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

Optimal. Leaf size=102 \[ -\frac{4 b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{4 b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}+\frac{\sin ^2(a+b x)}{d (c+d x)}-\frac{3 \cos ^2(a+b x)}{d (c+d x)} \]

[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

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Rubi [A]  time = 0.276183, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4431, 3313, 12, 3303, 3299, 3302} \[ -\frac{4 b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{4 b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}+\frac{\sin ^2(a+b x)}{d (c+d x)}-\frac{3 \cos ^2(a+b x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[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 4431

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]

Rule 3313

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 12

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

Rule 3303

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 3299

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 3302

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]

Rubi steps

\begin{align*} \int \frac{\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^2} \, dx &=\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 \text{Ci}\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]  time = 0.54519, size = 81, normalized size = 0.79 \[ -\frac{4 b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )+4 b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )+\frac{d (2 \cos (2 (a+b x))+1)}{c+d x}}{d^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.041, size = 169, normalized size = 1.7 \begin{align*}{\frac{1}{d \left ( dx+c \right ) }}+4\,{\frac{1}{b} \left ( 1/4\,{b}^{2} \left ( -2\,{\frac{\cos \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 2\,{\frac{-ad+bc}{d}} \right ) }-2\,{\frac{1}{d}{\it Ci} \left ( 2\,bx+2\,a+2\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 2\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) -1/2\,{\frac{{b}^{2}}{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.32883, size = 159, normalized size = 1.56 \begin{align*} -\frac{{\left (E_{2}\left (\frac{2 i \, b d x + 2 i \, b c}{d}\right ) + E_{2}\left (-\frac{2 i \, b d x + 2 i \, b c}{d}\right )\right )} \cos \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) -{\left (i \, E_{2}\left (\frac{2 i \, b d x + 2 i \, b c}{d}\right ) - i \, E_{2}\left (-\frac{2 i \, b d x + 2 i \, b c}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) + 1}{d^{2} x + c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Fricas [A]  time = 0.518525, size = 321, normalized size = 3.15 \begin{align*} -\frac{4 \, d \cos \left (b x + a\right )^{2} + 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 ) + 2 \,{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\right )} \operatorname{Ci}\left (-\frac{2 \,{\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right ) - d}{d^{3} x + c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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