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

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

[Out]

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

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Rubi [A]  time = 0.266386, 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} \[ \frac{b \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (\frac{b c}{d}+b x\right )}{4 d^2}+\frac{3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{b \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{4 d^2}-\frac{3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{\sin (a+b x)}{4 d (c+d x)}-\frac{\sin (3 a+3 b x)}{4 d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Cos[a - (b*c)/d]*CosIntegral[(b*c)/d + b*x])/(4*d^2) + (3*b*Cos[3*a - (3*b*c)/d]*CosIntegral[(3*b*c)/d + 3*
b*x])/(4*d^2) - Sin[a + b*x]/(4*d*(c + d*x)) - Sin[3*a + 3*b*x]/(4*d*(c + d*x)) - (b*Sin[a - (b*c)/d]*SinInteg
ral[(b*c)/d + b*x])/(4*d^2) - (3*b*Sin[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 ^2(a+b x) \sin (a+b x)}{(c+d x)^2} \, dx &=\int \left (\frac{\sin (a+b x)}{4 (c+d x)^2}+\frac{\sin (3 a+3 b x)}{4 (c+d x)^2}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\sin (a+b x)}{(c+d x)^2} \, dx+\frac{1}{4} \int \frac{\sin (3 a+3 b x)}{(c+d x)^2} \, dx\\ &=-\frac{\sin (a+b x)}{4 d (c+d x)}-\frac{\sin (3 a+3 b x)}{4 d (c+d x)}+\frac{b \int \frac{\cos (a+b x)}{c+d x} \, dx}{4 d}+\frac{(3 b) \int \frac{\cos (3 a+3 b x)}{c+d x} \, dx}{4 d}\\ &=-\frac{\sin (a+b x)}{4 d (c+d x)}-\frac{\sin (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{\cos \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{\cos \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{\sin \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{\sin \left (\frac{b c}{d}+b x\right )}{c+d x} \, dx}{4 d}\\ &=\frac{b \cos \left (a-\frac{b c}{d}\right ) \text{Ci}\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{Ci}\left (\frac{3 b c}{d}+3 b x\right )}{4 d^2}-\frac{\sin (a+b x)}{4 d (c+d x)}-\frac{\sin (3 a+3 b x)}{4 d (c+d x)}-\frac{b \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (\frac{b c}{d}+b x\right )}{4 d^2}-\frac{3 b \sin \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 = 1.13457, size = 139, normalized size = 0.83 \[ -\frac{-b \cos \left (a-\frac{b c}{d}\right ) \text{CosIntegral}\left (b \left (\frac{c}{d}+x\right )\right )-3 b \cos \left (3 a-\frac{3 b c}{d}\right ) \text{CosIntegral}\left (\frac{3 b (c+d x)}{d}\right )+b \sin \left (a-\frac{b c}{d}\right ) \text{Si}\left (b \left (\frac{c}{d}+x\right )\right )+3 b \sin \left (3 a-\frac{3 b c}{d}\right ) \text{Si}\left (\frac{3 b (c+d x)}{d}\right )+\frac{d \sin (a+b x)}{c+d x}+\frac{d \sin (3 (a+b x))}{c+d x}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 240, normalized size = 1.4 \begin{align*}{\frac{1}{b} \left ({\frac{{b}^{2}}{12} \left ( -3\,{\frac{\sin \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 ) \sin \left ( 3\,{\frac{-ad+bc}{d}} \right ) }+3\,{\frac{1}{d}{\it Ci} \left ( 3\,bx+3\,a+3\,{\frac{-ad+bc}{d}} \right ) \cos \left ( 3\,{\frac{-ad+bc}{d}} \right ) } \right ) } \right ) }+{\frac{{b}^{2}}{4} \left ( -{\frac{\sin \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 ) \sin \left ({\frac{-ad+bc}{d}} \right ) }+{\frac{1}{d}{\it Ci} \left ( bx+a+{\frac{-ad+bc}{d}} \right ) \cos \left ({\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)^2*sin(b*x+a)/(d*x+c)^2,x)

[Out]

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

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Maxima [C]  time = 1.72511, size = 405, normalized size = 2.41 \begin{align*} -\frac{b^{2}{\left (i \, E_{2}\left (\frac{i \, b c + i \,{\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, 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 ) + b^{2}{\left (i \, E_{2}\left (\frac{3 i \, b c + 3 i \,{\left (b x + a\right )} d - 3 i \, a d}{d}\right ) - i \, 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 (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 )} \sin \left (-\frac{b c - a d}{d}\right ) + 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 )} \sin \left (-\frac{3 \,{\left (b c - a d\right )}}{d}\right )}{8 \,{\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)^2*sin(b*x+a)/(d*x+c)^2,x, algorithm="maxima")

[Out]

-1/8*(b^2*(I*exp_integral_e(2, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_integral_e(2, -(I*b*c + I*(b*x + a)*
d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^2*(I*exp_integral_e(2, (3*I*b*c + 3*I*(b*x + a)*d - 3*I*a*d)/d) - I*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*(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))*sin(-(b*c - a*d)/d) + 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))*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 = 0.538292, size = 586, normalized size = 3.49 \begin{align*} -\frac{8 \, d \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + 6 \,{\left (b d x + b c\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 d x + b c\right )} \sin \left (-\frac{b c - a d}{d}\right ) \operatorname{Si}\left (\frac{b d x + b c}{d}\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 )} \cos \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 )} \cos \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)^2*sin(b*x+a)/(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/8*(8*d*cos(b*x + a)^2*sin(b*x + a) + 6*(b*d*x + b*c)*sin(-3*(b*c - a*d)/d)*sin_integral(3*(b*d*x + b*c)/d)
+ 2*(b*d*x + b*c)*sin(-(b*c - a*d)/d)*sin_integral((b*d*x + b*c)/d) - ((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))*cos(-(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))*cos(-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{\left (a + b x \right )} \cos ^{2}{\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)**2*sin(b*x+a)/(d*x+c)**2,x)

[Out]

Integral(sin(a + b*x)*cos(a + b*x)**2/(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 )^{2} \sin \left (b x + a\right )}{{\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)^2*sin(b*x+a)/(d*x+c)^2,x, algorithm="giac")

[Out]

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