Integrand size = 23, antiderivative size = 153 \[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\frac {92 c x \left (c+d x^3\right )^{11/12}}{405 a^2 \left (a+b x^3\right )^{5/4}}+\frac {4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}+\frac {253 c x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \left (c+d x^3\right )^{11/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{405 a^2 \left (a+b x^3\right )^{5/4}} \]
92/405*c*x*(d*x^3+c)^(11/12)/a^2/(b*x^3+a)^(5/4)+4/27*x*(d*x^3+c)^(23/12)/ a/(b*x^3+a)^(9/4)+253/405*c*x*(c*(b*x^3+a)/a/(d*x^3+c))^(5/4)*(d*x^3+c)^(1 1/12)*hypergeom([1/3, 5/4],[4/3],-(-a*d+b*c)*x^3/a/(d*x^3+c))/a^2/(b*x^3+a )^(5/4)
Time = 5.72 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\frac {c x \sqrt [4]{1+\frac {b x^3}{a}} \left (c+d x^3\right )^{11/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {13}{4},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{a^3 \sqrt [4]{a+b x^3} \left (1+\frac {d x^3}{c}\right )^{5/4}} \]
(c*x*(1 + (b*x^3)/a)^(1/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 13/4 , 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))])/(a^3*(a + b*x^3)^(1/4)*(1 + (d*x^3)/c)^(5/4))
Time = 0.26 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {903, 903, 905}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx\) |
\(\Big \downarrow \) 903 |
\(\displaystyle \frac {23 c \int \frac {\left (d x^3+c\right )^{11/12}}{\left (b x^3+a\right )^{9/4}}dx}{27 a}+\frac {4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}\) |
\(\Big \downarrow \) 903 |
\(\displaystyle \frac {23 c \left (\frac {11 c \int \frac {1}{\left (b x^3+a\right )^{5/4} \sqrt [12]{d x^3+c}}dx}{15 a}+\frac {4 x \left (c+d x^3\right )^{11/12}}{15 a \left (a+b x^3\right )^{5/4}}\right )}{27 a}+\frac {4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}\) |
\(\Big \downarrow \) 905 |
\(\displaystyle \frac {23 c \left (\frac {11 x \left (c+d x^3\right )^{11/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{5/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{15 a \left (a+b x^3\right )^{5/4}}+\frac {4 x \left (c+d x^3\right )^{11/12}}{15 a \left (a+b x^3\right )^{5/4}}\right )}{27 a}+\frac {4 x \left (c+d x^3\right )^{23/12}}{27 a \left (a+b x^3\right )^{9/4}}\) |
(4*x*(c + d*x^3)^(23/12))/(27*a*(a + b*x^3)^(9/4)) + (23*c*((4*x*(c + d*x^ 3)^(11/12))/(15*a*(a + b*x^3)^(5/4)) + (11*x*((c*(a + b*x^3))/(a*(c + d*x^ 3)))^(5/4)*(c + d*x^3)^(11/12)*Hypergeometric2F1[1/3, 5/4, 4/3, -(((b*c - a*d)*x^3)/(a*(c + d*x^3)))])/(15*a*(a + b*x^3)^(5/4))))/(27*a)
3.2.32.3.1 Defintions of rubi rules used
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] - Simp[ 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]
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]
\[\int \frac {\left (d \,x^{3}+c \right )^{\frac {23}{12}}}{\left (b \,x^{3}+a \right )^{\frac {13}{4}}}d x\]
\[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {23}{12}}}{{\left (b x^{3} + a\right )}^{\frac {13}{4}}} \,d x } \]
integral((b*x^3 + a)^(3/4)*(d*x^3 + c)^(23/12)/(b^4*x^12 + 4*a*b^3*x^9 + 6 *a^2*b^2*x^6 + 4*a^3*b*x^3 + a^4), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {23}{12}}}{{\left (b x^{3} + a\right )}^{\frac {13}{4}}} \,d x } \]
\[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {23}{12}}}{{\left (b x^{3} + a\right )}^{\frac {13}{4}}} \,d x } \]
Timed out. \[ \int \frac {\left (c+d x^3\right )^{23/12}}{\left (a+b x^3\right )^{13/4}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{23/12}}{{\left (b\,x^3+a\right )}^{13/4}} \,d x \]