3.5.37 \(\int \frac {\sqrt {x} (c+d x^2)^3}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=376 \[ \frac {(b c-a d)^2 (11 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}-\frac {(b c-a d)^2 (11 a d+b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}-\frac {(b c-a d)^2 (11 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{15/4}}+\frac {(b c-a d)^2 (11 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{5/4} b^{15/4}}-\frac {d x^{3/2} \left (11 a^2 d^2-21 a b c d+6 b^2 c^2\right )}{6 a b^3}-\frac {d^2 x^{7/2} (7 b c-11 a d)}{14 a b^2}+\frac {x^{3/2} \left (c+d x^2\right )^2 (b c-a d)}{2 a b \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.43, antiderivative size = 376, 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, 468, 570, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {d x^{3/2} \left (11 a^2 d^2-21 a b c d+6 b^2 c^2\right )}{6 a b^3}+\frac {(b c-a d)^2 (11 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}-\frac {(b c-a d)^2 (11 a d+b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}-\frac {(b c-a d)^2 (11 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{15/4}}+\frac {(b c-a d)^2 (11 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{5/4} b^{15/4}}-\frac {d^2 x^{7/2} (7 b c-11 a d)}{14 a b^2}+\frac {x^{3/2} \left (c+d x^2\right )^2 (b c-a d)}{2 a b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(d*(6*b^2*c^2 - 21*a*b*c*d + 11*a^2*d^2)*x^(3/2))/(6*a*b^3) - (d^2*(7*b*c - 11*a*d)*x^(7/2))/(14*a*b^2) + ((b
*c - a*d)*x^(3/2)*(c + d*x^2)^2)/(2*a*b*(a + b*x^2)) - ((b*c - a*d)^2*(b*c + 11*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/
4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(5/4)*b^(15/4)) + ((b*c - a*d)^2*(b*c + 11*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)
*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(5/4)*b^(15/4)) + ((b*c - a*d)^2*(b*c + 11*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)
*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(5/4)*b^(15/4)) - ((b*c - a*d)^2*(b*c + 11*a*d)*Log[Sqrt[a] + Sqrt
[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(8*Sqrt[2]*a^(5/4)*b^(15/4))

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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && 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 {\sqrt {x} \left (c+d x^2\right )^3}{\left (a+b x^2\right )^2} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^2 \left (c+d x^4\right )^3}{\left (a+b x^4\right )^2} \, dx,x,\sqrt {x}\right )\\ &=\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (c+d x^4\right ) \left (-c (b c+3 a d)+d (7 b c-11 a d) x^4\right )}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 a b}\\ &=\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}-\frac {\operatorname {Subst}\left (\int \left (\frac {d \left (6 b^2 c^2-21 a b c d+11 a^2 d^2\right ) x^2}{b^2}+\frac {d^2 (7 b c-11 a d) x^6}{b}-\frac {\left (b^3 c^3+9 a b^2 c^2 d-21 a^2 b c d^2+11 a^3 d^3\right ) x^2}{b^2 \left (a+b x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{2 a b}\\ &=-\frac {d \left (6 b^2 c^2-21 a b c d+11 a^2 d^2\right ) x^{3/2}}{6 a b^3}-\frac {d^2 (7 b c-11 a d) x^{7/2}}{14 a b^2}+\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}+\frac {\left ((b c-a d)^2 (b c+11 a d)\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 a b^3}\\ &=-\frac {d \left (6 b^2 c^2-21 a b c d+11 a^2 d^2\right ) x^{3/2}}{6 a b^3}-\frac {d^2 (7 b c-11 a d) x^{7/2}}{14 a b^2}+\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}-\frac {\left ((b c-a d)^2 (b c+11 a d)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a b^{7/2}}+\frac {\left ((b c-a d)^2 (b c+11 a d)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 a b^{7/2}}\\ &=-\frac {d \left (6 b^2 c^2-21 a b c d+11 a^2 d^2\right ) x^{3/2}}{6 a b^3}-\frac {d^2 (7 b c-11 a d) x^{7/2}}{14 a b^2}+\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}+\frac {\left ((b c-a d)^2 (b c+11 a d)\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 a b^4}+\frac {\left ((b c-a d)^2 (b c+11 a d)\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 a b^4}+\frac {\left ((b c-a d)^2 (b c+11 a d)\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} a^{5/4} b^{15/4}}+\frac {\left ((b c-a d)^2 (b c+11 a d)\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} a^{5/4} b^{15/4}}\\ &=-\frac {d \left (6 b^2 c^2-21 a b c d+11 a^2 d^2\right ) x^{3/2}}{6 a b^3}-\frac {d^2 (7 b c-11 a d) x^{7/2}}{14 a b^2}+\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}+\frac {(b c-a d)^2 (b c+11 a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}-\frac {(b c-a d)^2 (b c+11 a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}+\frac {\left ((b c-a d)^2 (b c+11 a d)\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} a^{5/4} b^{15/4}}-\frac {\left ((b c-a d)^2 (b c+11 a d)\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} a^{5/4} b^{15/4}}\\ &=-\frac {d \left (6 b^2 c^2-21 a b c d+11 a^2 d^2\right ) x^{3/2}}{6 a b^3}-\frac {d^2 (7 b c-11 a d) x^{7/2}}{14 a b^2}+\frac {(b c-a d) x^{3/2} \left (c+d x^2\right )^2}{2 a b \left (a+b x^2\right )}-\frac {(b c-a d)^2 (b c+11 a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{15/4}}+\frac {(b c-a d)^2 (b c+11 a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{5/4} b^{15/4}}+\frac {(b c-a d)^2 (b c+11 a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}-\frac {(b c-a d)^2 (b c+11 a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{5/4} b^{15/4}}\\ \end {align*}

