### 3.35 $$\int (d+e x^3)^{5/2} (a+b x^3+c x^6) \, dx$$

Optimal. Leaf size=396 $\frac{54\ 3^{3/4} \sqrt{2+\sqrt{3}} d^3 \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (667 a e^2-58 b d e+16 c d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt{3}\right )}{124729 e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}+\frac{2 x \left (d+e x^3\right )^{5/2} \left (667 a e^2-58 b d e+16 c d^2\right )}{11339 e^2}+\frac{30 d x \left (d+e x^3\right )^{3/2} \left (667 a e^2-58 b d e+16 c d^2\right )}{124729 e^2}+\frac{54 d^2 x \sqrt{d+e x^3} \left (667 a e^2-58 b d e+16 c d^2\right )}{124729 e^2}-\frac{2 x \left (d+e x^3\right )^{7/2} (8 c d-29 b e)}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}$

[Out]

(54*d^2*(16*c*d^2 - 58*b*d*e + 667*a*e^2)*x*Sqrt[d + e*x^3])/(124729*e^2) + (30*d*(16*c*d^2 - 58*b*d*e + 667*a
*e^2)*x*(d + e*x^3)^(3/2))/(124729*e^2) + (2*(16*c*d^2 - 58*b*d*e + 667*a*e^2)*x*(d + e*x^3)^(5/2))/(11339*e^2
) - (2*(8*c*d - 29*b*e)*x*(d + e*x^3)^(7/2))/(667*e^2) + (2*c*x^4*(d + e*x^3)^(7/2))/(29*e) + (54*3^(3/4)*Sqrt
[2 + Sqrt[3]]*d^3*(16*c*d^2 - 58*b*d*e + 667*a*e^2)*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x +
e^(2/3)*x^2)/((1 + Sqrt[3])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^(1/3)*x)/((1 +
Sqrt[3])*d^(1/3) + e^(1/3)*x)], -7 - 4*Sqrt[3]])/(124729*e^(7/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + S
qrt[3])*d^(1/3) + e^(1/3)*x)^2]*Sqrt[d + e*x^3])

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Rubi [A]  time = 0.41577, antiderivative size = 396, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {1411, 388, 195, 218} $\frac{2 x \left (d+e x^3\right )^{5/2} \left (667 a e^2-58 b d e+16 c d^2\right )}{11339 e^2}+\frac{30 d x \left (d+e x^3\right )^{3/2} \left (667 a e^2-58 b d e+16 c d^2\right )}{124729 e^2}+\frac{54 d^2 x \sqrt{d+e x^3} \left (667 a e^2-58 b d e+16 c d^2\right )}{124729 e^2}+\frac{54\ 3^{3/4} \sqrt{2+\sqrt{3}} d^3 \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (667 a e^2-58 b d e+16 c d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{e} x+\left (1-\sqrt{3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt{3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt{3}\right )}{124729 e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}-\frac{2 x \left (d+e x^3\right )^{7/2} (8 c d-29 b e)}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x^3)^(5/2)*(a + b*x^3 + c*x^6),x]

[Out]

(54*d^2*(16*c*d^2 - 58*b*d*e + 667*a*e^2)*x*Sqrt[d + e*x^3])/(124729*e^2) + (30*d*(16*c*d^2 - 58*b*d*e + 667*a
*e^2)*x*(d + e*x^3)^(3/2))/(124729*e^2) + (2*(16*c*d^2 - 58*b*d*e + 667*a*e^2)*x*(d + e*x^3)^(5/2))/(11339*e^2
) - (2*(8*c*d - 29*b*e)*x*(d + e*x^3)^(7/2))/(667*e^2) + (2*c*x^4*(d + e*x^3)^(7/2))/(29*e) + (54*3^(3/4)*Sqrt
[2 + Sqrt[3]]*d^3*(16*c*d^2 - 58*b*d*e + 667*a*e^2)*(d^(1/3) + e^(1/3)*x)*Sqrt[(d^(2/3) - d^(1/3)*e^(1/3)*x +
e^(2/3)*x^2)/((1 + Sqrt[3])*d^(1/3) + e^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*d^(1/3) + e^(1/3)*x)/((1 +
Sqrt[3])*d^(1/3) + e^(1/3)*x)], -7 - 4*Sqrt[3]])/(124729*e^(7/3)*Sqrt[(d^(1/3)*(d^(1/3) + e^(1/3)*x))/((1 + S
qrt[3])*d^(1/3) + e^(1/3)*x)^2]*Sqrt[d + e*x^3])

