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

Optimal. Leaf size=282 \[ -\frac{e^{3/2} (b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{3/2} d^{9/2}}+\frac{e \left (c+d x^2\right ) \left (-5 a^2 d^2-50 a b c d+79 b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 b d^4}+\frac{c^2 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac{\left (c+d x^2\right )^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 b d^2 e}-\frac{e \left (c+d x^2\right )^2 (a d+11 b c) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 d^4} \]

[Out]

(c^2*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^4 + ((79*b^2*c^2 - 50*a*b*c*d - 5*a^2*d^2)*e*Sqrt[(e*(
a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(48*b*d^4) - ((11*b*c + a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*
x^2)^2)/(24*d^4) + (((e*(a + b*x^2))/(c + d*x^2))^(5/2)*(c + d*x^2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(35*b^2*c^2
- 10*a*b*c*d - a^2*d^2)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^
(3/2)*d^(9/2))

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Rubi [A]  time = 0.383243, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {1960, 463, 455, 1157, 388, 208} \[ -\frac{e^{3/2} (b c-a d) \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{3/2} d^{9/2}}+\frac{e \left (c+d x^2\right ) \left (-5 a^2 d^2-50 a b c d+79 b^2 c^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{48 b d^4}+\frac{c^2 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac{\left (c+d x^2\right )^3 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2}}{6 b d^2 e}-\frac{e \left (c+d x^2\right )^2 (a d+11 b c) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{24 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/d^4 + ((79*b^2*c^2 - 50*a*b*c*d - 5*a^2*d^2)*e*Sqrt[(e*(
a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(48*b*d^4) - ((11*b*c + a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*
x^2)^2)/(24*d^4) + (((e*(a + b*x^2))/(c + d*x^2))^(5/2)*(c + d*x^2)^3)/(6*b*d^2*e) - ((b*c - a*d)*(35*b^2*c^2
- 10*a*b*c*d - a^2*d^2)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(16*b^
(3/2)*d^(9/2))

Rule 1960

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Den
ominator[p]}, Dist[(q*e*(b*c - a*d))/n, Subst[Int[(x^(q*(p + 1) - 1)*(-(a*e) + c*x^q)^(Simplify[(m + 1)/n] - 1
))/(b*e - d*x^q)^(Simplify[(m + 1)/n] + 1), x], x, ((e*(a + b*x^n))/(c + d*x^n))^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, m, n}, x] && FractionQ[p] && IntegerQ[Simplify[(m + 1)/n]]

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 455

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

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

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

Rule 208

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

Rubi steps

\begin{align*} \int x^5 \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^4 \left (-a e+c x^2\right )^2}{\left (b e-d x^2\right )^4} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{x^4 \left (-6 a^2 d^2 e^2+5 (b c e-a d e)^2+6 b c^2 d e x^2\right )}{\left (b e-d x^2\right )^3} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{6 b d^2}\\ &=-\frac{(11 b c+a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{-b d (b c-a d) (11 b c+a d) e^3-4 d^2 (b c-a d) (11 b c+a d) e^2 x^2-24 b c^2 d^3 e x^4}{\left (b e-d x^2\right )^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{24 b d^5}\\ &=\frac{\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac{(11 b c+a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}+\frac{(b c-a d) \operatorname{Subst}\left (\int \frac{-3 b d \left (19 b^2 c^2-10 a b c d-a^2 d^2\right ) e^3-48 b^2 c^2 d^2 e^2 x^2}{b e-d x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{48 b^2 d^5 e}\\ &=\frac{c^2 (b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac{\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac{(11 b c+a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{\left ((b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) e^2\right ) \operatorname{Subst}\left (\int \frac{1}{b e-d x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{16 b d^4}\\ &=\frac{c^2 (b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{d^4}+\frac{\left (79 b^2 c^2-50 a b c d-5 a^2 d^2\right ) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{48 b d^4}-\frac{(11 b c+a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{24 d^4}+\frac{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{5/2} \left (c+d x^2\right )^3}{6 b d^2 e}-\frac{(b c-a d) \left (35 b^2 c^2-10 a b c d-a^2 d^2\right ) e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{16 b^{3/2} d^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.569633, size = 294, normalized size = 1.04 \[ \frac{e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (b \sqrt{d} \sqrt{b c-a d} \left (a^2 b d \left (-100 c^2-35 c d x^2+17 d^2 x^4\right )+3 a^3 d^2 \left (c+d x^2\right )+a b^2 \left (-65 c^2 d x^2+105 c^3-52 c d^2 x^4+22 d^3 x^6\right )+b^3 x^2 \left (35 c^2 d x^2+105 c^3-14 c d^2 x^4+8 d^3 x^6\right )\right )-3 \sqrt{a+b x^2} (b c-a d)^2 \left (-a^2 d^2-10 a b c d+35 b^2 c^2\right ) \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )\right )}{48 b^2 d^{9/2} \left (a+b x^2\right ) \sqrt{b c-a d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(b*Sqrt[d]*Sqrt[b*c - a*d]*(3*a^3*d^2*(c + d*x^2) + a^2*b*d*(-100*c^2 - 3
5*c*d*x^2 + 17*d^2*x^4) + b^3*x^2*(105*c^3 + 35*c^2*d*x^2 - 14*c*d^2*x^4 + 8*d^3*x^6) + a*b^2*(105*c^3 - 65*c^
2*d*x^2 - 52*c*d^2*x^4 + 22*d^3*x^6)) - 3*(b*c - a*d)^2*(35*b^2*c^2 - 10*a*b*c*d - a^2*d^2)*Sqrt[a + b*x^2]*Sq
rt[(b*(c + d*x^2))/(b*c - a*d)]*ArcSinh[(Sqrt[d]*Sqrt[a + b*x^2])/Sqrt[b*c - a*d]]))/(48*b^2*d^(9/2)*Sqrt[b*c
- a*d]*(a + b*x^2))