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Mathematica [C]  time = 2.63, size = 355, normalized size = 0.94 \begin {gather*} \frac {95 a \left (a \left (77 a^2 \left (16875 c^3+50625 c^2 d x^2+50625 c d^2 x^4+15467 d^3 x^6\right )+22 a b x^2 \left (25931 c^3+77793 c^2 d x^2+87201 c d^2 x^4+28043 d^3 x^6\right )+b^2 x^4 \left (56099 c^3+79593 c^2 d x^2+79593 c d^2 x^4+26531 d^3 x^6\right )\right )-77 \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\frac {b x^2}{a}\right ) \left (a^3 \left (16875 c^3+50625 c^2 d x^2+50625 c d^2 x^4+15467 d^3 x^6\right )+a^2 b x^2 \left (14641 c^3+43923 c^2 d x^2+46611 c d^2 x^4+14641 d^3 x^6\right )+a b^2 x^4 \left (2401 c^3+6051 c^2 d x^2+7203 c d^2 x^4+2401 d^3 x^6\right )+b^3 x^6 \left (-101 c^3+81 c^2 d x^2+81 c d^2 x^4+27 d^3 x^6\right )\right )\right )-32768 b^4 x^8 \left (c+d x^2\right )^3 \, _5F_4\left (\frac {7}{4},2,2,2,2;1,1,1,\frac {23}{4};-\frac {b x^2}{a}\right )}{5617920 a^3 b^3 x^{9/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(95*a*(a*(77*a^2*(16875*c^3 + 50625*c^2*d*x^2 + 50625*c*d^2*x^4 + 15467*d^3*x^6) + b^2*x^4*(56099*c^3 + 79593*
c^2*d*x^2 + 79593*c*d^2*x^4 + 26531*d^3*x^6) + 22*a*b*x^2*(25931*c^3 + 77793*c^2*d*x^2 + 87201*c*d^2*x^4 + 280
43*d^3*x^6)) - 77*(b^3*x^6*(-101*c^3 + 81*c^2*d*x^2 + 81*c*d^2*x^4 + 27*d^3*x^6) + a*b^2*x^4*(2401*c^3 + 6051*
c^2*d*x^2 + 7203*c*d^2*x^4 + 2401*d^3*x^6) + a^2*b*x^2*(14641*c^3 + 43923*c^2*d*x^2 + 46611*c*d^2*x^4 + 14641*
d^3*x^6) + a^3*(16875*c^3 + 50625*c^2*d*x^2 + 50625*c*d^2*x^4 + 15467*d^3*x^6))*Hypergeometric2F1[3/4, 1, 7/4,
-((b*x^2)/a)]) - 32768*b^4*x^8*(c + d*x^2)^3*HypergeometricPFQ[{7/4, 2, 2, 2, 2}, {1, 1, 1, 23/4}, -((b*x^2)/
a)])/(5617920*a^3*b^3*x^(9/2))

