\(\int \frac {(a+b \sin (c+d x^3))^2}{x^2} \, dx\) [75]

Optimal result

Integrand size = 18, antiderivative size = 231 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^2} \, dx=\frac {-2 a^2-b^2}{2 x}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x}-\frac {a b d e^{i c} x^2 \Gamma \left (\frac {2}{3},-i d x^3\right )}{\left (-i d x^3\right )^{2/3}}-\frac {a b d e^{-i c} x^2 \Gamma \left (\frac {2}{3},i d x^3\right )}{\left (i d x^3\right )^{2/3}}+\frac {i b^2 d e^{2 i c} x^2 \Gamma \left (\frac {2}{3},-2 i d x^3\right )}{2\ 2^{2/3} \left (-i d x^3\right )^{2/3}}-\frac {i b^2 d e^{-2 i c} x^2 \Gamma \left (\frac {2}{3},2 i d x^3\right )}{2\ 2^{2/3} \left (i d x^3\right )^{2/3}}-\frac {2 a b \sin \left (c+d x^3\right )}{x} \] Output:

1/2*(-2*a^2-b^2)/x+1/2*b^2*cos(2*d*x^3+2*c)/x-a*b*d*exp(I*c)*x^2*GAMMA(2/3 
,-I*d*x^3)/(-I*d*x^3)^(2/3)-a*b*d*x^2*GAMMA(2/3,I*d*x^3)/exp(I*c)/(I*d*x^3 
)^(2/3)+1/4*I*b^2*d*exp(2*I*c)*x^2*GAMMA(2/3,-2*I*d*x^3)*2^(1/3)/(-I*d*x^3 
)^(2/3)-1/4*I*b^2*d*x^2*GAMMA(2/3,2*I*d*x^3)*2^(1/3)/exp(2*I*c)/(I*d*x^3)^ 
(2/3)-2*a*b*sin(d*x^3+c)/x
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^2} \, dx=\frac {-4 a^2 \left (d^2 x^6\right )^{2/3}-2 b^2 \left (d^2 x^6\right )^{2/3}+2 b^2 \left (d^2 x^6\right )^{2/3} \cos \left (2 \left (c+d x^3\right )\right )+\sqrt [3]{2} b^2 \left (i d x^3\right )^{5/3} \cos (2 c) \Gamma \left (\frac {2}{3},-2 i d x^3\right )+\sqrt [3]{2} b^2 \left (-i d x^3\right )^{5/3} \cos (2 c) \Gamma \left (\frac {2}{3},2 i d x^3\right )-4 i a b \left (-i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},i d x^3\right ) (\cos (c)-i \sin (c))+4 i a b \left (i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},-i d x^3\right ) (\cos (c)+i \sin (c))+i \sqrt [3]{2} b^2 \left (i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},-2 i d x^3\right ) \sin (2 c)-i \sqrt [3]{2} b^2 \left (-i d x^3\right )^{5/3} \Gamma \left (\frac {2}{3},2 i d x^3\right ) \sin (2 c)-8 a b \left (d^2 x^6\right )^{2/3} \sin \left (c+d x^3\right )}{4 x \left (d^2 x^6\right )^{2/3}} \] Input:

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

Output:

(-4*a^2*(d^2*x^6)^(2/3) - 2*b^2*(d^2*x^6)^(2/3) + 2*b^2*(d^2*x^6)^(2/3)*Co 
s[2*(c + d*x^3)] + 2^(1/3)*b^2*(I*d*x^3)^(5/3)*Cos[2*c]*Gamma[2/3, (-2*I)* 
d*x^3] + 2^(1/3)*b^2*((-I)*d*x^3)^(5/3)*Cos[2*c]*Gamma[2/3, (2*I)*d*x^3] - 
 (4*I)*a*b*((-I)*d*x^3)^(5/3)*Gamma[2/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (4 
*I)*a*b*(I*d*x^3)^(5/3)*Gamma[2/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c]) + I*2^( 
1/3)*b^2*(I*d*x^3)^(5/3)*Gamma[2/3, (-2*I)*d*x^3]*Sin[2*c] - I*2^(1/3)*b^2 
*((-I)*d*x^3)^(5/3)*Gamma[2/3, (2*I)*d*x^3]*Sin[2*c] - 8*a*b*(d^2*x^6)^(2/ 
3)*Sin[c + d*x^3])/(4*x*(d^2*x^6)^(2/3))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 229, 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.

