3.5.21 \(\int \frac {x^7 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\)

Optimal. Leaf size=336 \[ -\frac {\left (-a^2 d^2-3 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3} d^3}-\frac {\left (-a^2 d^2-3 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3} d^3}-\frac {c^{5/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^3}+\frac {c^{5/3} \sqrt [3]{b c-a d} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^3}+\frac {c^{5/3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^3}-\frac {x^2 \sqrt [3]{a+b x^3} (6 b c-a d)}{18 b d^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 d} \]

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Rubi [C]  time = 0.05, antiderivative size = 64, normalized size of antiderivative = 0.19, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {511, 510} \begin {gather*} \frac {x^8 \sqrt [3]{a+b x^3} F_1\left (\frac {8}{3};-\frac {1}{3},1;\frac {11}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{8 c \sqrt [3]{\frac {b x^3}{a}+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(x^7*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

(x^8*(a + b*x^3)^(1/3)*AppellF1[8/3, -1/3, 1, 11/3, -((b*x^3)/a), -((d*x^3)/c)])/(8*c*(1 + (b*x^3)/a)^(1/3))

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {x^7 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {x^7 \sqrt [3]{1+\frac {b x^3}{a}}}{c+d x^3} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {x^8 \sqrt [3]{a+b x^3} F_1\left (\frac {8}{3};-\frac {1}{3},1;\frac {11}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{8 c \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C]  time = 0.37, size = 226, normalized size = 0.67 \begin {gather*} \frac {5 c x^2 \left (a \left (\frac {b x^3}{a}+1\right )^{2/3} (6 b c-a d) \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {(a d-b c) x^3}{a \left (d x^3+c\right )}\right )+\left (a+b x^3\right ) \left (\frac {d x^3}{c}+1\right )^{2/3} \left (a d-6 b c+3 b d x^3\right )\right )-2 x^5 \left (\frac {b x^3}{a}+1\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3} \left (a^2 d^2+3 a b c d-9 b^2 c^2\right ) F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{90 b c d^2 \left (a+b x^3\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^7*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

(-2*(-9*b^2*c^2 + 3*a*b*c*d + a^2*d^2)*x^5*(1 + (b*x^3)/a)^(2/3)*(1 + (d*x^3)/c)^(2/3)*AppellF1[5/3, 2/3, 1, 8
/3, -((b*x^3)/a), -((d*x^3)/c)] + 5*c*x^2*((a + b*x^3)*(-6*b*c + a*d + 3*b*d*x^3)*(1 + (d*x^3)/c)^(2/3) + a*(6
*b*c - a*d)*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))]))/(90
*b*c*d^2*(a + b*x^3)^(2/3)*(1 + (d*x^3)/c)^(2/3))

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IntegrateAlgebraic [C]  time = 5.50, size = 613, normalized size = 1.82 \begin {gather*} \frac {\left (a^2 d^2+3 a b c d-9 b^2 c^2\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{27 b^{5/3} d^3}-\frac {\left (-a^2 d^2-3 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{9 \sqrt {3} b^{5/3} d^3}+\frac {\left (-a^2 d^2-3 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}+b^{2/3} x^2\right )}{54 b^{5/3} d^3}+\frac {i \left (\sqrt {3} c^{5/3} \sqrt [3]{b c-a d}+i c^{5/3} \sqrt [3]{b c-a d}\right ) \log \left (2 x \sqrt [3]{b c-a d}+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{6 d^3}-\frac {\sqrt {-1-i \sqrt {3}} c^{5/3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {3 x \sqrt [3]{b c-a d}}{\sqrt {3} x \sqrt [3]{b c-a d}-\sqrt {3} \sqrt [3]{c} \sqrt [3]{a+b x^3}-3 i \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{\sqrt {6} d^3}+\frac {\left (c^{5/3} \sqrt [3]{b c-a d}-i \sqrt {3} c^{5/3} \sqrt [3]{b c-a d}\right ) \log \left (\left (\sqrt {3}+i\right ) c^{2/3} \left (a+b x^3\right )^{2/3}+\sqrt [3]{c} \left (-\sqrt {3} x+i x\right ) \sqrt [3]{a+b x^3} \sqrt [3]{b c-a d}-2 i x^2 (b c-a d)^{2/3}\right )}{12 d^3}+\frac {\sqrt [3]{a+b x^3} \left (a d x^2-6 b c x^2+3 b d x^5\right )}{18 b d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^7*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

