\(\int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx\) [367]

Optimal result

Integrand size = 14, antiderivative size = 99 \[ \int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx=-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}-\frac {4 \cos \left (\frac {2 c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 c}{d}+2 x\right )}{d^3}+\frac {4 \cos (x) \sin (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {4 \sin \left (\frac {2 c}{d}\right ) \text {Si}\left (\frac {2 c}{d}+2 x\right )}{d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx=\frac {-8 \cos \left (\frac {2 c}{d}\right ) \operatorname {CosIntegral}\left (2 \left (\frac {c}{d}+x\right )\right )+\frac {d (-d-2 d \cos (2 x)+4 (c+d x) \sin (2 x))}{(c+d x)^2}-8 \sin \left (\frac {2 c}{d}\right ) \text {Si}\left (2 \left (\frac {c}{d}+x\right )\right )}{2 d^3} \] Input:

Integrate[(Csc[x]*Sin[3*x])/(c + d*x)^3,x]
 

Output:

(-8*Cos[(2*c)/d]*CosIntegral[2*(c/d + x)] + (d*(-d - 2*d*Cos[2*x] + 4*(c + 
 d*x)*Sin[2*x]))/(c + d*x)^2 - 8*Sin[(2*c)/d]*SinIntegral[2*(c/d + x)])/(2 
*d^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4931, 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.

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

\(\Big \downarrow \) 4931

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 \cos \left (\frac {2 c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 c}{d}+2 x\right )}{d^3}-\frac {4 \sin \left (\frac {2 c}{d}\right ) \text {Si}\left (\frac {2 c}{d}+2 x\right )}{d^3}+\frac {4 \sin (x) \cos (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}\)

Input:

Int[(Csc[x]*Sin[3*x])/(c + d*x)^3,x]
 

Output:

(-3*Cos[x]^2)/(2*d*(c + d*x)^2) - (4*Cos[(2*c)/d]*CosIntegral[(2*c)/d + 2* 
x])/d^3 + (4*Cos[x]*Sin[x])/(d^2*(c + d*x)) + Sin[x]^2/(2*d*(c + d*x)^2) - 
 (4*Sin[(2*c)/d]*SinIntegral[(2*c)/d + 2*x])/d^3
 

Defintions of rubi rules used

rule 2009

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

rule 4931

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + 
(d_.)*(x_)]^(q_.), x_Symbol] :> Int[ExpandTrigExpand[(e + f*x)^m*G[c + d*x] 
^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Member 
Q[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && E 
qQ[b*c - a*d, 0] && IGtQ[b/d, 1]
 

Maple [A] (verified)

Time = 2.01 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.11

method	result	size
default	\(-\frac {2}{\left (2 d x +2 c \right )^{2} d}-\frac {4 \cos \left (2 x \right )}{\left (2 d x +2 c \right )^{2} d}-\frac {4 \left (-\frac {\sin \left (2 x \right )}{\left (2 d x +2 c \right ) d}+\frac {\frac {\sin \left (\frac {2 c}{d}\right ) \operatorname {Si}\left (\frac {2 c}{d}+2 x \right )}{d}+\frac {\cos \left (\frac {2 c}{d}\right ) \operatorname {Ci}\left (\frac {2 c}{d}+2 x \right )}{d}}{d}\right )}{d}\)	\(110\)
risch	\(-\frac {1}{2 d \left (d x +c \right )^{2}}+\frac {2 \,{\mathrm e}^{\frac {2 i c}{d}} \operatorname {expIntegral}_{1}\left (2 i x +\frac {2 i c}{d}\right )}{d^{3}}+\frac {2 \,{\mathrm e}^{-\frac {2 i c}{d}} \operatorname {expIntegral}_{1}\left (-2 i x -\frac {2 i c}{d}\right )}{d^{3}}-\frac {\cos \left (2 x \right )}{d \left (d x +c \right )^{2}}+\frac {i \left (-4 i d x -4 i c \right ) \sin \left (2 x \right )}{2 \left (d x +c \right )^{2} d^{2}}\)	\(111\)

Input:

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

Output:

-2/(2*d*x+2*c)^2/d-4*cos(2*x)/(2*d*x+2*c)^2/d-4*(-sin(2*x)/(2*d*x+2*c)/d+( 
sin(2*c/d)*Si(2*c/d+2*x)/d+cos(2*c/d)*Ci(2*c/d+2*x)/d)/d)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.29 \[ \int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx=-\frac {4 \, d^{2} \cos \left (x\right )^{2} + 8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) - 8 \, {\left (d^{2} x + c d\right )} \cos \left (x\right ) \sin \left (x\right ) + 8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) - d^{2}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

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

Output:

