\(\int \frac {1}{\sqrt [4]{a+b x^3} (c+d x^3)^{13/12}} \, dx\) [126]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\frac {x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{c \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}} \]

[Out]

x*(c*(b*x^3+a)/a/(d*x^3+c))^(1/4)*hypergeom([1/4, 1/3],[4/3],-(-a*d+b*c)*x^3/a/(d*x^3+c))/c/(b*x^3+a)^(1/4)/(d
*x^3+c)^(1/12)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {388} \[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\frac {x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{c \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}} \]

[In]

Int[1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)),x]

[Out]

(x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(1/4)*Hypergeometric2F1[1/4, 1/3, 4/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)
))])/(c*(a + b*x^3)^(1/4)*(c + d*x^3)^(1/12))

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(c*(c*((a
+ b*x^n)/(a*(c + d*x^n))))^p*(c + d*x^n)^(1/n + p)))*Hypergeometric2F1[1/n, -p, 1 + 1/n, (-(b*c - a*d))*(x^n/(
a*(c + d*x^n)))], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x \sqrt [4]{\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}} \, _2F_1\left (\frac {1}{4},\frac {1}{3};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{c \sqrt [4]{a+b x^3} \sqrt [12]{c+d x^3}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 3.50 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\frac {x \sqrt [4]{1+\frac {b x^3}{a}} \left (1+\frac {d x^3}{c}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \]

[In]

Integrate[1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)),x]

[Out]

(x*(1 + (b*x^3)/a)^(1/4)*(1 + (d*x^3)/c)^(3/4)*Hypergeometric2F1[1/4, 1/3, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d
*x^3))])/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12))

Maple [F]

\[\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{4}} \left (d \,x^{3}+c \right )^{\frac {13}{12}}}d x\]

[In]

int(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x)

[Out]

int(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x)

Fricas [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{4}} {\left (d x^{3} + c\right )}^{\frac {13}{12}}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(3/4)*(d*x^3 + c)^(11/12)/(b*d^2*x^9 + (2*b*c*d + a*d^2)*x^6 + (b*c^2 + 2*a*c*d)*x^3 + a*
c^2), x)

Sympy [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\int \frac {1}{\sqrt [4]{a + b x^{3}} \left (c + d x^{3}\right )^{\frac {13}{12}}}\, dx \]

[In]

integrate(1/(b*x**3+a)**(1/4)/(d*x**3+c)**(13/12),x)

[Out]

Integral(1/((a + b*x**3)**(1/4)*(c + d*x**3)**(13/12)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{4}} {\left (d x^{3} + c\right )}^{\frac {13}{12}}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x, algorithm="maxima")

[Out]

integrate(1/((b*x^3 + a)^(1/4)*(d*x^3 + c)^(13/12)), x)

Giac [F]

\[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{4}} {\left (d x^{3} + c\right )}^{\frac {13}{12}}} \,d x } \]

[In]

integrate(1/(b*x^3+a)^(1/4)/(d*x^3+c)^(13/12),x, algorithm="giac")

[Out]

integrate(1/((b*x^3 + a)^(1/4)*(d*x^3 + c)^(13/12)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{a+b x^3} \left (c+d x^3\right )^{13/12}} \, dx=\int \frac {1}{{\left (b\,x^3+a\right )}^{1/4}\,{\left (d\,x^3+c\right )}^{13/12}} \,d x \]

[In]

int(1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)),x)

[Out]

int(1/((a + b*x^3)^(1/4)*(c + d*x^3)^(13/12)), x)