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

Optimal. Leaf size=340 \[ -\frac{4 c^{11/4} e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (57 a^2 d^2+b c (9 b c-38 a d)\right ) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right ),\frac{1}{2}\right )}{4389 d^{13/4} \sqrt{c+d x^2}}+\frac{8 c^2 e \sqrt{e x} \sqrt{c+d x^2} \left (57 a^2 d^2+b c (9 b c-38 a d)\right )}{4389 d^3}+\frac{2 (e x)^{5/2} \left (c+d x^2\right )^{3/2} \left (57 a^2 d^2+b c (9 b c-38 a d)\right )}{627 d^2 e}+\frac{4 c (e x)^{5/2} \sqrt{c+d x^2} \left (57 a^2 d^2+b c (9 b c-38 a d)\right )}{1463 d^2 e}-\frac{2 b (e x)^{5/2} \left (c+d x^2\right )^{5/2} (9 b c-38 a d)}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3} \]

[Out]

(8*c^2*(57*a^2*d^2 + b*c*(9*b*c - 38*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(4389*d^3) + (4*c*(57*a^2*d^2 + b*c*(9
*b*c - 38*a*d))*(e*x)^(5/2)*Sqrt[c + d*x^2])/(1463*d^2*e) + (2*(57*a^2*d^2 + b*c*(9*b*c - 38*a*d))*(e*x)^(5/2)
*(c + d*x^2)^(3/2))/(627*d^2*e) - (2*b*(9*b*c - 38*a*d)*(e*x)^(5/2)*(c + d*x^2)^(5/2))/(285*d^2*e) + (2*b^2*(e
*x)^(9/2)*(c + d*x^2)^(5/2))/(19*d*e^3) - (4*c^(11/4)*(57*a^2*d^2 + b*c*(9*b*c - 38*a*d))*e^(3/2)*(Sqrt[c] + S
qrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])],
1/2])/(4389*d^(13/4)*Sqrt[c + d*x^2])

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Rubi [A]  time = 0.323289, antiderivative size = 340, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {464, 459, 279, 321, 329, 220} \[ -\frac{4 c^{11/4} e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} \left (57 a^2 d^2+b c (9 b c-38 a d)\right ) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{4389 d^{13/4} \sqrt{c+d x^2}}+\frac{8 c^2 e \sqrt{e x} \sqrt{c+d x^2} \left (57 a^2 d^2+b c (9 b c-38 a d)\right )}{4389 d^3}+\frac{2 (e x)^{5/2} \left (c+d x^2\right )^{3/2} \left (57 a^2 d^2+b c (9 b c-38 a d)\right )}{627 d^2 e}+\frac{4 c (e x)^{5/2} \sqrt{c+d x^2} \left (57 a^2 d^2+b c (9 b c-38 a d)\right )}{1463 d^2 e}-\frac{2 b (e x)^{5/2} \left (c+d x^2\right )^{5/2} (9 b c-38 a d)}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*c^2*(57*a^2*d^2 + b*c*(9*b*c - 38*a*d))*e*Sqrt[e*x]*Sqrt[c + d*x^2])/(4389*d^3) + (4*c*(57*a^2*d^2 + b*c*(9
*b*c - 38*a*d))*(e*x)^(5/2)*Sqrt[c + d*x^2])/(1463*d^2*e) + (2*(57*a^2*d^2 + b*c*(9*b*c - 38*a*d))*(e*x)^(5/2)
*(c + d*x^2)^(3/2))/(627*d^2*e) - (2*b*(9*b*c - 38*a*d)*(e*x)^(5/2)*(c + d*x^2)^(5/2))/(285*d^2*e) + (2*b^2*(e
*x)^(9/2)*(c + d*x^2)^(5/2))/(19*d*e^3) - (4*c^(11/4)*(57*a^2*d^2 + b*c*(9*b*c - 38*a*d))*e^(3/2)*(Sqrt[c] + S
qrt[d]*x)*Sqrt[(c + d*x^2)/(Sqrt[c] + Sqrt[d]*x)^2]*EllipticF[2*ArcTan[(d^(1/4)*Sqrt[e*x])/(c^(1/4)*Sqrt[e])],
1/2])/(4389*d^(13/4)*Sqrt[c + d*x^2])

