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

Optimal. Leaf size=346 \[ \frac {(11 b c-3 a d) (b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} \sqrt [4]{c} d^{15/4}}-\frac {(11 b c-3 a d) (b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} \sqrt [4]{c} d^{15/4}}-\frac {(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} \sqrt [4]{c} d^{15/4}}+\frac {(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} \sqrt [4]{c} d^{15/4}}-\frac {x^{3/2} (11 b c-3 a d) (b c-a d)}{6 c d^3}+\frac {x^{7/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {2 b^2 x^{7/2}}{7 d^2} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 346, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {463, 459, 321, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {x^{7/2} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}-\frac {x^{3/2} (11 b c-3 a d) (b c-a d)}{6 c d^3}+\frac {(11 b c-3 a d) (b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} \sqrt [4]{c} d^{15/4}}-\frac {(11 b c-3 a d) (b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{8 \sqrt {2} \sqrt [4]{c} d^{15/4}}-\frac {(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} \sqrt [4]{c} d^{15/4}}+\frac {(11 b c-3 a d) (b c-a d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{4 \sqrt {2} \sqrt [4]{c} d^{15/4}}+\frac {2 b^2 x^{7/2}}{7 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((11*b*c - 3*a*d)*(b*c - a*d)*x^(3/2))/(6*c*d^3) + (2*b^2*x^(7/2))/(7*d^2) + ((b*c - a*d)^2*x^(7/2))/(2*c*d^2
*(c + d*x^2)) - ((11*b*c - 3*a*d)*(b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(1/4
)*d^(15/4)) + ((11*b*c - 3*a*d)*(b*c - a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(4*Sqrt[2]*c^(1/4)*
d^(15/4)) + ((11*b*c - 3*a*d)*(b*c - a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(8*Sqrt[
2]*c^(1/4)*d^(15/4)) - ((11*b*c - 3*a*d)*(b*c - a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x
])/(8*Sqrt[2]*c^(1/4)*d^(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 321

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 459

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

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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

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Mathematica [A]  time = 0.19, size = 337, normalized size = 0.97 \[ \frac {\frac {21 \sqrt {2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{\sqrt [4]{c}}-\frac {21 \sqrt {2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{\sqrt [4]{c}}-\frac {42 \sqrt {2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{\sqrt [4]{c}}+\frac {42 \sqrt {2} \left (3 a^2 d^2-14 a b c d+11 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{\sqrt [4]{c}}-448 b d^{3/4} x^{3/2} (b c-a d)-\frac {168 d^{3/4} x^{3/2} (b c-a d)^2}{c+d x^2}+96 b^2 d^{7/4} x^{7/2}}{336 d^{15/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-448*b*d^(3/4)*(b*c - a*d)*x^(3/2) + 96*b^2*d^(7/4)*x^(7/2) - (168*d^(3/4)*(b*c - a*d)^2*x^(3/2))/(c + d*x^2)
- (42*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[1 - (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(1/4) +
(42*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/c^(1/4) + (21
*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/c^(
1/4) - (21*Sqrt[2]*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[
d]*x])/c^(1/4))/(336*d^(15/4))

