3.203 \(\int (a+b x^3)^{3/2} (A+B x^3) \, dx\)

Optimal. Leaf size=299 \[ \frac{18\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (17 A b-2 a B) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{935 b^{4/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 x \left (a+b x^3\right )^{3/2} (17 A b-2 a B)}{187 b}+\frac{18 a x \sqrt{a+b x^3} (17 A b-2 a B)}{935 b}+\frac{2 B x \left (a+b x^3\right )^{5/2}}{17 b} \]

[Out]

(18*a*(17*A*b - 2*a*B)*x*Sqrt[a + b*x^3])/(935*b) + (2*(17*A*b - 2*a*B)*x*(a + b*x^3)^(3/2))/(187*b) + (2*B*x*
(a + b*x^3)^(5/2))/(17*b) + (18*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*(17*A*b - 2*a*B)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(
2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])
*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(935*b^(4/3)*Sqrt[(a^(1/3)*(a^(1/
3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])

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Rubi [A]  time = 0.123566, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {388, 195, 218} \[ \frac{18\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (17 A b-2 a B) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{935 b^{4/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 x \left (a+b x^3\right )^{3/2} (17 A b-2 a B)}{187 b}+\frac{18 a x \sqrt{a+b x^3} (17 A b-2 a B)}{935 b}+\frac{2 B x \left (a+b x^3\right )^{5/2}}{17 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/2)*(A + B*x^3),x]

[Out]

(18*a*(17*A*b - 2*a*B)*x*Sqrt[a + b*x^3])/(935*b) + (2*(17*A*b - 2*a*B)*x*(a + b*x^3)^(3/2))/(187*b) + (2*B*x*
(a + b*x^3)^(5/2))/(17*b) + (18*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*(17*A*b - 2*a*B)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(
2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])
*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*Sqrt[3]])/(935*b^(4/3)*Sqrt[(a^(1/3)*(a^(1/
3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[a + b*x^3])

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 (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx &=\frac{2 B x \left (a+b x^3\right )^{5/2}}{17 b}-\frac{\left (2 \left (-\frac{17 A b}{2}+a B\right )\right ) \int \left (a+b x^3\right )^{3/2} \, dx}{17 b}\\ &=\frac{2 (17 A b-2 a B) x \left (a+b x^3\right )^{3/2}}{187 b}+\frac{2 B x \left (a+b x^3\right )^{5/2}}{17 b}+\frac{(9 a (17 A b-2 a B)) \int \sqrt{a+b x^3} \, dx}{187 b}\\ &=\frac{18 a (17 A b-2 a B) x \sqrt{a+b x^3}}{935 b}+\frac{2 (17 A b-2 a B) x \left (a+b x^3\right )^{3/2}}{187 b}+\frac{2 B x \left (a+b x^3\right )^{5/2}}{17 b}+\frac{\left (27 a^2 (17 A b-2 a B)\right ) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{935 b}\\ &=\frac{18 a (17 A b-2 a B) x \sqrt{a+b x^3}}{935 b}+\frac{2 (17 A b-2 a B) x \left (a+b x^3\right )^{3/2}}{187 b}+\frac{2 B x \left (a+b x^3\right )^{5/2}}{17 b}+\frac{18\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 (17 A b-2 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt{3}\right )}{935 b^{4/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.056973, size = 77, normalized size = 0.26 \[ \frac{2 x \sqrt{a+b x^3} \left (B \left (a+b x^3\right )^2-\frac{a \left (a B-\frac{17 A b}{2}\right ) \, _2F_1\left (-\frac{3}{2},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{\sqrt{\frac{b x^3}{a}+1}}\right )}{17 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2)*(A + B*x^3),x]

[Out]

(2*x*Sqrt[a + b*x^3]*(B*(a + b*x^3)^2 - (a*((-17*A*b)/2 + a*B)*Hypergeometric2F1[-3/2, 1/3, 4/3, -((b*x^3)/a)]
)/Sqrt[1 + (b*x^3)/a]))/(17*b)

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Maple [B]  time = 0.014, size = 654, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(3/2)*(B*x^3+A),x)

[Out]

B*(2/17*b*x^7*(b*x^3+a)^(1/2)+40/187*a*x^4*(b*x^3+a)^(1/2)+54/935*a^2/b*x*(b*x^3+a)^(1/2)+36/935*I*a^3/b^2*3^(
1/2)*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)
*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(
1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*
(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^
(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))+A*(2/11*b*x^4*(b*x^3+a)^(1/2)+28/55*a*x*
(b*x^3+a)^(1/2)-18/55*I*a^2*3^(1/2)/b*(-a*b^2)^(1/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)
)*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3
)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a
)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1
/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A),x, algorithm="fricas")

[Out]

integral((B*b*x^6 + (B*a + A*b)*x^3 + A*a)*sqrt(b*x^3 + a), x)

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Sympy [A]  time = 4.0534, size = 170, normalized size = 0.57 \begin{align*} \frac{A a^{\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{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{4}{3}\right )} + \frac{A \sqrt{a} b 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{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{7}{3}\right )} + \frac{B a^{\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{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{7}{3}\right )} + \frac{B \sqrt{a} b 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{b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac{10}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(3/2)*(B*x**3+A),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(3/2)*(B*x^3+A),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)*(b*x^3 + a)^(3/2), x)