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IntegrateAlgebraic [A]  time = 0.52, size = 246, normalized size = 0.65 \begin {gather*} -\frac {(11 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} a^{5/4} b^{15/4}}-\frac {(11 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} a^{5/4} b^{15/4}}-\frac {x^{3/2} \left (77 a^3 d^3-147 a^2 b c d^2+44 a^2 b d^3 x^2+63 a b^2 c^2 d-84 a b^2 c d^2 x^2-12 a b^2 d^3 x^4-21 b^3 c^3\right )}{42 a b^3 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/42*(x^(3/2)*(-21*b^3*c^3 + 63*a*b^2*c^2*d - 147*a^2*b*c*d^2 + 77*a^3*d^3 - 84*a*b^2*c*d^2*x^2 + 44*a^2*b*d^
3*x^2 - 12*a*b^2*d^3*x^4))/(a*b^3*(a + b*x^2)) - ((-(b*c) + a*d)^2*(b*c + 11*a*d)*ArcTan[(Sqrt[a] - Sqrt[b]*x)
/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(4*Sqrt[2]*a^(5/4)*b^(15/4)) - ((-(b*c) + a*d)^2*(b*c + 11*a*d)*ArcTanh[(
Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(4*Sqrt[2]*a^(5/4)*b^(15/4))

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fricas [B]  time = 1.39, size = 2531, normalized size = 6.73

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3*x^(1/2)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

