3.375 \(\int \frac{\csc (a+b x) \sin (3 a+3 b x)}{(c+d x)^4} \, dx\)

Optimal. Leaf size=205 \[ \frac{8 b^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{8 b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac{2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac{4 b \sin (a+b x) \cos (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{\cos ^2(a+b x)}{d (c+d x)^3}-\frac{2 b^2}{3 d^3 (c+d x)} \]

[Out]

(-2*b^2)/(3*d^3*(c + d*x)) - Cos[a + b*x]^2/(d*(c + d*x)^3) + (2*b^2*Cos[a + b*x]^2)/(d^3*(c + d*x)) + (8*b^3*
CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/(3*d^4) + (4*b*Cos[a + b*x]*Sin[a + b*x])/(3*d^2*(c + d*x
)^2) + Sin[a + b*x]^2/(3*d*(c + d*x)^3) - (2*b^2*Sin[a + b*x]^2)/(3*d^3*(c + d*x)) + (8*b^3*Cos[2*a - (2*b*c)/
d]*SinIntegral[(2*b*c)/d + 2*b*x])/(3*d^4)

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Rubi [A]  time = 0.379893, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4431, 3314, 32, 3313, 12, 3303, 3299, 3302} \[ \frac{8 b^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}+\frac{8 b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}-\frac{2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac{2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac{4 b \sin (a+b x) \cos (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{\cos ^2(a+b x)}{d (c+d x)^3}-\frac{2 b^2}{3 d^3 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[a + b*x]*Sin[3*a + 3*b*x])/(c + d*x)^4,x]

[Out]

(-2*b^2)/(3*d^3*(c + d*x)) - Cos[a + b*x]^2/(d*(c + d*x)^3) + (2*b^2*Cos[a + b*x]^2)/(d^3*(c + d*x)) + (8*b^3*
CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/(3*d^4) + (4*b*Cos[a + b*x]*Sin[a + b*x])/(3*d^2*(c + d*x
)^2) + Sin[a + b*x]^2/(3*d*(c + d*x)^3) - (2*b^2*Sin[a + b*x]^2)/(3*d^3*(c + d*x)) + (8*b^3*Cos[2*a - (2*b*c)/
d]*SinIntegral[(2*b*c)/d + 2*b*x])/(3*d^4)

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 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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)^4} \, dx &=\int \left (\frac{3 \cos ^2(a+b x)}{(c+d x)^4}-\frac{\sin ^2(a+b x)}{(c+d x)^4}\right ) \, dx\\ &=3 \int \frac{\cos ^2(a+b x)}{(c+d x)^4} \, dx-\int \frac{\sin ^2(a+b x)}{(c+d x)^4} \, dx\\ &=-\frac{\cos ^2(a+b x)}{d (c+d x)^3}+\frac{4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{b^2 \int \frac{1}{(c+d x)^2} \, dx}{3 d^2}+\frac{\left (2 b^2\right ) \int \frac{\sin ^2(a+b x)}{(c+d x)^2} \, dx}{3 d^2}+\frac{b^2 \int \frac{1}{(c+d x)^2} \, dx}{d^2}-\frac{\left (2 b^2\right ) \int \frac{\cos ^2(a+b x)}{(c+d x)^2} \, dx}{d^2}\\ &=-\frac{2 b^2}{3 d^3 (c+d x)}-\frac{\cos ^2(a+b x)}{d (c+d x)^3}+\frac{2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac{4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac{\left (4 b^3\right ) \int \frac{\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{3 d^3}-\frac{\left (4 b^3\right ) \int -\frac{\sin (2 a+2 b x)}{2 (c+d x)} \, dx}{d^3}\\ &=-\frac{2 b^2}{3 d^3 (c+d x)}-\frac{\cos ^2(a+b x)}{d (c+d x)^3}+\frac{2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac{4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac{\left (2 b^3\right ) \int \frac{\sin (2 a+2 b x)}{c+d x} \, dx}{3 d^3}+\frac{\left (2 b^3\right ) \int \frac{\sin (2 a+2 b x)}{c+d x} \, dx}{d^3}\\ &=-\frac{2 b^2}{3 d^3 (c+d x)}-\frac{\cos ^2(a+b x)}{d (c+d x)^3}+\frac{2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac{4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac{\left (2 b^3 \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}{3 d^3}+\frac{\left (2 b^3 \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^3}+\frac{\left (2 b^3 \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}{3 d^3}+\frac{\left (2 b^3 \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^3}\\ &=-\frac{2 b^2}{3 d^3 (c+d x)}-\frac{\cos ^2(a+b x)}{d (c+d x)^3}+\frac{2 b^2 \cos ^2(a+b x)}{d^3 (c+d x)}+\frac{8 b^3 \text{Ci}\left (\frac{2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac{2 b c}{d}\right )}{3 d^4}+\frac{4 b \cos (a+b x) \sin (a+b x)}{3 d^2 (c+d x)^2}+\frac{\sin ^2(a+b x)}{3 d (c+d x)^3}-\frac{2 b^2 \sin ^2(a+b x)}{3 d^3 (c+d x)}+\frac{8 b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{3 d^4}\\ \end{align*}