Rule 1411

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Simp[(c*x^(n + 1)*
(d + e*x^n)^(q + 1))/(e*(n*(q + 2) + 1)), x] + Dist[1/(e*(n*(q + 2) + 1)), Int[(d + e*x^n)^q*(a*e*(n*(q + 2) +
1) - (c*d*(n + 1) - b*e*(n*(q + 2) + 1))*x^n), x], x] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && N
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

\begin{align*} \int \left (d+e x^3\right )^{5/2} \left (a+b x^3+c x^6\right ) \, dx &=\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}+\frac{2 \int \left (d+e x^3\right )^{5/2} \left (\frac{29 a e}{2}-\left (4 c d-\frac{29 b e}{2}\right ) x^3\right ) \, dx}{29 e}\\ &=-\frac{2 (8 c d-29 b e) x \left (d+e x^3\right )^{7/2}}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}-\frac{1}{667} \left (-667 a-\frac{2 d (8 c d-29 b e)}{e^2}\right ) \int \left (d+e x^3\right )^{5/2} \, dx\\ &=\frac{2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{5/2}}{11339}-\frac{2 (8 c d-29 b e) x \left (d+e x^3\right )^{7/2}}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}+\frac{\left (15 d \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right )\right ) \int \left (d+e x^3\right )^{3/2} \, dx}{11339}\\ &=\frac{30 d \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{3/2}}{124729}+\frac{2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{5/2}}{11339}-\frac{2 (8 c d-29 b e) x \left (d+e x^3\right )^{7/2}}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}+\frac{\left (135 d^2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right )\right ) \int \sqrt{d+e x^3} \, dx}{124729}\\ &=\frac{54 d^2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \sqrt{d+e x^3}}{124729}+\frac{30 d \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{3/2}}{124729}+\frac{2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{5/2}}{11339}-\frac{2 (8 c d-29 b e) x \left (d+e x^3\right )^{7/2}}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}+\frac{\left (81 d^3 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right )\right ) \int \frac{1}{\sqrt{d+e x^3}} \, dx}{124729}\\ &=\frac{54 d^2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \sqrt{d+e x^3}}{124729}+\frac{30 d \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{3/2}}{124729}+\frac{2 \left (667 a+\frac{2 d (8 c d-29 b e)}{e^2}\right ) x \left (d+e x^3\right )^{5/2}}{11339}-\frac{2 (8 c d-29 b e) x \left (d+e x^3\right )^{7/2}}{667 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{7/2}}{29 e}+\frac{54\ 3^{3/4} \sqrt{2+\sqrt{3}} d^3 \left (16 c d^2-58 b d e+667 a e^2\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right )|-7-4 \sqrt{3}\right )}{124729 e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}\\ \end{align*}

Mathematica [C]  time = 0.178911, size = 103, normalized size = 0.26 $\frac{x \sqrt{d+e x^3} \left (\frac{\, _2F_1\left (-\frac{5}{2},\frac{1}{3};\frac{4}{3};-\frac{e x^3}{d}\right ) \left (29 d^2 e (23 a e-2 b d)+16 c d^4\right )}{\sqrt{\frac{e x^3}{d}+1}}-2 \left (d+e x^3\right )^3 \left (-29 b e+8 c d-23 c e x^3\right )\right )}{667 e^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x^3)^(5/2)*(a + b*x^3 + c*x^6),x]

[Out]

(x*Sqrt[d + e*x^3]*(-2*(d + e*x^3)^3*(8*c*d - 29*b*e - 23*c*e*x^3) + ((16*c*d^4 + 29*d^2*e*(-2*b*d + 23*a*e))*
Hypergeometric2F1[-5/2, 1/3, 4/3, -((e*x^3)/d)])/Sqrt[1 + (e*x^3)/d]))/(667*e^2)

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Maple [B]  time = 0.161, size = 1070, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^3+d)^(5/2)*(c*x^6+b*x^3+a),x)