-1/168*(84*(a*b^4*x^2 + a^2*b^3)*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3
- 10017*a^4*b^8*c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8
*b^4*c^4*d^8 - 676588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641*a^12*d^12)/(a^
5*b^15))^(1/4)*arctan((sqrt((b^18*c^18 + 54*a*b^17*c^17*d + 1089*a^2*b^16*c^16*d^2 + 8976*a^3*b^15*c^15*d^3 +
5940*a^4*b^14*c^14*d^4 - 279576*a^5*b^13*c^13*d^5 - 338844*a^6*b^12*c^12*d^6 + 6001776*a^7*b^11*c^11*d^7 - 641
2626*a^8*b^10*c^10*d^8 - 62165180*a^9*b^9*c^9*d^9 + 294333534*a^10*b^8*c^8*d^10 - 671362704*a^11*b^7*c^7*d^11
+ 974580036*a^12*b^6*c^6*d^12 - 971334936*a^13*b^5*c^5*d^13 + 678512340*a^14*b^4*c^4*d^14 - 328575984*a^15*b^3
*c^3*d^15 + 105546969*a^16*b^2*c^2*d^16 - 20292426*a^17*b*c*d^17 + 1771561*a^18*d^18)*x - (a^3*b^19*c^12 + 36*
a^4*b^18*c^11*d + 402*a^5*b^17*c^10*d^2 + 692*a^6*b^16*c^9*d^3 - 10017*a^7*b^15*c^8*d^4 - 5688*a^8*b^14*c^7*d^
5 + 160188*a^9*b^13*c^6*d^6 - 486648*a^10*b^12*c^5*d^7 + 746703*a^11*b^11*c^4*d^8 - 676588*a^12*b^10*c^3*d^9 +
368082*a^13*b^9*c^2*d^10 - 111804*a^14*b^8*c*d^11 + 14641*a^15*b^7*d^12)*sqrt(-(b^12*c^12 + 36*a*b^11*c^11*d
+ 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3 - 10017*a^4*b^8*c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*
c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8*b^4*c^4*d^8 - 676588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10
- 111804*a^11*b*c*d^11 + 14641*a^12*d^12)/(a^5*b^15)))*a*b^4*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^
10*d^2 + 692*a^3*b^9*c^9*d^3 - 10017*a^4*b^8*c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*
a^7*b^5*c^5*d^7 + 746703*a^8*b^4*c^4*d^8 - 676588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c
*d^11 + 14641*a^12*d^12)/(a^5*b^15))^(1/4) - (a*b^13*c^9 + 27*a^2*b^12*c^8*d + 180*a^3*b^11*c^7*d^2 - 372*a^4*
b^10*c^6*d^3 - 3186*a^5*b^9*c^5*d^4 + 13194*a^6*b^8*c^4*d^5 - 21372*a^7*b^7*c^3*d^6 + 17820*a^8*b^6*c^2*d^7 -
7623*a^9*b^5*c*d^8 + 1331*a^10*b^4*d^9)*sqrt(x)*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*
a^3*b^9*c^9*d^3 - 10017*a^4*b^8*c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d
^7 + 746703*a^8*b^4*c^4*d^8 - 676588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641
*a^12*d^12)/(a^5*b^15))^(1/4))/(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3 - 1
0017*a^4*b^8*c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8*b^4
*c^4*d^8 - 676588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641*a^12*d^12)) - 21*(
a*b^4*x^2 + a^2*b^3)*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3 - 10017*a^4
*b^8*c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8*b^4*c^4*d^8
- 676588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641*a^12*d^12)/(a^5*b^15))^(1/
4)*log(a^4*b^11*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3 - 10017*a^4*b^8*
c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8*b^4*c^4*d^8 - 67
6588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641*a^12*d^12)/(a^5*b^15))^(3/4) +
(b^9*c^9 + 27*a*b^8*c^8*d + 180*a^2*b^7*c^7*d^2 - 372*a^3*b^6*c^6*d^3 - 3186*a^4*b^5*c^5*d^4 + 13194*a^5*b^4*c
^4*d^5 - 21372*a^6*b^3*c^3*d^6 + 17820*a^7*b^2*c^2*d^7 - 7623*a^8*b*c*d^8 + 1331*a^9*d^9)*sqrt(x)) + 21*(a*b^4
*x^2 + a^2*b^3)*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3 - 10017*a^4*b^8*
c^8*d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8*b^4*c^4*d^8 - 67
6588*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641*a^12*d^12)/(a^5*b^15))^(1/4)*lo
g(-a^4*b^11*(-(b^12*c^12 + 36*a*b^11*c^11*d + 402*a^2*b^10*c^10*d^2 + 692*a^3*b^9*c^9*d^3 - 10017*a^4*b^8*c^8*
d^4 - 5688*a^5*b^7*c^7*d^5 + 160188*a^6*b^6*c^6*d^6 - 486648*a^7*b^5*c^5*d^7 + 746703*a^8*b^4*c^4*d^8 - 676588
*a^9*b^3*c^3*d^9 + 368082*a^10*b^2*c^2*d^10 - 111804*a^11*b*c*d^11 + 14641*a^12*d^12)/(a^5*b^15))^(3/4) + (b^9
*c^9 + 27*a*b^8*c^8*d + 180*a^2*b^7*c^7*d^2 - 372*a^3*b^6*c^6*d^3 - 3186*a^4*b^5*c^5*d^4 + 13194*a^5*b^4*c^4*d
^5 - 21372*a^6*b^3*c^3*d^6 + 17820*a^7*b^2*c^2*d^7 - 7623*a^8*b*c*d^8 + 1331*a^9*d^9)*sqrt(x)) - 4*(12*a*b^2*d
^3*x^5 + 4*(21*a*b^2*c*d^2 - 11*a^2*b*d^3)*x^3 + 7*(3*b^3*c^3 - 9*a*b^2*c^2*d + 21*a^2*b*c*d^2 - 11*a^3*d^3)*x
)*sqrt(x))/(a*b^4*x^2 + a^2*b^3)

