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

Optimal. Leaf size=99 \[ -\frac {4 \cos \left (\frac {2 c}{d}\right ) \text {Ci}\left (\frac {2 c}{d}+2 x\right )}{d^3}-\frac {4 \sin \left (\frac {2 c}{d}\right ) \text {Si}\left (\frac {2 c}{d}+2 x\right )}{d^3}+\frac {4 \sin (x) \cos (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {3 \cos ^2(x)}{2 d (c+d x)^2} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4431, 3314, 31, 3312, 3303, 3299, 3302} \[ -\frac {4 \cos \left (\frac {2 c}{d}\right ) \text {CosIntegral}\left (\frac {2 c}{d}+2 x\right )}{d^3}-\frac {4 \sin \left (\frac {2 c}{d}\right ) \text {Si}\left (\frac {2 c}{d}+2 x\right )}{d^3}+\frac {4 \sin (x) \cos (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {3 \cos ^2(x)}{2 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[x]*Sin[3*x])/(c + d*x)^3,x]

[Out]

(-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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 3314

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

Rubi steps

\begin {align*} \int \frac {\csc (x) \sin (3 x)}{(c+d x)^3} \, dx &=\int \left (\frac {3 \cos ^2(x)}{(c+d x)^3}-\frac {\sin ^2(x)}{(c+d x)^3}\right ) \, dx\\ &=3 \int \frac {\cos ^2(x)}{(c+d x)^3} \, dx-\int \frac {\sin ^2(x)}{(c+d x)^3} \, dx\\ &=-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}+\frac {4 \cos (x) \sin (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {\int \frac {1}{c+d x} \, dx}{d^2}+\frac {2 \int \frac {\sin ^2(x)}{c+d x} \, dx}{d^2}+\frac {3 \int \frac {1}{c+d x} \, dx}{d^2}-\frac {6 \int \frac {\cos ^2(x)}{c+d x} \, dx}{d^2}\\ &=-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}+\frac {2 \log (c+d x)}{d^3}+\frac {4 \cos (x) \sin (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}+\frac {2 \int \left (\frac {1}{2 (c+d x)}-\frac {\cos (2 x)}{2 (c+d x)}\right ) \, dx}{d^2}-\frac {6 \int \left (\frac {1}{2 (c+d x)}+\frac {\cos (2 x)}{2 (c+d x)}\right ) \, dx}{d^2}\\ &=-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}+\frac {4 \cos (x) \sin (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {\int \frac {\cos (2 x)}{c+d x} \, dx}{d^2}-\frac {3 \int \frac {\cos (2 x)}{c+d x} \, dx}{d^2}\\ &=-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}+\frac {4 \cos (x) \sin (x)}{d^2 (c+d x)}+\frac {\sin ^2(x)}{2 d (c+d x)^2}-\frac {\cos \left (\frac {2 c}{d}\right ) \int \frac {\cos \left (\frac {2 c}{d}+2 x\right )}{c+d x} \, dx}{d^2}-\frac {\left (3 \cos \left (\frac {2 c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 c}{d}+2 x\right )}{c+d x} \, dx}{d^2}-\frac {\sin \left (\frac {2 c}{d}\right ) \int \frac {\sin \left (\frac {2 c}{d}+2 x\right )}{c+d x} \, dx}{d^2}-\frac {\left (3 \sin \left (\frac {2 c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 c}{d}+2 x\right )}{c+d x} \, dx}{d^2}\\ &=-\frac {3 \cos ^2(x)}{2 d (c+d x)^2}-\frac {4 \cos \left (\frac {2 c}{d}\right ) \text {Ci}\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}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 77, normalized size = 0.78 \[ \frac {-8 \cos \left (\frac {2 c}{d}\right ) \text {Ci}\left (2 \left (\frac {c}{d}+x\right )\right )-8 \sin \left (\frac {2 c}{d}\right ) \text {Si}\left (2 \left (\frac {c}{d}+x\right )\right )+\frac {d (4 \sin (2 x) (c+d x)-2 d \cos (2 x)-d)}{(c+d x)^2}}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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)

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fricas [A]  time = 0.44, size = 158, normalized size = 1.60 \[ -\frac {4 \, d^{2} \cos \relax (x)^{2} - 8 \, {\left (d^{2} x + c d\right )} \cos \relax (x) \sin \relax (x) + 8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) - d^{2} + 4 \, {\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (d x + c\right )}}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right )}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*d^2*cos(x)^2 - 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 + 4*((d^2*x^2 + 2*c*d*x + c^2)*cos_integral(2*(d*x + c)/d) + (d^2*x^2 + 2*c*d*x + c^2)*cos_i
ntegral(-2*(d*x + c)/d))*cos(2*c/d))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

