\(\int \frac {\cos ^4(a+b x)}{x^3} \, dx\) [28]

Optimal result

Integrand size = 12, antiderivative size = 90 \[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=-\frac {\cos ^4(a+b x)}{2 x^2}-b^2 \cos (2 a) \operatorname {CosIntegral}(2 b x)-b^2 \cos (4 a) \operatorname {CosIntegral}(4 b x)+\frac {2 b \cos ^3(a+b x) \sin (a+b x)}{x}+b^2 \sin (2 a) \text {Si}(2 b x)+b^2 \sin (4 a) \text {Si}(4 b x) \] Output:

-1/2*cos(b*x+a)^4/x^2-b^2*cos(2*a)*Ci(2*b*x)-b^2*cos(4*a)*Ci(4*b*x)+2*b*co 
s(b*x+a)^3*sin(b*x+a)/x+b^2*sin(2*a)*Si(2*b*x)+b^2*sin(4*a)*Si(4*b*x)
 

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=-\frac {3+4 \cos (2 (a+b x))+\cos (4 (a+b x))+16 b^2 x^2 \cos (2 a) \operatorname {CosIntegral}(2 b x)+16 b^2 x^2 \cos (4 a) \operatorname {CosIntegral}(4 b x)-8 b x \sin (2 (a+b x))-4 b x \sin (4 (a+b x))-16 b^2 x^2 \sin (2 a) \text {Si}(2 b x)-16 b^2 x^2 \sin (4 a) \text {Si}(4 b x)}{16 x^2} \] Input:

Integrate[Cos[a + b*x]^4/x^3,x]
 

Output:

-1/16*(3 + 4*Cos[2*(a + b*x)] + Cos[4*(a + b*x)] + 16*b^2*x^2*Cos[2*a]*Cos 
Integral[2*b*x] + 16*b^2*x^2*Cos[4*a]*CosIntegral[4*b*x] - 8*b*x*Sin[2*(a 
+ b*x)] - 4*b*x*Sin[4*(a + b*x)] - 16*b^2*x^2*Sin[2*a]*SinIntegral[2*b*x] 
- 16*b^2*x^2*Sin[4*a]*SinIntegral[4*b*x])/x^2
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.53, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 3795, 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.

\(\displaystyle \int \frac {\cos ^4(a+b x)}{x^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )^4}{x^3}dx\)

\(\Big \downarrow \) 3795

\(\displaystyle -8 b^2 \int \frac {\cos ^4(a+b x)}{x}dx+6 b^2 \int \frac {\cos ^2(a+b x)}{x}dx-\frac {\cos ^4(a+b x)}{2 x^2}+\frac {2 b \sin (a+b x) \cos ^3(a+b x)}{x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 6 b^2 \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )^2}{x}dx-8 b^2 \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )^4}{x}dx-\frac {\cos ^4(a+b x)}{2 x^2}+\frac {2 b \sin (a+b x) \cos ^3(a+b x)}{x}\)

\(\Big \downarrow \) 3793

\(\displaystyle 6 b^2 \int \left (\frac {\cos (2 a+2 b x)}{2 x}+\frac {1}{2 x}\right )dx-8 b^2 \int \left (\frac {\cos (2 a+2 b x)}{2 x}+\frac {\cos (4 a+4 b x)}{8 x}+\frac {3}{8 x}\right )dx-\frac {\cos ^4(a+b x)}{2 x^2}+\frac {2 b \sin (a+b x) \cos ^3(a+b x)}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 6 b^2 \left (\frac {1}{2} \cos (2 a) \operatorname {CosIntegral}(2 b x)-\frac {1}{2} \sin (2 a) \text {Si}(2 b x)+\frac {\log (x)}{2}\right )-8 b^2 \left (\frac {1}{2} \cos (2 a) \operatorname {CosIntegral}(2 b x)+\frac {1}{8} \cos (4 a) \operatorname {CosIntegral}(4 b x)-\frac {1}{2} \sin (2 a) \text {Si}(2 b x)-\frac {1}{8} \sin (4 a) \text {Si}(4 b x)+\frac {3 \log (x)}{8}\right )-\frac {\cos ^4(a+b x)}{2 x^2}+\frac {2 b \sin (a+b x) \cos ^3(a+b x)}{x}\)