Mathematica [A]  time = 1.12737, size = 125, normalized size = 0.61 \[ \frac{8 b^3 \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )+\frac{d \left (\cos (2 (a+b x)) \left (4 b^2 (c+d x)^2-2 d^2\right )+d (2 b (c+d x) \sin (2 (a+b x))-d)\right )}{(c+d x)^3}+8 b^3 \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )}{3 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[a + b*x]*Sin[3*a + 3*b*x])/(c + d*x)^4,x]

[Out]

(8*b^3*CosIntegral[(2*b*(c + d*x))/d]*Sin[2*a - (2*b*c)/d] + (d*((-2*d^2 + 4*b^2*(c + d*x)^2)*Cos[2*(a + b*x)]
+ d*(-d + 2*b*(c + d*x)*Sin[2*(a + b*x)])))/(c + d*x)^3 + 8*b^3*Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*
x))/d])/(3*d^4)

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Maple [A]  time = 0.039, size = 243, normalized size = 1.2 \begin{align*}{\frac{1}{3\,d \left ( dx+c \right ) ^{3}}}+4\,{\frac{1}{b} \left ( 1/4\,{b}^{4} \left ( -2/3\,{\frac{\cos \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}d}}-2/3\,{\frac{1}{d} \left ( -{\frac{\sin \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{2}d}}+{\frac{1}{d} \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 ) } \right ) } \right ) -1/6\,{\frac{{b}^{4}}{ \left ( \left ( bx+a \right ) d-ad+bc \right ) ^{3}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)^4,x)

[Out]

1/3/d/(d*x+c)^3+4/b*(1/4*b^4*(-2/3*cos(2*b*x+2*a)/((b*x+a)*d-a*d+b*c)^3/d-2/3*(-sin(2*b*x+2*a)/((b*x+a)*d-a*d+
b*c)^2/d+(-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)/d)/d)-1/6*b^4/((b*x+a)*d-a*d+b*c)^3/d)

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Maxima [C]  time = 1.51092, size = 190, normalized size = 0.93 \begin{align*} -\frac{3 \,{\left (E_{4}\left (\frac{2 i \, b d x + 2 i \, b c}{d}\right ) + E_{4}\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 (3 i \, E_{4}\left (\frac{2 i \, b d x + 2 i \, b c}{d}\right ) - 3 i \, E_{4}\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}{3 \,{\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} \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)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.556165, size = 748, normalized size = 3.65 \begin{align*} -\frac{4 \, b^{2} d^{3} x^{2} + 8 \, b^{2} c d^{2} x + 4 \, b^{2} c^{2} d - d^{3} - 4 \,{\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{2} - 4 \,{\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 8 \,{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\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 ) - 4 \,{\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\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 )}{3 \,{\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \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)^4,x, algorithm="fricas")

[Out]

-1/3*(4*b^2*d^3*x^2 + 8*b^2*c*d^2*x + 4*b^2*c^2*d - d^3 - 4*(2*b^2*d^3*x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d - d^3
)*cos(b*x + a)^2 - 4*(b*d^3*x + b*c*d^2)*cos(b*x + a)*sin(b*x + a) - 8*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*
c^2*d*x + b^3*c^3)*cos(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) - 4*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 +
3*b^3*c^2*d*x + b^3*c^3)*cos_integral(2*(b*d*x + b*c)/d) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b
^3*c^3)*cos_integral(-2*(b*d*x + b*c)/d))*sin(-2*(b*c - a*d)/d))/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^
4)

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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)**4,x)

[Out]

Timed out

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Giac [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)^4,x, algorithm="giac")

[Out]

Timed out