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

Optimal. Leaf size=254 \[ -\frac {b^{4/3} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d}-\frac {b^{4/3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} d}-\frac {(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^{4/3} d}+\frac {(b c-a d)^{4/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{4/3} d}+\frac {(b c-a d)^{4/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{4/3} d}-\frac {a \sqrt [3]{a+b x^3}}{c x} \]

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Rubi [C]  time = 0.06, antiderivative size = 63, normalized size of antiderivative = 0.25, 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 \sqrt [3]{a+b x^3} F_1\left (-\frac {1}{3};-\frac {4}{3},1;\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c x \sqrt [3]{\frac {b x^3}{a}+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.55, size = 161, normalized size = 0.63 \begin {gather*} \frac {2 b^2 c x^6 \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-\frac {5 a x^3 \left (\frac {b x^3}{a}+1\right )^{2/3} (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 )}{\left (\frac {d x^3}{c}+1\right )^{2/3}}-10 a c \left (a+b x^3\right )}{10 c^2 x \left (a+b x^3\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-10*a*c*(a + b*x^3) + 2*b^2*c*x^6*(1 + (b*x^3)/a)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -((b*x^3)/a), -((d*x^3)/c)
] - (5*a*(-2*b*c + a*d)*x^3*(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))])/(1 + (d*x^3)/c)^(2/3))/(10*c^2*x*(a + b*x^3)^(2/3))

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IntegrateAlgebraic [C]  time = 5.54, size = 509, normalized size = 2.00 \begin {gather*} -\frac {b^{4/3} \log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{3 d}-\frac {b^{4/3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2 \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{\sqrt {3} d}+\frac {b^{4/3} \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 )}{6 d}+\frac {i \left (\sqrt {3} (b c-a d)^{4/3}+i (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 c^{4/3} d}-\frac {\sqrt {\frac {1}{6} \left (-1-i \sqrt {3}\right )} (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 )}{c^{4/3} d}+\frac {\left ((b c-a d)^{4/3}-i \sqrt {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 c^{4/3} d}-\frac {a \sqrt [3]{a+b x^3}}{c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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}}}{{\left (d x^{3} + c\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}}}{\left (d \,x^{3}+c \right ) x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((b*x^3+a)^(4/3)/x^2/(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}}}{{\left (d x^{3} + c\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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