### 3.36 $$\int (d+e x^3)^{3/2} (a+b x^3+c x^6) \, dx$$

Optimal. Leaf size=356 $\frac{18\ 3^{3/4} \sqrt{2+\sqrt{3}} d^2 \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (391 a e^2-46 b d e+16 c d^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt{3}\right )}{21505 e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}+\frac{2 x \left (d+e x^3\right )^{3/2} \left (391 a e^2-46 b d e+16 c d^2\right )}{4301 e^2}+\frac{18 d x \sqrt{d+e x^3} \left (391 a e^2-46 b d e+16 c d^2\right )}{21505 e^2}-\frac{2 x \left (d+e x^3\right )^{5/2} (8 c d-23 b e)}{391 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}$

[Out]

(18*d*(16*c*d^2 - 46*b*d*e + 391*a*e^2)*x*Sqrt[d + e*x^3])/(21505*e^2) + (2*(16*c*d^2 - 46*b*d*e + 391*a*e^2)*
x*(d + e*x^3)^(3/2))/(4301*e^2) - (2*(8*c*d - 23*b*e)*x*(d + e*x^3)^(5/2))/(391*e^2) + (2*c*x^4*(d + e*x^3)^(5
/2))/(23*e) + (18*3^(3/4)*Sqrt[2 + Sqrt[3]]*d^2*(16*c*d^2 - 46*b*d*e + 391*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]])/(21505*e^(7/3)*Sqrt[(d^(1/3)*(
d^(1/3) + e^(1/3)*x))/((1 + Sqrt[3])*d^(1/3) + e^(1/3)*x)^2]*Sqrt[d + e*x^3])

________________________________________________________________________________________

Rubi [A]  time = 0.311484, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {1411, 388, 195, 218} $\frac{2 x \left (d+e x^3\right )^{3/2} \left (391 a e^2-46 b d e+16 c d^2\right )}{4301 e^2}+\frac{18 d x \sqrt{d+e x^3} \left (391 a e^2-46 b d e+16 c d^2\right )}{21505 e^2}+\frac{18\ 3^{3/4} \sqrt{2+\sqrt{3}} d^2 \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt{\frac{d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (391 a e^2-46 b d e+16 c d^2\right ) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{e} x+\left (1-\sqrt{3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt{3}\right ) \sqrt [3]{d}}\right )|-7-4 \sqrt{3}\right )}{21505 e^{7/3} \sqrt{\frac{\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt{d+e x^3}}-\frac{2 x \left (d+e x^3\right )^{5/2} (8 c d-23 b e)}{391 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(18*d*(16*c*d^2 - 46*b*d*e + 391*a*e^2)*x*Sqrt[d + e*x^3])/(21505*e^2) + (2*(16*c*d^2 - 46*b*d*e + 391*a*e^2)*
x*(d + e*x^3)^(3/2))/(4301*e^2) - (2*(8*c*d - 23*b*e)*x*(d + e*x^3)^(5/2))/(391*e^2) + (2*c*x^4*(d + e*x^3)^(5
/2))/(23*e) + (18*3^(3/4)*Sqrt[2 + Sqrt[3]]*d^2*(16*c*d^2 - 46*b*d*e + 391*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]])/(21505*e^(7/3)*Sqrt[(d^(1/3)*(
d^(1/3) + e^(1/3)*x))/((1 + Sqrt[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 )^{3/2} \left (a+b x^3+c x^6\right ) \, dx &=\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}+\frac{2 \int \left (d+e x^3\right )^{3/2} \left (\frac{23 a e}{2}-\left (4 c d-\frac{23 b e}{2}\right ) x^3\right ) \, dx}{23 e}\\ &=-\frac{2 (8 c d-23 b e) x \left (d+e x^3\right )^{5/2}}{391 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}-\frac{1}{391} \left (-391 a-\frac{2 d (8 c d-23 b e)}{e^2}\right ) \int \left (d+e x^3\right )^{3/2} \, dx\\ &=\frac{2 \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right ) x \left (d+e x^3\right )^{3/2}}{4301}-\frac{2 (8 c d-23 b e) x \left (d+e x^3\right )^{5/2}}{391 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}+\frac{\left (9 d \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right )\right ) \int \sqrt{d+e x^3} \, dx}{4301}\\ &=\frac{18 d \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right ) x \sqrt{d+e x^3}}{21505}+\frac{2 \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right ) x \left (d+e x^3\right )^{3/2}}{4301}-\frac{2 (8 c d-23 b e) x \left (d+e x^3\right )^{5/2}}{391 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}+\frac{\left (27 d^2 \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right )\right ) \int \frac{1}{\sqrt{d+e x^3}} \, dx}{21505}\\ &=\frac{18 d \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right ) x \sqrt{d+e x^3}}{21505}+\frac{2 \left (391 a+\frac{2 d (8 c d-23 b e)}{e^2}\right ) x \left (d+e x^3\right )^{3/2}}{4301}-\frac{2 (8 c d-23 b e) x \left (d+e x^3\right )^{5/2}}{391 e^2}+\frac{2 c x^4 \left (d+e x^3\right )^{5/2}}{23 e}+\frac{18\ 3^{3/4} \sqrt{2+\sqrt{3}} d^2 \left (16 c d^2-46 b d e+391 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 )}{21505 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 = 0.153007, size = 101, normalized size = 0.28 $\frac{x \sqrt{d+e x^3} \left (\frac{\, _2F_1\left (-\frac{3}{2},\frac{1}{3};\frac{4}{3};-\frac{e x^3}{d}\right ) \left (23 d e (17 a e-2 b d)+16 c d^3\right )}{\sqrt{\frac{e x^3}{d}+1}}-2 \left (d+e x^3\right )^2 \left (-23 b e+8 c d-17 c e x^3\right )\right )}{391 e^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(x*Sqrt[d + e*x^3]*(-2*(d + e*x^3)^2*(8*c*d - 23*b*e - 17*c*e*x^3) + ((16*c*d^3 + 23*d*e*(-2*b*d + 17*a*e))*Hy
pergeometric2F1[-3/2, 1/3, 4/3, -((e*x^3)/d)])/Sqrt[1 + (e*x^3)/d]))/(391*e^2)