Input:

Int[Cos[a + b*x]^4/x^3,x]
 

Output:

-1/2*Cos[a + b*x]^4/x^2 + (2*b*Cos[a + b*x]^3*Sin[a + b*x])/x + 6*b^2*((Co 
s[2*a]*CosIntegral[2*b*x])/2 + Log[x]/2 - (Sin[2*a]*SinIntegral[2*b*x])/2) 
 - 8*b^2*((Cos[2*a]*CosIntegral[2*b*x])/2 + (Cos[4*a]*CosIntegral[4*b*x])/ 
8 + (3*Log[x])/8 - (Sin[2*a]*SinIntegral[2*b*x])/2 - (Sin[4*a]*SinIntegral 
[4*b*x])/8)
 

Defintions of rubi rules used

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]
 

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\cos \left (4 b x +4 a \right )}{16 b^{2} x^{2}}+\frac {\sin \left (4 b x +4 a \right )}{4 b x}+\sin \left (4 a \right ) \operatorname {Si}\left (4 b x \right )-\cos \left (4 a \right ) \operatorname {Ci}\left (4 b x \right )-\frac {\cos \left (2 b x +2 a \right )}{4 b^{2} x^{2}}+\frac {\sin \left (2 b x +2 a \right )}{2 b x}+\sin \left (2 a \right ) \operatorname {Si}\left (2 b x \right )-\cos \left (2 a \right ) \operatorname {Ci}\left (2 b x \right )-\frac {3}{16 x^{2} b^{2}}\right )\)	\(124\)
default	\(b^{2} \left (-\frac {\cos \left (4 b x +4 a \right )}{16 b^{2} x^{2}}+\frac {\sin \left (4 b x +4 a \right )}{4 b x}+\sin \left (4 a \right ) \operatorname {Si}\left (4 b x \right )-\cos \left (4 a \right ) \operatorname {Ci}\left (4 b x \right )-\frac {\cos \left (2 b x +2 a \right )}{4 b^{2} x^{2}}+\frac {\sin \left (2 b x +2 a \right )}{2 b x}+\sin \left (2 a \right ) \operatorname {Si}\left (2 b x \right )-\cos \left (2 a \right ) \operatorname {Ci}\left (2 b x \right )-\frac {3}{16 x^{2} b^{2}}\right )\)	\(124\)
risch	\(-\frac {8 i {\mathrm e}^{-2 i a} \pi \,\operatorname {csgn}\left (b x \right ) b^{2} x^{2}+8 i \pi \,\operatorname {csgn}\left (b x \right ) {\mathrm e}^{-4 i a} b^{2} x^{2}-16 i {\mathrm e}^{-2 i a} \operatorname {Si}\left (2 b x \right ) b^{2} x^{2}-16 i \operatorname {Si}\left (4 b x \right ) {\mathrm e}^{-4 i a} b^{2} x^{2}-8 \,{\mathrm e}^{-2 i a} \operatorname {expIntegral}_{1}\left (-2 i b x \right ) b^{2} x^{2}-8 \,{\mathrm e}^{2 i a} \operatorname {expIntegral}_{1}\left (-2 i b x \right ) x^{2} b^{2}-8 \,{\mathrm e}^{4 i a} \operatorname {expIntegral}_{1}\left (-4 i b x \right ) x^{2} b^{2}-8 \,\operatorname {expIntegral}_{1}\left (-4 i b x \right ) {\mathrm e}^{-4 i a} b^{2} x^{2}-4 x \sin \left (4 b x +4 a \right ) b -8 x \sin \left (2 b x +2 a \right ) b +\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3}{16 x^{2}}\)	\(210\)