-1/2*(4*d^2*cos(x)^2 + 8*(d^2*x^2 + 2*c*d*x + c^2)*cos(2*c/d)*cos_integral 
(2*(d*x + c)/d) - 8*(d^2*x + c*d)*cos(x)*sin(x) + 8*(d^2*x^2 + 2*c*d*x + c 
^2)*sin(2*c/d)*sin_integral(2*(d*x + c)/d) - d^2)/(d^5*x^2 + 2*c*d^4*x + c 
^2*d^3)
 

Sympy [F]

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

integrate(csc(x)*sin(3*x)/(d*x+c)**3,x)
 

Output:

Integral(sin(3*x)*csc(x)/(c + d*x)**3, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.69 \[ \int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx=-\frac {{\left (E_{3}\left (\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right )^{3} - {\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right )\right )} \sin \left (\frac {2 \, c}{d}\right )^{3} + {\left ({\left (E_{3}\left (\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right ) + 1\right )} \sin \left (\frac {2 \, c}{d}\right )^{2} + {\left (E_{3}\left (\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right ) + \cos \left (\frac {2 \, c}{d}\right )^{2} - {\left ({\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right )^{2} + i \, E_{3}\left (\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, d x - i \, c\right )}}{d}\right )\right )} \sin \left (\frac {2 \, c}{d}\right )}{2 \, {\left ({\left (\cos \left (\frac {2 \, c}{d}\right )^{2} + \sin \left (\frac {2 \, c}{d}\right )^{2}\right )} d^{3} x^{2} + 2 \, {\left (c \cos \left (\frac {2 \, c}{d}\right )^{2} + c \sin \left (\frac {2 \, c}{d}\right )^{2}\right )} d^{2} x + {\left (c^{2} \cos \left (\frac {2 \, c}{d}\right )^{2} + c^{2} \sin \left (\frac {2 \, c}{d}\right )^{2}\right )} d\right )}} \] Input:

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

Output:

-1/2*((exp_integral_e(3, 2*(-I*d*x - I*c)/d) + exp_integral_e(3, -2*(-I*d* 
x - I*c)/d))*cos(2*c/d)^3 - (I*exp_integral_e(3, 2*(-I*d*x - I*c)/d) - I*e 
xp_integral_e(3, -2*(-I*d*x - I*c)/d))*sin(2*c/d)^3 + ((exp_integral_e(3, 
2*(-I*d*x - I*c)/d) + exp_integral_e(3, -2*(-I*d*x - I*c)/d))*cos(2*c/d) + 
 1)*sin(2*c/d)^2 + (exp_integral_e(3, 2*(-I*d*x - I*c)/d) + exp_integral_e 
(3, -2*(-I*d*x - I*c)/d))*cos(2*c/d) + cos(2*c/d)^2 - ((I*exp_integral_e(3 
, 2*(-I*d*x - I*c)/d) - I*exp_integral_e(3, -2*(-I*d*x - I*c)/d))*cos(2*c/ 
d)^2 + I*exp_integral_e(3, 2*(-I*d*x - I*c)/d) - I*exp_integral_e(3, -2*(- 
I*d*x - I*c)/d))*sin(2*c/d))/((cos(2*c/d)^2 + sin(2*c/d)^2)*d^3*x^2 + 2*(c 
*cos(2*c/d)^2 + c*sin(2*c/d)^2)*d^2*x + (c^2*cos(2*c/d)^2 + c^2*sin(2*c/d) 
^2)*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (95) = 190\).

Time = 0.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 2.03 \[ \int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx=-\frac {8 \, d^{2} x^{2} \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 8 \, d^{2} x^{2} \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 16 \, c d x \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 16 \, c d x \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 8 \, c^{2} \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) - 4 \, d^{2} x \sin \left (2 \, x\right ) + 8 \, c^{2} \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 2 \, d^{2} \cos \left (2 \, x\right ) - 4 \, c d \sin \left (2 \, x\right ) + d^{2}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:

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

Output:

-1/2*(8*d^2*x^2*cos(2*c/d)*cos_integral(2*(d*x + c)/d) + 8*d^2*x^2*sin(2*c 
/d)*sin_integral(2*(d*x + c)/d) + 16*c*d*x*cos(2*c/d)*cos_integral(2*(d*x 
+ c)/d) + 16*c*d*x*sin(2*c/d)*sin_integral(2*(d*x + c)/d) + 8*c^2*cos(2*c/ 
d)*cos_integral(2*(d*x + c)/d) - 4*d^2*x*sin(2*x) + 8*c^2*sin(2*c/d)*sin_i 
ntegral(2*(d*x + c)/d) + 2*d^2*cos(2*x) - 4*c*d*sin(2*x) + d^2)/(d^5*x^2 + 
 2*c*d^4*x + c^2*d^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(csc(x)*sin(3*x)/(d*x+c)^3,x)
 

Output:

int(csc(x)*sin(3*x)/(d*x+c)^3,x)