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

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

[Out]

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

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Rubi [A]  time = 0.0289201, 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} \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{1}{3};\frac{8}{3},3;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2 c^3 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 8/3, 3, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a^2*c^3*(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 )^3} \, 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 )^3} \, 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},3;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a^2 c^3 \left (a+b x^3\right )^{2/3}}\\ \end{align*}

Mathematica [B]  time = 1.63346, size = 515, normalized size = 8.31 \[ \frac{x \left (b d x^3 \left (\frac{b x^3}{a}+1\right )^{2/3} \left (-110 a^2 b c d^2+25 a^3 d^3-171 a b^2 c^2 d+36 b^3 c^3\right ) F_1\left (\frac{4}{3};\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+\frac{4 c \left (\frac{4 a c \left (c+d x^3\right ) \left (540 a^2 b^2 c^2 d^2-235 a^3 b c d^3+50 a^4 d^4-171 a b^3 c^3 d+36 b^4 c^4\right ) F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{4 a c F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )-x^3 \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )}+\frac{5 a^3 b^2 d^3 x^3 \left (-50 c^2-36 c d x^3+5 d^2 x^6\right )-a^2 b^3 c d \left (378 c^2 d x^3+189 c^3+314 c d^2 x^6+110 d^3 x^9\right )+5 a^4 b d^3 \left (-25 c^2-6 c d x^3+10 d^2 x^6\right )+5 a^5 d^4 \left (8 c+5 d x^3\right )+9 a b^4 c^2 \left (6 c-19 d x^3\right ) \left (c+d x^3\right )^2+36 b^5 c^3 x^3 \left (c+d x^3\right )^2}{a+b x^3}\right )}{\left (c+d x^3\right )^2}\right )}{360 a^2 c^3 \left (a+b x^3\right )^{2/3} (b c-a d)^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(b*d*(36*b^3*c^3 - 171*a*b^2*c^2*d - 110*a^2*b*c*d^2 + 25*a^3*d^3)*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)] + (4*c*((36*b^5*c^3*x^3*(c + d*x^3)^2 + 9*a*b^4*c^2*(6*c - 19*d*x^3)*
(c + d*x^3)^2 + 5*a^5*d^4*(8*c + 5*d*x^3) + 5*a^3*b^2*d^3*x^3*(-50*c^2 - 36*c*d*x^3 + 5*d^2*x^6) + 5*a^4*b*d^3
*(-25*c^2 - 6*c*d*x^3 + 10*d^2*x^6) - a^2*b^3*c*d*(189*c^3 + 378*c^2*d*x^3 + 314*c*d^2*x^6 + 110*d^3*x^9))/(a
+ b*x^3) + (4*a*c*(36*b^4*c^4 - 171*a*b^3*c^3*d + 540*a^2*b^2*c^2*d^2 - 235*a^3*b*c*d^3 + 50*a^4*d^4)*(c + d*x
^3)*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(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)]))))/(c + d*x^3)^2))/(360*a^2*c^3*(b*c - a*d)^4*(a + b*x^3)^(2/3))

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Maple [F]  time = 0.45, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d{x}^{3}+c \right ) ^{3}} \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)^3,x)

[Out]

int(1/(b*x^3+a)^(8/3)/(d*x^3+c)^3,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 )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

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

[Out]

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