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

Optimal. Leaf size=235 \[ \frac {9 b^2 \sin \left (6 a-\frac {6 b c}{d}\right ) \text {Ci}\left (\frac {6 b c}{d}+6 b x\right )}{16 d^3}-\frac {3 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{16 d^3}-\frac {3 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{16 d^3}+\frac {9 b^2 \cos \left (6 a-\frac {6 b c}{d}\right ) \text {Si}\left (\frac {6 b c}{d}+6 b x\right )}{16 d^3}-\frac {3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac {3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac {3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac {\sin (6 a+6 b x)}{64 d (c+d x)^2} \]

[Out]

-3/32*b*cos(2*b*x+2*a)/d^2/(d*x+c)+3/32*b*cos(6*b*x+6*a)/d^2/(d*x+c)-3/16*b^2*cos(2*a-2*b*c/d)*Si(2*b*c/d+2*b*
x)/d^3+9/16*b^2*cos(6*a-6*b*c/d)*Si(6*b*c/d+6*b*x)/d^3+9/16*b^2*Ci(6*b*c/d+6*b*x)*sin(6*a-6*b*c/d)/d^3-3/16*b^
2*Ci(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^3-3/64*sin(2*b*x+2*a)/d/(d*x+c)^2+1/64*sin(6*b*x+6*a)/d/(d*x+c)^2

