3.66 \(\int \sqrt [3]{a+b x^3} (c+d x^3) \, dx\)

Optimal. Leaf size=82 \[ \frac{x \sqrt [3]{a+b x^3} (5 b c-a d) \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{5 b \sqrt [3]{\frac{b x^3}{a}+1}}+\frac{d x \left (a+b x^3\right )^{4/3}}{5 b} \]

[Out]

(d*x*(a + b*x^3)^(4/3))/(5*b) + ((5*b*c - a*d)*x*(a + b*x^3)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)
/a)])/(5*b*(1 + (b*x^3)/a)^(1/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0210526, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {388, 246, 245} \[ \frac{x \sqrt [3]{a+b x^3} (5 b c-a d) \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{5 b \sqrt [3]{\frac{b x^3}{a}+1}}+\frac{d x \left (a+b x^3\right )^{4/3}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)*(c + d*x^3),x]

[Out]

(d*x*(a + b*x^3)^(4/3))/(5*b) + ((5*b*c - a*d)*x*(a + b*x^3)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)
/a)])/(5*b*(1 + (b*x^3)/a)^(1/3))

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 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sqrt [3]{a+b x^3} \left (c+d x^3\right ) \, dx &=\frac{d x \left (a+b x^3\right )^{4/3}}{5 b}-\frac{(-5 b c+a d) \int \sqrt [3]{a+b x^3} \, dx}{5 b}\\ &=\frac{d x \left (a+b x^3\right )^{4/3}}{5 b}-\frac{\left ((-5 b c+a d) \sqrt [3]{a+b x^3}\right ) \int \sqrt [3]{1+\frac{b x^3}{a}} \, dx}{5 b \sqrt [3]{1+\frac{b x^3}{a}}}\\ &=\frac{d x \left (a+b x^3\right )^{4/3}}{5 b}+\frac{(5 b c-a d) x \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{5 b \sqrt [3]{1+\frac{b x^3}{a}}}\\ \end{align*}

Mathematica [A]  time = 0.0669086, size = 72, normalized size = 0.88 \[ \frac{x \sqrt [3]{a+b x^3} \left (\frac{(5 b c-a d) \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{4}{3};-\frac{b x^3}{a}\right )}{\sqrt [3]{\frac{b x^3}{a}+1}}+d \left (a+b x^3\right )\right )}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(1/3)*(c + d*x^3),x]

[Out]

(x*(a + b*x^3)^(1/3)*(d*(a + b*x^3) + ((5*b*c - a*d)*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)])/(1 + (b*
x^3)/a)^(1/3)))/(5*b)

________________________________________________________________________________________

Maple [F]  time = 0.208, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{b{x}^{3}+a} \left ( d{x}^{3}+c \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)*(d*x^3+c),x)

[Out]

int((b*x^3+a)^(1/3)*(d*x^3+c),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)*(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)*(d*x^3 + c), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)*(d*x^3+c),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(1/3)*(d*x^3 + c), x)

________________________________________________________________________________________

Sympy [C]  time = 2.0371, size = 82, normalized size = 1. \begin{align*} \frac{\sqrt [3]{a} c x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \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{\sqrt [3]{a} d x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)*(d*x**3+c),x)

[Out]

a**(1/3)*c*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + a**(1/3)*d*x**4*
gamma(4/3)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{3} + a\right )}^{\frac{1}{3}}{\left (d x^{3} + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)*(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*(d*x^3 + c), x)