Integrand size = 18, antiderivative size = 285 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=\frac {-2 a^2-b^2}{8 x^4}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}-\frac {3 i a b d^2 e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac {3 i a b d^2 e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac {3 b^2 d^2 e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {3 b^2 d^2 e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x} \]
1/8*(-2*a^2-b^2)/x^4-3/2*a*b*d*cos(d*x^3+c)/x+1/8*b^2*cos(2*d*x^3+2*c)/x^4 -3/4*I*a*b*d^2*exp(I*c)*x^2*GAMMA(2/3,-I*d*x^3)/(-I*d*x^3)^(2/3)+3/4*I*a*b *d^2*x^2*GAMMA(2/3,I*d*x^3)/exp(I*c)/(I*d*x^3)^(2/3)-3/8*b^2*d^2*exp(2*I*c )*x^2*GAMMA(2/3,-2*I*d*x^3)*2^(1/3)/(-I*d*x^3)^(2/3)-3/8*b^2*d^2*x^2*GAMMA (2/3,2*I*d*x^3)*2^(1/3)/exp(2*I*c)/(I*d*x^3)^(2/3)-1/2*a*b*sin(d*x^3+c)/x^ 4-3/4*b^2*d*sin(2*d*x^3+2*c)/x
Time = 1.84 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=-\frac {2 a^2+b^2+12 a b d x^3 \cos \left (c+d x^3\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )-3 \sqrt [3]{2} b^2 \left (i d x^3\right )^{4/3} \cos (2 c) \Gamma \left (\frac {2}{3},2 i d x^3\right )+6 i a b \left (i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))+6 i a b \left (i d x^3\right )^{2/3} \sqrt [3]{d^2 x^6} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))-3 \sqrt [3]{2} b^2 \left (-i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))+3 i \sqrt [3]{2} b^2 \left (i d x^3\right )^{4/3} \Gamma \left (\frac {2}{3},2 i d x^3\right ) \sin (2 c)+4 a b \sin \left (c+d x^3\right )+6 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )}{8 x^4} \]
-1/8*(2*a^2 + b^2 + 12*a*b*d*x^3*Cos[c + d*x^3] - b^2*Cos[2*(c + d*x^3)] - 3*2^(1/3)*b^2*(I*d*x^3)^(4/3)*Cos[2*c]*Gamma[2/3, (2*I)*d*x^3] + (6*I)*a* b*(I*d*x^3)^(4/3)*Gamma[2/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (6*I)*a*b*(I*d *x^3)^(2/3)*(d^2*x^6)^(1/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) - 3 *2^(1/3)*b^2*((-I)*d*x^3)^(4/3)*Gamma[2/3, (-2*I)*d*x^3]*(Cos[2*c] + I*Sin [2*c]) + (3*I)*2^(1/3)*b^2*(I*d*x^3)^(4/3)*Gamma[2/3, (2*I)*d*x^3]*Sin[2*c ] + 4*a*b*Sin[c + d*x^3] + 6*b^2*d*x^3*Sin[2*(c + d*x^3)])/x^4
Time = 0.46 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3884, 6, 2009}
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.