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fricas [B]  time = 0.60, size = 1733, normalized size = 5.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/168*(84*(d^4*x^2 + c*d^3)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^
5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8
)/(c*d^15))^(1/4)*arctan((sqrt((1771561*b^12*c^12 - 13528284*a*b^11*c^11*d + 45943458*a^2*b^10*c^10*d^2 - 9149
2940*a^3*b^9*c^9*d^3 + 118659255*a^4*b^8*c^8*d^4 - 105323064*a^5*b^7*c^7*d^5 + 65490076*a^6*b^6*c^6*d^6 - 2872
4472*a^7*b^5*c^5*d^7 + 8825895*a^8*b^4*c^4*d^8 - 1855980*a^9*b^3*c^3*d^9 + 254178*a^10*b^2*c^2*d^10 - 20412*a^
11*b*c*d^11 + 729*a^12*d^12)*x - (14641*b^8*c^9*d^7 - 74536*a*b^7*c^8*d^8 + 158268*a^2*b^6*c^7*d^9 - 181720*a^
3*b^5*c^6*d^10 + 122566*a^4*b^4*c^5*d^11 - 49560*a^5*b^3*c^4*d^12 + 11772*a^6*b^2*c^3*d^13 - 1512*a^7*b*c^2*d^
14 + 81*a^8*c*d^15)*sqrt(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3
+ 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*
d^15)))*d^4*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3 + 122566*a^
4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*d^15))^(1/4)
- (1331*b^6*c^6*d^4 - 5082*a*b^5*c^5*d^5 + 7557*a^2*b^4*c^4*d^6 - 5516*a^3*b^3*c^3*d^7 + 2061*a^4*b^2*c^2*d^8
- 378*a^5*b*c*d^9 + 27*a^6*d^10)*sqrt(x)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 1817
20*a^3*b^5*c^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7
+ 81*a^8*d^8)/(c*d^15))^(1/4))/(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c
^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^
8)) - 21*(d^4*x^2 + c*d^3)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*
d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/
(c*d^15))^(1/4)*log(c*d^11*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*
d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/
(c*d^15))^(3/4) + (1331*b^6*c^6 - 5082*a*b^5*c^5*d + 7557*a^2*b^4*c^4*d^2 - 5516*a^3*b^3*c^3*d^3 + 2061*a^4*b^
2*c^2*d^4 - 378*a^5*b*c*d^5 + 27*a^6*d^6)*sqrt(x)) + 21*(d^4*x^2 + c*d^3)*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*d
+ 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a^
6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*d^15))^(1/4)*log(-c*d^11*(-(14641*b^8*c^8 - 74536*a*b^7*c^7*
d + 158268*a^2*b^6*c^6*d^2 - 181720*a^3*b^5*c^5*d^3 + 122566*a^4*b^4*c^4*d^4 - 49560*a^5*b^3*c^3*d^5 + 11772*a
^6*b^2*c^2*d^6 - 1512*a^7*b*c*d^7 + 81*a^8*d^8)/(c*d^15))^(3/4) + (1331*b^6*c^6 - 5082*a*b^5*c^5*d + 7557*a^2*
b^4*c^4*d^2 - 5516*a^3*b^3*c^3*d^3 + 2061*a^4*b^2*c^2*d^4 - 378*a^5*b*c*d^5 + 27*a^6*d^6)*sqrt(x)) - 4*(12*b^2
*d^2*x^5 - 4*(11*b^2*c*d - 14*a*b*d^2)*x^3 - 7*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*x)*sqrt(x))/(d^4*x^2 + c*
d^3)

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giac [A]  time = 0.48, size = 413, normalized size = 1.19 \[ -\frac {b^{2} c^{2} x^{\frac {3}{2}} - 2 \, a b c d x^{\frac {3}{2}} + a^{2} d^{2} x^{\frac {3}{2}}}{2 \, {\left (d x^{2} + c\right )} d^{3}} + \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{8 \, c d^{6}} + \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{8 \, c d^{6}} - \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{16 \, c d^{6}} + \frac {\sqrt {2} {\left (11 \, \left (c d^{3}\right )^{\frac {3}{4}} b^{2} c^{2} - 14 \, \left (c d^{3}\right )^{\frac {3}{4}} a b c d + 3 \, \left (c d^{3}\right )^{\frac {3}{4}} a^{2} d^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {c}{d}\right )^{\frac {1}{4}} + x + \sqrt {\frac {c}{d}}\right )}{16 \, c d^{6}} + \frac {2 \, {\left (3 \, b^{2} d^{12} x^{\frac {7}{2}} - 14 \, b^{2} c d^{11} x^{\frac {3}{2}} + 14 \, a b d^{12} x^{\frac {3}{2}}\right )}}{21 \, d^{14}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*c^2*x^(3/2) - 2*a*b*c*d*x^(3/2) + a^2*d^2*x^(3/2))/((d*x^2 + c)*d^3) + 1/8*sqrt(2)*(11*(c*d^3)^(3/4)
*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqr
t(x))/(c/d)^(1/4))/(c*d^6) + 1/8*sqrt(2)*(11*(c*d^3)^(3/4)*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4
)*a^2*d^2)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^(1/4))/(c*d^6) - 1/16*sqrt(2)*(11*(c*d^
3)^(3/4)*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4)*a^2*d^2)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + s
qrt(c/d))/(c*d^6) + 1/16*sqrt(2)*(11*(c*d^3)^(3/4)*b^2*c^2 - 14*(c*d^3)^(3/4)*a*b*c*d + 3*(c*d^3)^(3/4)*a^2*d^
2)*log(-sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(c*d^6) + 2/21*(3*b^2*d^12*x^(7/2) - 14*b^2*c*d^11*x^(3/2
) + 14*a*b*d^12*x^(3/2))/d^14