Input:

int(cos(b*x+a)^4/x^3,x,method=_RETURNVERBOSE)
 

Output:

b^2*(-1/16*cos(4*b*x+4*a)/b^2/x^2+1/4*sin(4*b*x+4*a)/b/x+sin(4*a)*Si(4*b*x 
)-cos(4*a)*Ci(4*b*x)-1/4*cos(2*b*x+2*a)/b^2/x^2+1/2*sin(2*b*x+2*a)/b/x+sin 
(2*a)*Si(2*b*x)-cos(2*a)*Ci(2*b*x)-3/16/x^2/b^2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=-\frac {2 \, b^{2} x^{2} \cos \left (4 \, a\right ) \operatorname {Ci}\left (4 \, b x\right ) + 2 \, b^{2} x^{2} \cos \left (2 \, a\right ) \operatorname {Ci}\left (2 \, b x\right ) - 4 \, b x \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) - 2 \, b^{2} x^{2} \sin \left (4 \, a\right ) \operatorname {Si}\left (4 \, b x\right ) - 2 \, b^{2} x^{2} \sin \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x\right ) + \cos \left (b x + a\right )^{4}}{2 \, x^{2}} \] Input:

integrate(cos(b*x+a)^4/x^3,x, algorithm="fricas")
 

Output:

-1/2*(2*b^2*x^2*cos(4*a)*cos_integral(4*b*x) + 2*b^2*x^2*cos(2*a)*cos_inte 
gral(2*b*x) - 4*b*x*cos(b*x + a)^3*sin(b*x + a) - 2*b^2*x^2*sin(4*a)*sin_i 
ntegral(4*b*x) - 2*b^2*x^2*sin(2*a)*sin_integral(2*b*x) + cos(b*x + a)^4)/ 
x^2
 

Sympy [F]

\[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=\int \frac {\cos ^{4}{\left (a + b x \right )}}{x^{3}}\, dx \] Input:

integrate(cos(b*x+a)**4/x**3,x)
 

Output:

Integral(cos(a + b*x)**4/x**3, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 790, normalized size of antiderivative = 8.78 \[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(b*x+a)^4/x^3,x, algorithm="maxima")
 

Output:

-1/32*(((exp_integral_e(3, 4*I*b*x) + exp_integral_e(3, -4*I*b*x))*cos(2*a 
)^2 + (exp_integral_e(3, 4*I*b*x) + exp_integral_e(3, -4*I*b*x))*sin(2*a)^ 
2)*cos(4*a)^3 + ((-I*exp_integral_e(3, 4*I*b*x) + I*exp_integral_e(3, -4*I 
*b*x))*cos(2*a)^2 + (-I*exp_integral_e(3, 4*I*b*x) + I*exp_integral_e(3, - 
4*I*b*x))*sin(2*a)^2)*sin(4*a)^3 + 2*(2*(exp_integral_e(3, 2*I*b*x) + exp_ 
integral_e(3, -2*I*b*x))*cos(2*a)^3 + 2*(-I*exp_integral_e(3, 2*I*b*x) + I 
*exp_integral_e(3, -2*I*b*x))*sin(2*a)^3 + (2*(exp_integral_e(3, 2*I*b*x) 
+ exp_integral_e(3, -2*I*b*x))*cos(2*a) + 3)*sin(2*a)^2 + 2*(exp_integral_ 
e(3, 2*I*b*x) + exp_integral_e(3, -2*I*b*x))*cos(2*a) + 3*cos(2*a)^2 + 2*( 
(-I*exp_integral_e(3, 2*I*b*x) + I*exp_integral_e(3, -2*I*b*x))*cos(2*a)^2 
 - I*exp_integral_e(3, 2*I*b*x) + I*exp_integral_e(3, -2*I*b*x))*sin(2*a)) 
*cos(4*a)^2 + (4*(exp_integral_e(3, 2*I*b*x) + exp_integral_e(3, -2*I*b*x) 
)*cos(2*a)^3 + 4*(-I*exp_integral_e(3, 2*I*b*x) + I*exp_integral_e(3, -2*I 
*b*x))*sin(2*a)^3 + 2*(2*(exp_integral_e(3, 2*I*b*x) + exp_integral_e(3, - 
2*I*b*x))*cos(2*a) + 3)*sin(2*a)^2 + ((exp_integral_e(3, 4*I*b*x) + exp_in 
tegral_e(3, -4*I*b*x))*cos(2*a)^2 + (exp_integral_e(3, 4*I*b*x) + exp_inte 
gral_e(3, -4*I*b*x))*sin(2*a)^2)*cos(4*a) + 4*(exp_integral_e(3, 2*I*b*x) 
+ exp_integral_e(3, -2*I*b*x))*cos(2*a) + 6*cos(2*a)^2 + 4*((-I*exp_integr 
al_e(3, 2*I*b*x) + I*exp_integral_e(3, -2*I*b*x))*cos(2*a)^2 - I*exp_integ 
ral_e(3, 2*I*b*x) + I*exp_integral_e(3, -2*I*b*x))*sin(2*a))*sin(4*a)^2...
 

