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

Optimal. Leaf size=334 \[ -\frac {\left (2 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{2/3} d^3}-\frac {\left (2 a^2 d^2-12 a b c d+9 b^2 c^2\right ) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{2/3} d^3}-\frac {c^{2/3} (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 d^3}+\frac {c^{2/3} (b c-a d)^{4/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^3}+\frac {c^{2/3} (b c-a d)^{4/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d^3}-\frac {x^2 \sqrt [3]{a+b x^3} (6 b c-7 a d)}{18 d^2}+\frac {b x^5 \sqrt [3]{a+b x^3}}{6 d} \]

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Rubi [C]  time = 0.06, antiderivative size = 65, 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 {a x^5 \sqrt [3]{a+b x^3} F_1\left (\frac {5}{3};-\frac {4}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c \sqrt [3]{\frac {b x^3}{a}+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x^5*(a + b*x^3)^(1/3)*AppellF1[5/3, -4/3, 1, 8/3, -((b*x^3)/a), -((d*x^3)/c)])/(5*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^4 \left (a+b x^3\right )^{4/3}}{c+d x^3} \, dx &=\frac {\left (a \sqrt [3]{a+b x^3}\right ) \int \frac {x^4 \left (1+\frac {b x^3}{a}\right )^{4/3}}{c+d x^3} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {a x^5 \sqrt [3]{a+b x^3} F_1\left (\frac {5}{3};-\frac {4}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C]  time = 0.31, size = 225, normalized size = 0.67 \begin {gather*} \frac {2 x^5 \left (\frac {b x^3}{a}+1\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3} \left (2 a^2 d^2-12 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 )+5 c x^2 \left (a \left (\frac {b x^3}{a}+1\right )^{2/3} (6 b c-7 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 (7 a d-6 b c+3 b d x^3\right )\right )}{90 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^4*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

(2*(9*b^2*c^2 - 12*a*b*c*d + 2*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 + 7*a*d + 3*b*d*x^3)*(1 + (d*x^3)/c)^(2/3) + a
*(6*b*c - 7*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*c*d^2*(a + b*x^3)^(2/3)*(1 + (d*x^3)/c)^(2/3))

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IntegrateAlgebraic [C]  time = 9.96, size = 611, normalized size = 1.83 \begin {gather*} \frac {\left (-2 a^2 d^2+12 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^{2/3} d^3}-\frac {\left (2 a^2 d^2-12 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^{2/3} d^3}+\frac {\left (2 a^2 d^2-12 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^{2/3} d^3}+\frac {i \left (\sqrt {3} c^{2/3} (b c-a d)^{4/3}+i c^{2/3} (b c-a d)^{4/3}\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 {\frac {1}{6} \left (-1-i \sqrt {3}\right )} c^{2/3} (b c-a d)^{4/3} \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 )}{d^3}+\frac {\left (c^{2/3} (b c-a d)^{4/3}-i \sqrt {3} c^{2/3} (b c-a d)^{4/3}\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 (7 a d x^2-6 b c x^2+3 b d x^5\right )}{18 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^3)^(1/3)*(-6*b*c*x^2 + 7*a*d*x^2 + 3*b*d*x^5))/(18*d^2) - ((9*b^2*c^2 - 12*a*b*c*d + 2*a^2*d^2)*ArcT
an[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/(9*Sqrt[3]*b^(2/3)*d^3) - (Sqrt[(-1 - I*Sqrt[3])/6]
*c^(2/3)*(b*c - a*d)^(4/3)*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))])/d^3 + ((-9*b^2*c^2 + 12*a*b*c*d - 2*a^2*d^2)*Log[-(b^(1/3)*x
) + (a + b*x^3)^(1/3)])/(27*b^(2/3)*d^3) + ((I/6)*(I*c^(2/3)*(b*c - a*d)^(4/3) + Sqrt[3]*c^(2/3)*(b*c - a*d)^(
4/3))*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 - 12*a*b*c*d +
2*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^(2/3)*d^3) + ((c^(2/3)*(
b*c - a*d)^(4/3) - I*Sqrt[3]*c^(2/3)*(b*c - a*d)^(4/3))*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 = 2.05, size = 550, normalized size = 1.65 \begin {gather*} \frac {2 \, \sqrt {3} {\left (9 \, b^{3} c^{2} - 12 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \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 )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) - 18 \, \sqrt {3} {\left (b^{3} c - a b^{2} d\right )} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} \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 ) - 2 \, {\left (9 \, b^{2} c^{2} - 12 \, a b c d + 2 \, 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} - 12 \, a b c d + 2 \, 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 ) - 18 \, {\left (b^{3} c - a b^{2} d\right )} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} \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^{3} c - a b^{2} d\right )} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} \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 ) + 3 \, {\left (3 \, b^{3} d^{2} x^{5} - {\left (6 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*(2*sqrt(3)*(9*b^3*c^2 - 12*a*b^2*c*d + 2*a^2*b*d^2)*sqrt(-(-b^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-b^2)^(1/3)
*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x)) - 18*sqrt(3)*(b^3*c - a*b^2*d)*(
-b*c^3 + a*c^2*d)^(1/3)*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)) - 2*(9*b^2*c^2 - 12*a*b*c*d + 2*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 - 12*a*b*c*d + 2*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) - 18*(b^3*c - a*b^2*d)*(-b*c^3 + a*c^2*d)^(1/3)*log(((b*x^3
+ a)^(1/3)*c + (-b*c^3 + a*c^2*d)^(1/3)*x)/x) + 9*(b^3*c - a*b^2*d)*(-b*c^3 + a*c^2*d)^(1/3)*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) + 3*(3*b^3*d^
2*x^5 - (6*b^3*c*d - 7*a*b^2*d^2)*x^2)*(b*x^3 + a)^(1/3))/(b^2*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 {4}{3}} x^{4}}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} x^{4}}{d \,x^{3}+c}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^4*(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 \begin {gather*} \int \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}} x^{4}}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a)^(4/3)*x^4/(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^4\,{\left (b\,x^3+a\right )}^{4/3}}{d\,x^3+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x^4*(a + b*x^3)^(4/3))/(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^{4} \left (a + b x^{3}\right )^{\frac {4}{3}}}{c + d x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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