3.80 \(\int (a+b x^3)^{4/3} (c+d x^3)^2 \, dx\)

Optimal. Leaf size=133 \[ \frac {a x \sqrt [3]{a+b x^3} \left (2 a^2 d^2-11 a b c d+44 b^2 c^2\right ) \, _2F_1\left (-\frac {4}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{44 b^2 \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3} (7 b c-2 a d)}{44 b^2}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b} \]

[Out]

1/44*d*(-2*a*d+7*b*c)*x*(b*x^3+a)^(7/3)/b^2+1/11*d*x*(b*x^3+a)^(7/3)*(d*x^3+c)/b+1/44*a*(2*a^2*d^2-11*a*b*c*d+
44*b^2*c^2)*x*(b*x^3+a)^(1/3)*hypergeom([-4/3, 1/3],[4/3],-b*x^3/a)/b^2/(1+b*x^3/a)^(1/3)

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Rubi [A]  time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {416, 388, 246, 245} \[ \frac {a x \sqrt [3]{a+b x^3} \left (2 a^2 d^2-11 a b c d+44 b^2 c^2\right ) \, _2F_1\left (-\frac {4}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{44 b^2 \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{7/3} (7 b c-2 a d)}{44 b^2}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(4/3)*(c + d*x^3)^2,x]

[Out]

(d*(7*b*c - 2*a*d)*x*(a + b*x^3)^(7/3))/(44*b^2) + (d*x*(a + b*x^3)^(7/3)*(c + d*x^3))/(11*b) + (a*(44*b^2*c^2
- 11*a*b*c*d + 2*a^2*d^2)*x*(a + b*x^3)^(1/3)*Hypergeometric2F1[-4/3, 1/3, 4/3, -((b*x^3)/a)])/(44*b^2*(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]

Rule 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rubi steps

\begin {align*} \int \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )^2 \, dx &=\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}+\frac {\int \left (a+b x^3\right )^{4/3} \left (c (11 b c-a d)+2 d (7 b c-2 a d) x^3\right ) \, dx}{11 b}\\ &=\frac {d (7 b c-2 a d) x \left (a+b x^3\right )^{7/3}}{44 b^2}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}-\frac {(2 a d (7 b c-2 a d)-8 b c (11 b c-a d)) \int \left (a+b x^3\right )^{4/3} \, dx}{88 b^2}\\ &=\frac {d (7 b c-2 a d) x \left (a+b x^3\right )^{7/3}}{44 b^2}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}-\frac {\left (a (2 a d (7 b c-2 a d)-8 b c (11 b c-a d)) \sqrt [3]{a+b x^3}\right ) \int \left (1+\frac {b x^3}{a}\right )^{4/3} \, dx}{88 b^2 \sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {d (7 b c-2 a d) x \left (a+b x^3\right )^{7/3}}{44 b^2}+\frac {d x \left (a+b x^3\right )^{7/3} \left (c+d x^3\right )}{11 b}+\frac {a \left (44 b^2 c^2-11 a b c d+2 a^2 d^2\right ) 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 )}{44 b^2 \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [A]  time = 5.19, size = 171, normalized size = 1.29 \[ \frac {x \left (2 a^2 \left (\frac {b x^3}{a}+1\right )^{2/3} \left (2 a^2 d^2-11 a b c d+44 b^2 c^2\right ) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )-\left (a+b x^3\right ) \left (4 a^3 d^2-2 a^2 b d \left (11 c+d x^3\right )-3 a b^2 \left (44 c^2+33 c d x^3+10 d^2 x^6\right )-b^3 x^3 \left (44 c^2+55 c d x^3+20 d^2 x^6\right )\right )\right )}{220 b^2 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(4/3)*(c + d*x^3)^2,x]

[Out]

(x*(-((a + b*x^3)*(4*a^3*d^2 - 2*a^2*b*d*(11*c + d*x^3) - 3*a*b^2*(44*c^2 + 33*c*d*x^3 + 10*d^2*x^6) - b^3*x^3
*(44*c^2 + 55*c*d*x^3 + 20*d^2*x^6))) + 2*a^2*(44*b^2*c^2 - 11*a*b*c*d + 2*a^2*d^2)*(1 + (b*x^3)/a)^(2/3)*Hype
rgeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)]))/(220*b^2*(a + b*x^3)^(2/3))

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b d^{2} x^{9} + {\left (2 \, b c d + a d^{2}\right )} x^{6} + {\left (b c^{2} + 2 \, a c d\right )} x^{3} + a c^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)*(d*x^3+c)^2,x, algorithm="fricas")

[Out]

integral((b*d^2*x^9 + (2*b*c*d + a*d^2)*x^6 + (b*c^2 + 2*a*c*d)*x^3 + a*c^2)*(b*x^3 + a)^(1/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)*(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(4/3)*(d*x^3 + c)^2, x)

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)*(d*x^3+c)^2,x)

[Out]

int((b*x^3+a)^(4/3)*(d*x^3+c)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)*(d*x^3+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)*(d*x^3 + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (b\,x^3+a\right )}^{4/3}\,{\left (d\,x^3+c\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(4/3)*(c + d*x^3)^2,x)

[Out]

int((a + b*x^3)^(4/3)*(c + d*x^3)^2, x)

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sympy [C]  time = 7.06, size = 270, normalized size = 2.03 \[ \frac {a^{\frac {4}{3}} c^{2} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \frac {2 a^{\frac {4}{3}} c d x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {a^{\frac {4}{3}} d^{2} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {\sqrt [3]{a} b c^{2} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {2 \sqrt [3]{a} b c d x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} + \frac {\sqrt [3]{a} b d^{2} x^{10} \Gamma \left (\frac {10}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {10}{3} \\ \frac {13}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {13}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(4/3)*(d*x**3+c)**2,x)

[Out]

a**(4/3)*c**2*x*gamma(1/3)*hyper((-1/3, 1/3), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(4/3)) + 2*a**(4/3)*c*
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**(4/3)*d**2*x**7*gam
ma(7/3)*hyper((-1/3, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3)) + a**(1/3)*b*c**2*x**4*gamma(4/3
)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(7/3)) + 2*a**(1/3)*b*c*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)) + a**(1/3)*b*d**2*x**10*gamma(10/3)*hyper((-1
/3, 10/3), (13/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(13/3))

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