________________________________________________________________________________________

Rubi [A]  time = 0.35, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4406, 3297, 3303, 3299, 3302} \[ \frac {9 b^2 \sin \left (6 a-\frac {6 b c}{d}\right ) \text {CosIntegral}\left (\frac {6 b c}{d}+6 b x\right )}{16 d^3}-\frac {3 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{16 d^3}-\frac {3 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{16 d^3}+\frac {9 b^2 \cos \left (6 a-\frac {6 b c}{d}\right ) \text {Si}\left (\frac {6 b c}{d}+6 b x\right )}{16 d^3}-\frac {3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac {3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac {3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac {\sin (6 a+6 b x)}{64 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*Cos[2*a + 2*b*x])/(32*d^2*(c + d*x)) + (3*b*Cos[6*a + 6*b*x])/(32*d^2*(c + d*x)) + (9*b^2*CosIntegral[(6
*b*c)/d + 6*b*x]*Sin[6*a - (6*b*c)/d])/(16*d^3) - (3*b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/
(16*d^3) - (3*Sin[2*a + 2*b*x])/(64*d*(c + d*x)^2) + Sin[6*a + 6*b*x]/(64*d*(c + d*x)^2) - (3*b^2*Cos[2*a - (2
*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/(16*d^3) + (9*b^2*Cos[6*a - (6*b*c)/d]*SinIntegral[(6*b*c)/d + 6*b*x]
)/(16*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 ^3(a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac {3 \sin (2 a+2 b x)}{32 (c+d x)^3}-\frac {\sin (6 a+6 b x)}{32 (c+d x)^3}\right ) \, dx\\ &=-\left (\frac {1}{32} \int \frac {\sin (6 a+6 b x)}{(c+d x)^3} \, dx\right )+\frac {3}{32} \int \frac {\sin (2 a+2 b x)}{(c+d x)^3} \, dx\\ &=-\frac {3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac {\sin (6 a+6 b x)}{64 d (c+d x)^2}+\frac {(3 b) \int \frac {\cos (2 a+2 b x)}{(c+d x)^2} \, dx}{32 d}-\frac {(3 b) \int \frac {\cos (6 a+6 b x)}{(c+d x)^2} \, dx}{32 d}\\ &=-\frac {3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac {3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac {3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac {\sin (6 a+6 b x)}{64 d (c+d x)^2}-\frac {\left (3 b^2\right ) \int \frac {\sin (2 a+2 b x)}{c+d x} \, dx}{16 d^2}+\frac {\left (9 b^2\right ) \int \frac {\sin (6 a+6 b x)}{c+d x} \, dx}{16 d^2}\\ &=-\frac {3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac {3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}-\frac {3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac {\sin (6 a+6 b x)}{64 d (c+d x)^2}+\frac {\left (9 b^2 \cos \left (6 a-\frac {6 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {6 b c}{d}+6 b x\right )}{c+d x} \, dx}{16 d^2}-\frac {\left (3 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}{16 d^2}+\frac {\left (9 b^2 \sin \left (6 a-\frac {6 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {6 b c}{d}+6 b x\right )}{c+d x} \, dx}{16 d^2}-\frac {\left (3 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}{16 d^2}\\ &=-\frac {3 b \cos (2 a+2 b x)}{32 d^2 (c+d x)}+\frac {3 b \cos (6 a+6 b x)}{32 d^2 (c+d x)}+\frac {9 b^2 \text {Ci}\left (\frac {6 b c}{d}+6 b x\right ) \sin \left (6 a-\frac {6 b c}{d}\right )}{16 d^3}-\frac {3 b^2 \text {Ci}\left (\frac {2 b c}{d}+2 b x\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{16 d^3}-\frac {3 \sin (2 a+2 b x)}{64 d (c+d x)^2}+\frac {\sin (6 a+6 b x)}{64 d (c+d x)^2}-\frac {3 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{16 d^3}+\frac {9 b^2 \cos \left (6 a-\frac {6 b c}{d}\right ) \text {Si}\left (\frac {6 b c}{d}+6 b x\right )}{16 d^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.04, size = 239, normalized size = 1.02 \[ \frac {6 b^2 (c+d x)^2 \left (6 \sin \left (6 a-\frac {6 b c}{d}\right ) \text {Ci}\left (\frac {6 b (c+d x)}{d}\right )-2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )-2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )+6 \cos \left (6 a-\frac {6 b c}{d}\right ) \text {Si}\left (\frac {6 b (c+d x)}{d}\right )\right )-3 d \cos (2 b x) (2 b \cos (2 a) (c+d x)+d \sin (2 a))+d \cos (6 b x) (6 b \cos (6 a) (c+d x)+d \sin (6 a))+3 d \sin (2 b x) (2 b \sin (2 a) (c+d x)-d \cos (2 a))+d \sin (6 b x) (d \cos (6 a)-6 b \sin (6 a) (c+d x))}{64 d^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*Cos[2*b*x]*(2*b*(c + d*x)*Cos[2*a] + d*Sin[2*a]) + d*Cos[6*b*x]*(6*b*(c + d*x)*Cos[6*a] + d*Sin[6*a]) +
3*d*(-(d*Cos[2*a]) + 2*b*(c + d*x)*Sin[2*a])*Sin[2*b*x] + d*(d*Cos[6*a] - 6*b*(c + d*x)*Sin[6*a])*Sin[6*b*x] +
6*b^2*(c + d*x)^2*(6*CosIntegral[(6*b*(c + d*x))/d]*Sin[6*a - (6*b*c)/d] - 2*CosIntegral[(2*b*(c + d*x))/d]*S
in[2*a - (2*b*c)/d] - 2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*(c + d*x))/d] + 6*Cos[6*a - (6*b*c)/d]*SinIntegr
al[(6*b*(c + d*x))/d]))/(64*d^3*(c + d*x)^2)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 434, normalized size = 1.85 \[ \frac {96 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{6} - 144 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{4} + 48 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2} + 18 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {6 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {6 \, {\left (b d x + b c\right )}}{d}\right ) - 6 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + 16 \, {\left (d^{2} \cos \left (b x + a\right )^{5} - d^{2} \cos \left (b x + a\right )^{3}\right )} \sin \left (b x + a\right ) - 3 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 9 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {6 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {6 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {6 \, {\left (b c - a d\right )}}{d}\right )}{32 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(96*(b*d^2*x + b*c*d)*cos(b*x + a)^6 - 144*(b*d^2*x + b*c*d)*cos(b*x + a)^4 + 48*(b*d^2*x + b*c*d)*cos(b*
x + a)^2 + 18*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-6*(b*c - a*d)/d)*sin_integral(6*(b*d*x + b*c)/d) - 6*
(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) + 16*(d^2*cos(b*x
+ a)^5 - d^2*cos(b*x + a)^3)*sin(b*x + a) - 3*((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) + 9*((b
^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(6*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*co
s_integral(-6*(b*d*x + b*c)/d))*sin(-6*(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)^3/(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.02, size = 329, normalized size = 1.40 \[ \frac {-\frac {b^{3} \left (-\frac {3 \sin \left (6 b x +6 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {18 \cos \left (6 b x +6 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {18 \left (\frac {6 \Si \left (6 b x +6 a +\frac {-6 d a +6 c b}{d}\right ) \cos \left (\frac {-6 d a +6 c b}{d}\right )}{d}-\frac {6 \Ci \left (6 b x +6 a +\frac {-6 d a +6 c b}{d}\right ) \sin \left (\frac {-6 d a +6 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{192}+\frac {3 b^{3} \left (-\frac {\sin \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right )^{2} d}+\frac {-\frac {2 \cos \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}}{d}\right )}{64}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/192*b^3*(-3*sin(6*b*x+6*a)/((b*x+a)*d-d*a+c*b)^2/d+3*(-6*cos(6*b*x+6*a)/((b*x+a)*d-d*a+c*b)/d-6*(6*Si(
6*b*x+6*a+6*(-a*d+b*c)/d)*cos(6*(-a*d+b*c)/d)/d-6*Ci(6*b*x+6*a+6*(-a*d+b*c)/d)*sin(6*(-a*d+b*c)/d)/d)/d)/d)+3/
64*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 = 1.43 \[ \frac {b^{3} {\left (-3 i \, E_{3}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + 3 i \, E_{3}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (i \, E_{3}\left (\frac {6 i \, b c + 6 i \, {\left (b x + a\right )} d - 6 i \, a d}{d}\right ) - i \, E_{3}\left (-\frac {6 i \, b c + 6 i \, {\left (b x + a\right )} d - 6 i \, a d}{d}\right )\right )} \cos \left (-\frac {6 \, {\left (b c - a d\right )}}{d}\right ) - 3 \, b^{3} {\left (E_{3}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{3}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (E_{3}\left (\frac {6 i \, b c + 6 i \, {\left (b x + a\right )} d - 6 i \, a d}{d}\right ) + E_{3}\left (-\frac {6 i \, b c + 6 i \, {\left (b x + a\right )} d - 6 i \, a d}{d}\right )\right )} \sin \left (-\frac {6 \, {\left (b c - a d\right )}}{d}\right )}{64 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(b^3*(-3*I*exp_integral_e(3, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + 3*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, (6*I*b*c + 6*I*(b*x + a)*d -
6*I*a*d)/d) - I*exp_integral_e(3, -(6*I*b*c + 6*I*(b*x + a)*d - 6*I*a*d)/d))*cos(-6*(b*c - a*d)/d) - 3*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, (6*I*b*c + 6*I*(b*x + a)*d - 6*I*a*d)/d) + exp_integr
al_e(3, -(6*I*b*c + 6*I*(b*x + a)*d - 6*I*a*d)/d))*sin(-6*(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 \[ \int \frac {{\cos \left (a+b\,x\right )}^3\,{\sin \left (a+b\,x\right )}^3}{{\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________