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

Optimal. Leaf size=71 \[ \frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d}-\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}+\frac {\log (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4431, 3312, 3303, 3299, 3302} \[ \frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d}-\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}+\frac {\log (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[2*a - (2*b*c)/d]*CosIntegral[(2*b*c)/d + 2*b*x])/d + Log[c + d*x]/d - (2*Sin[2*a - (2*b*c)/d]*SinIntegr
al[(2*b*c)/d + 2*b*x])/d

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]

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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 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]

Rubi steps

\begin {align*} \int \frac {\csc (a+b x) \sin (3 a+3 b x)}{c+d x} \, dx &=\int \left (\frac {3 \cos ^2(a+b x)}{c+d x}-\frac {\sin ^2(a+b x)}{c+d x}\right ) \, dx\\ &=3 \int \frac {\cos ^2(a+b x)}{c+d x} \, dx-\int \frac {\sin ^2(a+b x)}{c+d x} \, dx\\ &=3 \int \left (\frac {1}{2 (c+d x)}+\frac {\cos (2 a+2 b x)}{2 (c+d x)}\right ) \, dx-\int \left (\frac {1}{2 (c+d x)}-\frac {\cos (2 a+2 b x)}{2 (c+d x)}\right ) \, dx\\ &=\frac {\log (c+d x)}{d}+\frac {1}{2} \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx+\frac {3}{2} \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx\\ &=\frac {\log (c+d x)}{d}+\frac {1}{2} \cos \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx+\frac {1}{2} \left (3 \cos \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-\frac {1}{2} \sin \left (2 a-\frac {2 b c}{d}\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx-\frac {1}{2} \left (3 \sin \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\\ &=\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d}+\frac {\log (c+d x)}{d}-\frac {2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 63, normalized size = 0.89 \[ \frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )-2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+\log (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 85, normalized size = 1.20 \[ \frac {{\left (\operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + \log \left (d x + c\right )}{d} \]

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),x, algorithm="fricas")

[Out]

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

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giac [C]  time = 0.32, size = 1118, normalized size = 15.75 \[ \text {result too large to display} \]

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),x, algorithm="giac")

[Out]

(log(abs(d*x + c))*tan(1/2*a)^4*tan(b*c/d)^2 - real_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)^4*tan(b*c/d
)^2 - real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^4*tan(b*c/d)^2 + 2*imag_part(cos_integral(2*b*x + 2
*b*c/d))*tan(1/2*a)^4*tan(b*c/d) - 2*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^4*tan(b*c/d) + 4*sin
_integral(2*(b*d*x + b*c)/d)*tan(1/2*a)^4*tan(b*c/d) - 4*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)^3
*tan(b*c/d)^2 + 4*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^3*tan(b*c/d)^2 - 8*sin_integral(2*(b*d*
x + b*c)/d)*tan(1/2*a)^3*tan(b*c/d)^2 + log(abs(d*x + c))*tan(1/2*a)^4 + real_part(cos_integral(2*b*x + 2*b*c/
d))*tan(1/2*a)^4 + real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^4 - 8*real_part(cos_integral(2*b*x + 2
*b*c/d))*tan(1/2*a)^3*tan(b*c/d) - 8*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^3*tan(b*c/d) + 2*log
(abs(d*x + c))*tan(1/2*a)^2*tan(b*c/d)^2 + 6*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)^2*tan(b*c/d)^
2 + 6*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^2*tan(b*c/d)^2 + 4*imag_part(cos_integral(2*b*x + 2
*b*c/d))*tan(1/2*a)^3 - 4*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a)^3 + 8*sin_integral(2*(b*d*x + b
*c)/d)*tan(1/2*a)^3 - 12*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)^2*tan(b*c/d) + 12*imag_part(cos_i
ntegral(-2*b*x - 2*b*c/d))*tan(1/2*a)^2*tan(b*c/d) - 24*sin_integral(2*(b*d*x + b*c)/d)*tan(1/2*a)^2*tan(b*c/d
) + 4*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)*tan(b*c/d)^2 - 4*imag_part(cos_integral(-2*b*x - 2*b
*c/d))*tan(1/2*a)*tan(b*c/d)^2 + 8*sin_integral(2*(b*d*x + b*c)/d)*tan(1/2*a)*tan(b*c/d)^2 + 2*log(abs(d*x + c
))*tan(1/2*a)^2 - 6*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)^2 - 6*real_part(cos_integral(-2*b*x -
2*b*c/d))*tan(1/2*a)^2 + 8*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(1/2*a)*tan(b*c/d) + 8*real_part(cos_in
tegral(-2*b*x - 2*b*c/d))*tan(1/2*a)*tan(b*c/d) + log(abs(d*x + c))*tan(b*c/d)^2 - real_part(cos_integral(2*b*
x + 2*b*c/d))*tan(b*c/d)^2 - real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d)^2 - 4*imag_part(cos_integral
(2*b*x + 2*b*c/d))*tan(1/2*a) + 4*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(1/2*a) - 8*sin_integral(2*(b*d
*x + b*c)/d)*tan(1/2*a) + 2*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d) - 2*imag_part(cos_integral(-2*
b*x - 2*b*c/d))*tan(b*c/d) + 4*sin_integral(2*(b*d*x + b*c)/d)*tan(b*c/d) + log(abs(d*x + c)) + real_part(cos_
integral(2*b*x + 2*b*c/d)) + real_part(cos_integral(-2*b*x - 2*b*c/d)))/(d*tan(1/2*a)^4*tan(b*c/d)^2 + d*tan(1
/2*a)^4 + 2*d*tan(1/2*a)^2*tan(b*c/d)^2 + 2*d*tan(1/2*a)^2 + d*tan(b*c/d)^2 + d)

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maple [A]  time = 0.04, size = 116, normalized size = 1.63 \[ -\frac {\ln \left (d x +c \right )}{d}+\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}+\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}+\frac {2 \ln \left (\left (b x +a \right ) d -d a +c b \right )}{d} \]

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),x)

[Out]

-ln(d*x+c)/d+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)*cos(2*(-a*d+b
*c)/d)/d+2*ln((b*x+a)*d-d*a+c*b)/d

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maxima [C]  time = 0.41, size = 117, normalized size = 1.65 \[ -\frac {{\left (E_{1}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) + E_{1}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (i \, E_{1}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) - i \, E_{1}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \log \left (d x + c\right )}{d} \]

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),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (3\,a+3\,b\,x\right )}{\sin \left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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),x)

[Out]

Timed out

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