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

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

[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)

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Rubi [A]  time = 0.301291, antiderivative size = 168, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 1.43532, size = 139, normalized size = 0.83 \[ -\frac{-3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )+b \sin \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )+b \cos \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )-3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )+\frac{d \cos (a+b x)}{c+d x}-\frac{d \cos (3 (a+b x))}{c+d x}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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*Co
s[3*a - (3*b*c)/d]*SinIntegral[(3*b*(c + d*x))/d])/(4*d^2)

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Maple [A]  time = 0.023, size = 242, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{2}}{4} \left ( -{\frac{\cos \left ( bx+a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-{\frac{1}{d} \left ({\frac{1}{d}{\it Si} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \cos \left ({\frac{-ad+bc}{d}} \right ) }-{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \sin \left ({\frac{-ad+bc}{d}} \right ) } \right ) } \right ) }-{\frac{{b}^{2}}{12} \left ( -3\,{\frac{\cos \left ( 3\,bx+3\,a \right ) }{ \left ( \left ( bx+a \right ) d-ad+bc \right ) d}}-3\,{\frac{1}{d} \left ( 3\,{\frac{1}{d}{\it Si} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 3\,{\frac{-ad+bc}{d}} \right ) }-3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \sin \left ( 3\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}

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-a*d+b*c)/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-a*d+b*c)/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 = 1.77077, size = 408, normalized size = 2.43 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 0.54075, size = 593, normalized size = 3.53 \begin{align*} \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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{2}{\left (a + b x \right )} \cos{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right ) \sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{2}}\,{d x} \end{align*}

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]

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