\(\int \frac {(c+d x^3)^{5/12}}{(a+b x^3)^{7/4}} \, dx\) [129]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\frac {4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}+\frac {5 x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{9 a \left (a+b x^3\right )^{3/4}} \]

[Out]

4/9*x*(d*x^3+c)^(5/12)/a/(b*x^3+a)^(3/4)+5/9*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))/a/(b*x^3+a)^(3/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 121, 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} \[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\frac {5 x \left (c+d x^3\right )^{5/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{9 a \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}} \]

[In]

Int[(c + d*x^3)^(5/12)/(a + b*x^3)^(7/4),x]

[Out]

(4*x*(c + d*x^3)^(5/12))/(9*a*(a + b*x^3)^(3/4)) + (5*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)))])/(9*a*(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 \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}+\frac {(5 c) \int \frac {1}{\left (a+b x^3\right )^{3/4} \left (c+d x^3\right )^{7/12}} \, dx}{9 a} \\ & = \frac {4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}+\frac {5 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 )}{9 a \left (a+b x^3\right )^{3/4}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 5.54 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.74 \[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{4},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{a \left (a+b x^3\right )^{3/4} \left (1+\frac {d x^3}{c}\right )^{3/4}} \]

[In]

Integrate[(c + d*x^3)^(5/12)/(a + b*x^3)^(7/4),x]

[Out]

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

Maple [F]

\[\int \frac {\left (d \,x^{3}+c \right )^{\frac {5}{12}}}{\left (b \,x^{3}+a \right )^{\frac {7}{4}}}d x\]

[In]

int((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x)

[Out]

int((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x)

Fricas [F]

\[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {5}{12}}}{{\left (b x^{3} + a\right )}^{\frac {7}{4}}} \,d x } \]

[In]

integrate((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\int \frac {\left (c + d x^{3}\right )^{\frac {5}{12}}}{\left (a + b x^{3}\right )^{\frac {7}{4}}}\, dx \]

[In]

integrate((d*x**3+c)**(5/12)/(b*x**3+a)**(7/4),x)

[Out]

Integral((c + d*x**3)**(5/12)/(a + b*x**3)**(7/4), x)

Maxima [F]

\[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {5}{12}}}{{\left (b x^{3} + a\right )}^{\frac {7}{4}}} \,d x } \]

[In]

integrate((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^3 + c)^(5/12)/(b*x^3 + a)^(7/4), x)

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {5}{12}}}{{\left (b x^{3} + a\right )}^{\frac {7}{4}}} \,d x } \]

[In]

integrate((d*x^3+c)^(5/12)/(b*x^3+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^(5/12)/(b*x^3 + a)^(7/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^{5/12}}{\left (a+b x^3\right )^{7/4}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{5/12}}{{\left (b\,x^3+a\right )}^{7/4}} \,d x \]

[In]

int((c + d*x^3)^(5/12)/(a + b*x^3)^(7/4),x)

[Out]

int((c + d*x^3)^(5/12)/(a + b*x^3)^(7/4), x)