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giac [A]  time = 0.50, size = 516, normalized size = 1.37 \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 )} a b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} + 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 21 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 11 \, \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^{2} b^{6}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} + 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 21 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 11 \, \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^{2} b^{6}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} + 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 21 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 11 \, \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^{2} b^{6}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} + 9 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d - 21 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} + 11 \, \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^{2} b^{6}} + \frac {2 \, {\left (3 \, b^{12} d^{3} x^{\frac {7}{2}} + 21 \, b^{12} c d^{2} x^{\frac {3}{2}} - 14 \, a b^{11} d^{3} x^{\frac {3}{2}}\right )}}{21 \, b^{14}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3*x^(1/2)/(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)*a*b^3) +
1/8*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 + 9*(a*b^3)^(3/4)*a*b^2*c^2*d - 21*(a*b^3)^(3/4)*a^2*b*c*d^2 + 11*(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^2*b^6) + 1/8*sqrt(2)*((a*b^
3)^(3/4)*b^3*c^3 + 9*(a*b^3)^(3/4)*a*b^2*c^2*d - 21*(a*b^3)^(3/4)*a^2*b*c*d^2 + 11*(a*b^3)^(3/4)*a^3*d^3)*arct
an(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a^2*b^6) - 1/16*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3
+ 9*(a*b^3)^(3/4)*a*b^2*c^2*d - 21*(a*b^3)^(3/4)*a^2*b*c*d^2 + 11*(a*b^3)^(3/4)*a^3*d^3)*log(sqrt(2)*sqrt(x)*
(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^6) + 1/16*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 + 9*(a*b^3)^(3/4)*a*b^2*c^2*d - 2
1*(a*b^3)^(3/4)*a^2*b*c*d^2 + 11*(a*b^3)^(3/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2
*b^6) + 2/21*(3*b^12*d^3*x^(7/2) + 21*b^12*c*d^2*x^(3/2) - 14*a*b^11*d^3*x^(3/2))/b^14

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maple [B]  time = 0.02, size = 706, normalized size = 1.88 \begin {gather*} \frac {2 d^{3} x^{\frac {7}{2}}}{7 b^{2}}-\frac {a^{2} d^{3} x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {3 a c \,d^{2} 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 ) a}-\frac {3 c^{2} d \,x^{\frac {3}{2}}}{2 \left (b \,x^{2}+a \right ) b}-\frac {4 a \,d^{3} x^{\frac {3}{2}}}{3 b^{3}}+\frac {2 c \,d^{2} x^{\frac {3}{2}}}{b^{2}}+\frac {11 \sqrt {2}\, a^{2} 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^{4}}+\frac {11 \sqrt {2}\, a^{2} 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^{4}}+\frac {11 \sqrt {2}\, a^{2} 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^{4}}-\frac {21 \sqrt {2}\, a 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^{3}}-\frac {21 \sqrt {2}\, a 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^{3}}-\frac {21 \sqrt {2}\, a 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^{3}}+\frac {\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}} a b}+\frac {\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}} a b}+\frac {\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}} a b}+\frac {9 \sqrt {2}\, 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^{2}}+\frac {9 \sqrt {2}\, 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^{2}}+\frac {9 \sqrt {2}\, 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^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^3*x^(1/2)/(b*x^2+a)^2,x)

[Out]

2/7*d^3/b^2*x^(7/2)-4/3*d^3/b^3*x^(3/2)*a+2*d^2/b^2*x^(3/2)*c-1/2/b^3*a^2*x^(3/2)/(b*x^2+a)*d^3+3/2/b^2*a*x^(3
/2)/(b*x^2+a)*c*d^2-3/2/b*x^(3/2)/(b*x^2+a)*c^2*d+1/2/a*x^(3/2)/(b*x^2+a)*c^3+11/8/b^4*a^2/(a/b)^(1/4)*2^(1/2)
*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)*d^3-21/8/b^3*a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1
)*c*d^2+9/8/b^2/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)*c^2*d+1/8/b/a/(a/b)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)*c^3+11/8/b^4*a^2/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)
*d^3-21/8/b^3*a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)*c*d^2+9/8/b^2/(a/b)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)*c^2*d+1/8/b/a/(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)*c^
3+11/16/b^4*a^2/(a/b)^(1/4)*2^(1/2)*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)))*d^3-21/16/b^3*a/(a/b)^(1/4)*2^(1/2)*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)))*c*d^2+9/16/b^2/(a/b)^(1/4)*2^(1/2)*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)))*c^2*d+1/16/b/a/(a/b)^(1/4)*2^(1/2)*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)))*c^3

