3.20 \(\int \frac {\cos (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=221 \[ -\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2} \]

[Out]

9/8*b^2*Ci(3*b*c/d+3*b*x)*cos(3*a-3*b*c/d)/d^3-1/8*b^2*Ci(b*c/d+b*x)*cos(a-b*c/d)/d^3-1/8*cos(b*x+a)/d/(d*x+c)
^2+1/8*cos(3*b*x+3*a)/d/(d*x+c)^2-9/8*b^2*Si(3*b*c/d+3*b*x)*sin(3*a-3*b*c/d)/d^3+1/8*b^2*Si(b*c/d+b*x)*sin(a-b
*c/d)/d^3+1/8*b*sin(b*x+a)/d^2/(d*x+c)-3/8*b*sin(3*b*x+3*a)/d^2/(d*x+c)

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Rubi [A]  time = 0.36, antiderivative size = 221, 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} \[ -\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[a + b*x]*Sin[a + b*x]^2)/(c + d*x)^3,x]

[Out]

-Cos[a + b*x]/(8*d*(c + d*x)^2) + Cos[3*a + 3*b*x]/(8*d*(c + d*x)^2) - (b^2*Cos[a - (b*c)/d]*CosIntegral[(b*c)
/d + b*x])/(8*d^3) + (9*b^2*Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*b*x])/(8*d^3) + (b*Sin[a + b*x])/(8
*d^2*(c + d*x)) - (3*b*Sin[3*a + 3*b*x])/(8*d^2*(c + d*x)) + (b^2*Sin[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])
/(8*d^3) - (9*b^2*Sin[3*a - (3*b*c)/d]*SinIntegral[(3*b*c)/d + 3*b*x])/(8*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 (a+b x) \sin ^2(a+b x)}{(c+d x)^3} \, dx &=\int \left (\frac {\cos (a+b x)}{4 (c+d x)^3}-\frac {\cos (3 a+3 b x)}{4 (c+d x)^3}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\cos (a+b x)}{(c+d x)^3} \, dx-\frac {1}{4} \int \frac {\cos (3 a+3 b x)}{(c+d x)^3} \, dx\\ &=-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2}-\frac {b \int \frac {\sin (a+b x)}{(c+d x)^2} \, dx}{8 d}+\frac {(3 b) \int \frac {\sin (3 a+3 b x)}{(c+d x)^2} \, dx}{8 d}\\ &=-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}-\frac {b^2 \int \frac {\cos (a+b x)}{c+d x} \, dx}{8 d^2}+\frac {\left (9 b^2\right ) \int \frac {\cos (3 a+3 b x)}{c+d x} \, dx}{8 d^2}\\ &=-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}+\frac {\left (9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{8 d^2}-\frac {\left (b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{8 d^2}-\frac {\left (9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {3 b c}{d}+3 b x\right )}{c+d x} \, dx}{8 d^2}+\frac {\left (b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{8 d^2}\\ &=-\frac {\cos (a+b x)}{8 d (c+d x)^2}+\frac {\cos (3 a+3 b x)}{8 d (c+d x)^2}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{8 d^3}+\frac {9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}+\frac {b \sin (a+b x)}{8 d^2 (c+d x)}-\frac {3 b \sin (3 a+3 b x)}{8 d^2 (c+d x)}+\frac {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{8 d^3}-\frac {9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{8 d^3}\\ \end {align*}

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Mathematica [A]  time = 2.16, size = 183, normalized size = 0.83 \[ \frac {b^2 \left (-\cos \left (a-\frac {b c}{d}\right )\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+9 b^2 \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )+b^2 \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )-9 b^2 \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )+\frac {d (b (c+d x) \sin (a+b x)-d \cos (a+b x))}{(c+d x)^2}+\frac {d (d \cos (3 (a+b x))-3 b (c+d x) \sin (3 (a+b x)))}{(c+d x)^2}}{8 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[a + b*x]*Sin[a + b*x]^2)/(c + d*x)^3,x]

[Out]

