3.5.90 \(\int \frac {1}{x^3 (a+b x^3)^{4/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=229 \[ -\frac {\left (a+b x^3\right )^{2/3} (3 b c-a d)}{2 a^2 c x^2 (b c-a d)}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{5/3} (b c-a d)^{4/3}}-\frac {d^2 \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{5/3} (b c-a d)^{4/3}}+\frac {d^2 \tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{5/3} (b c-a d)^{4/3}}+\frac {b}{a x^2 \sqrt [3]{a+b x^3} (b c-a d)} \]

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Rubi [C]  time = 1.29, antiderivative size = 542, normalized size of antiderivative = 2.37, 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 {-9 c^2 x^6 (b c-a d)^2 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-9 d^2 x^{12} (b c-a d)^2 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-18 c d x^9 (b c-a d)^2 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-126 c^2 d^2 x^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+126 c^2 d^2 x^6 \left (a+b x^3\right )^2-15 c^2 x^6 (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-168 c^3 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-28 c^4 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+168 c^3 d x^3 \left (a+b x^3\right )^2+28 c^4 \left (a+b x^3\right )^2-27 d^2 x^{12} (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-42 c d x^9 (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )}{14 c^4 x^5 \left (a+b x^3\right )^{7/3} (b c-a d)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[1/(x^3*(a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(28*c^4*(a + b*x^3)^2 + 168*c^3*d*x^3*(a + b*x^3)^2 + 126*c^2*d^2*x^6*(a + b*x^3)^2 - 28*c^4*(a + b*x^3)^2*Hyp
ergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 168*c^3*d*x^3*(a + b*x^3)^2*Hypergeometric2F1
[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 126*c^2*d^2*x^6*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/3
, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 15*c^2*(b*c - a*d)^2*x^6*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x
^3)/(c*(a + b*x^3))] - 42*c*d*(b*c - a*d)^2*x^9*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^
3))] - 27*d^2*(b*c - a*d)^2*x^12*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 9*c^2*(b
*c - a*d)^2*x^6*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 18*c*d*(b*c - a
*d)^2*x^9*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] - 9*d^2*(b*c - a*d)^2*x
^12*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(14*c^4*(b*c - a*d)*x^5*(a +
b*x^3)^(7/3))

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 {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx &=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \int \frac {1}{x^3 \left (1+\frac {b x^3}{a}\right )^{4/3} \left (c+d x^3\right )} \, dx}{a \sqrt [3]{a+b x^3}}\\ &=\frac {28 c^4 \left (a+b x^3\right )^2+168 c^3 d x^3 \left (a+b x^3\right )^2+126 c^2 d^2 x^6 \left (a+b x^3\right )^2-28 c^4 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-168 c^3 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-126 c^2 d^2 x^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-15 c^2 (b c-a d)^2 x^6 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-42 c d (b c-a d)^2 x^9 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-27 d^2 (b c-a d)^2 x^{12} \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-9 c^2 (b c-a d)^2 x^6 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-18 c d (b c-a d)^2 x^9 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )-9 d^2 (b c-a d)^2 x^{12} \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (a+b x^3\right )}\right )}{14 c^4 (b c-a d) x^5 \left (a+b x^3\right )^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.83, size = 542, normalized size = 2.37 \begin {gather*} \frac {9 c^2 x^6 (b c-a d)^2 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+9 d^2 x^{12} (b c-a d)^2 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+18 c d x^9 (b c-a d)^2 \, _3F_2\left (2,2,\frac {7}{3};1,\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+28 c^4 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-28 c^4 \left (a+b x^3\right )^2+168 c^3 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-168 c^3 d x^3 \left (a+b x^3\right )^2+126 c^2 d^2 x^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-126 c^2 d^2 x^6 \left (a+b x^3\right )^2+15 c^2 x^6 (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+27 d^2 x^{12} (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+42 c d x^9 (b c-a d)^2 \, _2F_1\left (2,\frac {7}{3};\frac {10}{3};\frac {(b c-a d) x^3}{c \left (b x^3+a\right )}\right )}{14 c^4 x^5 \left (a+b x^3\right )^{7/3} (a d-b c)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^3*(a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(-28*c^4*(a + b*x^3)^2 - 168*c^3*d*x^3*(a + b*x^3)^2 - 126*c^2*d^2*x^6*(a + b*x^3)^2 + 28*c^4*(a + b*x^3)^2*Hy
pergeometric2F1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 168*c^3*d*x^3*(a + b*x^3)^2*Hypergeometric2F
1[1/3, 1, 4/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 126*c^2*d^2*x^6*(a + b*x^3)^2*Hypergeometric2F1[1/3, 1, 4/
3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 15*c^2*(b*c - a*d)^2*x^6*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*
x^3)/(c*(a + b*x^3))] + 42*c*d*(b*c - a*d)^2*x^9*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x
^3))] + 27*d^2*(b*c - a*d)^2*x^12*Hypergeometric2F1[2, 7/3, 10/3, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 9*c^2*(
b*c - a*d)^2*x^6*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 18*c*d*(b*c -
a*d)^2*x^9*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))] + 9*d^2*(b*c - a*d)^2*
x^12*HypergeometricPFQ[{2, 2, 7/3}, {1, 10/3}, ((b*c - a*d)*x^3)/(c*(a + b*x^3))])/(14*c^4*(-(b*c) + a*d)*x^5*
(a + b*x^3)^(7/3))

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IntegrateAlgebraic [C]  time = 3.19, size = 397, normalized size = 1.73 \begin {gather*} \frac {-a^2 d+a b c-a b d x^3+3 b^2 c x^3}{2 a^2 c x^2 \sqrt [3]{a+b x^3} (a d-b c)}+\frac {\left (d^2+i \sqrt {3} d^2\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^{5/3} (b c-a d)^{4/3}}-\frac {\sqrt {\frac {1}{6} \left (-1+i \sqrt {3}\right )} d^2 \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^{5/3} (b c-a d)^{4/3}}-\frac {i \left (\sqrt {3} d^2-i d^2\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^{5/3} (b c-a d)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^3*(a + b*x^3)^(4/3)*(c + d*x^3)),x]

[Out]

(a*b*c - a^2*d + 3*b^2*c*x^3 - a*b*d*x^3)/(2*a^2*c*(-(b*c) + a*d)*x^2*(a + b*x^3)^(1/3)) - (Sqrt[(-1 + I*Sqrt[
3])/6]*d^2*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) - Sqr
t[3]*c^(1/3)*(a + b*x^3)^(1/3))])/(c^(5/3)*(b*c - a*d)^(4/3)) + ((d^2 + I*Sqrt[3]*d^2)*Log[2*(b*c - a*d)^(1/3)
*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(6*c^(5/3)*(b*c - a*d)^(4/3)) - ((I/12)*((-I)*d^2 + Sqrt[3]*d
^2)*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 + Sq
rt[3])*c^(2/3)*(a + b*x^3)^(2/3)])/(c^(5/3)*(b*c - a*d)^(4/3))

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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(1/x^3/(b*x^3+a)^(4/3)/(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 {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)*x^3), x)

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {4}{3}} \left (d \,x^{3}+c \right ) x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x)

[Out]

int(1/x^3/(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 {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^3+a)^(4/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(4/3)*(d*x^3 + c)*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^3)^(4/3)*(c + d*x^3)),x)

[Out]

int(1/(x^3*(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 {1}{x^{3} \left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**3+a)**(4/3)/(d*x**3+c),x)

[Out]

Integral(1/(x**3*(a + b*x**3)**(4/3)*(c + d*x**3)), x)

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