\(\int \frac {(a+b \sin (c+d x^3))^2}{x^6} \, dx\) [80]

Optimal result

Integrand size = 18, antiderivative size = 277 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=\frac {-2 a^2-b^2}{10 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {3 i a b d^2 e^{i c} x \Gamma \left (\frac {1}{3},-i d x^3\right )}{10 \sqrt [3]{-i d x^3}}+\frac {3 i a b d^2 e^{-i c} x \Gamma \left (\frac {1}{3},i d x^3\right )}{10 \sqrt [3]{i d x^3}}-\frac {3 b^2 d^2 e^{2 i c} x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {3 b^2 d^2 e^{-2 i c} x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{i d x^3}}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2} \] Output:

1/10*(-2*a^2-b^2)/x^5-3/5*a*b*d*cos(d*x^3+c)/x^2+1/10*b^2*cos(2*d*x^3+2*c) 
/x^5-3/10*I*a*b*d^2*exp(I*c)*x*GAMMA(1/3,-I*d*x^3)/(-I*d*x^3)^(1/3)+3/10*I 
*a*b*d^2*x*GAMMA(1/3,I*d*x^3)/exp(I*c)/(I*d*x^3)^(1/3)-3/20*b^2*d^2*exp(2* 
I*c)*x*GAMMA(1/3,-2*I*d*x^3)*2^(2/3)/(-I*d*x^3)^(1/3)-3/20*b^2*d^2*x*GAMMA 
(1/3,2*I*d*x^3)*2^(2/3)/exp(2*I*c)/(I*d*x^3)^(1/3)-2/5*a*b*sin(d*x^3+c)/x^ 
5-3/10*b^2*d*sin(2*d*x^3+2*c)/x^2
 

Mathematica [A] (verified)

Time = 3.36 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=-\frac {4 a^2+2 b^2+12 a b d x^3 \cos \left (c+d x^3\right )-2 b^2 \cos \left (2 \left (c+d x^3\right )\right )-3\ 2^{2/3} b^2 \left (i d x^3\right )^{5/3} \cos (2 c) \Gamma \left (\frac {1}{3},2 i d x^3\right )+6 i a b \left (i d x^3\right )^{5/3} \Gamma \left (\frac {1}{3},i d x^3\right ) (\cos (c)-i \sin (c))+6 i a b \sqrt [3]{i d x^3} \left (d^2 x^6\right )^{2/3} \Gamma \left (\frac {1}{3},-i d x^3\right ) (\cos (c)+i \sin (c))-3\ 2^{2/3} b^2 \left (-i d x^3\right )^{5/3} \Gamma \left (\frac {1}{3},-2 i d x^3\right ) (\cos (2 c)+i \sin (2 c))+3 i 2^{2/3} b^2 \left (i d x^3\right )^{5/3} \Gamma \left (\frac {1}{3},2 i d x^3\right ) \sin (2 c)+8 a b \sin \left (c+d x^3\right )+6 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )}{20 x^5} \] Input:

Integrate[(a + b*Sin[c + d*x^3])^2/x^6,x]
 

Output:

-1/20*(4*a^2 + 2*b^2 + 12*a*b*d*x^3*Cos[c + d*x^3] - 2*b^2*Cos[2*(c + d*x^ 
3)] - 3*2^(2/3)*b^2*(I*d*x^3)^(5/3)*Cos[2*c]*Gamma[1/3, (2*I)*d*x^3] + (6* 
I)*a*b*(I*d*x^3)^(5/3)*Gamma[1/3, I*d*x^3]*(Cos[c] - I*Sin[c]) + (6*I)*a*b 
*(I*d*x^3)^(1/3)*(d^2*x^6)^(2/3)*Gamma[1/3, (-I)*d*x^3]*(Cos[c] + I*Sin[c] 
) - 3*2^(2/3)*b^2*((-I)*d*x^3)^(5/3)*Gamma[1/3, (-2*I)*d*x^3]*(Cos[2*c] + 
I*Sin[2*c]) + (3*I)*2^(2/3)*b^2*(I*d*x^3)^(5/3)*Gamma[1/3, (2*I)*d*x^3]*Si 
n[2*c] + 8*a*b*Sin[c + d*x^3] + 6*b^2*d*x^3*Sin[2*(c + d*x^3)])/x^5
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 275, 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.

