Integrand size = 18, antiderivative size = 227 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {-2 a^2-b^2}{4 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{4 x^2}-\frac {a b d e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{2 \sqrt [3]{-i d x^3}}-\frac {a b d e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{2 \sqrt [3]{i d x^3}}+\frac {i b^2 d e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {i b^2 d e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{i d x^3}}-\frac {a b \sin \left (c+d x^3\right )}{x^2} \]
1/4*(-2*a^2-b^2)/x^2+1/4*b^2*cos(2*d*x^3+2*c)/x^2-1/2*a*b*d*exp(I*c)*x*GAM MA(1/3,-I*d*x^3)/(-I*d*x^3)^(1/3)-1/2*a*b*d*x*GAMMA(1/3,I*d*x^3)/exp(I*c)/ (I*d*x^3)^(1/3)+1/8*I*b^2*d*exp(2*I*c)*x*GAMMA(1/3,-2*I*d*x^3)*2^(2/3)/(-I *d*x^3)^(1/3)-1/8*I*b^2*d*x*GAMMA(1/3,2*I*d*x^3)*2^(2/3)/exp(2*I*c)/(I*d*x ^3)^(1/3)-a*b*sin(d*x^3+c)/x^2
Time = 0.53 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {-2 a^2-b^2}{4 x^2}+\frac {b^2 \cos (2 c) \cos \left (2 d x^3\right )}{4 x^2}+\frac {3}{2} a b d \cos (c) \left (-\frac {x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}\right )-\frac {a b \cos \left (d x^3\right ) \sin (c)}{x^2}+\frac {3}{2} i a b d \left (-\frac {x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}+\frac {x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}\right ) \sin (c)-\frac {b^2 \left (i d x^3\right )^{2/3} \Gamma \left (\frac {1}{3},2 i d x^3\right ) (\cos (2 c)-i \sin (2 c))}{4 \sqrt [3]{2} x^2}+\frac {i b^2 d x \Gamma \left (\frac {1}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))}{4 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {a b \cos (c) \sin \left (d x^3\right )}{x^2}-\frac {b^2 \sin (2 c) \sin \left (2 d x^3\right )}{4 x^2} \]
(-2*a^2 - b^2)/(4*x^2) + (b^2*Cos[2*c]*Cos[2*d*x^3])/(4*x^2) + (3*a*b*d*Co s[c]*(-1/3*(x*Gamma[1/3, (-I)*d*x^3])/((-I)*d*x^3)^(1/3) - (x*Gamma[1/3, I *d*x^3])/(3*(I*d*x^3)^(1/3))))/2 - (a*b*Cos[d*x^3]*Sin[c])/x^2 + ((3*I)/2) *a*b*d*(-1/3*(x*Gamma[1/3, (-I)*d*x^3])/((-I)*d*x^3)^(1/3) + (x*Gamma[1/3, I*d*x^3])/(3*(I*d*x^3)^(1/3)))*Sin[c] - (b^2*(I*d*x^3)^(2/3)*Gamma[1/3, ( 2*I)*d*x^3]*(Cos[2*c] - I*Sin[2*c]))/(4*2^(1/3)*x^2) + ((I/4)*b^2*d*x*Gamm a[1/3, (-2*I)*d*x^3]*(Cos[2*c] + I*Sin[2*c]))/(2^(1/3)*((-I)*d*x^3)^(1/3)) - (a*b*Cos[c]*Sin[d*x^3])/x^2 - (b^2*Sin[2*c]*Sin[2*d*x^3])/(4*x^2)
Time = 0.34 (sec) , antiderivative size = 225, 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.
