3.143 \(\int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=231 \[ -\frac {b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2} \]

[Out]

-1/4*b*cos(2*b*x+2*a)/d^2/(d*x+c)-1/4*b*cos(4*b*x+4*a)/d^2/(d*x+c)-1/2*b^2*cos(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/
d^3-b^2*cos(4*a-4*b*c/d)*Si(4*b*c/d+4*b*x)/d^3-b^2*Ci(4*b*c/d+4*b*x)*sin(4*a-4*b*c/d)/d^3-1/2*b^2*Ci(2*b*c/d+2
*b*x)*sin(2*a-2*b*c/d)/d^3-1/8*sin(2*b*x+2*a)/d/(d*x+c)^2-1/16*sin(4*b*x+4*a)/d/(d*x+c)^2

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4406, 3297, 3303, 3299, 3302} \[ -\frac {b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {CosIntegral}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[2*a + 2*b*x])/(4*d^2*(c + d*x)) - (b*Cos[4*a + 4*b*x])/(4*d^2*(c + d*x)) - (b^2*CosIntegral[(4*b*c)/d
+ 4*b*x]*Sin[4*a - (4*b*c)/d])/d^3 - (b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/(2*d^3) - Sin[2
*a + 2*b*x]/(8*d*(c + d*x)^2) - Sin[4*a + 4*b*x]/(16*d*(c + d*x)^2) - (b^2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2
*b*c)/d + 2*b*x])/(2*d^3) - (b^2*Cos[4*a - (4*b*c)/d]*SinIntegral[(4*b*c)/d + 4*b*x])/d^3

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^3(a+b x) \sin (a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac {\sin (2 a+2 b x)}{4 (c+d x)^3}+\frac {\sin (4 a+4 b x)}{8 (c+d x)^3}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sin (4 a+4 b x)}{(c+d x)^3} \, dx+\frac {1}{4} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3} \, dx\\ &=-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}+\frac {b \int \frac {\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{4 d}+\frac {b \int \frac {\cos (4 a+4 b x)}{(c+d x)^2} \, dx}{4 d}\\ &=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac {b^2 \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{2 d^2}-\frac {b^2 \int \frac {\sin (4 a+4 b x)}{c+d x} \, dx}{d^2}\\ &=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac {\left (b^2 \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx}{d^2}-\frac {\left (b^2 \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}{2 d^2}-\frac {\left (b^2 \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {4 b c}{d}+4 b x\right )}{c+d x} \, dx}{d^2}-\frac {\left (b^2 \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}{2 d^2}\\ &=-\frac {b \cos (2 a+2 b x)}{4 d^2 (c+d x)}-\frac {b \cos (4 a+4 b x)}{4 d^2 (c+d x)}-\frac {b^2 \text {Ci}\left (\frac {4 b c}{d}+4 b x\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{d^3}-\frac {b^2 \text {Ci}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{2 d^3}-\frac {\sin (2 a+2 b x)}{8 d (c+d x)^2}-\frac {\sin (4 a+4 b x)}{16 d (c+d x)^2}-\frac {b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{2 d^3}-\frac {b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b c}{d}+4 b x\right )}{d^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.75, size = 197, normalized size = 0.85 \[ -\frac {16 b^2 \sin \left (4 a-\frac {4 b c}{d}\right ) \text {Ci}\left (\frac {4 b (c+d x)}{d}\right )+8 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )+8 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+16 b^2 \cos \left (4 a-\frac {4 b c}{d}\right ) \text {Si}\left (\frac {4 b (c+d x)}{d}\right )+\frac {2 d (2 b (c+d x) \cos (2 (a+b x))+d \sin (2 (a+b x)))}{(c+d x)^2}+\frac {d (4 b (c+d x) \cos (4 (a+b x))+d \sin (4 (a+b x)))}{(c+d x)^2}}{16 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(16*b^2*CosIntegral[(4*b*(c + d*x))/d]*Sin[4*a - (4*b*c)/d] + 8*b^2*CosIntegral[(2*b*(c + d*x))/d]*Sin[2
*a - (2*b*c)/d] + (2*d*(2*b*(c + d*x)*Cos[2*(a + b*x)] + d*Sin[2*(a + b*x)]))/(c + d*x)^2 + (d*(4*b*(c + d*x)*
Cos[4*(a + b*x)] + d*Sin[4*(a + b*x)]))/(c + d*x)^2 + 8*b^2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d
] + 16*b^2*Cos[4*a - (4*b*c)/d]*SinIntegral[(4*b*(c + d*x))/d])/d^3

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 397, normalized size = 1.72 \[ -\frac {2 \, d^{2} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right ) + 8 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} - 6 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 4 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {4 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {4 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.02, size = 329, normalized size = 1.42 \[ \frac {\frac {b^{3} \left (-\frac {2 \sin \left (4 b x +4 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {8 \cos \left (4 b x +4 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {8 \left (\frac {4 \Si \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \cos \left (\frac {-4 d a +4 c b}{d}\right )}{d}-\frac {4 \Ci \left (4 b x +4 a +\frac {-4 d a +4 c b}{d}\right ) \sin \left (\frac {-4 d a +4 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{32}+\frac {b^{3} \left (-\frac {\sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {2 \cos \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{8}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/32*b^3*(-2*sin(4*b*x+4*a)/((b*x+a)*d-d*a+c*b)^2/d+2*(-4*cos(4*b*x+4*a)/((b*x+a)*d-d*a+c*b)/d-4*(4*Si(4*
b*x+4*a+4*(-a*d+b*c)/d)*cos(4*(-a*d+b*c)/d)/d-4*Ci(4*b*x+4*a+4*(-a*d+b*c)/d)*sin(4*(-a*d+b*c)/d)/d)/d)/d)+1/8*
b^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))

________________________________________________________________________________________

maxima [C]  time = 0.69, size = 336, normalized size = 1.45 \[ -\frac {b^{3} {\left (2 i \, E_{3}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 2 i \, E_{3}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (i \, E_{3}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{3} {\left (E_{3}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{3}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (E_{3}\left (\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right ) + E_{3}\left (-\frac {4 i \, b c + 4 i \, {\left (b x + a\right )} d - 4 i \, a d}{d}\right )\right )} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )}{16 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(b^3*(2*I*exp_integral_e(3, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) - 2*I*exp_integral_e(3, -(2*I*b*c +
2*I*(b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) + b^3*(I*exp_integral_e(3, (4*I*b*c + 4*I*(b*x + a)*d -
4*I*a*d)/d) - I*exp_integral_e(3, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*cos(-4*(b*c - a*d)/d) + 2*b^3*(ex
p_integral_e(3, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(3, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I
*a*d)/d))*sin(-2*(b*c - a*d)/d) + b^3*(exp_integral_e(3, (4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d) + exp_integr
al_e(3, -(4*I*b*c + 4*I*(b*x + a)*d - 4*I*a*d)/d))*sin(-4*(b*c - a*d)/d))/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a
)^2*d^3 + a^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*b)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________