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

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

[Out]

-2/3*b^2/d^3/(d*x+c)-cos(b*x+a)^2/d/(d*x+c)^3+2*b^2*cos(b*x+a)^2/d^3/(d*x+c)+8/3*b^3*cos(2*a-2*b*c/d)*Si(2*b*c
/d+2*b*x)/d^4+8/3*b^3*Ci(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^4+4/3*b*cos(b*x+a)*sin(b*x+a)/d^2/(d*x+c)^2+1/3*sin
(b*x+a)^2/d/(d*x+c)^3-2/3*b^2*sin(b*x+a)^2/d^3/(d*x+c)

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Rubi [A]  time = 0.38, antiderivative size = 205, normalized size of antiderivative = 1.00, 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 12

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

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 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 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 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 (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*}

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Mathematica [A]  time = 1.14, size = 125, normalized size = 0.61 \[ \frac {8 b^3 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )+8 b^3 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+\frac {d \left (\cos (2 (a+b x)) \left (4 b^2 (c+d x)^2-2 d^2\right )+d (2 b (c+d x) \sin (2 (a+b x))-d)\right )}{(c+d x)^3}}{3 d^4} \]

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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fricas [A]  time = 0.49, size = 343, normalized size = 1.67 \[ -\frac {4 \, b^{2} d^{3} x^{2} + 8 \, b^{2} c d^{2} x + 4 \, b^{2} c^{2} d - d^{3} - 4 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \cos \left (b x + a\right )^{2} - 4 \, {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 8 \, {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) - 4 \, {\left ({\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{3 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \]

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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giac [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)^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.04, size = 243, normalized size = 1.19 \[ \frac {1}{3 d \left (d x +c \right )^{3}}+\frac {b^{4} \left (-\frac {2 \cos \left (2 b x +2 a \right )}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}-\frac {2 \left (-\frac {\sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {2 \cos \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{3 d}\right )-\frac {2 b^{4}}{3 \left (\left (b x +a \right ) d -d a +c b \right )^{3} d}}{b} \]

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

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maxima [C]  time = 0.45, size = 141, normalized size = 0.69 \[ -\frac {3 \, {\left (E_{4}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) + E_{4}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - {\left (3 i \, E_{4}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) - 3 i \, E_{4}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 1}{3 \, {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (3\,a+3\,b\,x\right )}{\sin \left (a+b\,x\right )\,{\left (c+d\,x\right )}^4} \,d x \]

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)^4),x)

[Out]

int(sin(3*a + 3*b*x)/(sin(a + b*x)*(c + d*x)^4), 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)**4,x)

[Out]

Timed out

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