Rule 464

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

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 279

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

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 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int (e x)^{3/2} \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx &=\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}+\frac{2 \int (e x)^{3/2} \left (c+d x^2\right )^{3/2} \left (\frac{19 a^2 d}{2}-\frac{1}{2} b (9 b c-38 a d) x^2\right ) \, dx}{19 d}\\ &=-\frac{2 b (9 b c-38 a d) (e x)^{5/2} \left (c+d x^2\right )^{5/2}}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}+\frac{1}{57} \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) \int (e x)^{3/2} \left (c+d x^2\right )^{3/2} \, dx\\ &=\frac{2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{627 e}-\frac{2 b (9 b c-38 a d) (e x)^{5/2} \left (c+d x^2\right )^{5/2}}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}+\frac{1}{209} \left (2 c \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right )\right ) \int (e x)^{3/2} \sqrt{c+d x^2} \, dx\\ &=\frac{4 c \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{1463 e}+\frac{2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{627 e}-\frac{2 b (9 b c-38 a d) (e x)^{5/2} \left (c+d x^2\right )^{5/2}}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}+\frac{\left (4 c^2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right )\right ) \int \frac{(e x)^{3/2}}{\sqrt{c+d x^2}} \, dx}{1463}\\ &=\frac{8 c^2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{4389 d}+\frac{4 c \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{1463 e}+\frac{2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{627 e}-\frac{2 b (9 b c-38 a d) (e x)^{5/2} \left (c+d x^2\right )^{5/2}}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}-\frac{\left (4 c^3 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) e^2\right ) \int \frac{1}{\sqrt{e x} \sqrt{c+d x^2}} \, dx}{4389 d}\\ &=\frac{8 c^2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{4389 d}+\frac{4 c \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{1463 e}+\frac{2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{627 e}-\frac{2 b (9 b c-38 a d) (e x)^{5/2} \left (c+d x^2\right )^{5/2}}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}-\frac{\left (8 c^3 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+\frac{d x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{4389 d}\\ &=\frac{8 c^2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) e \sqrt{e x} \sqrt{c+d x^2}}{4389 d}+\frac{4 c \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \sqrt{c+d x^2}}{1463 e}+\frac{2 \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) (e x)^{5/2} \left (c+d x^2\right )^{3/2}}{627 e}-\frac{2 b (9 b c-38 a d) (e x)^{5/2} \left (c+d x^2\right )^{5/2}}{285 d^2 e}+\frac{2 b^2 (e x)^{9/2} \left (c+d x^2\right )^{5/2}}{19 d e^3}-\frac{4 c^{11/4} \left (57 a^2+\frac{b c (9 b c-38 a d)}{d^2}\right ) e^{3/2} \left (\sqrt{c}+\sqrt{d} x\right ) \sqrt{\frac{c+d x^2}{\left (\sqrt{c}+\sqrt{d} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt{e x}}{\sqrt [4]{c} \sqrt{e}}\right )|\frac{1}{2}\right )}{4389 d^{5/4} \sqrt{c+d x^2}}\\ \end{align*}

Mathematica [C]  time = 0.263317, size = 259, normalized size = 0.76 \[ \frac{(e x)^{3/2} \left (\frac{2 \sqrt{x} \left (c+d x^2\right ) \left (285 a^2 d^2 \left (4 c^2+13 c d x^2+7 d^2 x^4\right )+38 a b d \left (12 c^2 d x^2-20 c^3+119 c d^2 x^4+77 d^3 x^6\right )+3 b^2 \left (28 c^2 d^2 x^4-36 c^3 d x^2+60 c^4+539 c d^3 x^6+385 d^4 x^8\right )\right )}{5 d^3}-\frac{8 i c^3 x \sqrt{\frac{c}{d x^2}+1} \left (57 a^2 d^2-38 a b c d+9 b^2 c^2\right ) \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}{\sqrt{x}}\right ),-1\right )}{d^3 \sqrt{\frac{i \sqrt{c}}{\sqrt{d}}}}\right )}{4389 x^{3/2} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(3/2)*((2*Sqrt[x]*(c + d*x^2)*(285*a^2*d^2*(4*c^2 + 13*c*d*x^2 + 7*d^2*x^4) + 38*a*b*d*(-20*c^3 + 12*c^
2*d*x^2 + 119*c*d^2*x^4 + 77*d^3*x^6) + 3*b^2*(60*c^4 - 36*c^3*d*x^2 + 28*c^2*d^2*x^4 + 539*c*d^3*x^6 + 385*d^
4*x^8)))/(5*d^3) - ((8*I)*c^3*(9*b^2*c^2 - 38*a*b*c*d + 57*a^2*d^2)*Sqrt[1 + c/(d*x^2)]*x*EllipticF[I*ArcSinh[
Sqrt[(I*Sqrt[c])/Sqrt[d]]/Sqrt[x]], -1])/(Sqrt[(I*Sqrt[c])/Sqrt[d]]*d^3)))/(4389*x^(3/2)*Sqrt[c + d*x^2])