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maxima [A]  time = 2.51, size = 309, normalized size = 0.82 \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 (a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {2 \, {\left (3 \, b d^{3} x^{\frac {7}{2}} + 7 \, {\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} x^{\frac {3}{2}}\right )}}{21 \, b^{3}} + \frac {{\left (b^{3} c^{3} + 9 \, a b^{2} c^{2} d - 21 \, a^{2} b c d^{2} + 11 \, 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 \, a b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^3*x^(1/2)/(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)/(a*b^4*x^2 + a^2*b^3) + 2/21*(3*b*d^3*x^(7/2)
+ 7*(3*b*c*d^2 - 2*a*d^3)*x^(3/2))/b^3 + 1/16*(b^3*c^3 + 9*a*b^2*c^2*d - 21*a^2*b*c*d^2 + 11*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)))/(a*b^3)

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mupad [B]  time = 0.23, size = 616, normalized size = 1.64 \begin {gather*} \frac {2\,d^3\,x^{7/2}}{7\,b^2}-x^{3/2}\,\left (\frac {4\,a\,d^3}{3\,b^3}-\frac {2\,c\,d^2}{b^2}\right )-\frac {x^{3/2}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )}{2\,a\,\left (b^4\,x^2+a\,b^3\right )}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (11\,a\,d+b\,c\right )\,\left (121\,a^6\,d^6-462\,a^5\,b\,c\,d^5+639\,a^4\,b^2\,c^2\,d^4-356\,a^3\,b^3\,c^3\,d^3+39\,a^2\,b^4\,c^4\,d^2+18\,a\,b^5\,c^5\,d+b^6\,c^6\right )}{{\left (-a\right )}^{1/4}\,\left (1331\,a^9\,d^9-7623\,a^8\,b\,c\,d^8+17820\,a^7\,b^2\,c^2\,d^7-21372\,a^6\,b^3\,c^3\,d^6+13194\,a^5\,b^4\,c^4\,d^5-3186\,a^4\,b^5\,c^5\,d^4-372\,a^3\,b^6\,c^6\,d^3+180\,a^2\,b^7\,c^7\,d^2+27\,a\,b^8\,c^8\,d+b^9\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (11\,a\,d+b\,c\right )}{4\,{\left (-a\right )}^{5/4}\,b^{15/4}}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,{\left (a\,d-b\,c\right )}^2\,\left (11\,a\,d+b\,c\right )\,\left (121\,a^6\,d^6-462\,a^5\,b\,c\,d^5+639\,a^4\,b^2\,c^2\,d^4-356\,a^3\,b^3\,c^3\,d^3+39\,a^2\,b^4\,c^4\,d^2+18\,a\,b^5\,c^5\,d+b^6\,c^6\right )\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}\,\left (1331\,a^9\,d^9-7623\,a^8\,b\,c\,d^8+17820\,a^7\,b^2\,c^2\,d^7-21372\,a^6\,b^3\,c^3\,d^6+13194\,a^5\,b^4\,c^4\,d^5-3186\,a^4\,b^5\,c^5\,d^4-372\,a^3\,b^6\,c^6\,d^3+180\,a^2\,b^7\,c^7\,d^2+27\,a\,b^8\,c^8\,d+b^9\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^2\,\left (11\,a\,d+b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{5/4}\,b^{15/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/2)*(c + d*x^2)^3)/(a + b*x^2)^2,x)

[Out]