(-(b^2*Cos[a - (b*c)/d]*CosIntegral[b*(c/d + x)]) + 9*b^2*Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*(c + d*x))/d]
+ (d*(-(d*Cos[a + b*x]) + b*(c + d*x)*Sin[a + b*x]))/(c + d*x)^2 + (d*(d*Cos[3*(a + b*x)] - 3*b*(c + d*x)*Sin[
3*(a + b*x)]))/(c + d*x)^2 + b^2*Sin[a - (b*c)/d]*SinIntegral[b*(c/d + x)] - 9*b^2*Sin[3*a - (3*b*c)/d]*SinInt
egral[(3*b*(c + d*x))/d])/(8*d^3)

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fricas [A]  time = 0.55, size = 399, normalized size = 1.81 \[ \frac {8 \, d^{2} \cos \left (b x + a\right )^{3} - 8 \, d^{2} \cos \left (b x + a\right ) - 18 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + 9 \, {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 8 \, {\left (b d^{2} x + b c d - 3 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{16 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x, algorithm="fricas")

[Out]

1/16*(8*d^2*cos(b*x + a)^3 - 8*d^2*cos(b*x + a) - 18*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*sin(-3*(b*c - a*d)/
d)*sin_integral(3*(b*d*x + b*c)/d) + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*sin(-(b*c - a*d)/d)*sin_integral(
(b*d*x + b*c)/d) - ((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral((b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2
*c*d*x + b^2*c^2)*cos_integral(-(b*d*x + b*c)/d))*cos(-(b*c - a*d)/d) + 9*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^
2)*cos_integral(3*(b*d*x + b*c)/d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-3*(b*d*x + b*c)/d))*c
os(-3*(b*c - a*d)/d) + 8*(b*d^2*x + b*c*d - 3*(b*d^2*x + b*c*d)*cos(b*x + a)^2)*sin(b*x + a))/(d^5*x^2 + 2*c*d
^4*x + c^2*d^3)

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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(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.02, size = 311, normalized size = 1.41 \[ \frac {\frac {b^{3} \left (-\frac {\cos \left (b x +a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {-\frac {\sin \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}+\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}\right )}{4}-\frac {b^{3} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{2 \left (\left (b x +a \right ) d -d a +c b \right )^{2} d}-\frac {3 \left (-\frac {3 \sin \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {9 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}+\frac {9 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}}{d}\right )}{2 d}\right )}{12}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/4*b^3*(-1/2*cos(b*x+a)/((b*x+a)*d-d*a+c*b)^2/d-1/2*(-sin(b*x+a)/((b*x+a)*d-d*a+c*b)/d+(Si(b*x+a+(-a*d+b
*c)/d)*sin((-a*d+b*c)/d)/d+Ci(b*x+a+(-a*d+b*c)/d)*cos((-a*d+b*c)/d)/d)/d)/d)-1/12*b^3*(-3/2*cos(3*b*x+3*a)/((b
*x+a)*d-d*a+c*b)^2/d-3/2*(-3*sin(3*b*x+3*a)/((b*x+a)*d-d*a+c*b)/d+3*(3*Si(3*b*x+3*a+3*(-a*d+b*c)/d)*sin(3*(-a*
d+b*c)/d)/d+3*Ci(3*b*x+3*a+3*(-a*d+b*c)/d)*cos(3*(-a*d+b*c)/d)/d)/d)/d))

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maxima [C]  time = 0.68, size = 337, normalized size = 1.52 \[ -\frac {8192 \, b^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - 8192 \, b^{3} {\left (E_{3}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{3}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{3} {\left (-8192 i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 8192 i \, E_{3}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + b^{3} {\left (8192 i \, E_{3}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - 8192 i \, E_{3}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{65536 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)^2/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/65536*(8192*b^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(3, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) - 8192*b^3*(exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d)
+ exp_integral_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b^3*(-8192*I*exp_integ
ral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 8192*I*exp_integral_e(3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*si
n(-(b*c - a*d)/d) + b^3*(8192*I*exp_integral_e(3, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - 8192*I*exp_integr
al_e(3, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(b*x+a)**2/(d*x+c)**3,x)

[Out]

Integral(sin(a + b*x)**2*cos(a + b*x)/(c + d*x)**3, x)

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