\(\int \frac {\sqrt [4]{a+b x^3}}{(c+d x^3)^{19/12}} \, dx\) [125]

Optimal result

Integrand size = 23, antiderivative size = 122 \[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\frac {4 x \sqrt [4]{a+b x^3}}{7 c \left (c+d x^3\right )^{7/12}}+\frac {3 a x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{7 c^2 \left (a+b x^3\right )^{3/4}} \]

[Out]

4/7*x*(b*x^3+a)^(1/4)/c/(d*x^3+c)^(7/12)+3/7*a*x*(c*(b*x^3+a)/a/(d*x^3+c))^(3/4)*(d*x^3+c)^(5/12)*hypergeom([1
/3, 3/4],[4/3],-(-a*d+b*c)*x^3/a/(d*x^3+c))/c^2/(b*x^3+a)^(3/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {386, 388} \[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\frac {3 a x \left (c+d x^3\right )^{5/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{7 c^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \sqrt [4]{a+b x^3}}{7 c \left (c+d x^3\right )^{7/12}} \]

[In]

Int[(a + b*x^3)^(1/4)/(c + d*x^3)^(19/12),x]

[Out]

(4*x*(a + b*x^3)^(1/4))/(7*c*(c + d*x^3)^(7/12)) + (3*a*x*((c*(a + b*x^3))/(a*(c + d*x^3)))^(3/4)*(c + d*x^3)^
(5/12)*Hypergeometric2F1[1/3, 3/4, 4/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)))])/(7*c^2*(a + b*x^3)^(3/4))

Rule 386

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

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 {4 x \sqrt [4]{a+b x^3}}{7 c \left (c+d x^3\right )^{7/12}}+\frac {(3 a) \int \frac {1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{7 c} \\ & = \frac {4 x \sqrt [4]{a+b x^3}}{7 c \left (c+d x^3\right )^{7/12}}+\frac {3 a x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \, _2F_1\left (\frac {1}{3},\frac {3}{4};\frac {4}{3};-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{7 c^2 \left (a+b x^3\right )^{3/4}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 3.78 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\frac {x \sqrt [4]{a+b x^3} \sqrt [4]{1+\frac {d x^3}{c}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{3},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{c \sqrt [4]{1+\frac {b x^3}{a}} \left (c+d x^3\right )^{7/12}} \]

[In]

Integrate[(a + b*x^3)^(1/4)/(c + d*x^3)^(19/12),x]

[Out]

(x*(a + b*x^3)^(1/4)*(1 + (d*x^3)/c)^(1/4)*Hypergeometric2F1[-1/4, 1/3, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^
3))])/(c*(1 + (b*x^3)/a)^(1/4)*(c + d*x^3)^(7/12))

Maple [F]

\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{4}}}{\left (d \,x^{3}+c \right )^{\frac {19}{12}}}d x\]

[In]

int((b*x^3+a)^(1/4)/(d*x^3+c)^(19/12),x)

[Out]

int((b*x^3+a)^(1/4)/(d*x^3+c)^(19/12),x)

Fricas [F]

\[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{4}}}{{\left (d x^{3} + c\right )}^{\frac {19}{12}}} \,d x } \]

[In]

integrate((b*x^3+a)^(1/4)/(d*x^3+c)^(19/12),x, algorithm="fricas")

[Out]

integral((b*x^3 + a)^(1/4)*(d*x^3 + c)^(5/12)/(d^2*x^6 + 2*c*d*x^3 + c^2), x)

Sympy [F]

\[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\int \frac {\sqrt [4]{a + b x^{3}}}{\left (c + d x^{3}\right )^{\frac {19}{12}}}\, dx \]

[In]

integrate((b*x**3+a)**(1/4)/(d*x**3+c)**(19/12),x)

[Out]

Integral((a + b*x**3)**(1/4)/(c + d*x**3)**(19/12), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{4}}}{{\left (d x^{3} + c\right )}^{\frac {19}{12}}} \,d x } \]

[In]

integrate((b*x^3+a)^(1/4)/(d*x^3+c)^(19/12),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/4)/(d*x^3 + c)^(19/12), x)

Giac [F]

\[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{4}}}{{\left (d x^{3} + c\right )}^{\frac {19}{12}}} \,d x } \]

[In]

integrate((b*x^3+a)^(1/4)/(d*x^3+c)^(19/12),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/4)/(d*x^3 + c)^(19/12), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{a+b x^3}}{\left (c+d x^3\right )^{19/12}} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/4}}{{\left (d\,x^3+c\right )}^{19/12}} \,d x \]

[In]

int((a + b*x^3)^(1/4)/(c + d*x^3)^(19/12),x)

[Out]

int((a + b*x^3)^(1/4)/(c + d*x^3)^(19/12), x)