\(\int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx\) [14]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx=\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}-\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}-\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3} \] Output:

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

Mathematica [A] (verified)

Time = 1.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx=-\frac {-2 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d \left (d \sin ^2(a+b x)+b (c+d x) \sin (2 (a+b x))\right )}{(c+d x)^2}+2 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{2 d^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3795, 16, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3795

\(\displaystyle -\frac {2 b^2 \int \frac {\sin ^2(a+b x)}{c+d x}dx}{d^2}+\frac {b^2 \int \frac {1}{c+d x}dx}{d^2}-\frac {b \sin (a+b x) \cos (a+b x)}{d^2 (c+d x)}-\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}\)

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3793

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(b^2*Log[c + d*x])/d^3 - (b*Cos[a + b*x]*Sin[a + b*x])/(d^2*(c + d*x)) - S 
in[a + b*x]^2/(2*d*(c + d*x)^2) - (2*b^2*(-1/2*(Cos[2*a - (2*b*c)/d]*CosIn 
tegral[(2*b*c)/d + 2*b*x])/d + Log[c + d*x]/(2*d) + (Sin[2*a - (2*b*c)/d]* 
SinIntegral[(2*b*c)/d + 2*b*x])/(2*d)))/d^2
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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]
 

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.71

method	result	size
derivativedivides	\(\frac {-\frac {b^{3}}{4 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}-\frac {b^{3} \left (-\frac {\cos \left (2 b x +2 a \right )}{\left (-a d +b c +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {4 \,\operatorname {Si}\left (-2 b x -2 a -\frac {2 \left (-a d +b c \right )}{d}\right ) \sin \left (\frac {-2 a d +2 b c}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 b c}{d}\right ) \cos \left (\frac {-2 a d +2 b c}{d}\right )}{d}}{d}}{d}\right )}{4}}{b}\)	\(193\)
default	\(\frac {-\frac {b^{3}}{4 \left (-a d +b c +d \left (b x +a \right )\right )^{2} d}-\frac {b^{3} \left (-\frac {\cos \left (2 b x +2 a \right )}{\left (-a d +b c +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-a d +b c +d \left (b x +a \right )\right ) d}+\frac {-\frac {4 \,\operatorname {Si}\left (-2 b x -2 a -\frac {2 \left (-a d +b c \right )}{d}\right ) \sin \left (\frac {-2 a d +2 b c}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 b c}{d}\right ) \cos \left (\frac {-2 a d +2 b c}{d}\right )}{d}}{d}}{d}\right )}{4}}{b}\)	\(193\)
risch	\(-\frac {b^{2} {\mathrm e}^{-\frac {2 i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -b c \right )}{d}\right )}{2 d^{3}}-\frac {b^{2} {\mathrm e}^{\frac {2 i \left (a d -b c \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i b c \right )}{d}\right )}{2 d^{3}}-\frac {1}{4 d \left (d x +c \right )^{2}}-\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (2 b x +2 a \right )}{8 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}+\frac {i \left (4 i b^{3} d^{3} x^{3}+12 i b^{3} c \,d^{2} x^{2}+12 i b^{3} c^{2} d x +4 i b^{3} c^{3}\right ) \sin \left (2 b x +2 a \right )}{8 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}\)	\(290\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.60 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx=\frac {d^{2} \cos \left (b x + a\right )^{2} + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - d^{2}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

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

Output:

1/2*(d^2*cos(b*x + a)^2 + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-2*( 
b*c - a*d)/d)*cos_integral(2*(b*d*x + b*c)/d) - 2*(b*d^2*x + b*c*d)*cos(b* 
x + a)*sin(b*x + a) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*sin(-2*(b*c 
- a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) - d^2)/(d^5*x^2 + 2*c*d^4*x + c^ 
2*d^3)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.82 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx=\frac {b^{3} {\left (E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, b c - i \, {\left (b x + a\right )} d + i \, a d\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{3}}{4 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \] Input:

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

Output:

1/4*(b^3*(exp_integral_e(3, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) + exp_in 
tegral_e(3, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*cos(-2*(b*c - a*d)/d) 
+ b^3*(I*exp_integral_e(3, 2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d) - I*exp_i 
ntegral_e(3, -2*(-I*b*c - I*(b*x + a)*d + I*a*d)/d))*sin(-2*(b*c - a*d)/d) 
 - b^3)/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 
 - a*d^3)*(b*x + a))*b)
 

Giac [C] (verification not implemented)

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

Time = 0.54 (sec) , antiderivative size = 5141, normalized size of antiderivative = 45.50 \[ \int \frac {\sin ^2(a+b x)}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(b^2*d^2*x^2*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a 
)^2*tan(b*c/d)^2 + b^2*d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*t 
an(b*x)^2*tan(a)^2*tan(b*c/d)^2 - 2*b^2*d^2*x^2*imag_part(cos_integral(2*b 
*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b^2*d^2*x^2*imag_part(co 
s_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 4*b^2*d^2*x 
^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b^2* 
d^2*x^2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c 
/d)^2 - 2*b^2*d^2*x^2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2 
*tan(a)*tan(b*c/d)^2 + 4*b^2*d^2*x^2*sin_integral(2*(b*d*x + b*c)/d)*tan(b 
*x)^2*tan(a)*tan(b*c/d)^2 + 2*b^2*c*d*x*real_part(cos_integral(2*b*x + 2*b 
*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + 2*b^2*c*d*x*real_part(cos_integr 
al(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b^2*d^2*x^2*real_ 
part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 - b^2*d^2*x^2*real 
_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 + 4*b^2*d^2*x^2* 
real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) + 4* 
b^2*d^2*x^2*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*ta 
n(b*c/d) - 4*b^2*c*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2 
*tan(a)^2*tan(b*c/d) + 4*b^2*c*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d 
))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 8*b^2*c*d*x*sin_integral(2*(b*d*x + b* 
c)/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d) - b^2*d^2*x^2*real_part(cos_integr...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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