3*c^3*d^3 + 191*a^5*b^2*c^2*d^4 - 110*a^6
*b*c*d^5))/((-a)^(1/4)*(125*a^10*d^9 - a*b^9*c^9 + 21*a^2*b^8*c^8*d - 180*a^3*b^7*c^7*d^2 + 820*a^4*b^6*c^6*d^
3 - 2190*a^5*b^5*c^5*d^4 + 3606*a^6*b^4*c^4*d^5 - 3716*a^7*b^3*c^3*d^6 + 2340*a^8*b^2*c^2*d^7 - 825*a^9*b*c*d^
8)))*(a*d - b*c)^2*(5*a*d - b*c))/(4*(-a)^(1/4)*b^(19/4)) - (atan((b^(1/4)*x^(1/2)*(a*d - b*c)^2*(5*a*d - b*c)
*(25*a^7*d^6 + a*b^6*c^6 - 14*a^2*b^5*c^5*d + 71*a^3*b^4*c^4*d^2 - 164*a^4*b^3*c^3*d^3 + 191*a^5*b^2*c^2*d^4 -
110*a^6*b*c*d^5)*1i)/((-a)^(1/4)*(125*a^10*d^9 - a*b^9*c^9 + 21*a^2*b^8*c^8*d - 180*a^3*b^7*c^7*d^2 + 820*a^4
*b^6*c^6*d^3 - 2190*a^5*b^5*c^5*d^4 + 3606*a^6*b^4*c^4*d^5 - 3716*a^7*b^3*c^3*d^6 + 2340*a^8*b^2*c^2*d^7 - 825
*a^9*b*c*d^8)))*(a*d - b*c)^2*(5*a*d - b*c)*3i)/(4*(-a)^(1/4)*b^(19/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(d*x**2+c)**3/(b*x**2+a)**2,x)

[Out]

Timed out

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