3.1.81 \(\int \sqrt [3]{a+b x^3} (c+d x^3)^2 \, dx\) [81]

Optimal. Leaf size=131 \[ \frac {d (11 b c-4 a d) x \left (a+b x^3\right )^{4/3}}{40 b^2}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}+\frac {\left (10 b^2 c^2-4 a b c d+a^2 d^2\right ) x \sqrt [3]{a+b x^3} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{10 b^2 \sqrt [3]{1+\frac {b x^3}{a}}} \]

[Out]

1/40*d*(-4*a*d+11*b*c)*x*(b*x^3+a)^(4/3)/b^2+1/8*d*x*(b*x^3+a)^(4/3)*(d*x^3+c)/b+1/10*(a^2*d^2-4*a*b*c*d+10*b^
2*c^2)*x*(b*x^3+a)^(1/3)*hypergeom([-1/3, 1/3],[4/3],-b*x^3/a)/b^2/(1+b*x^3/a)^(1/3)

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Rubi [A]
time = 0.05, antiderivative size = 131, 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 = {427, 396, 252, 251} \begin {gather*} \frac {x \sqrt [3]{a+b x^3} \left (a^2 d^2-4 a b c d+10 b^2 c^2\right ) \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{10 b^2 \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {d x \left (a+b x^3\right )^{4/3} (11 b c-4 a d)}{40 b^2}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(1/3)*(c + d*x^3)^2,x]

[Out]

(d*(11*b*c - 4*a*d)*x*(a + b*x^3)^(4/3))/(40*b^2) + (d*x*(a + b*x^3)^(4/3)*(c + d*x^3))/(8*b) + ((10*b^2*c^2 -
 4*a*b*c*d + a^2*d^2)*x*(a + b*x^3)^(1/3)*Hypergeometric2F1[-1/3, 1/3, 4/3, -((b*x^3)/a)])/(10*b^2*(1 + (b*x^3
)/a)^(1/3))

Rule 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 396

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 427

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 \sqrt [3]{a+b x^3} \left (c+d x^3\right )^2 \, dx &=\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}+\frac {\int \sqrt [3]{a+b x^3} \left (c (8 b c-a d)+d (11 b c-4 a d) x^3\right ) \, dx}{8 b}\\ &=\frac {d (11 b c-4 a d) x \left (a+b x^3\right )^{4/3}}{40 b^2}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}+\frac {\left (10 b^2 c^2-4 a b c d+a^2 d^2\right ) \int \sqrt [3]{a+b x^3} \, dx}{10 b^2}\\ &=\frac {d (11 b c-4 a d) x \left (a+b x^3\right )^{4/3}}{40 b^2}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}+\frac {\left (\left (10 b^2 c^2-4 a b c d+a^2 d^2\right ) \sqrt [3]{a+b x^3}\right ) \int \sqrt [3]{1+\frac {b x^3}{a}} \, dx}{10 b^2 \sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {d (11 b c-4 a d) x \left (a+b x^3\right )^{4/3}}{40 b^2}+\frac {d x \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )}{8 b}+\frac {\left (10 b^2 c^2-4 a b c d+a^2 d^2\right ) 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 )}{10 b^2 \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [A]
time = 7.69, size = 179, normalized size = 1.37 \begin {gather*} \frac {x \sqrt [3]{a+b x^3} \left (20 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \Gamma \left (-\frac {1}{3}\right ) \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {10}{3};-\frac {b x^3}{a}\right )-3 b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \Gamma \left (\frac {2}{3}\right ) \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {13}{3};-\frac {b x^3}{a}\right )-9 b x^3 \left (c+d x^3\right )^2 \Gamma \left (\frac {2}{3}\right ) \, _3F_2\left (\frac {2}{3},\frac {4}{3},2;1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{280 a \sqrt [3]{1+\frac {b x^3}{a}} \Gamma \left (-\frac {1}{3}\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(1/3)*(c + d*x^3)^2,x]

[Out]

(x*(a + b*x^3)^(1/3)*(20*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Gamma[-1/3]*Hypergeometric2F1[-1/3, 1/3, 10/3, -((
b*x^3)/a)] - 3*b*x^3*(11*c^2 + 16*c*d*x^3 + 5*d^2*x^6)*Gamma[2/3]*Hypergeometric2F1[2/3, 4/3, 13/3, -((b*x^3)/
a)] - 9*b*x^3*(c + d*x^3)^2*Gamma[2/3]*HypergeometricPFQ[{2/3, 4/3, 2}, {1, 13/3}, -((b*x^3)/a)]))/(280*a*(1 +
 (b*x^3)/a)^(1/3)*Gamma[-1/3])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (d \,x^{3}+c \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)*(d*x^3+c)^2,x)

[Out]

int((b*x^3+a)^(1/3)*(d*x^3+c)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)*(d*x^3+c)^2,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)*(d*x^3 + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)*(d*x^3+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.81, size = 131, normalized size = 1.00 \begin {gather*} \frac {\sqrt [3]{a} 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 \sqrt [3]{a} 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 {\sqrt [3]{a} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)*(d*x**3+c)**2,x)

[Out]

a**(1/3)*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**(1/3)*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**(1/3)*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))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)*(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*(d*x^3 + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)*(c + d*x^3)^2,x)

[Out]

int((a + b*x^3)^(1/3)*(c + d*x^3)^2, x)

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