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

Optimal. Leaf size=277 \[ \frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 \sqrt [3]{c} d^2}+\frac {\sqrt [3]{b} (3 b c-4 a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 d^2}-\frac {(b c-a d)^{4/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} d^2}+\frac {\sqrt [3]{b} (3 b c-4 a d) \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{3 \sqrt {3} d^2}-\frac {(b c-a d)^{4/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} d^2}+\frac {b x^2 \sqrt [3]{a+b x^3}}{3 d} \]

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Rubi [C]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 0.23, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {511, 510} \begin {gather*} \frac {a x^2 \sqrt [3]{a+b x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c \sqrt [3]{\frac {b x^3}{a}+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x^2*(a + b*x^3)^(1/3)*AppellF1[2/3, -4/3, 1, 5/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*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 \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 \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^2 \sqrt [3]{a+b x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C]  time = 0.32, size = 198, normalized size = 0.71 \begin {gather*} \frac {2 b x^5 \left (\frac {b x^3}{a}+1\right )^{2/3} \left (\frac {d x^3}{c}+1\right )^{2/3} (4 a d-3 b c) 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 x^2 \left (a \left (\frac {b x^3}{a}+1\right )^{2/3} (3 a d-2 b c) \, _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 )+2 b c \left (a+b x^3\right ) \left (\frac {d x^3}{c}+1\right )^{2/3}\right )}{30 c d \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*(a + b*x^3)^(4/3))/(c + d*x^3),x]

[Out]

(2*b*(-3*b*c + 4*a*d)*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*x^2*(2*b*c*(a + b*x^3)*(1 + (d*x^3)/c)^(2/3) + a*(-2*b*c + 3*a*d)*(1 + (b*x^3)/a)^(2/3)*Hyp
ergeometric2F1[2/3, 2/3, 5/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))]))/(30*c*d*(a + b*x^3)^(2/3)*(1 + (d*x^3)/c
)^(2/3))

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IntegrateAlgebraic [C]  time = 7.07, size = 552, normalized size = 1.99 \begin {gather*} \frac {\left (3 b^{4/3} c-4 a \sqrt [3]{b} d\right ) \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{9 d^2}+\frac {\left (3 b^{4/3} c-4 a \sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{3 \sqrt {3} d^2}+\frac {\left (4 a \sqrt [3]{b} d-3 b^{4/3} c\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 )}{18 d^2}+\frac {i \left (\sqrt {3} (b c-a d)^{4/3}+i (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 \sqrt [3]{c} d^2}+\frac {\left ((b c-a d)^{4/3}-i \sqrt {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 \sqrt [3]{c} d^2}+\frac {\sqrt {-1-i \sqrt {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 )}{\sqrt {6} \sqrt [3]{c} d^2}+\frac {b x^2 \sqrt [3]{a+b x^3}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^2*(a + b*x^3)^(1/3))/(3*d) + ((3*b^(4/3)*c - 4*a*b^(1/3)*d)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a
+ b*x^3)^(1/3))])/(3*Sqrt[3]*d^2) + (Sqrt[-1 - I*Sqrt[3]]*(b*c - a*d)^(4/3)*ArcTan[(3*(b*c - a*d)^(1/3)*x)/(Sq
rt[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]*c^
(1/3)*d^2) + ((3*b^(4/3)*c - 4*a*b^(1/3)*d)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/(9*d^2) + (((b*c - a*d)^(4/
3) - I*Sqrt[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)])/(6*c
^(1/3)*d^2) + ((-3*b^(4/3)*c + 4*a*b^(1/3)*d)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3
)])/(18*d^2) + ((I/12)*(I*(b*c - a*d)^(4/3) + Sqrt[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)])/(c^(1/
3)*d^2)

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fricas [A]  time = 0.71, size = 396, normalized size = 1.43 \begin {gather*} \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b d x^{2} - 6 \, \sqrt {3} {\left (b c - a d\right )} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left (b c - a d\right )} x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c \left (\frac {b c - a d}{c}\right )^{\frac {2}{3}}}{3 \, {\left (b c - a d\right )} x}\right ) + 2 \, \sqrt {3} {\left (3 \, b c - 4 \, a d\right )} \left (-b\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}}}{3 \, b x}\right ) - 2 \, {\left (3 \, b c - 4 \, a d\right )} \left (-b\right )^{\frac {1}{3}} \log \left (\frac {\left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) - 6 \, {\left (b c - a d\right )} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \log \left (-\frac {x \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} - {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + {\left (3 \, b c - 4 \, a d\right )} \left (-b\right )^{\frac {1}{3}} \log \left (\frac {\left (-b\right )^{\frac {2}{3}} x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 3 \, {\left (b c - a d\right )} \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} \log \left (\frac {x^{2} \left (\frac {b c - a d}{c}\right )^{\frac {2}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} x \left (\frac {b c - a d}{c}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{18 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(6*(b*x^3 + a)^(1/3)*b*d*x^2 - 6*sqrt(3)*(b*c - a*d)*((b*c - a*d)/c)^(1/3)*arctan(-1/3*(sqrt(3)*(b*c - a*
d)*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*c*((b*c - a*d)/c)^(2/3))/((b*c - a*d)*x)) + 2*sqrt(3)*(3*b*c - 4*a*d)*(-b)^
(1/3)*arctan(1/3*(sqrt(3)*b*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-b)^(2/3))/(b*x)) - 2*(3*b*c - 4*a*d)*(-b)^(1/3)*
log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - 6*(b*c - a*d)*((b*c - a*d)/c)^(1/3)*log(-(x*((b*c - a*d)/c)^(1/3)
- (b*x^3 + a)^(1/3))/x) + (3*b*c - 4*a*d)*(-b)^(1/3)*log(((-b)^(2/3)*x^2 - (b*x^3 + a)^(1/3)*(-b)^(1/3)*x + (b
*x^3 + a)^(2/3))/x^2) + 3*(b*c - a*d)*((b*c - a*d)/c)^(1/3)*log((x^2*((b*c - a*d)/c)^(2/3) + (b*x^3 + a)^(1/3)
*x*((b*c - a*d)/c)^(1/3) + (b*x^3 + a)^(2/3))/x^2))/d^2

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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}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}} x}{d \,x^{3}+c}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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}{d x^{3} + c}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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