________________________________________________________________________________________

Maple [B]  time = 0.029, size = 1010, normalized size = 2.8 \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)^(3/2)*(c*x^6+b*x^3+a),x)

[Out]

c*(2/23*e*x^10*(e*x^3+d)^(1/2)+52/391*d*x^7*(e*x^3+d)^(1/2)+54/4301*d^2/e*x^4*(e*x^3+d)^(1/2)-432/21505*d^3/e^
2*x*(e*x^3+d)^(1/2)-288/21505*I*d^4/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*(2/17*e*x^7*(e*x^3+d)^(1/2)+40/187*d*x^4*(e*x^3+d)^(1/2)+54/935*d^2/e*x*(e*x^3+d)^(1/2)+36/935*I*d^3/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*(2/11*e*x^4*(e*x^3+d)^(1/2)+28/55*d
*x*(e*x^3+d)^(1/2)-18/55*I*d^2*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)))

________________________________________________________________________________________

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{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 5.8478, size = 257, normalized size = 0.72 \begin{align*} \frac{a d^{\frac{3}{2}} x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{3} \\ \frac{4}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} + \frac{a \sqrt{d} e x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \Gamma \left (\frac{7}{3}\right )} + \frac{b d^{\frac{3}{2}} x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{4}{3} \\ \frac{7}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \Gamma \left (\frac{7}{3}\right )} + \frac{b \sqrt{d} e x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \Gamma \left (\frac{10}{3}\right )} + \frac{c d^{\frac{3}{2}} x^{7} \Gamma \left (\frac{7}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{3} \\ \frac{10}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \Gamma \left (\frac{10}{3}\right )} + \frac{c \sqrt{d} e x^{10} \Gamma \left (\frac{10}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{10}{3} \\ \frac{13}{3} \end{matrix}\middle |{\frac{e x^{3} e^{i \pi }}{d}} \right )}}{3 \Gamma \left (\frac{13}{3}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**(3/2)*x*gamma(1/3)*hyper((-1/2, 1/3), (4/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(4/3)) + a*sqrt(d)*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)) + b*d**(3/2)*x**4*gamma(4/3)*h
yper((-1/2, 4/3), (7/3,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(7/3)) + b*sqrt(d)*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)) + c*d**(3/2)*x**7*gamma(7/3)*hyper((-1/2, 7/3), (10/3
,), e*x**3*exp_polar(I*pi)/d)/(3*gamma(10/3)) + c*sqrt(d)*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))

________________________________________________________________________________________

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{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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