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

\(\Big \downarrow \) 3884

\(\displaystyle \int \left (\frac {a^2}{x^2}+\frac {2 a b \sin \left (c+d x^3\right )}{x^2}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^2}+\frac {b^2}{2 x^2}\right )dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \left (\frac {a^2+\frac {b^2}{2}}{x^2}+\frac {2 a b \sin \left (c+d x^3\right )}{x^2}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*(2*a^2 + b^2)/x + (b^2*Cos[2*c + 2*d*x^3])/(2*x) - (a*b*d*E^(I*c)*x^2 
*Gamma[2/3, (-I)*d*x^3])/((-I)*d*x^3)^(2/3) - (a*b*d*x^2*Gamma[2/3, I*d*x^ 
3])/(E^(I*c)*(I*d*x^3)^(2/3)) + ((I/2)*b^2*d*E^((2*I)*c)*x^2*Gamma[2/3, (- 
2*I)*d*x^3])/(2^(2/3)*((-I)*d*x^3)^(2/3)) - ((I/2)*b^2*d*x^2*Gamma[2/3, (2 
*I)*d*x^3])/(2^(2/3)*E^((2*I)*c)*(I*d*x^3)^(2/3)) - (2*a*b*Sin[c + d*x^3]) 
/x
 

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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^2} \, dx=\frac {4 \, b^{2} \cos \left (d x^{3} + c\right )^{2} - 8 \, a b \sin \left (d x^{3} + c\right ) - {\left (b^{2} x \cos \left (2 \, c\right ) - i \, b^{2} x \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, 2 i \, d x^{3}\right ) - 4 \, {\left (-i \, a b x \cos \left (c\right ) - a b x \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, i \, d x^{3}\right ) - 4 \, {\left (i \, a b x \cos \left (c\right ) - a b x \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -i \, d x^{3}\right ) - {\left (b^{2} x \cos \left (2 \, c\right ) + i \, b^{2} x \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -2 i \, d x^{3}\right ) - 4 \, a^{2} - 4 \, b^{2}}{4 \, x} \] Input:

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

Output:

1/4*(4*b^2*cos(d*x^3 + c)^2 - 8*a*b*sin(d*x^3 + c) - (b^2*x*cos(2*c) - I*b 
^2*x*sin(2*c))*(2*I*d)^(1/3)*gamma(2/3, 2*I*d*x^3) - 4*(-I*a*b*x*cos(c) - 
a*b*x*sin(c))*(I*d)^(1/3)*gamma(2/3, I*d*x^3) - 4*(I*a*b*x*cos(c) - a*b*x* 
sin(c))*(-I*d)^(1/3)*gamma(2/3, -I*d*x^3) - (b^2*x*cos(2*c) + I*b^2*x*sin( 
2*c))*(-2*I*d)^(1/3)*gamma(2/3, -2*I*d*x^3) - 4*a^2 - 4*b^2)/x
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^2} \, dx=-\frac {\left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {1}{3}, i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {1}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) + {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {1}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {1}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b}{6 \, x} + \frac {{\left (2^{\frac {1}{3}} \left (d x^{3}\right )^{\frac {1}{3}} {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {1}{3}, 2 i \, d x^{3}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {1}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} - 12\right )} b^{2}}{24 \, x} - \frac {a^{2}}{x} \] Input:

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

Output:

-1/6*(d*x^3)^(1/3)*(((I*sqrt(3) - 1)*gamma(-1/3, I*d*x^3) + (-I*sqrt(3) - 
1)*gamma(-1/3, -I*d*x^3))*cos(c) + ((sqrt(3) + I)*gamma(-1/3, I*d*x^3) + ( 
sqrt(3) - I)*gamma(-1/3, -I*d*x^3))*sin(c))*a*b/x + 1/24*(2^(1/3)*(d*x^3)^ 
(1/3)*(((sqrt(3) + I)*gamma(-1/3, 2*I*d*x^3) + (sqrt(3) - I)*gamma(-1/3, - 
2*I*d*x^3))*cos(2*c) - ((I*sqrt(3) - 1)*gamma(-1/3, 2*I*d*x^3) + (-I*sqrt( 
3) - 1)*gamma(-1/3, -2*I*d*x^3))*sin(2*c)) - 12)*b^2/x - a^2/x
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^2} \, dx=\frac {\left (\int \frac {\sin \left (d \,x^{3}+c \right )^{2}}{x^{2}}d x \right ) b^{2} x +12 \left (\int \frac {x}{\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}+1}d x \right ) a b d x -2 \left (\int \frac {1}{x^{2}}d x \right ) b^{2} x -2 \sin \left (d \,x^{3}+c \right ) a b -a^{2}-3 a b d \,x^{3}-2 b^{2}}{x} \] Input:

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

Output:

(int(sin(c + d*x**3)**2/x**2,x)*b**2*x + 12*int(x/(tan((c + d*x**3)/2)**2 
+ 1),x)*a*b*d*x - 2*int(1/x**2,x)*b**2*x - 2*sin(c + d*x**3)*a*b - a**2 - 
3*a*b*d*x**3 - 2*b**2)/x