3.120 \(\int \frac{1}{(a+b x^3)^{5/3} (c+d x^3)^3} \, dx\)

Optimal. Leaf size=62 \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{1}{3};\frac{5}{3},3;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a c^3 \left (a+b x^3\right )^{2/3}} \]

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 5/3, 3, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*c^3*(a + b*x^3)^(2/3))

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Rubi [A]  time = 0.0291526, 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} \[ \frac{x \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{1}{3};\frac{5}{3},3;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a c^3 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(5/3)*(c + d*x^3)^3),x]

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*AppellF1[1/3, 5/3, 3, 4/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*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 )^{5/3} \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 )^{5/3} \left (c+d x^3\right )^3} \, dx}{a \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{5}{3},3;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a c^3 \left (a+b x^3\right )^{2/3}}\\ \end{align*}

Mathematica [B]  time = 0.916664, size = 531, normalized size = 8.56 \[ \frac{\frac{b d x^4 \left (\frac{b x^3}{a}+1\right )^{2/3} \left (5 a^2 d^2-16 a b c d-9 b^2 c^2\right ) F_1\left (\frac{4}{3};\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{(a d-b c)^3}-\frac{4 c \left (x^4 \left (a^2 b d^2 \left (19 c^2+8 c d x^3-5 d^2 x^6\right )-a^3 d^3 \left (8 c+5 d x^3\right )+a b^2 c d^2 x^3 \left (19 c+16 d x^3\right )+9 b^3 c^2 \left (c+d x^3\right )^2\right ) \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )+4 a c x \left (a^2 b d^2 \left (-54 c^2-43 c d x^3+5 d^2 x^6\right )+3 a^3 d^3 \left (6 c+5 d x^3\right )+a b^2 c d \left (54 c^2+35 c d x^3-16 d^2 x^6\right )-9 b^3 c^2 \left (2 c^2+3 c d x^3+d^2 x^6\right )\right ) F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )}{\left (c+d x^3\right )^2 (b c-a d)^3 \left (4 a c F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )-x^3 \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )\right )}}{72 a c^3 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^3)^(5/3)*(c + d*x^3)^3),x]

[Out]

((b*d*(-9*b^2*c^2 - 16*a*b*c*d + 5*a^2*d^2)*x^4*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a),
-((d*x^3)/c)])/(-(b*c) + a*d)^3 - (4*c*(4*a*c*x*(3*a^3*d^3*(6*c + 5*d*x^3) + a*b^2*c*d*(54*c^2 + 35*c*d*x^3 -
16*d^2*x^6) - 9*b^3*c^2*(2*c^2 + 3*c*d*x^3 + d^2*x^6) + a^2*b*d^2*(-54*c^2 - 43*c*d*x^3 + 5*d^2*x^6))*AppellF
1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + x^4*(9*b^3*c^2*(c + d*x^3)^2 - a^3*d^3*(8*c + 5*d*x^3) + a*b
^2*c*d^2*x^3*(19*c + 16*d*x^3) + a^2*b*d^2*(19*c^2 + 8*c*d*x^3 - 5*d^2*x^6))*(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)])))/((b*c - a*d)^3*
(c + d*x^3)^2*(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)]))))/(72*a*c^3
*(a + b*x^3)^(2/3))

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Maple [F]  time = 0.429, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( d{x}^{3}+c \right ) ^{3}} \left ( b{x}^{3}+a \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(5/3)/(d*x^3+c)^3,x)

[Out]

int(1/(b*x^3+a)^(5/3)/(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{5}{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)^(5/3)/(d*x^3+c)^3,x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(5/3)*(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)^(5/3)/(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)**(5/3)/(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{5}{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)^(5/3)/(d*x^3+c)^3,x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)^3), x)