3.97 \(\int \frac{1}{(a+b x^3)^{8/3} (c+d x^3)} \, dx\)

Optimal. Leaf size=62 \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{1}{3};\frac{8}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2 c \left (a+b x^3\right )^{2/3}} \]

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 8/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a^2*c*(a + b*x^3)^(2/3))

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Rubi [A]  time = 0.0288902, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {430, 429} \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{1}{3};\frac{8}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2 c \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(8/3)*(c + d*x^3)),x]

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 8/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a^2*c*(a + b*x^3)^(2/3))

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^3\right )^{8/3} \left (c+d x^3\right )} \, dx &=\frac{\left (1+\frac{b x^3}{a}\right )^{2/3} \int \frac{1}{\left (1+\frac{b x^3}{a}\right )^{8/3} \left (c+d x^3\right )} \, dx}{a^2 \left (a+b x^3\right )^{2/3}}\\ &=\frac{x \left (1+\frac{b x^3}{a}\right )^{2/3} F_1\left (\frac{1}{3};\frac{8}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2 c \left (a+b x^3\right )^{2/3}}\\ \end{align*}

Mathematica [B]  time = 0.7847, size = 429, normalized size = 6.92 \[ -\frac{x \left (\frac{4 \left (4 a c \left (-a^2 b d \left (20 c+d x^3\right )+10 a^3 d^2+a b^2 \left (10 c^2-12 c d x^3-9 d^2 x^6\right )+4 b^3 c x^3 \left (2 c+d x^3\right )\right ) F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+b x^3 \left (c+d x^3\right ) \left (11 a^2 d+a b \left (9 d x^3-6 c\right )-4 b^2 c x^3\right ) \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )\right )}{\left (a+b x^3\right ) \left (c+d x^3\right ) \left (x^3 \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )-4 a c F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )}+\frac{b d x^3 \left (\frac{b x^3}{a}+1\right )^{2/3} (9 a d-4 b c) F_1\left (\frac{4}{3};\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c}\right )}{40 a^2 \left (a+b x^3\right )^{2/3} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^3)^(8/3)*(c + d*x^3)),x]

[Out]

-(x*((b*d*(-4*b*c + 9*a*d)*x^3*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/c
+ (4*(4*a*c*(10*a^3*d^2 + 4*b^3*c*x^3*(2*c + d*x^3) - a^2*b*d*(20*c + d*x^3) + a*b^2*(10*c^2 - 12*c*d*x^3 - 9
*d^2*x^6))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + b*x^3*(c + d*x^3)*(11*a^2*d - 4*b^2*c*x^3
+ a*b*(-6*c + 9*d*x^3))*(3*a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3, 5/
3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))/((a + b*x^3)*(c + d*x^3)*(-4*a*c*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3
)/a), -((d*x^3)/c)] + x^3*(3*a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3,
5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)])))))/(40*a^2*(b*c - a*d)^2*(a + b*x^3)^(2/3))

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Maple [F]  time = 0.408, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{d{x}^{3}+c} \left ( b{x}^{3}+a \right ) ^{-{\frac{8}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(8/3)/(d*x^3+c),x)

[Out]

int(1/(b*x^3+a)^(8/3)/(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{8}{3}}{\left (d x^{3} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(8/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(8/3)*(d*x^3 + c)), 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/(b*x^3+a)^(8/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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/(b*x**3+a)**(8/3)/(d*x**3+c),x)

[Out]

Timed out

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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{8}{3}}{\left (d x^{3} + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(8/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(8/3)*(d*x^3 + c)), x)