Integrand size = 23, antiderivative size = 136 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b \cos (a+b x) \sin (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}+\frac {4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3} \] Output:
-3/2*cos(b*x+a)^2/d/(d*x+c)^2-4*b^2*cos(2*a-2*b*c/d)*Ci(2*b*c/d+2*b*x)/d^3 +4*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+4*b^2* sin(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/d^3
Time = 0.71 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {8 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d (d+2 d \cos (2 (a+b x))-4 b (c+d x) \sin (2 (a+b x)))}{(c+d x)^2}-8 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{2 d^3} \] Input:
Integrate[(Csc[a + b*x]*Sin[3*a + 3*b*x])/(c + d*x)^3,x]
Output:
-1/2*(8*b^2*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*(c + d*x))/d] + (d*(d + 2*d*Cos[2*(a + b*x)] - 4*b*(c + d*x)*Sin[2*(a + b*x)]))/(c + d*x)^2 - 8*b^ 2*Sin[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d])/d^3
Time = 0.57 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
\(\displaystyle \int \frac {\sin (3 a+3 b x) \csc (a+b x)}{(c+d x)^3} \, dx\) |
\(\Big \downarrow \) 4931 |
\(\displaystyle \int \left (\frac {3 \cos ^2(a+b x)}{(c+d x)^3}-\frac {\sin ^2(a+b x)}{(c+d x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}+\frac {4 b \sin (a+b x) \cos (a+b x)}{d^2 (c+d x)}+\frac {\sin ^2(a+b x)}{2 d (c+d x)^2}-\frac {3 \cos ^2(a+b x)}{2 d (c+d x)^2}\) |
Input:
Int[(Csc[a + b*x]*Sin[3*a + 3*b*x])/(c + d*x)^3,x]
Output:
(-3*Cos[a + b*x]^2)/(2*d*(c + d*x)^2) - (4*b^2*Cos[2*a - (2*b*c)/d]*CosInt egral[(2*b*c)/d + 2*b*x])/d^3 + (4*b*Cos[a + b*x]*Sin[a + b*x])/(d^2*(c + d*x)) + Sin[a + b*x]^2/(2*d*(c + d*x)^2) + (4*b^2*Sin[2*a - (2*b*c)/d]*Sin Integral[(2*b*c)/d + 2*b*x])/d^3
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]
Time = 1.76 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.52
method | result | size |
default | \(\frac {1}{2 d \left (d x +c \right )^{2}}+\frac {b^{3} \left (-\frac {\cos \left (2 b x +2 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right )^{2} d}-\frac {-\frac {2 \sin \left (2 b x +2 a \right )}{\left (-a d +c b +d \left (b x +a \right )\right ) d}+\frac {\frac {4 \,\operatorname {Si}\left (2 b x +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}+\frac {4 \,\operatorname {Ci}\left (2 b x +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}}{d}}{d}\right )-\frac {b^{3}}{\left (-a d +c b +d \left (b x +a \right )\right )^{2} d}}{b}\) | \(207\) |
risch | \(-\frac {1}{2 d \left (d x +c \right )^{2}}+\frac {2 b^{2} {\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{d^{3}}+\frac {2 b^{2} {\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {expIntegral}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{d^{3}}+\frac {\left (-2 b^{2} d^{3} x^{2}-4 b^{2} c \,d^{2} x -2 b^{2} c^{2} d \right ) \cos \left (2 b x +2 a \right )}{2 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}-\frac {i \left (4 i b^{3} d^{3} x^{3}+12 i b^{3} c \,d^{2} x^{2}+12 i b^{3} c^{2} d x +4 i c^{3} b^{3}\right ) \sin \left (2 b x +2 a \right )}{2 d^{2} \left (x^{2} d^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right ) \left (d x +c \right )^{2}}\) | \(290\) |
Input:
int(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)^3,x,method=_RETURNVERBOSE)
Output:
1/2/d/(d*x+c)^2+4/b*(1/4*b^3*(-cos(2*b*x+2*a)/(-a*d+c*b+d*(b*x+a))^2/d-(-2 *sin(2*b*x+2*a)/(-a*d+c*b+d*(b*x+a))/d+2*(2*Si(2*b*x+2*a+2*(-a*d+b*c)/d)*s in(2*(-a*d+b*c)/d)/d+2*Ci(2*b*x+2*a+2*(-a*d+b*c)/d)*cos(2*(-a*d+b*c)/d)/d) /d)/d)-1/4*b^3/(-a*d+c*b+d*(b*x+a))^2/d)