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Maple [B]  time = 0.032, size = 1027, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(-12*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^4*a*b*d^3+60*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*
(b*d)^(1/2)*x^4*b^2*c*d^2+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)
^(1/2))*x^2*a^3*d^4+27*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2
))*x^2*a^2*b*c*d^3-135*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2
))*x^2*a*b^2*c^2*d^2+105*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1
/2))*x^2*b^3*c^3*d-16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*x^2*b*d^2-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(1/2)*(b*d)^(1/2)*x^2*a^2*d^3+108*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*a*b*c*d^2-54*(b*d*x^4+
a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b^2*c^2*d+3*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)
*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*c*d^3+27*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^
(1/2)+a*d+b*c)/(b*d)^(1/2))*a^2*b*c^2*d^2-135*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1
/2)+a*d+b*c)/(b*d)^(1/2))*a*b^2*c^3*d+105*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)+
a*d+b*c)/(b*d)^(1/2))*b^3*c^4+96*((d*x^2+c)*(b*x^2+a))^(1/2)*(b*d)^(1/2)*a*b*c^2*d-96*((d*x^2+c)*(b*x^2+a))^(1
/2)*(b*d)^(1/2)*b^2*c^3-16*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(b*d)^(1/2)*b*c*d-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*
c)^(1/2)*(b*d)^(1/2)*a^2*c*d^2+120*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*a*b*c^2*d-114*(b*d*x^4+a*d*
x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b^2*c^3)/d^4/b*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(3/2)/(b*d)^(1/2)/((d*x^2+
c)*(b*x^2+a))^(1/2)/(b*x^2+a)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 17.1062, size = 1175, normalized size = 4.17 \begin{align*} \left [\frac{3 \,{\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} e \sqrt{\frac{e}{b d}} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} e x^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \,{\left (2 \, b^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} +{\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{b d}}\right ) + 4 \,{\left (8 \, b^{2} d^{3} e x^{6} - 14 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} e x^{4} +{\left (35 \, b^{2} c^{2} d - 38 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} e x^{2} +{\left (105 \, b^{2} c^{3} - 100 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{192 \, b d^{4}}, \frac{3 \,{\left (35 \, b^{3} c^{3} - 45 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} e \sqrt{-\frac{e}{b d}} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{e}{b d}}}{2 \,{\left (b e x^{2} + a e\right )}}\right ) + 2 \,{\left (8 \, b^{2} d^{3} e x^{6} - 14 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} e x^{4} +{\left (35 \, b^{2} c^{2} d - 38 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} e x^{2} +{\left (105 \, b^{2} c^{3} - 100 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{96 \, b d^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)*e*sqrt(e/(b*d))*log(8*b^2*d^2*e*x^4 + 8*(b^2
*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*e - 4*(2*b^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*
c*d^2 + a*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(e/(b*d))) + 4*(8*b^2*d^3*e*x^6 - 14*(b^2*c*d^2 -
a*b*d^3)*e*x^4 + (35*b^2*c^2*d - 38*a*b*c*d^2 + 3*a^2*d^3)*e*x^2 + (105*b^2*c^3 - 100*a*b*c^2*d + 3*a^2*c*d^2)
*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b*d^4), 1/96*(3*(35*b^3*c^3 - 45*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3
)*e*sqrt(-e/(b*d))*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^
2 + a*e)) + 2*(8*b^2*d^3*e*x^6 - 14*(b^2*c*d^2 - a*b*d^3)*e*x^4 + (35*b^2*c^2*d - 38*a*b*c*d^2 + 3*a^2*d^3)*e*
x^2 + (105*b^2*c^3 - 100*a*b*c^2*d + 3*a^2*c*d^2)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b*d^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}\right )^{\frac{3}{2}} x^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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