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx\)

\(\Big \downarrow \) 3884

\(\displaystyle \int \left (\frac {a^2}{x^6}+\frac {2 a b \sin \left (c+d x^3\right )}{x^6}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^6}+\frac {b^2}{2 x^6}\right )dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \left (\frac {a^2+\frac {b^2}{2}}{x^6}+\frac {2 a b \sin \left (c+d x^3\right )}{x^6}-\frac {b^2 \cos \left (2 c+2 d x^3\right )}{2 x^6}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2+b^2}{10 x^5}-\frac {3 i a b e^{i c} d^2 x \Gamma \left (\frac {1}{3},-i d x^3\right )}{10 \sqrt [3]{-i d x^3}}+\frac {3 i a b e^{-i c} d^2 x \Gamma \left (\frac {1}{3},i d x^3\right )}{10 \sqrt [3]{i d x^3}}-\frac {2 a b \sin \left (c+d x^3\right )}{5 x^5}-\frac {3 a b d \cos \left (c+d x^3\right )}{5 x^2}-\frac {3 b^2 e^{2 i c} d^2 x \Gamma \left (\frac {1}{3},-2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{-i d x^3}}-\frac {3 b^2 e^{-2 i c} d^2 x \Gamma \left (\frac {1}{3},2 i d x^3\right )}{10 \sqrt [3]{2} \sqrt [3]{i d x^3}}+\frac {b^2 \cos \left (2 c+2 d x^3\right )}{10 x^5}-\frac {3 b^2 d \sin \left (2 c+2 d x^3\right )}{10 x^2}\)

Input:

Int[(a + b*Sin[c + d*x^3])^2/x^6,x]
 

Output:

-1/10*(2*a^2 + b^2)/x^5 - (3*a*b*d*Cos[c + d*x^3])/(5*x^2) + (b^2*Cos[2*c 
+ 2*d*x^3])/(10*x^5) - (((3*I)/10)*a*b*d^2*E^(I*c)*x*Gamma[1/3, (-I)*d*x^3 
])/((-I)*d*x^3)^(1/3) + (((3*I)/10)*a*b*d^2*x*Gamma[1/3, I*d*x^3])/(E^(I*c 
)*(I*d*x^3)^(1/3)) - (3*b^2*d^2*E^((2*I)*c)*x*Gamma[1/3, (-2*I)*d*x^3])/(1 
0*2^(1/3)*((-I)*d*x^3)^(1/3)) - (3*b^2*d^2*x*Gamma[1/3, (2*I)*d*x^3])/(10* 
2^(1/3)*E^((2*I)*c)*(I*d*x^3)^(1/3)) - (2*a*b*Sin[c + d*x^3])/(5*x^5) - (3 
*b^2*d*Sin[2*c + 2*d*x^3])/(10*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]

\[\int \frac {{\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}}{x^{6}}d x\]

Input:

int((a+b*sin(d*x^3+c))^2/x^6,x)
 

Output:

int((a+b*sin(d*x^3+c))^2/x^6,x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=-\frac {12 \, a b d x^{3} \cos \left (d x^{3} + c\right ) - 4 \, b^{2} \cos \left (d x^{3} + c\right )^{2} + 3 \, {\left (-i \, b^{2} d x^{5} \cos \left (2 \, c\right ) - b^{2} d x^{5} \sin \left (2 \, c\right )\right )} \left (2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, 2 i \, d x^{3}\right ) - 6 \, {\left (a b d x^{5} \cos \left (c\right ) - i \, a b d x^{5} \sin \left (c\right )\right )} \left (i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, i \, d x^{3}\right ) - 6 \, {\left (a b d x^{5} \cos \left (c\right ) + i \, a b d x^{5} \sin \left (c\right )\right )} \left (-i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -i \, d x^{3}\right ) + 3 \, {\left (i \, b^{2} d x^{5} \cos \left (2 \, c\right ) - b^{2} d x^{5} \sin \left (2 \, c\right )\right )} \left (-2 i \, d\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -2 i \, d x^{3}\right ) + 4 \, a^{2} + 4 \, b^{2} + 4 \, {\left (3 \, b^{2} d x^{3} \cos \left (d x^{3} + c\right ) + 2 \, a b\right )} \sin \left (d x^{3} + c\right )}{20 \, x^{5}} \] Input:

integrate((a+b*sin(d*x^3+c))^2/x^6,x, algorithm="fricas")
 

Output:

