3.65 \(\int (a+b x^3)^{4/3} (c+d x^3) \, dx\)

Optimal. Leaf size=83 \[ \frac {a x \sqrt [3]{a+b x^3} (8 b c-a d) \, _2F_1\left (-\frac {4}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3}}{8 b} \]

[Out]

1/8*d*x*(b*x^3+a)^(7/3)/b+1/8*a*(-a*d+8*b*c)*x*(b*x^3+a)^(1/3)*hypergeom([-4/3, 1/3],[4/3],-b*x^3/a)/b/(1+b*x^
3/a)^(1/3)

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Rubi [A]  time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.00, 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} \[ \frac {a x \sqrt [3]{a+b x^3} (8 b c-a d) \, _2F_1\left (-\frac {4}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(4/3)*(c + d*x^3),x]

[Out]

(d*x*(a + b*x^3)^(7/3))/(8*b) + (a*(8*b*c - a*d)*x*(a + b*x^3)^(1/3)*Hypergeometric2F1[-4/3, 1/3, 4/3, -((b*x^
3)/a)])/(8*b*(1 + (b*x^3)/a)^(1/3))

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])

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 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]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right ) \, dx &=\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}-\frac {(-8 b c+a d) \int \left (a+b x^3\right )^{4/3} \, dx}{8 b}\\ &=\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}-\frac {\left (a (-8 b c+a d) \sqrt [3]{a+b x^3}\right ) \int \left (1+\frac {b x^3}{a}\right )^{4/3} \, dx}{8 b \sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {d x \left (a+b x^3\right )^{7/3}}{8 b}+\frac {a (8 b c-a d) x \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac {4}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 75, normalized size = 0.90 \[ \frac {x \sqrt [3]{a+b x^3} \left (d \left (a+b x^3\right )^2-\frac {a (a d-8 b c) \, _2F_1\left (-\frac {4}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{\sqrt [3]{\frac {b x^3}{a}+1}}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(4/3)*(c + d*x^3),x]

[Out]

(x*(a + b*x^3)^(1/3)*(d*(a + b*x^3)^2 - (a*(-8*b*c + a*d)*Hypergeometric2F1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(1
+ (b*x^3)/a)^(1/3)))/(8*b)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b d x^{6} + {\left (b c + a d\right )} x^{3} + a c\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)*(d*x^3+c),x, algorithm="fricas")

[Out]

integral((b*d*x^6 + (b*c + a*d)*x^3 + a*c)*(b*x^3 + a)^(1/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)*(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(4/3)*(d*x^3 + c), x)

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)*(d*x^3+c),x)

[Out]

int((b*x^3+a)^(4/3)*(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)*(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)*(d*x^3 + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(4/3)*(c + d*x^3),x)

[Out]

int((a + b*x^3)^(4/3)*(c + d*x^3), x)

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sympy [C]  time = 7.40, size = 170, normalized size = 2.05 \[ \frac {a^{\frac {4}{3}} c x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {a^{\frac {4}{3}} d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} b c x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {\sqrt [3]{a} b d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(4/3)*(d*x**3+c),x)

[Out]

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

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