\(\int \frac {(c \sin ^3(a+b x))^{2/3}}{x^3} \, dx\) [341]

Optimal result

Integrand size = 18, antiderivative size = 119 \[ \int \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x^3} \, dx=-\frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{2 x^2}-\frac {b \cot (a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}}{x}+b^2 \cos (2 a) \operatorname {CosIntegral}(2 b x) \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3}-b^2 \csc ^2(a+b x) \sin (2 a) \left (c \sin ^3(a+b x)\right )^{2/3} \text {Si}(2 b x) \] Output:

-1/2*(c*sin(b*x+a)^3)^(2/3)/x^2-b*cot(b*x+a)*(c*sin(b*x+a)^3)^(2/3)/x+b^2* 
cos(2*a)*Ci(2*b*x)*csc(b*x+a)^2*(c*sin(b*x+a)^3)^(2/3)-b^2*csc(b*x+a)^2*si 
n(2*a)*(c*sin(b*x+a)^3)^(2/3)*Si(2*b*x)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x^3} \, dx=\frac {\csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \left (-1+\cos (2 (a+b x))+4 b^2 x^2 \cos (2 a) \operatorname {CosIntegral}(2 b x)-2 b x \sin (2 (a+b x))-4 b^2 x^2 \sin (2 a) \text {Si}(2 b x)\right )}{4 x^2} \] Input:

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

Output:

(Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(-1 + Cos[2*(a + b*x)] + 4*b^2*x^ 
2*Cos[2*a]*CosIntegral[2*b*x] - 2*b*x*Sin[2*(a + b*x)] - 4*b^2*x^2*Sin[2*a 
]*SinIntegral[2*b*x]))/(4*x^2)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {7271, 3042, 3795, 14, 3042, 3793, 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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3795

\(\displaystyle \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \left (-2 b^2 \int \frac {\sin ^2(a+b x)}{x}dx+b^2 \int \frac {1}{x}dx-\frac {\sin ^2(a+b x)}{2 x^2}-\frac {b \sin (a+b x) \cos (a+b x)}{x}\right )\)

\(\Big \downarrow \) 14

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

\(\displaystyle \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \left (-2 b^2 \int \left (\frac {1}{2 x}-\frac {\cos (2 a+2 b x)}{2 x}\right )dx-\frac {\sin ^2(a+b x)}{2 x^2}-\frac {b \sin (a+b x) \cos (a+b x)}{x}+b^2 \log (x)\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \csc ^2(a+b x) \left (c \sin ^3(a+b x)\right )^{2/3} \left (-2 b^2 \left (-\frac {1}{2} \cos (2 a) \operatorname {CosIntegral}(2 b x)+\frac {1}{2} \sin (2 a) \text {Si}(2 b x)+\frac {\log (x)}{2}\right )-\frac {\sin ^2(a+b x)}{2 x^2}-\frac {b \sin (a+b x) \cos (a+b x)}{x}+b^2 \log (x)\right )\)

Input:

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

Output:

Csc[a + b*x]^2*(c*Sin[a + b*x]^3)^(2/3)*(b^2*Log[x] - (b*Cos[a + b*x]*Sin[ 
a + b*x])/x - Sin[a + b*x]^2/(2*x^2) - 2*b^2*(-1/2*(Cos[2*a]*CosIntegral[2 
*b*x]) + Log[x]/2 + (Sin[2*a]*SinIntegral[2*b*x])/2))
 

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 3795

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), x] + (-Simp[ 
b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(d^2*(m + 1) 
*(m + 2))), x] + Simp[b^2*f^2*n*((n - 1)/(d^2*(m + 1)*(m + 2)))   Int[(c + 
d*x)^(m + 2)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[f^2*(n^2/(d^2*(m + 1)* 
(m + 2)))   Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, 
c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
 