Time = 0.09 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.34 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {4 \, d^{2} \cos \left (b x + a\right )^{2} + 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - 8 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 8 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - d^{2}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \] Input:
integrate(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)^3,x, algorithm="fricas")
Output:
-1/2*(4*d^2*cos(b*x + a)^2 + 8*(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) - 8*(b*d^2*x + b*c*d)*cos (b*x + a)*sin(b*x + a) - 8*(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)
Timed out. \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)**3,x)
Output:
Timed out
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.97 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {2 \, {\left (E_{3}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + E_{3}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left (i \, E_{3}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) - i \, E_{3}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 1}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} \] Input:
integrate(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)^3,x, algorithm="maxima")
Output:
-1/2*(2*(exp_integral_e(3, 2*(-I*b*d*x - I*b*c)/d) + exp_integral_e(3, -2* (-I*b*d*x - I*b*c)/d))*cos(-2*(b*c - a*d)/d) + 2*(I*exp_integral_e(3, 2*(- I*b*d*x - I*b*c)/d) - I*exp_integral_e(3, -2*(-I*b*d*x - I*b*c)/d))*sin(-2 *(b*c - a*d)/d) + 1)/(d^3*x^2 + 2*c*d^2*x + c^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 704 vs. \(2 (132) = 264\).
Time = 0.20 (sec) , antiderivative size = 704, normalized size of antiderivative = 5.18 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx =\text {Too large to display} \] Input:
integrate(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)^3,x, algorithm="giac")
Output:
-1/2*(8*b^5*c^2*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) + 16*(b*x + a)*b^4*c*d*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) - 16*a*b^4*c*d*cos(-2*(b*c - a*d)/d)*cos_integral(2 *(b*c + (b*x + a)*d - a*d)/d) + 8*(b*x + a)^2*b^3*d^2*cos(-2*(b*c - a*d)/d )*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) - 16*(b*x + a)*a*b^3*d^2*cos (-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) + 8*a^2*b^3 *d^2*cos(-2*(b*c - a*d)/d)*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d) + 8 *b^5*c^2*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d ) + 16*(b*x + a)*b^4*c*d*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 16*a*b^4*c*d*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b* c + (b*x + a)*d - a*d)/d) + 8*(b*x + a)^2*b^3*d^2*sin(-2*(b*c - a*d)/d)*si n_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 16*(b*x + a)*a*b^3*d^2*sin(-2 *(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 8*a^2*b^3*d ^2*sin(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 4* b^4*c*d*sin(2*b*x + 2*a) - 4*(b*x + a)*b^3*d^2*sin(2*b*x + 2*a) + 4*a*b^3* d^2*sin(2*b*x + 2*a) + 2*b^3*d^2*cos(2*b*x + 2*a) + b^3*d^2)/((b^2*c^2*d^3 + 2*(b*x + a)*b*c*d^4 - 2*a*b*c*d^4 + (b*x + a)^2*d^5 - 2*(b*x + a)*a*d^5 + a^2*d^5)*b)
Timed out. \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )}{\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^3} \,d x \] Input:
int(sin(3*a + 3*b*x)/(sin(a + b*x)*(c + d*x)^3),x)
Output:
int(sin(3*a + 3*b*x)/(sin(a + b*x)*(c + d*x)^3), x)
\[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\csc \left (b x +a \right ) \sin \left (3 b x +3 a \right )}{\left (d x +c \right )^{3}}d x \] Input:
int(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)^3,x)
Output:
int(csc(b*x+a)*sin(3*b*x+3*a)/(d*x+c)^3,x)