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giac [B]  time = 2.73, size = 201, normalized size = 2.03 \[ -\frac {8 \, d^{2} x^{2} \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 8 \, d^{2} x^{2} \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 16 \, c d x \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 16 \, c d x \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 8 \, c^{2} \cos \left (\frac {2 \, c}{d}\right ) \operatorname {Ci}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) - 4 \, d^{2} x \sin \left (2 \, x\right ) + 8 \, c^{2} \sin \left (\frac {2 \, c}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (d x + c\right )}}{d}\right ) + 2 \, d^{2} \cos \left (2 \, x\right ) - 4 \, c d \sin \left (2 \, x\right ) + d^{2}}{2 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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_integral(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)

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maple [A]  time = 0.06, size = 104, normalized size = 1.05 \[ -\frac {\cos \left (2 x \right )}{\left (d x +c \right )^{2} d}-\frac {-\frac {2 \sin \left (2 x \right )}{\left (d x +c \right ) d}+\frac {\frac {4 \Si \left (\frac {2 c}{d}+2 x \right ) \sin \left (\frac {2 c}{d}\right )}{d}+\frac {4 \Ci \left (\frac {2 c}{d}+2 x \right ) \cos \left (\frac {2 c}{d}\right )}{d}}{d}}{d}-\frac {1}{2 d \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.42, size = 362, normalized size = 3.66 \[ -\frac {2 \, {\left (E_{3}\left (\frac {2 i \, d x + 2 i \, c}{d}\right ) + E_{3}\left (-\frac {2 i \, d x + 2 i \, c}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right )^{3} + {\left (2 i \, E_{3}\left (\frac {2 i \, d x + 2 i \, c}{d}\right ) - 2 i \, E_{3}\left (-\frac {2 i \, d x + 2 i \, c}{d}\right )\right )} \sin \left (\frac {2 \, c}{d}\right )^{3} + 2 \, {\left ({\left (E_{3}\left (\frac {2 i \, d x + 2 i \, c}{d}\right ) + E_{3}\left (-\frac {2 i \, d x + 2 i \, c}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right ) + 1\right )} \sin \left (\frac {2 \, c}{d}\right )^{2} + 2 \, {\left (E_{3}\left (\frac {2 i \, d x + 2 i \, c}{d}\right ) + E_{3}\left (-\frac {2 i \, d x + 2 i \, c}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right ) + 2 \, \cos \left (\frac {2 \, c}{d}\right )^{2} + {\left ({\left (2 i \, E_{3}\left (\frac {2 i \, d x + 2 i \, c}{d}\right ) - 2 i \, E_{3}\left (-\frac {2 i \, d x + 2 i \, c}{d}\right )\right )} \cos \left (\frac {2 \, c}{d}\right )^{2} + 2 i \, E_{3}\left (\frac {2 i \, d x + 2 i \, c}{d}\right ) - 2 i \, E_{3}\left (-\frac {2 i \, d x + 2 i \, c}{d}\right )\right )} \sin \left (\frac {2 \, c}{d}\right )}{4 \, {\left ({\left (\cos \left (\frac {2 \, c}{d}\right )^{2} + \sin \left (\frac {2 \, c}{d}\right )^{2}\right )} d^{3} x^{2} + 2 \, {\left (c \cos \left (\frac {2 \, c}{d}\right )^{2} + c \sin \left (\frac {2 \, c}{d}\right )^{2}\right )} d^{2} x + {\left (c^{2} \cos \left (\frac {2 \, c}{d}\right )^{2} + c^{2} \sin \left (\frac {2 \, c}{d}\right )^{2}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*(exp_integral_e(3, (2*I*d*x + 2*I*c)/d) + exp_integral_e(3, -(2*I*d*x + 2*I*c)/d))*cos(2*c/d)^3 + (2*I
*exp_integral_e(3, (2*I*d*x + 2*I*c)/d) - 2*I*exp_integral_e(3, -(2*I*d*x + 2*I*c)/d))*sin(2*c/d)^3 + 2*((exp_
integral_e(3, (2*I*d*x + 2*I*c)/d) + exp_integral_e(3, -(2*I*d*x + 2*I*c)/d))*cos(2*c/d) + 1)*sin(2*c/d)^2 + 2
*(exp_integral_e(3, (2*I*d*x + 2*I*c)/d) + exp_integral_e(3, -(2*I*d*x + 2*I*c)/d))*cos(2*c/d) + 2*cos(2*c/d)^
2 + ((2*I*exp_integral_e(3, (2*I*d*x + 2*I*c)/d) - 2*I*exp_integral_e(3, -(2*I*d*x + 2*I*c)/d))*cos(2*c/d)^2 +
2*I*exp_integral_e(3, (2*I*d*x + 2*I*c)/d) - 2*I*exp_integral_e(3, -(2*I*d*x + 2*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(3*x)/(sin(x)*(c + d*x)^3),x)

[Out]

int(sin(3*x)/(sin(x)*(c + d*x)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (3 x \right )} \csc {\relax (x )}}{\left (c + d x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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