\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx\) |
\(\Big \downarrow \) 3884 |
\(\displaystyle \int \left (\frac {a^2}{x^3}+\frac {2 a b \sin \left (c+d x^3\right )}{x^3}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^3}+\frac {b^2}{2 x^3}\right )dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \left (\frac {a^2+\frac {b^2}{2}}{x^3}+\frac {2 a b \sin \left (c+d x^3\right )}{x^3}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^2+b^2}{4 x^2}-\frac {a b e^{i c} d x \Gamma \left (\frac {1}{3},-i d x^3\right )}{2 \sqrt [3]{-i d x^3}}-\frac {a b e^{-i c} d x \Gamma \left (\frac {1}{3},i d x^3\right )}{2 \sqrt [3]{i d x^3}}-\frac {a b \sin \left (c+d x^3\right )}{x^2}+\frac {i b^2 e^{2 i c} d x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {i b^2 e^{-2 i c} d x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{i d x^3}}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{4 x^2}\) |
-1/4*(2*a^2 + b^2)/x^2 + (b^2*Cos[2*c + 2*d*x^3])/(4*x^2) - (a*b*d*E^(I*c) *x*Gamma[1/3, (-I)*d*x^3])/(2*((-I)*d*x^3)^(1/3)) - (a*b*d*x*Gamma[1/3, I* d*x^3])/(2*E^(I*c)*(I*d*x^3)^(1/3)) + ((I/4)*b^2*d*E^((2*I)*c)*x*Gamma[1/3 , (-2*I)*d*x^3])/(2^(1/3)*((-I)*d*x^3)^(1/3)) - ((I/4)*b^2*d*x*Gamma[1/3, (2*I)*d*x^3])/(2^(1/3)*E^((2*I)*c)*(I*d*x^3)^(1/3)) - (a*b*Sin[c + d*x^3]) /x^2
3.1.79.3.1 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]
\[\int \frac {{\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}}{x^{3}}d x\]
Time = 0.12 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {4 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 8 \, a b \sin \left (d x^{3} + c\right ) - {\left (b^{2} x^{2} \cos \left (2 \, c\right ) - i \, b^{2} x^{2} \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) - 4 \, {\left (-i \, a b x^{2} \cos \left (c\right ) - a b x^{2} \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 4 \, {\left (i \, a b x^{2} \cos \left (c\right ) - a b x^{2} \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) - {\left (b^{2} x^{2} \cos \left (2 \, c\right ) + i \, b^{2} x^{2} \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right ) - 4 \, a^{2} - 4 \, b^{2}}{8 \, x^{2}} \]
1/8*(4*b^2*cos(d*x^3 + c)^2 - 8*a*b*sin(d*x^3 + c) - (b^2*x^2*cos(2*c) - I *b^2*x^2*sin(2*c))*(2*I*d)^(2/3)*gamma(1/3, 2*I*d*x^3) - 4*(-I*a*b*x^2*cos (c) - a*b*x^2*sin(c))*(I*d)^(2/3)*gamma(1/3, I*d*x^3) - 4*(I*a*b*x^2*cos(c ) - a*b*x^2*sin(c))*(-I*d)^(2/3)*gamma(1/3, -I*d*x^3) - (b^2*x^2*cos(2*c) + I*b^2*x^2*sin(2*c))*(-2*I*d)^(2/3)*gamma(1/3, -2*I*d*x^3) - 4*a^2 - 4*b^ 2)/x^2
\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{3}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {\left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {2}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {2}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {2}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {2}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b}{6 \, x^{2}} - \frac {{\left (2^{\frac {2}{3}} \left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} + 6\right )} b^{2}}{24 \, x^{2}} - \frac {a^{2}}{2 \, x^{2}} \]
1/6*(d*x^3)^(2/3)*(((sqrt(3) - I)*gamma(-2/3, I*d*x^3) + (sqrt(3) + I)*gam ma(-2/3, -I*d*x^3))*cos(c) - ((I*sqrt(3) + 1)*gamma(-2/3, I*d*x^3) + (-I*s qrt(3) + 1)*gamma(-2/3, -I*d*x^3))*sin(c))*a*b/x^2 - 1/24*(2^(2/3)*(d*x^3) ^(2/3)*(((-I*sqrt(3) - 1)*gamma(-2/3, 2*I*d*x^3) + (I*sqrt(3) - 1)*gamma(- 2/3, -2*I*d*x^3))*cos(2*c) - ((sqrt(3) - I)*gamma(-2/3, 2*I*d*x^3) + (sqrt (3) + I)*gamma(-2/3, -2*I*d*x^3))*sin(2*c)) + 6)*b^2/x^2 - 1/2*a^2/x^2
\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\int { \frac {{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^3} \,d x \]