\(\int \frac {(a+b \sin (c+d x^3))^2}{x^3} \, dx\) [79]

Optimal result

Integrand size = 18, antiderivative size = 227 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {-2 a^2-b^2}{4 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{4 x^2}-\frac {a b d e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{2 \sqrt [3]{-i d x^3}}-\frac {a b d e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{2 \sqrt [3]{i d x^3}}+\frac {i b^2 d e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {i b^2 d e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{i d x^3}}-\frac {a b \sin \left (c+d x^3\right )}{x^2} \] Output:

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
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {-2 a^2-b^2}{4 x^2}+\frac {b^2 \cos (2 c) \cos \left (2 d x^3\right )}{4 x^2}+\frac {3}{2} a b d \cos (c) \left (-\frac {x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}-\frac {x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}\right )-\frac {a b \cos \left (d x^3\right ) \sin (c)}{x^2}+\frac {3}{2} i a b d \left (-\frac {x \Gamma \left (\frac {1}{3},-i d x^3\right )}{3 \sqrt [3]{-i d x^3}}+\frac {x \Gamma \left (\frac {1}{3},i d x^3\right )}{3 \sqrt [3]{i d x^3}}\right ) \sin (c)-\frac {b^2 \left (i d x^3\right )^{2/3} \Gamma \left (\frac {1}{3},2 i d x^3\right ) (\cos (2 c)-i \sin (2 c))}{4 \sqrt [3]{2} x^2}+\frac {i b^2 d x \Gamma \left (\frac {1}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))}{4 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {a b \cos (c) \sin \left (d x^3\right )}{x^2}-\frac {b^2 \sin (2 c) \sin \left (2 d x^3\right )}{4 x^2} \] Input:

Integrate[(a + b*Sin[c + d*x^3])^2/x^3,x]
 

Output:

(-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)
 

Rubi [A] (verified)

Time = 0.36 (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

\(\displaystyle -\frac {2 a^2+b^2}{4 x^2}-\frac {a b e^{i c} d x \Gamma \left (\frac {1}{3},-i d x^3\right )}{2 \sqrt [3]{-i d x^3}}-\frac {a b e^{-i c} d x \Gamma \left (\frac {1}{3},i d x^3\right )}{2 \sqrt [3]{i d x^3}}-\frac {a b \sin \left (c+d x^3\right )}{x^2}+\frac {i b^2 e^{2 i c} d x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {i b^2 e^{-2 i c} d x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{4 \sqrt [3]{2} \sqrt [3]{i d x^3}}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{4 x^2}\)

Input:

Int[(a + b*Sin[c + d*x^3])^2/x^3,x]
 

Output:

-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
 

Defintions of rubi rules used

rule 6

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3884

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]
 

Maple [F]

\[\int \frac {{\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}}{x^{3}}d x\]

Input:

int((a+b*sin(d*x^3+c))^2/x^3,x)
 

Output:

int((a+b*sin(d*x^3+c))^2/x^3,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {4 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 8 \, a b \sin \left (d x^{3} + c\right ) - {\left (b^{2} x^{2} \cos \left (2 \, c\right ) - i \, b^{2} x^{2} \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) - 4 \, {\left (-i \, a b x^{2} \cos \left (c\right ) - a b x^{2} \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 4 \, {\left (i \, a b x^{2} \cos \left (c\right ) - a b x^{2} \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) - {\left (b^{2} x^{2} \cos \left (2 \, c\right ) + i \, b^{2} x^{2} \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right ) - 4 \, a^{2} - 4 \, b^{2}}{8 \, x^{2}} \] Input:

integrate((a+b*sin(d*x^3+c))^2/x^3,x, algorithm="fricas")
 

Output:

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
 

Sympy [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{3}}\, dx \] Input:

integrate((a+b*sin(d*x**3+c))**2/x**3,x)
 

Output:

Integral((a + b*sin(c + d*x**3))**2/x**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {\left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {2}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {2}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {2}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (-\frac {2}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b}{6 \, x^{2}} - \frac {{\left (2^{\frac {2}{3}} \left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {2}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {2}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} + 6\right )} b^{2}}{24 \, x^{2}} - \frac {a^{2}}{2 \, x^{2}} \] Input:

integrate((a+b*sin(d*x^3+c))^2/x^3,x, algorithm="maxima")
 

Output:

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
 

Giac [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\int { \frac {{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{3}} \,d x } \] Input:

integrate((a+b*sin(d*x^3+c))^2/x^3,x, algorithm="giac")
 

Output:

integrate((b*sin(d*x^3 + c) + a)^2/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^3} \,d x \] Input:

int((a + b*sin(c + d*x^3))^2/x^3,x)
 

Output:

int((a + b*sin(c + d*x^3))^2/x^3, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^3} \, dx=\frac {2 \left (\int \frac {\sin \left (d \,x^{3}+c \right )^{2}}{x^{3}}d x \right ) b^{2} x^{2}+4 \left (\int \frac {\sin \left (d \,x^{3}+c \right )}{x^{3}}d x \right ) a b \,x^{2}-a^{2}}{2 x^{2}} \] Input:

int((a+b*sin(d*x^3+c))^2/x^3,x)
 

Output:

(2*int(sin(c + d*x**3)**2/x**3,x)*b**2*x**2 + 4*int(sin(c + d*x**3)/x**3,x 
)*a*b*x**2 - a**2)/(2*x**2)