(2*d^3*x^(7/2))/(7*b^2) - x^(3/2)*((4*a*d^3)/(3*b^3) - (2*c*d^2)/b^2) - (x^(3/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^2*
c^2*d - 3*a^2*b*c*d^2))/(2*a*(a*b^3 + b^4*x^2)) - (atan((b^(1/4)*x^(1/2)*(a*d - b*c)^2*(11*a*d + b*c)*(121*a^6
*d^6 + b^6*c^6 + 39*a^2*b^4*c^4*d^2 - 356*a^3*b^3*c^3*d^3 + 639*a^4*b^2*c^2*d^4 + 18*a*b^5*c^5*d - 462*a^5*b*c
*d^5))/((-a)^(1/4)*(1331*a^9*d^9 + b^9*c^9 + 180*a^2*b^7*c^7*d^2 - 372*a^3*b^6*c^6*d^3 - 3186*a^4*b^5*c^5*d^4
+ 13194*a^5*b^4*c^4*d^5 - 21372*a^6*b^3*c^3*d^6 + 17820*a^7*b^2*c^2*d^7 + 27*a*b^8*c^8*d - 7623*a^8*b*c*d^8)))
*(a*d - b*c)^2*(11*a*d + b*c))/(4*(-a)^(5/4)*b^(15/4)) - (atan((b^(1/4)*x^(1/2)*(a*d - b*c)^2*(11*a*d + b*c)*(
121*a^6*d^6 + b^6*c^6 + 39*a^2*b^4*c^4*d^2 - 356*a^3*b^3*c^3*d^3 + 639*a^4*b^2*c^2*d^4 + 18*a*b^5*c^5*d - 462*
a^5*b*c*d^5)*1i)/((-a)^(1/4)*(1331*a^9*d^9 + b^9*c^9 + 180*a^2*b^7*c^7*d^2 - 372*a^3*b^6*c^6*d^3 - 3186*a^4*b^
5*c^5*d^4 + 13194*a^5*b^4*c^4*d^5 - 21372*a^6*b^3*c^3*d^6 + 17820*a^7*b^2*c^2*d^7 + 27*a*b^8*c^8*d - 7623*a^8*
b*c*d^8)))*(a*d - b*c)^2*(11*a*d + b*c)*1i)/(4*(-a)^(5/4)*b^(15/4))

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sympy [A]  time = 136.46, size = 173, normalized size = 0.46 \begin {gather*} - \frac {4 a d^{3} x^{\frac {3}{2}}}{3 b^{3}} - \frac {2 x^{\frac {3}{2}} \left (a d - b c\right )^{3}}{4 a^{2} b^{3} + 4 a b^{4} x^{2}} + \frac {2 c d^{2} x^{\frac {3}{2}}}{b^{2}} + \frac {2 d^{3} x^{\frac {7}{2}}}{7 b^{2}} + \frac {6 d \left (a d - b c\right )^{2} \operatorname {RootSum} {\left (256 t^{4} a b^{3} + 1, \left (t \mapsto t \log {\left (64 t^{3} a b^{2} + \sqrt {x} \right )} \right )\right )}}{b^{3}} - \frac {2 \left (a d - b c\right )^{3} \operatorname {RootSum} {\left (65536 t^{4} a^{5} b^{3} + 1, \left (t \mapsto t \log {\left (4096 t^{3} a^{4} b^{2} + \sqrt {x} \right )} \right )\right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**3*x**(1/2)/(b*x**2+a)**2,x)

[Out]

-4*a*d**3*x**(3/2)/(3*b**3) - 2*x**(3/2)*(a*d - b*c)**3/(4*a**2*b**3 + 4*a*b**4*x**2) + 2*c*d**2*x**(3/2)/b**2
+ 2*d**3*x**(7/2)/(7*b**2) + 6*d*(a*d - b*c)**2*RootSum(256*_t**4*a*b**3 + 1, Lambda(_t, _t*log(64*_t**3*a*b*
*2 + sqrt(x))))/b**3 - 2*(a*d - b*c)**3*RootSum(65536*_t**4*a**5*b**3 + 1, Lambda(_t, _t*log(4096*_t**3*a**4*b
**2 + sqrt(x))))/b**3

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