Integrand size = 23, antiderivative size = 153 \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\frac {68 c x \left (c+d x^3\right )^{5/12}}{189 a^2 \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}+\frac {85 c x \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (c+d x^3\right )}\right )}{189 a^2 \left (a+b x^3\right )^{3/4}} \]
68/189*c*x*(d*x^3+c)^(5/12)/a^2/(b*x^3+a)^(3/4)+4/21*x*(d*x^3+c)^(17/12)/a /(b*x^3+a)^(7/4)+85/189*c*x*(c*(b*x^3+a)/a/(d*x^3+c))^(3/4)*(d*x^3+c)^(5/1 2)*hypergeom([1/3, 3/4],[4/3],-(-a*d+b*c)*x^3/a/(d*x^3+c))/a^2/(b*x^3+a)^( 3/4)
Time = 5.70 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.59 \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\frac {c x \left (1+\frac {b x^3}{a}\right )^{3/4} \left (c+d x^3\right )^{5/12} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{4},\frac {4}{3},\frac {(-b c+a d) x^3}{a \left (c+d x^3\right )}\right )}{a^2 \left (a+b x^3\right )^{3/4} \left (1+\frac {d x^3}{c}\right )^{3/4}} \]
(c*x*(1 + (b*x^3)/a)^(3/4)*(c + d*x^3)^(5/12)*Hypergeometric2F1[1/3, 11/4, 4/3, ((-(b*c) + a*d)*x^3)/(a*(c + d*x^3))])/(a^2*(a + b*x^3)^(3/4)*(1 + ( d*x^3)/c)^(3/4))
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.
\(\displaystyle \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx\) |
\(\Big \downarrow \) 903 |
\(\displaystyle \frac {17 c \int \frac {\left (d x^3+c\right )^{5/12}}{\left (b x^3+a\right )^{7/4}}dx}{21 a}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}\) |
\(\Big \downarrow \) 903 |
\(\displaystyle \frac {17 c \left (\frac {5 c \int \frac {1}{\left (b x^3+a\right )^{3/4} \left (d x^3+c\right )^{7/12}}dx}{9 a}+\frac {4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}\right )}{21 a}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}\) |
\(\Big \downarrow \) 905 |
\(\displaystyle \frac {17 c \left (\frac {5 x \left (c+d x^3\right )^{5/12} \left (\frac {c \left (a+b x^3\right )}{a \left (c+d x^3\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {4}{3},-\frac {(b c-a d) x^3}{a \left (d x^3+c\right )}\right )}{9 a \left (a+b x^3\right )^{3/4}}+\frac {4 x \left (c+d x^3\right )^{5/12}}{9 a \left (a+b x^3\right )^{3/4}}\right )}{21 a}+\frac {4 x \left (c+d x^3\right )^{17/12}}{21 a \left (a+b x^3\right )^{7/4}}\) |
(4*x*(c + d*x^3)^(17/12))/(21*a*(a + b*x^3)^(7/4)) + (17*c*((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))))/(21*a)
3.2.31.3.1 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]
\[\int \frac {\left (d \,x^{3}+c \right )^{\frac {17}{12}}}{\left (b \,x^{3}+a \right )^{\frac {11}{4}}}d x\]
\[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {17}{12}}}{{\left (b x^{3} + a\right )}^{\frac {11}{4}}} \,d x } \]
integral((b*x^3 + a)^(1/4)*(d*x^3 + c)^(17/12)/(b^3*x^9 + 3*a*b^2*x^6 + 3* a^2*b*x^3 + a^3), x)
Timed out. \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {17}{12}}}{{\left (b x^{3} + a\right )}^{\frac {11}{4}}} \,d x } \]
\[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {17}{12}}}{{\left (b x^{3} + a\right )}^{\frac {11}{4}}} \,d x } \]
Timed out. \[ \int \frac {\left (c+d x^3\right )^{17/12}}{\left (a+b x^3\right )^{11/4}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^{17/12}}{{\left (b\,x^3+a\right )}^{11/4}} \,d x \]