3.119 \(\int \frac {1}{(a+b x^3)^{2/3} (c+d x^3)^3} \, dx\)

Optimal. Leaf size=59 \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \left (a+b x^3\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, 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} \[ \frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} F_1\left (\frac {1}{3};\frac {2}{3},3;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c^3 \left (a+b x^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^3)^(2/3)*(c + d*x^3)^3),x]

[Out]

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

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])

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])

Rubi steps

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

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Mathematica [B]  time = 0.77, size = 442, normalized size = 7.49 \[ \frac {x \left (\frac {4 c \left (d x^3 \left (a^2 d \left (8 c+5 d x^3\right )+a b \left (-13 c^2-2 c d x^3+5 d^2 x^6\right )-b^2 c x^3 \left (13 c+10 d x^3\right )\right ) \left (3 a d F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )-4 a c \left (3 a^2 d^2 \left (6 c+5 d x^3\right )+a b d \left (-36 c^2-25 c d x^3+5 d^2 x^6\right )+b^2 c \left (18 c^2+5 c d x^3-10 d^2 x^6\right )\right ) F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}{\left (c+d x^3\right )^2 \left (x^3 \left (3 a d F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )-4 a c F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}+5 b d x^3 \left (\frac {b x^3}{a}+1\right )^{2/3} (a d-2 b c) F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}{72 c^3 \left (a+b x^3\right )^{2/3} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*x^3)^(2/3)*(c + d*x^3)^3),x]

[Out]

(x*(5*b*d*(-2*b*c + 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)] + (4
*c*(-4*a*c*(3*a^2*d^2*(6*c + 5*d*x^3) + b^2*c*(18*c^2 + 5*c*d*x^3 - 10*d^2*x^6) + a*b*d*(-36*c^2 - 25*c*d*x^3
+ 5*d^2*x^6))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + d*x^3*(a^2*d*(8*c + 5*d*x^3) - b^2*c*x^
3*(13*c + 10*d*x^3) + a*b*(-13*c^2 - 2*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)])))/((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*c^3*(b*c - a*d)^2*(a + b*x^3)^(2/
3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(2/3)/(d*x^3+c)^3,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(2/3)/(d*x^3+c)^3,x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*(d*x^3 + c)^3), x)

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maple [F]  time = 0.59, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (d \,x^{3}+c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^3+a)^(2/3)/(d*x^3+c)^3,x)

[Out]

int(1/(b*x^3+a)^(2/3)/(d*x^3+c)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} {\left (d x^{3} + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^3+a)^(2/3)/(d*x^3+c)^3,x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(2/3)*(d*x^3 + c)^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (b\,x^3+a\right )}^{2/3}\,{\left (d\,x^3+c\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^3)^(2/3)*(c + d*x^3)^3),x)

[Out]

int(1/((a + b*x^3)^(2/3)*(c + d*x^3)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{3}\right )^{\frac {2}{3}} \left (c + d x^{3}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**3+a)**(2/3)/(d*x**3+c)**3,x)

[Out]

Integral(1/((a + b*x**3)**(2/3)*(c + d*x**3)**3), x)

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