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

Optimal. Leaf size=168 \[ \frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{4 d^2}-\frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {\cos (a+b x)}{4 d (c+d x)}+\frac {\cos (3 a+3 b x)}{4 d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {CosIntegral}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{4 d^2}-\frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{4 d^2}+\frac {3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b c}{d}+3 b x\right )}{4 d^2}-\frac {\cos (a+b x)}{4 d (c+d x)}+\frac {\cos (3 a+3 b x)}{4 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 1.37, size = 139, normalized size = 0.83 \[ -\frac {-3 b \sin \left (3 a-\frac {3 b c}{d}\right ) \text {Ci}\left (\frac {3 b (c+d x)}{d}\right )+b \sin \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )+b \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )-3 b \cos \left (3 a-\frac {3 b c}{d}\right ) \text {Si}\left (\frac {3 b (c+d x)}{d}\right )+\frac {d \cos (a+b x)}{c+d x}-\frac {d \cos (3 (a+b x))}{c+d x}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 236, normalized size = 1.40 \[ \frac {8 \, d \cos \left (b x + a\right )^{3} + 6 \, {\left (b d x + b c\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) - 2 \, {\left (b d x + b c\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - 8 \, d \cos \left (b x + a\right ) - {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + 3 \, {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {3 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {3 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{8 \, {\left (d^{3} x + c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 242, normalized size = 1.44 \[ \frac {\frac {b^{2} \left (-\frac {\cos \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}-\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}}{d}\right )}{4}-\frac {b^{2} \left (-\frac {3 \cos \left (3 b x +3 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {3 \left (\frac {3 \Si \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \cos \left (\frac {-3 d a +3 c b}{d}\right )}{d}-\frac {3 \Ci \left (3 b x +3 a +\frac {-3 d a +3 c b}{d}\right ) \sin \left (\frac {-3 d a +3 c b}{d}\right )}{d}\right )}{d}\right )}{12}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.52, size = 302, normalized size = 1.80 \[ -\frac {8192 \, b^{2} {\left (E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - 8192 \, b^{2} {\left (E_{2}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) + E_{2}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + b^{2} {\left (-8192 i \, E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + 8192 i \, E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right ) + b^{2} {\left (8192 i \, E_{2}\left (\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - 8192 i \, E_{2}\left (-\frac {3 i \, b c + 3 i \, {\left (b x + a\right )} d - 3 i \, a d}{d}\right )\right )} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )}{65536 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/65536*(8192*b^2*(exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_e(2, -(I*b*c + I*(b*x
+ a)*d - I*a*d)/d))*cos(-(b*c - a*d)/d) - 8192*b^2*(exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d)
+ exp_integral_e(2, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*cos(-3*(b*c - a*d)/d) + b^2*(-8192*I*exp_integ
ral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + 8192*I*exp_integral_e(2, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*si
n(-(b*c - a*d)/d) + b^2*(8192*I*exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - 8192*I*exp_integr
al_e(2, -(3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d))*sin(-3*(b*c - a*d)/d))/((b*c*d + (b*x + a)*d^2 - a*d^2)*b)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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