[Out]

c*(2/29*e^2*x^13*(e*x^3+d)^(1/2)+122/667*d*e*x^10*(e*x^3+d)^(1/2)+1562/11339*d^2*x^7*(e*x^3+d)^(1/2)+810/12472
9*d^3/e*x^4*(e*x^3+d)^(1/2)-1296/124729*d^4/e^2*x*(e*x^3+d)^(1/2)-864/124729*I*d^5/e^3*3^(1/2)*(-d*e^2)^(1/3)*
(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1
/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e
*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1
/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^
2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))+b*(2/23*e^2*x^10*(e*x^3+d)^(1/2)+98/391*d*e*x^7*(e*x^3+d)^(1
/2)+974/4301*d^2*x^4*(e*x^3+d)^(1/2)+162/4301*d^3/e*x*(e*x^3+d)^(1/2)+108/4301*I*d^4/e^2*3^(1/2)*(-d*e^2)^(1/3
)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^
(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)
/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^
(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*
e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3)))^(1/2)))+a*(2/17*e^2*x^7*(e*x^3+d)^(1/2)+74/187*d*e*x^4*(e*x^3+d)^(
1/2)+106/187*d^2*x*(e*x^3+d)^(1/2)-54/187*I*d^3*3^(1/2)/e*(-d*e^2)^(1/3)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1
/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(1/3))^(1/2)*((x-1/e*(-d*e^2)^(1/3))/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(
1/2)/e*(-d*e^2)^(1/3)))^(1/2)*(-I*(x+1/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3^(1/2)*e/(-d*e^2)^(
1/3))^(1/2)/(e*x^3+d)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/e*(-d*e^2)^(1/3)-1/2*I*3^(1/2)/e*(-d*e^2)^(1/3))*3
^(1/2)*e/(-d*e^2)^(1/3))^(1/2),(I*3^(1/2)/e*(-d*e^2)^(1/3)/(-3/2/e*(-d*e^2)^(1/3)+1/2*I*3^(1/2)/e*(-d*e^2)^(1/
3)))^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{6} + b x^{3} + a\right )}{\left (e x^{3} + d\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^3+d)^(5/2)*(c*x^6+b*x^3+a),x, algorithm="maxima")

[Out]

integrate((c*x^6 + b*x^3 + a)*(e*x^3 + d)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c e^{2} x^{12} +{\left (2 \, c d e + b e^{2}\right )} x^{9} +{\left (c d^{2} + 2 \, b d e + a e^{2}\right )} x^{6} +{\left (b d^{2} + 2 \, a d e\right )} x^{3} + a d^{2}\right )} \sqrt{e x^{3} + d}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^3+d)^(5/2)*(c*x^6+b*x^3+a),x, algorithm="fricas")

[Out]

integral((c*e^2*x^12 + (2*c*d*e + b*e^2)*x^9 + (c*d^2 + 2*b*d*e + a*e^2)*x^6 + (b*d^2 + 2*a*d*e)*x^3 + a*d^2)*
sqrt(e*x^3 + d), x)

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Sympy [A]  time = 10.0827, size = 400, normalized size = 1.01 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**3+d)**(5/2)*(c*x**6+b*x**3+a),x)

[Out]

a*d**(5/2)*x*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(4/3)) + 2*a*d**(3/2)*e*x
**4*gamma(4/3)*hyper((-1/2, 4/3), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(7/3)) + a*sqrt(d)*e**2*x**7*gamma
(7/3)*hyper((-1/2, 7/3), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(10/3)) + b*d**(5/2)*x**4*gamma(4/3)*hyper
((-1/2, 4/3), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(7/3)) + 2*b*d**(3/2)*e*x**7*gamma(7/3)*hyper((-1/2, 7
/3), (10/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(10/3)) + b*sqrt(d)*e**2*x**10*gamma(10/3)*hyper((-1/2, 10/3),
(13/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(13/3)) + c*d**(5/2)*x**7*gamma(7/3)*hyper((-1/2, 7/3), (10/3,), e
*x**3*exp_polar(I*pi)/d)/(3*gamma(10/3)) + 2*c*d**(3/2)*e*x**10*gamma(10/3)*hyper((-1/2, 10/3), (13/3,), e*x**
3*exp_polar(I*pi)/d)/(3*gamma(13/3)) + c*sqrt(d)*e**2*x**13*gamma(13/3)*hyper((-1/2, 13/3), (16/3,), e*x**3*ex
p_polar(I*pi)/d)/(3*gamma(16/3))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{6} + b x^{3} + a\right )}{\left (e x^{3} + d\right )}^{\frac{5}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^3+d)^(5/2)*(c*x^6+b*x^3+a),x, algorithm="giac")

[Out]

integrate((c*x^6 + b*x^3 + a)*(e*x^3 + d)^(5/2), x)