\(\int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx\) [373]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx=-\frac {3 \cos ^2(a+b x)}{d (c+d x)}-\frac {4 b \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^2}+\frac {\sin ^2(a+b x)}{d (c+d x)}-\frac {4 b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4516, 3394, 12, 3384, 3380, 3383} \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx=-\frac {4 b \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}-\frac {4 b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {\sin ^2(a+b x)}{d (c+d x)}-\frac {3 \cos ^2(a+b x)}{d (c+d x)} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3394

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

Rule 4516

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] && M
emberQ[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && EqQ[b*c - a*d, 0] && IGtQ[b/d,
1]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 \cos ^2(a+b x)}{(c+d x)^2}-\frac {\sin ^2(a+b x)}{(c+d x)^2}\right ) \, dx \\ & = 3 \int \frac {\cos ^2(a+b x)}{(c+d x)^2} \, dx-\int \frac {\sin ^2(a+b x)}{(c+d x)^2} \, dx \\ & = -\frac {3 \cos ^2(a+b x)}{d (c+d x)}+\frac {\sin ^2(a+b x)}{d (c+d x)}-\frac {(2 b) \int \frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{d}+\frac {(6 b) \int -\frac {\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{d} \\ & = -\frac {3 \cos ^2(a+b x)}{d (c+d x)}+\frac {\sin ^2(a+b x)}{d (c+d x)}-\frac {b \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{d}-\frac {(3 b) \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{d} \\ & = -\frac {3 \cos ^2(a+b x)}{d (c+d x)}+\frac {\sin ^2(a+b x)}{d (c+d x)}-\frac {\left (b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}-\frac {\left (3 b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}-\frac {\left (b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}-\frac {\left (3 b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d} \\ & = -\frac {3 \cos ^2(a+b x)}{d (c+d x)}-\frac {4 b \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{d^2}+\frac {\sin ^2(a+b x)}{d (c+d x)}-\frac {4 b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx=-\frac {\frac {d (1+2 \cos (2 (a+b x)))}{c+d x}+4 b \operatorname {CosIntegral}\left (\frac {2 b (c+d x)}{d}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )+4 b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )}{d^2} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.97 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.52

method	result	size
risch	\(-\frac {1}{d \left (d x +c \right )}+\frac {2 i b \,{\mathrm e}^{-\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (2 i b x +2 i a -\frac {2 i \left (a d -c b \right )}{d}\right )}{d^{2}}-\frac {2 i b \,{\mathrm e}^{\frac {2 i \left (a d -c b \right )}{d}} \operatorname {Ei}_{1}\left (-2 i b x -2 i a -\frac {2 \left (-i a d +i c b \right )}{d}\right )}{d^{2}}-\frac {\left (-2 d x b -2 c b \right ) \cos \left (2 x b +2 a \right )}{d \left (-d x b -c b \right ) \left (d x +c \right )}\)	\(155\)
default	\(\frac {1}{d \left (d x +c \right )}+\frac {b^{2} \left (-\frac {2 \cos \left (2 x b +2 a \right )}{\left (-a d +c b +d \left (x b +a \right )\right ) d}-\frac {2 \left (\frac {2 \,\operatorname {Si}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \cos \left (\frac {-2 a d +2 c b}{d}\right )}{d}-\frac {2 \,\operatorname {Ci}\left (2 x b +2 a +\frac {-2 a d +2 c b}{d}\right ) \sin \left (\frac {-2 a d +2 c b}{d}\right )}{d}\right )}{d}\right )-\frac {2 b^{2}}{\left (-a d +c b +d \left (x b +a \right )\right ) d}}{b}\)	\(169\)

[In]

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

[Out]

-1/d/(d*x+c)+2*I*b/d^2*exp(-2*I*(a*d-b*c)/d)*Ei(1,2*I*b*x+2*I*a-2*I*(a*d-b*c)/d)-2*I*b/d^2*exp(2*I*(a*d-b*c)/d
)*Ei(1,-2*I*x*b-2*I*a-2*(-I*a*d+I*c*b)/d)-1/d*(-2*b*d*x-2*b*c)/(-b*d*x-b*c)/(d*x+c)*cos(2*b*x+2*a)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx=-\frac {4 \, d \cos \left (b x + a\right )^{2} + 4 \, {\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left (b d x + b c\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - d}{d^{3} x + c d^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.18 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx=-\frac {{\left (E_{2}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + E_{2}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (-i \, E_{2}\left (\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right ) + i \, E_{2}\left (-\frac {2 \, {\left (-i \, b d x - i \, b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 1}{d^{2} x + c d} \]

[In]

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

[Out]

-((exp_integral_e(2, 2*(-I*b*d*x - I*b*c)/d) + exp_integral_e(2, -2*(-I*b*d*x - I*b*c)/d))*cos(-2*(b*c - a*d)/
d) - (-I*exp_integral_e(2, 2*(-I*b*d*x - I*b*c)/d) + I*exp_integral_e(2, -2*(-I*b*d*x - I*b*c)/d))*sin(-2*(b*c
- a*d)/d) + 1)/(d^2*x + c*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 308 vs. \(2 (102) = 204\).

Time = 0.31 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.02 \[ \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx=-\frac {4 \, b^{3} c \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left (b x + a\right )} b^{2} d \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, a b^{2} d \operatorname {Ci}\left (\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, b^{3} c \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) - 4 \, {\left (b x + a\right )} b^{2} d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 4 \, a b^{2} d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left (b c + {\left (b x + a\right )} d - a d\right )}}{d}\right ) + 2 \, b^{2} d \cos \left (2 \, b x + 2 \, a\right ) + b^{2} d}{{\left (b c d^{2} + {\left (b x + a\right )} d^{3} - a d^{3}\right )} b} \]

[In]

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

[Out]

-(4*b^3*c*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) + 4*(b*x + a)*b^2*d*cos_integral(2
*(b*c + (b*x + a)*d - a*d)/d)*sin(-2*(b*c - a*d)/d) - 4*a*b^2*d*cos_integral(2*(b*c + (b*x + a)*d - a*d)/d)*si
n(-2*(b*c - a*d)/d) - 4*b^3*c*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) - 4*(b*x + a)
*b^2*d*cos(-2*(b*c - a*d)/d)*sin_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 4*a*b^2*d*cos(-2*(b*c - a*d)/d)*si
n_integral(-2*(b*c + (b*x + a)*d - a*d)/d) + 2*b^2*d*cos(2*b*x + 2*a) + b^2*d)/((b*c*d^2 + (b*x + a)*d^3 - a*d
^3)*b)

Mupad [F(-1)]

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

[In]

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

[Out]

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