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Maple [A]  time = 0.026, size = 489, normalized size = 1.4 \begin{align*} -{\frac{2\,e}{21945\,x{d}^{4}}\sqrt{ex} \left ( -1155\,{x}^{11}{b}^{2}{d}^{6}-2926\,{x}^{9}ab{d}^{6}-2772\,{x}^{9}{b}^{2}c{d}^{5}-1995\,{x}^{7}{a}^{2}{d}^{6}-7448\,{x}^{7}abc{d}^{5}-1701\,{x}^{7}{b}^{2}{c}^{2}{d}^{4}+570\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){a}^{2}{c}^{3}{d}^{2}-380\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ) ab{c}^{4}d+90\,\sqrt{-cd}\sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{2}\sqrt{{\frac{-dx+\sqrt{-cd}}{\sqrt{-cd}}}}\sqrt{-{\frac{dx}{\sqrt{-cd}}}}{\it EllipticF} \left ( \sqrt{{\frac{dx+\sqrt{-cd}}{\sqrt{-cd}}}},1/2\,\sqrt{2} \right ){b}^{2}{c}^{5}-5700\,{x}^{5}{a}^{2}c{d}^{5}-4978\,{x}^{5}ab{c}^{2}{d}^{4}+24\,{x}^{5}{b}^{2}{c}^{3}{d}^{3}-4845\,{x}^{3}{a}^{2}{c}^{2}{d}^{4}+304\,{x}^{3}ab{c}^{3}{d}^{3}-72\,{x}^{3}{b}^{2}{c}^{4}{d}^{2}-1140\,x{a}^{2}{c}^{3}{d}^{3}+760\,xab{c}^{4}{d}^{2}-180\,x{b}^{2}{c}^{5}d \right ){\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21945*e/x*(e*x)^(1/2)/(d*x^2+c)^(1/2)*(-1155*x^11*b^2*d^6-2926*x^9*a*b*d^6-2772*x^9*b^2*c*d^5-1995*x^7*a^2*
d^6-7448*x^7*a*b*c*d^5-1701*x^7*b^2*c^2*d^4+570*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*(
(-d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^
(1/2),1/2*2^(1/2))*a^2*c^3*d^2-380*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^
(1/2))/(-c*d)^(1/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(
1/2))*a*b*c^4*d+90*(-c*d)^(1/2)*((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2)*2^(1/2)*((-d*x+(-c*d)^(1/2))/(-c*d)^(1
/2))^(1/2)*(-x/(-c*d)^(1/2)*d)^(1/2)*EllipticF(((d*x+(-c*d)^(1/2))/(-c*d)^(1/2))^(1/2),1/2*2^(1/2))*b^2*c^5-57
00*x^5*a^2*c*d^5-4978*x^5*a*b*c^2*d^4+24*x^5*b^2*c^3*d^3-4845*x^3*a^2*c^2*d^4+304*x^3*a*b*c^3*d^3-72*x^3*b^2*c
^4*d^2-1140*x*a^2*c^3*d^3+760*x*a*b*c^4*d^2-180*x*b^2*c^5*d)/d^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} d e x^{7} +{\left (b^{2} c + 2 \, a b d\right )} e x^{5} + a^{2} c e x +{\left (2 \, a b c + a^{2} d\right )} e x^{3}\right )} \sqrt{d x^{2} + c} \sqrt{e x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*d*e*x^7 + (b^2*c + 2*a*b*d)*e*x^5 + a^2*c*e*x + (2*a*b*c + a^2*d)*e*x^3)*sqrt(d*x^2 + c)*sqrt(e*
x), x)

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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((e*x)**(3/2)*(b*x**2+a)**2*(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 (b x^{2} + a\right )}^{2}{\left (d x^{2} + c\right )}^{\frac{3}{2}} \left (e x\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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