-1/20*(12*a*b*d*x^3*cos(d*x^3 + c) - 4*b^2*cos(d*x^3 + c)^2 + 3*(-I*b^2*d* 
x^5*cos(2*c) - b^2*d*x^5*sin(2*c))*(2*I*d)^(2/3)*gamma(1/3, 2*I*d*x^3) - 6 
*(a*b*d*x^5*cos(c) - I*a*b*d*x^5*sin(c))*(I*d)^(2/3)*gamma(1/3, I*d*x^3) - 
 6*(a*b*d*x^5*cos(c) + I*a*b*d*x^5*sin(c))*(-I*d)^(2/3)*gamma(1/3, -I*d*x^ 
3) + 3*(I*b^2*d*x^5*cos(2*c) - b^2*d*x^5*sin(2*c))*(-2*I*d)^(2/3)*gamma(1/ 
3, -2*I*d*x^3) + 4*a^2 + 4*b^2 + 4*(3*b^2*d*x^3*cos(d*x^3 + c) + 2*a*b)*si 
n(d*x^3 + c))/x^5
 

Sympy [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=\int \frac {\left (a + b \sin {\left (c + d x^{3} \right )}\right )^{2}}{x^{6}}\, dx \] Input:

integrate((a+b*sin(d*x**3+c))**2/x**6,x)
 

Output:

Integral((a + b*sin(c + d*x**3))**2/x**6, x)
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=-\frac {\left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \cos \left (c\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {5}{3}, i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {5}{3}, -i \, d x^{3}\right )\right )} \sin \left (c\right )\right )} a b d}{6 \, x^{2}} - \frac {{\left (5 \cdot 2^{\frac {2}{3}} \left (d x^{3}\right )^{\frac {2}{3}} {\left ({\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {5}{3}, 2 i \, d x^{3}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {5}{3}, -2 i \, d x^{3}\right )\right )} \cos \left (2 \, c\right ) + {\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, 2 i \, d x^{3}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {5}{3}, -2 i \, d x^{3}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{3} + 6\right )} b^{2}}{60 \, x^{5}} - \frac {a^{2}}{5 \, x^{5}} \] Input:

integrate((a+b*sin(d*x^3+c))^2/x^6,x, algorithm="maxima")
 

Output:

-1/6*(d*x^3)^(2/3)*(((-I*sqrt(3) - 1)*gamma(-5/3, I*d*x^3) + (I*sqrt(3) - 
1)*gamma(-5/3, -I*d*x^3))*cos(c) - ((sqrt(3) - I)*gamma(-5/3, I*d*x^3) + ( 
sqrt(3) + I)*gamma(-5/3, -I*d*x^3))*sin(c))*a*b*d/x^2 - 1/60*(5*2^(2/3)*(d 
*x^3)^(2/3)*(((sqrt(3) - I)*gamma(-5/3, 2*I*d*x^3) + (sqrt(3) + I)*gamma(- 
5/3, -2*I*d*x^3))*cos(2*c) + ((-I*sqrt(3) - 1)*gamma(-5/3, 2*I*d*x^3) + (I 
*sqrt(3) - 1)*gamma(-5/3, -2*I*d*x^3))*sin(2*c))*d*x^3 + 6)*b^2/x^5 - 1/5* 
a^2/x^5
 

Giac [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=\int { \frac {{\left (b \sin \left (d x^{3} + c\right ) + a\right )}^{2}}{x^{6}} \,d x } \] Input:

integrate((a+b*sin(d*x^3+c))^2/x^6,x, algorithm="giac")
 

Output:

integrate((b*sin(d*x^3 + c) + a)^2/x^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=\int \frac {{\left (a+b\,\sin \left (d\,x^3+c\right )\right )}^2}{x^6} \,d x \] Input:

int((a + b*sin(c + d*x^3))^2/x^6,x)
 

Output:

int((a + b*sin(c + d*x^3))^2/x^6, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \sin \left (c+d x^3\right )\right )^2}{x^6} \, dx=\frac {5 \left (\int \frac {\sin \left (d \,x^{3}+c \right )^{2}}{x^{6}}d x \right ) b^{2} x^{5}+10 \left (\int \frac {\sin \left (d \,x^{3}+c \right )}{x^{6}}d x \right ) a b \,x^{5}-a^{2}}{5 x^{5}} \] Input:

int((a+b*sin(d*x^3+c))^2/x^6,x)
 

Output:

(5*int(sin(c + d*x**3)**2/x**6,x)*b**2*x**5 + 10*int(sin(c + d*x**3)/x**6, 
x)*a*b*x**5 - a**2)/(5*x**5)