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maple [A]  time = 0.02, size = 523, normalized size = 1.51 \[ \frac {2 b^{2} x^{\frac {7}{2}}}{7 d^{2}}-\frac {a^{2} x^{\frac {3}{2}}}{2 \left (d \,x^{2}+c \right ) d}+\frac {a b c \,x^{\frac {3}{2}}}{\left (d \,x^{2}+c \right ) d^{2}}-\frac {b^{2} c^{2} x^{\frac {3}{2}}}{2 \left (d \,x^{2}+c \right ) d^{3}}+\frac {4 a b \,x^{\frac {3}{2}}}{3 d^{2}}-\frac {4 b^{2} c \,x^{\frac {3}{2}}}{3 d^{3}}+\frac {3 \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {3 \sqrt {2}\, a^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}+\frac {3 \sqrt {2}\, a^{2} \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{16 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{2}}-\frac {7 \sqrt {2}\, a b c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}}-\frac {7 \sqrt {2}\, a b c \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{4 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}}-\frac {7 \sqrt {2}\, a b c \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{3}}+\frac {11 \sqrt {2}\, b^{2} c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{4}}+\frac {11 \sqrt {2}\, b^{2} c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{8 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{4}}+\frac {11 \sqrt {2}\, b^{2} c^{2} \ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {c}{d}}}\right )}{16 \left (\frac {c}{d}\right )^{\frac {1}{4}} d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.42, size = 272, normalized size = 0.79 \[ -\frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{\frac {3}{2}}}{2 \, {\left (d^{4} x^{2} + c d^{3}\right )}} + \frac {{\left (11 \, b^{2} c^{2} - 14 \, a b c d + 3 \, a^{2} d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} + 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} - 2 \, \sqrt {d} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} \sqrt {x} + \sqrt {d} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{16 \, d^{3}} + \frac {2 \, {\left (3 \, b^{2} d x^{\frac {7}{2}} - 14 \, {\left (b^{2} c - a b d\right )} x^{\frac {3}{2}}\right )}}{21 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^(3/2)/(d^4*x^2 + c*d^3) + 1/16*(11*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2)*(2
*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)
*sqrt(d))*sqrt(d)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) - 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*
sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt(d)) - sqrt(2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))
/(c^(1/4)*d^(3/4)) + sqrt(2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(1/4)*d^(3/4)))/d^
3 + 2/21*(3*b^2*d*x^(7/2) - 14*(b^2*c - a*b*d)*x^(3/2))/d^3

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mupad [B]  time = 0.22, size = 160, normalized size = 0.46 \[ \frac {2\,b^2\,x^{7/2}}{7\,d^2}-\frac {x^{3/2}\,\left (\frac {a^2\,d^2}{2}-a\,b\,c\,d+\frac {b^2\,c^2}{2}\right )}{d^4\,x^2+c\,d^3}-x^{3/2}\,\left (\frac {4\,b^2\,c}{3\,d^3}-\frac {4\,a\,b}{3\,d^2}\right )+\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}}{{\left (-c\right )}^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (3\,a\,d-11\,b\,c\right )}{4\,{\left (-c\right )}^{1/4}\,d^{15/4}}+\frac {\mathrm {atan}\left (\frac {d^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-c\right )}^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (3\,a\,d-11\,b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{1/4}\,d^{15/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^2*x^(7/2))/(7*d^2) - (x^(3/2)*((a^2*d^2)/2 + (b^2*c^2)/2 - a*b*c*d))/(c*d^3 + d^4*x^2) - x^(3/2)*((4*b^2*
c)/(3*d^3) - (4*a*b)/(3*d^2)) + (atan((d^(1/4)*x^(1/2))/(-c)^(1/4))*(a*d - b*c)*(3*a*d - 11*b*c))/(4*(-c)^(1/4
)*d^(15/4)) + (atan((d^(1/4)*x^(1/2)*1i)/(-c)^(1/4))*(a*d - b*c)*(3*a*d - 11*b*c)*1i)/(4*(-c)^(1/4)*d^(15/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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