3.5.52 \(\int \frac {(a+b x^3)^{4/3}}{x^8 (c+d x^3)} \, dx\)

Optimal. Leaf size=250 \[ -\frac {\sqrt [3]{a+b x^3} \left (28 a^2 d^2-35 a b c d+4 b^2 c^2\right )}{28 a c^3 x}-\frac {d (b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^{10/3}}+\frac {d (b c-a d)^{4/3} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{10/3}}+\frac {d (b c-a d)^{4/3} \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{10/3}}-\frac {\sqrt [3]{a+b x^3} (8 b c-7 a d)}{28 c^2 x^4}-\frac {a \sqrt [3]{a+b x^3}}{7 c x^7} \]

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Rubi [C]  time = 0.51, antiderivative size = 169, normalized size of antiderivative = 0.68, 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 {12 c x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) (b c-a d) \, _2F_1\left (-\frac {1}{3},2;\frac {2}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-\left (4 c-3 d x^3\right ) \left (c \left (a+b x^3\right ) \left (a \left (c-4 d x^3\right )+5 b c x^3\right )-2 x^6 (b c-a d)^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )\right )}{28 c^4 x^7 \left (a+b x^3\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(4/3)/(x^8*(c + d*x^3)),x]

[Out]

(12*c*(b*c - a*d)*x^3*(a + b*x^3)*(c + d*x^3)*Hypergeometric2F1[-1/3, 2, 2/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3)
)] - (4*c - 3*d*x^3)*(c*(a + b*x^3)*(5*b*c*x^3 + a*(c - 4*d*x^3)) - 2*(b*c - a*d)^2*x^6*Hypergeometric2F1[2/3,
1, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))]))/(28*c^4*x^7*(a + b*x^3)^(2/3))

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^8 \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^8 \left (c+d x^3\right )} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {12 c (b c-a d) x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \, _2F_1\left (-\frac {1}{3},2;\frac {2}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-\left (4 c-3 d x^3\right ) \left (c \left (a+b x^3\right ) \left (5 b c x^3+a \left (c-4 d x^3\right )\right )-2 (b c-a d)^2 x^6 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )\right )}{28 c^4 x^7 \left (a+b x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.60, size = 179, normalized size = 0.72 \begin {gather*} -\frac {a \left (\frac {b x^3}{a}+1\right ) \left (12 c x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) (a d-b c) \, _2F_1\left (-\frac {1}{3},2;\frac {2}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+\left (4 c-3 d x^3\right ) \left (c \left (a+b x^3\right ) \left (a \left (c-4 d x^3\right )+5 b c x^3\right )-2 x^6 (b c-a d)^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )\right )\right )}{28 c^4 x^7 \left (a+b x^3\right )^{5/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(4/3)/(x^8*(c + d*x^3)),x]

[Out]

-1/28*(a*(1 + (b*x^3)/a)*(12*c*(-(b*c) + a*d)*x^3*(a + b*x^3)*(c + d*x^3)*Hypergeometric2F1[-1/3, 2, 2/3, ((b*
c - a*d)*x^3)/(c*(a + b*x^3))] + (4*c - 3*d*x^3)*(c*(a + b*x^3)*(5*b*c*x^3 + a*(c - 4*d*x^3)) - 2*(b*c - a*d)^
2*x^6*Hypergeometric2F1[2/3, 1, 5/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])))/(c^4*x^7*(a + b*x^3)^(5/3))

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IntegrateAlgebraic [C]  time = 3.20, size = 434, normalized size = 1.74 \begin {gather*} \frac {\sqrt [3]{a+b x^3} \left (-4 a^2 c^2+7 a^2 c d x^3-28 a^2 d^2 x^6-8 a b c^2 x^3+35 a b c d x^6-4 b^2 c^2 x^6\right )}{28 a c^3 x^7}+\frac {i \left (\sqrt {3} d (b c-a d)^{4/3}+i d (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^{10/3}}-\frac {\sqrt {\frac {1}{6} \left (-1-i \sqrt {3}\right )} d (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^{10/3}}+\frac {\left (d (b c-a d)^{4/3}-i \sqrt {3} d (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^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^3)^(4/3)/(x^8*(c + d*x^3)),x]

[Out]

((a + b*x^3)^(1/3)*(-4*a^2*c^2 - 8*a*b*c^2*x^3 + 7*a^2*c*d*x^3 - 4*b^2*c^2*x^6 + 35*a*b*c*d*x^6 - 28*a^2*d^2*x
^6))/(28*a*c^3*x^7) - (Sqrt[(-1 - I*Sqrt[3])/6]*d*(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^(10/3) + ((I/6)*(
I*d*(b*c - a*d)^(4/3) + Sqrt[3]*d*(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^(10/3) + ((d*(b*c - a*d)^(4/3) - I*Sqrt[3]*d*(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^(10/3))

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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^8/(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^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^8), 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^{8}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(4/3)/x^8/(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^{8}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^8/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^8), 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^8\,\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^8*(c + d*x^3)),x)

[Out]

int((a + b*x^3)^(4/3)/(x^8*(c + d*x^3)), x)

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sympy [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**8/(d*x**3+c),x)

[Out]

Timed out

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