((a + b*x^3)^(1/3)*(-6*b*c*x^2 + a*d*x^2 + 3*b*d*x^5))/(18*b*d^2) - ((9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*ArcTan[
(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/(9*Sqrt[3]*b^(5/3)*d^3) - (Sqrt[-1 - I*Sqrt[3]]*c^(5/3
)*(b*c - a*d)^(1/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I)*c^(1/3)*(a + b*x^3)^(1
/3) - Sqrt[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(Sqrt[6]*d^3) + ((-9*b^2*c^2 + 3*a*b*c*d + a^2*d^2)*Log[-(b^(1/3)*x
) + (a + b*x^3)^(1/3)])/(27*b^(5/3)*d^3) + ((I/6)*(I*c^(5/3)*(b*c - a*d)^(1/3) + Sqrt[3]*c^(5/3)*(b*c - a*d)^(
1/3))*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/d^3 + ((9*b^2*c^2 - 3*a*b*c*d -
a^2*d^2)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/(54*b^(5/3)*d^3) + ((c^(5/3)*(b*c
- a*d)^(1/3) - I*Sqrt[3]*c^(5/3)*(b*c - a*d)^(1/3))*Log[(-2*I)*(b*c - a*d)^(2/3)*x^2 + c^(1/3)*(b*c - a*d)^(1
/3)*(I*x - Sqrt[3]*x)*(a + b*x^3)^(1/3) + (I + Sqrt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(12*d^3)

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fricas [A]  time = 1.95, size = 494, normalized size = 1.47 \begin {gather*} \frac {18 \, \sqrt {3} {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} b^{3} c \arctan \left (-\frac {\sqrt {3} {\left (b c^{2} - a c d\right )} x + 2 \, \sqrt {3} {\left (b c^{3} - a c^{2} d\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{3 \, {\left (b c^{2} - a c d\right )} x}\right ) + 18 \, {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} b^{3} c \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} c - {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} x}{x}\right ) - 9 \, {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} b^{3} c \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} + {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x + {\left (b c^{3} - a c^{2} d\right )}^{\frac {2}{3}} x^{2}}{x^{2}}\right ) + 2 \, \sqrt {3} {\left (9 \, b^{3} c^{2} - 3 \, a b^{2} c d - a^{2} b d^{2}\right )} {\left (b^{2}\right )}^{\frac {1}{6}} \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) - 2 \, {\left (9 \, b^{2} c^{2} - 3 \, a b c d - a^{2} d^{2}\right )} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + {\left (9 \, b^{2} c^{2} - 3 \, a b c d - a^{2} d^{2}\right )} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (3 \, b^{3} d^{2} x^{5} - {\left (6 \, b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

1/54*(18*sqrt(3)*(b*c^3 - a*c^2*d)^(1/3)*b^3*c*arctan(-1/3*(sqrt(3)*(b*c^2 - a*c*d)*x + 2*sqrt(3)*(b*c^3 - a*c
^2*d)^(2/3)*(b*x^3 + a)^(1/3))/((b*c^2 - a*c*d)*x)) + 18*(b*c^3 - a*c^2*d)^(1/3)*b^3*c*log(((b*x^3 + a)^(1/3)*
c - (b*c^3 - a*c^2*d)^(1/3)*x)/x) - 9*(b*c^3 - a*c^2*d)^(1/3)*b^3*c*log(((b*x^3 + a)^(2/3)*c^2 + (b*c^3 - a*c^
2*d)^(1/3)*(b*x^3 + a)^(1/3)*c*x + (b*c^3 - a*c^2*d)^(2/3)*x^2)/x^2) + 2*sqrt(3)*(9*b^3*c^2 - 3*a*b^2*c*d - a^
2*b*d^2)*(b^2)^(1/6)*arctan(1/3*(sqrt(3)*(b^2)^(1/3)*b*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6
)/(b^2*x)) - 2*(9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + (
9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*(b^2)^(2/3)*log(((b^2)^(1/3)*b*x^2 + (b*x^3 + a)^(1/3)*(b^2)^(2/3)*x + (b*x^3
+ a)^(2/3)*b)/x^2) + 3*(3*b^3*d^2*x^5 - (6*b^3*c*d - a*b^2*d^2)*x^2)*(b*x^3 + a)^(1/3))/(b^3*d^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{7}}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*x^7/(d*x^3 + c), x)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{7}}{d \,x^{3}+c}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{7}}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)*x^7/(d*x^3 + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^7\,{\left (b\,x^3+a\right )}^{1/3}}{d\,x^3+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(a + b*x^3)^(1/3))/(c + d*x^3),x)

[Out]

int((x^7*(a + b*x^3)^(1/3))/(c + d*x^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**7*(a + b*x**3)**(1/3)/(c + d*x**3), x)

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