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

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

[Out]

b/(a*(b*c - a*d)*x^2*(a + b*x^3)^(1/3)) - ((3*b*c - a*d)*(a + b*x^3)^(2/3))/(2*a^2*c*(b*c - a*d)*x^2) + (d^2*A
rcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(Sqrt[3]*c^(5/3)*(b*c - a*d)^(4/3))
+ (d^2*Log[c + d*x^3])/(6*c^(5/3)*(b*c - a*d)^(4/3)) - (d^2*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1
/3)])/(2*c^(5/3)*(b*c - a*d)^(4/3))

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Rubi [C]  time = 1.28743, 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} \[ \frac{-9 c^2 x^6 (b c-a d)^2 \, _3F_2\left (2,2,\frac{7}{3};1,\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-9 d^2 x^{12} (b c-a d)^2 \, _3F_2\left (2,2,\frac{7}{3};1,\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-18 c d x^9 (b c-a d)^2 \, _3F_2\left (2,2,\frac{7}{3};1,\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-126 c^2 d^2 x^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+126 c^2 d^2 x^6 \left (a+b x^3\right )^2-15 c^2 x^6 (b c-a d)^2 \, _2F_1\left (2,\frac{7}{3};\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-168 c^3 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-28 c^4 \left (a+b x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+168 c^3 d x^3 \left (a+b x^3\right )^2+28 c^4 \left (a+b x^3\right )^2-27 d^2 x^{12} (b c-a d)^2 \, _2F_1\left (2,\frac{7}{3};\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-42 c d x^9 (b c-a d)^2 \, _2F_1\left (2,\frac{7}{3};\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )}{14 c^4 x^5 \left (a+b x^3\right )^{7/3} (b c-a d)} \]

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 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])

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])

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*}

Mathematica [C]  time = 0.686156, size = 542, normalized size = 2.37 \[ \frac{9 c^2 x^6 (b c-a d)^2 \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{3}\right \},\left \{1,\frac{10}{3}\right \},\frac{x^3 (b c-a d)}{c \left (a+b x^3\right )}\right )+9 d^2 x^{12} (b c-a d)^2 \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{3}\right \},\left \{1,\frac{10}{3}\right \},\frac{x^3 (b c-a d)}{c \left (a+b x^3\right )}\right )+18 c d x^9 (b c-a d)^2 \text{HypergeometricPFQ}\left (\left \{2,2,\frac{7}{3}\right \},\left \{1,\frac{10}{3}\right \},\frac{x^3 (b c-a d)}{c \left (a+b x^3\right )}\right )+126 c^2 d^2 x^6 \left (a+b x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-126 c^2 d^2 x^6 \left (a+b x^3\right )^2+15 c^2 x^6 (b c-a d)^2 \, _2F_1\left (2,\frac{7}{3};\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+168 c^3 d x^3 \left (a+b x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+28 c^4 \left (a+b x^3\right )^2 \, _2F_1\left (\frac{1}{3},1;\frac{4}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )-168 c^3 d x^3 \left (a+b x^3\right )^2-28 c^4 \left (a+b x^3\right )^2+27 d^2 x^{12} (b c-a d)^2 \, _2F_1\left (2,\frac{7}{3};\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )+42 c d x^9 (b c-a d)^2 \, _2F_1\left (2,\frac{7}{3};\frac{10}{3};\frac{(b c-a d) x^3}{c \left (b x^3+a\right )}\right )}{14 c^4 x^5 \left (a+b x^3\right )^{7/3} (a d-b c)} \]

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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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( d{x}^{3}+c \right ) } \left ( b{x}^{3}+a \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{4}{3}}{\left (d x^{3} + c\right )} x^{3}}\,{d x} \end{align*}

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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}^{\frac{4}{3}}{\left (d x^{3} + c\right )} x^{3}}\,{d x} \end{align*}

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)