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx\) |
| \(\Big \downarrow \) 3884 |
| \(\displaystyle \int \left (\frac {a^2}{x^5}+\frac {2 a b \sin \left (c+d x^3\right )}{x^5}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^5}+\frac {b^2}{2 x^5}\right )dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \left (\frac {a^2+\frac {b^2}{2}}{x^5}+\frac {2 a b \sin \left (c+d x^3\right )}{x^5}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^5}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 a^2+b^2}{8 x^4}-\frac {3 i a b e^{i c} d^2 x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{4 \left (-i d x^3\right )^{2/3}}+\frac {3 i a b e^{-i c} d^2 x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{4 \left (i d x^3\right )^{2/3}}-\frac {3 a b d \cos \left (c+d x^3\right )}{2 x}-\frac {a b \sin \left (c+d x^3\right )}{2 x^4}-\frac {3 b^2 e^{2 i c} d^2 x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{4\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {3 b^2 e^{-2 i c} d^2 x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{4\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{4 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{8 x^4}\) |
-1/8*(2*a^2 + b^2)/x^4 - (3*a*b*d*Cos[c + d*x^3])/(2*x) + (b^2*Cos[2*c + 2 *d*x^3])/(8*x^4) - (((3*I)/4)*a*b*d^2*E^(I*c)*x^2*Gamma[2/3, (-I)*d*x^3])/ ((-I)*d*x^3)^(2/3) + (((3*I)/4)*a*b*d^2*x^2*Gamma[2/3, I*d*x^3])/(E^(I*c)* (I*d*x^3)^(2/3)) - (3*b^2*d^2*E^((2*I)*c)*x^2*Gamma[2/3, (-2*I)*d*x^3])/(4 *2^(2/3)*((-I)*d*x^3)^(2/3)) - (3*b^2*d^2*x^2*Gamma[2/3, (2*I)*d*x^3])/(4* 2^(2/3)*E^((2*I)*c)*(I*d*x^3)^(2/3)) - (a*b*Sin[c + d*x^3])/(2*x^4) - (3*b ^2*d*Sin[2*c + 2*d*x^3])/(4*x)
3.1.76.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x _Symbol] :> Int[ExpandTrigReduce[(e*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]
\[\int \frac {{\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}}{x^{5}}d x\]
Time = 0.12 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=-\frac {12 \, a b d x^{3} \cos \left (d x^{3} + c\right ) - 2 \, b^{2} \cos \left (d x^{3} + c\right )^{2} + 3 \, {\left (-i \, b^{2} d x^{4} \cos \left (2 \, c\right ) - b^{2} d x^{4} \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) - 6 \, {\left (a b d x^{4} \cos \left (c\right ) - i \, a b d x^{4} \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) - 6 \, {\left (a b d x^{4} \cos \left (c\right ) + i \, a b d x^{4} \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) + 3 \, {\left (i \, b^{2} d x^{4} \cos \left (2 \, c\right ) - b^{2} d x^{4} \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right ) + 2 \, a^{2} + 2 \, b^{2} + 4 \, {\left (3 \, b^{2} d x^{3} \cos \left (d x^{3} + c\right ) + a b\right )} \sin \left (d x^{3} + c\right )}{8 \, x^{4}} \]
-1/8*(12*a*b*d*x^3*cos(d*x^3 + c) - 2*b^2*cos(d*x^3 + c)^2 + 3*(-I*b^2*d*x ^4*cos(2*c) - b^2*d*x^4*sin(2*c))*(2*I*d)^(1/3)*gamma(2/3, 2*I*d*x^3) - 6* (a*b*d*x^4*cos(c) - I*a*b*d*x^4*sin(c))*(I*d)^(1/3)*gamma(2/3, I*d*x^3) - 6*(a*b*d*x^4*cos(c) + I*a*b*d*x^4*sin(c))*(-I*d)^(1/3)*gamma(2/3, -I*d*x^3 ) + 3*(I*b^2*d*x^4*cos(2*c) - b^2*d*x^4*sin(2*c))*(-2*I*d)^(1/3)*gamma(2/3 , -2*I*d*x^3) + 2*a^2 + 2*b^2 + 4*(3*b^2*d*x^3*cos(d*x^3 + c) + a*b)*sin(d *x^3 + c))/x^4
\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{5}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=\frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b d}{6 \, x} - \frac {{\left (2 \cdot 2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {4}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {4}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} + 3\right )} b^{2}}{24 \, x^{4}} - \frac {a^{2}}{4 \, x^{4}} \]
1/6*(d*x^3)^(1/3)*(((sqrt(3) + I)*gamma(-4/3, I*d*x^3) + (sqrt(3) - I)*gam ma(-4/3, -I*d*x^3))*cos(c) - ((I*sqrt(3) - 1)*gamma(-4/3, I*d*x^3) + (-I*s qrt(3) - 1)*gamma(-4/3, -I*d*x^3))*sin(c))*a*b*d/x - 1/24*(2*2^(1/3)*(d*x^ 3)^(1/3)*(((-I*sqrt(3) + 1)*gamma(-4/3, 2*I*d*x^3) + (I*sqrt(3) + 1)*gamma (-4/3, -2*I*d*x^3))*cos(2*c) - ((sqrt(3) + I)*gamma(-4/3, 2*I*d*x^3) + (sq rt(3) - I)*gamma(-4/3, -2*I*d*x^3))*sin(2*c))*d*x^3 + 3)*b^2/x^4 - 1/4*a^2 /x^4
\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=\int { \frac {{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^5} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^5} \,d x \]