\(\int \frac {1}{(a+b x^3)^{5/3} (c+d x^3)^2} \, dx\) [107]

Optimal result

Integrand size = 21, antiderivative size = 62 \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{3},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^2 \left (a+b x^3\right )^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {441, 440} \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{3},\frac {5}{3},2,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^2 \left (a+b x^3\right )^{2/3}} \]

[In]

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

[Out]

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

Rule 440

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 441

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*} \text {integral}& = \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 )^2} \, 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},2;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a c^2 \left (a+b x^3\right )^{2/3}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(386\) vs. \(2(62)=124\).

Time = 10.53 (sec) , antiderivative size = 386, normalized size of antiderivative = 6.23 \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\frac {x \left (b d (3 b c+2 a d) x^3 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {c \left (16 a c \left (6 a^2 d^2+2 a b d \left (-6 c+d x^3\right )+3 b^2 c \left (2 c+d x^3\right )\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-4 x^3 \left (2 a^2 d^2+2 a b d^2 x^3+3 b^2 c \left (c+d x^3\right )\right ) \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (4 a c \operatorname {AppellF1}\left (\frac {1}{3},\frac {2}{3},1,\frac {4}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-x^3 \left (3 a d \operatorname {AppellF1}\left (\frac {4}{3},\frac {2}{3},2,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{3},1,\frac {7}{3},-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )\right )}\right )}{24 a c^2 (b c-a d)^2 \left (a+b x^3\right )^{2/3}} \]

[In]

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

[Out]

(x*(b*d*(3*b*c + 2*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*
(16*a*c*(6*a^2*d^2 + 2*a*b*d*(-6*c + d*x^3) + 3*b^2*c*(2*c + d*x^3))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a),
-((d*x^3)/c)] - 4*x^3*(2*a^2*d^2 + 2*a*b*d^2*x^3 + 3*b^2*c*(c + 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)])))/((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)])))))/(24*a*c^2*(b*c - a*d)^2*(a +
b*x^3)^(2/3))

Maple [F]

\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {5}{3}} \left (d \,x^{3}+c \right )^{2}}d x\]

[In]

int(1/(b*x^3+a)^(5/3)/(d*x^3+c)^2,x)

[Out]

int(1/(b*x^3+a)^(5/3)/(d*x^3+c)^2,x)

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\text {Timed out} \]

[In]

integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{\left (a + b x^{3}\right )^{\frac {5}{3}} \left (c + d x^{3}\right )^{2}}\, dx \]

[In]

integrate(1/(b*x**3+a)**(5/3)/(d*x**3+c)**2,x)

[Out]

Integral(1/((a + b*x**3)**(5/3)*(c + d*x**3)**2), x)

Maxima [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}^{2}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)^2), x)

Giac [F]

\[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {5}{3}} {\left (d x^{3} + c\right )}^{2}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(5/3)/(d*x^3+c)^2,x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(5/3)*(d*x^3 + c)^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b x^3\right )^{5/3} \left (c+d x^3\right )^2} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{5/3}\,{\left (d\,x^3+c\right )}^2} \,d x \]

[In]

int(1/((a + b*x^3)^(5/3)*(c + d*x^3)^2),x)

[Out]

int(1/((a + b*x^3)^(5/3)*(c + d*x^3)^2), x)