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.39 (sec) , antiderivative size = 3920, normalized size of antiderivative = 43.56 \[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=\text {Too large to display} \] Input:

integrate(cos(b*x+a)^4/x^3,x, algorithm="giac")
 

Output:

1/8*(4*b^2*x^2*real_part(cos_integral(4*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan( 
2*a)^2*tan(a)^2 + 4*b^2*x^2*real_part(cos_integral(2*b*x))*tan(2*b*x)^2*ta 
n(b*x)^2*tan(2*a)^2*tan(a)^2 + 4*b^2*x^2*real_part(cos_integral(-2*b*x))*t 
an(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2 + 4*b^2*x^2*real_part(cos_integ 
ral(-4*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a)^2 + 8*b^2*x^2*imag_ 
part(cos_integral(2*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*tan(a) - 8*b^ 
2*x^2*imag_part(cos_integral(-2*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2*t 
an(a) + 16*b^2*x^2*sin_integral(2*b*x)*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2* 
tan(a) + 8*b^2*x^2*imag_part(cos_integral(4*b*x))*tan(2*b*x)^2*tan(b*x)^2* 
tan(2*a)*tan(a)^2 - 8*b^2*x^2*imag_part(cos_integral(-4*b*x))*tan(2*b*x)^2 
*tan(b*x)^2*tan(2*a)*tan(a)^2 + 16*b^2*x^2*sin_integral(4*b*x)*tan(2*b*x)^ 
2*tan(b*x)^2*tan(2*a)*tan(a)^2 + 4*b^2*x^2*real_part(cos_integral(4*b*x))* 
tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2 - 4*b^2*x^2*real_part(cos_integral(2*b* 
x))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2 - 4*b^2*x^2*real_part(cos_integral( 
-2*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2 + 4*b^2*x^2*real_part(cos_inte 
gral(-4*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(2*a)^2 - 4*b^2*x^2*real_part(cos 
_integral(4*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(a)^2 + 4*b^2*x^2*real_part(c 
os_integral(2*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(a)^2 + 4*b^2*x^2*real_part 
(cos_integral(-2*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(a)^2 - 4*b^2*x^2*real_p 
art(cos_integral(-4*b*x))*tan(2*b*x)^2*tan(b*x)^2*tan(a)^2 + 4*b^2*x^2*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^4(a+b x)}{x^3} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^4}{x^3} \,d x \] Input:

int(cos(a + b*x)^4/x^3,x)
 

Output:

int(cos(a + b*x)^4/x^3, x)
 

Reduce [F]

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

int(cos(b*x+a)^4/x^3,x)
 

Output:

int(cos(a + b*x)**4/x**3,x)