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.50 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {\left (i c \,{\mathrm e}^{-3 i \left (b x +a \right )} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}\right )^{\frac {2}{3}} \left (4 \,\operatorname {expIntegral}_{1}\left (-2 i b x \right ) {\mathrm e}^{2 i \left (b x +2 a \right )} x^{2} b^{2}+4 \,{\mathrm e}^{2 i b x} \operatorname {expIntegral}_{1}\left (2 i b x \right ) x^{2} b^{2}-2 i {\mathrm e}^{4 i \left (b x +a \right )} x b +2 i b x +2 \,{\mathrm e}^{2 i \left (b x +a \right )}-{\mathrm e}^{4 i \left (b x +a \right )}-1\right )}{8 \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2} x^{2}}\)	\(137\)

Input:

int((c*sin(b*x+a)^3)^(2/3)/x^3,x,method=_RETURNVERBOSE)
 

Output:

1/8*(I*c*exp(-3*I*(b*x+a))*(exp(2*I*(b*x+a))-1)^3)^(2/3)*(4*Ei(1,-2*I*b*x) 
*exp(2*I*(b*x+2*a))*x^2*b^2+4*exp(2*I*b*x)*Ei(1,2*I*b*x)*x^2*b^2-2*I*exp(4 
*I*(b*x+a))*x*b+2*I*b*x+2*exp(2*I*(b*x+a))-exp(4*I*(b*x+a))-1)/(exp(2*I*(b 
*x+a))-1)^2/x^2
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x^3} \, dx=-\frac {{\left (b^{2} x^{2} {\rm Ei}\left (2 i \, b x\right ) e^{\left (2 i \, a\right )} + b^{2} x^{2} {\rm Ei}\left (-2 i \, b x\right ) e^{\left (-2 i \, a\right )} - 2 \, b x \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \cos \left (b x + a\right )^{2} - 1\right )} \left (-{\left (c \cos \left (b x + a\right )^{2} - c\right )} \sin \left (b x + a\right )\right )^{\frac {2}{3}}}{2 \, {\left (x^{2} \cos \left (b x + a\right )^{2} - x^{2}\right )}} \] Input:

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

Output:

-1/2*(b^2*x^2*Ei(2*I*b*x)*e^(2*I*a) + b^2*x^2*Ei(-2*I*b*x)*e^(-2*I*a) - 2* 
b*x*cos(b*x + a)*sin(b*x + a) + cos(b*x + a)^2 - 1)*(-(c*cos(b*x + a)^2 - 
c)*sin(b*x + a))^(2/3)/(x^2*cos(b*x + a)^2 - x^2)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.49 \[ \int \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x^3} \, dx=-\frac {{\left ({\left ({\left (-i \, \sqrt {3} + 1\right )} E_{3}\left (2 i \, b x\right ) + {\left (i \, \sqrt {3} + 1\right )} E_{3}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right )^{3} - {\left ({\left (\sqrt {3} + i\right )} E_{3}\left (2 i \, b x\right ) + {\left (\sqrt {3} - i\right )} E_{3}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right )^{3} + {\left ({\left ({\left (-i \, \sqrt {3} + 1\right )} E_{3}\left (2 i \, b x\right ) + {\left (i \, \sqrt {3} + 1\right )} E_{3}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) - 2\right )} \sin \left (2 \, a\right )^{2} + {\left ({\left (i \, \sqrt {3} + 1\right )} E_{3}\left (2 i \, b x\right ) + {\left (-i \, \sqrt {3} + 1\right )} E_{3}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right ) - 2 \, \cos \left (2 \, a\right )^{2} - {\left ({\left ({\left (\sqrt {3} + i\right )} E_{3}\left (2 i \, b x\right ) + {\left (\sqrt {3} - i\right )} E_{3}\left (-2 i \, b x\right )\right )} \cos \left (2 \, a\right )^{2} - {\left (\sqrt {3} - i\right )} E_{3}\left (2 i \, b x\right ) - {\left (\sqrt {3} + i\right )} E_{3}\left (-2 i \, b x\right )\right )} \sin \left (2 \, a\right )\right )} b^{2} c^{\frac {2}{3}}}{16 \, {\left (a^{2} \cos \left (2 \, a\right )^{2} + a^{2} \sin \left (2 \, a\right )^{2} + {\left (b x + a\right )}^{2} {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} - 2 \, {\left (a \cos \left (2 \, a\right )^{2} + a \sin \left (2 \, a\right )^{2}\right )} {\left (b x + a\right )}\right )}} \] Input:

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

Output:

-1/16*(((-I*sqrt(3) + 1)*exp_integral_e(3, 2*I*b*x) + (I*sqrt(3) + 1)*exp_ 
integral_e(3, -2*I*b*x))*cos(2*a)^3 - ((sqrt(3) + I)*exp_integral_e(3, 2*I 
*b*x) + (sqrt(3) - I)*exp_integral_e(3, -2*I*b*x))*sin(2*a)^3 + (((-I*sqrt 
(3) + 1)*exp_integral_e(3, 2*I*b*x) + (I*sqrt(3) + 1)*exp_integral_e(3, -2 
*I*b*x))*cos(2*a) - 2)*sin(2*a)^2 + ((I*sqrt(3) + 1)*exp_integral_e(3, 2*I 
*b*x) + (-I*sqrt(3) + 1)*exp_integral_e(3, -2*I*b*x))*cos(2*a) - 2*cos(2*a 
)^2 - (((sqrt(3) + I)*exp_integral_e(3, 2*I*b*x) + (sqrt(3) - I)*exp_integ 
ral_e(3, -2*I*b*x))*cos(2*a)^2 - (sqrt(3) - I)*exp_integral_e(3, 2*I*b*x) 
- (sqrt(3) + I)*exp_integral_e(3, -2*I*b*x))*sin(2*a))*b^2*c^(2/3)/(a^2*co 
s(2*a)^2 + a^2*sin(2*a)^2 + (b*x + a)^2*(cos(2*a)^2 + sin(2*a)^2) - 2*(a*c 
os(2*a)^2 + a*sin(2*a)^2)*(b*x + a))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c \sin ^3(a+b x)\right )^{2/3}}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(c**(2/3)*(2*cos(a + b*x)*sin(a + b*x)*tan((a + b*x)/2)**4*b*x + 4*cos(a + 
 b*x)*sin(a + b*x)*tan((a + b*x)/2)**2*b*x + 2*cos(a + b*x)*sin(a + b*x)*b 
*x - 8*cos(a + b*x)*tan((a + b*x)/2)**4 - 16*cos(a + b*x)*tan((a + b*x)/2) 
**2 - 8*cos(a + b*x) - 48*int(tan((a + b*x)/2)**2/(tan((a + b*x)/2)**4*x + 
 2*tan((a + b*x)/2)**2*x + x),x)*tan((a + b*x)/2)**4*b**2*x**2 - 96*int(ta 
n((a + b*x)/2)**2/(tan((a + b*x)/2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x) 
*tan((a + b*x)/2)**2*b**2*x**2 - 48*int(tan((a + b*x)/2)**2/(tan((a + b*x) 
/2)**4*x + 2*tan((a + b*x)/2)**2*x + x),x)*b**2*x**2 + 6*log(x)*tan((a + b 
*x)/2)**4*b**2*x**2 + 12*log(x)*tan((a + b*x)/2)**2*b**2*x**2 + 6*log(x)*b 
**2*x**2 + sin(a + b*x)**2*tan((a + b*x)/2)**4 + 2*sin(a + b*x)**2*tan((a 
+ b*x)/2)**2 + sin(a + b*x)**2 + 8*sin(a + b*x)*tan((a + b*x)/2)**4*b*x + 
16*sin(a + b*x)*tan((a + b*x)/2)**2*b*x + 8*sin(a + b*x)*b*x - 8*tan((a + 
b*x)/2)**4 - 16*tan((a + b*x)/2)**2 - 32*tan((a + b*x)/2)*b*x + 8))/(6*x** 
2*(tan((a + b*x)/2)**4 + 2*tan((a + b*x)/2)**2 + 1))