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

Optimal. Leaf size=81 \[ \frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}+\frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{\sin ^2(a+b x)}{d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.139409, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3313, 12, 3303, 3299, 3302} \[ \frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}+\frac{b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b c}{d}+2 b x\right )}{d^2}-\frac{\sin ^2(a+b x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(c + d*x)^2,x]

[Out]

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

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.398654, size = 75, normalized size = 0.93 \[ \frac{b \sin \left (2 a-\frac{2 b c}{d}\right ) \text{CosIntegral}\left (\frac{2 b (c+d x)}{d}\right )+b \cos \left (2 a-\frac{2 b c}{d}\right ) \text{Si}\left (\frac{2 b (c+d x)}{d}\right )-\frac{d \sin ^2(a+b x)}{c+d x}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 156, normalized size = 1.9 \begin{align*}{\frac{1}{b} \left ( -{\frac{{b}^{2}}{ \left ( 2\, \left ( bx+a \right ) d-2\,da+2\,cb \right ) d}}-{\frac{{b}^{2}}{4} \left ( -2\,{\frac{\cos \left ( 2\,bx+2\,a \right ) }{ \left ( \left ( bx+a \right ) d-da+cb \right ) d}}-2\,{\frac{1}{d} \left ( 2\,{\frac{1}{d}{\it Si} \left ( 2\,bx+2\,a+2\,{\frac{-da+cb}{d}} \right ) \cos \left ( 2\,{\frac{-da+cb}{d}} \right ) }-2\,{\frac{1}{d}{\it Ci} \left ( 2\,bx+2\,a+2\,{\frac{-da+cb}{d}} \right ) \sin \left ( 2\,{\frac{-da+cb}{d}} \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.32235, size = 231, normalized size = 2.85 \begin{align*} \frac{16 \, b^{2}{\left (E_{2}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{2}\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^{2}{\left (16 i \, E_{2}\left (\frac{2 i \, b c + 2 i \,{\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 16 i \, E_{2}\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 ) - 32 \, b^{2}}{64 \,{\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(sin(b*x+a)^2/(d*x+c)^2,x, algorithm="maxima")

[Out]

1/64*(16*b^2*(exp_integral_e(2, (2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d) + exp_integral_e(2, -(2*I*b*c + 2*I*(
b*x + a)*d - 2*I*a*d)/d))*cos(-2*(b*c - a*d)/d) - b^2*(16*I*exp_integral_e(2, (2*I*b*c + 2*I*(b*x + a)*d - 2*I
*a*d)/d) - 16*I*exp_integral_e(2, -(2*I*b*c + 2*I*(b*x + a)*d - 2*I*a*d)/d))*sin(-2*(b*c - a*d)/d) - 32*b^2)/(
(b*c*d + (b*x + a)*d^2 - a*d^2)*b)

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Fricas [A]  time = 1.81944, size = 325, normalized size = 4.01 \begin{align*} \frac{2 \, d \cos \left (b x + a\right )^{2} + 2 \,{\left (b d x + b c\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 ) +{\left ({\left (b d x + b c\right )} \operatorname{Ci}\left (\frac{2 \,{\left (b d x + b c\right )}}{d}\right ) +{\left (b d x + b c\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 ) - 2 \, d}{2 \,{\left (d^{3} x + c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d*cos(b*x + a)^2 + 2*(b*d*x + b*c)*cos(-2*(b*c - a*d)/d)*sin_integral(2*(b*d*x + b*c)/d) + ((b*d*x + b*
c)*cos_integral(2*(b*d*x + b*c)/d) + (b*d*x + b*c)*cos_integral(-2*(b*d*x + b*c)/d))*sin(-2*(b*c - a*d)/d) - 2
*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 )}}{\left (c + d x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.34291, size = 3976, normalized size = 49.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b*d*x*imag_part(cos_int
egral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + 2*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2
*tan(a)^2*tan(b*c/d)^2 + 2*b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) + 2*b
*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - 2*b*d*x*real_part(cos_integral
(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 - 2*b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)
^2*tan(a)*tan(b*c/d)^2 + b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b*c*i
mag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + 2*b*c*sin_integral(2*(b*d*x + b*c)
/d)*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 - b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 + b*
d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 - 2*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*ta
n(b*x)^2*tan(a)^2 + 4*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) - 4*b*d*x*im
ag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) + 8*b*d*x*sin_integral(2*(b*d*x + b*c)/d)
*tan(b*x)^2*tan(a)*tan(b*c/d) + 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d)
+ 2*b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2*tan(b*c/d) - b*d*x*imag_part(cos_integra
l(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 + b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b
*c/d)^2 - 2*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(b*c/d)^2 - 2*b*c*real_part(cos_integral(2*b*x
 + 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d)^2 - 2*b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a
)*tan(b*c/d)^2 + b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 - b*d*x*imag_part(cos_in
tegral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 + 2*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*tan(a)^2*tan(b*c/d)^
2 + 2*b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a) + 2*b*d*x*real_part(cos_integral(-2*b*x
 - 2*b*c/d))*tan(b*x)^2*tan(a) - b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)^2 + b*c*imag_p
art(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)^2 - 2*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan
(a)^2 - 2*b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d) - 2*b*d*x*real_part(cos_integra
l(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d) + 4*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a)*
tan(b*c/d) - 4*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a)*tan(b*c/d) + 8*b*c*sin_integral
(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(a)*tan(b*c/d) + 2*b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*t
an(b*c/d) + 2*b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d) - b*c*imag_part(cos_integral
(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d)^2 + b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/
d)^2 - 2*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2*tan(b*c/d)^2 - 2*b*d*x*real_part(cos_integral(2*b*x +
2*b*c/d))*tan(a)*tan(b*c/d)^2 - 2*b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d)^2 + b*c*im
ag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d)^2 - b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*t
an(a)^2*tan(b*c/d)^2 + 2*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(a)^2*tan(b*c/d)^2 + b*d*x*imag_part(cos_integ
ral(2*b*x + 2*b*c/d))*tan(b*x)^2 - b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2 + 2*b*d*x*sin_in
tegral(2*(b*d*x + b*c)/d)*tan(b*x)^2 + 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(a) + 2*b*
c*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(a) - b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))
*tan(a)^2 + b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2 - 2*b*d*x*sin_integral(2*(b*d*x + b*c)/d)
*tan(a)^2 - 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2*tan(b*c/d) - 2*b*c*real_part(cos_integra
l(-2*b*x - 2*b*c/d))*tan(b*x)^2*tan(b*c/d) + 4*b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d
) - 4*b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d) + 8*b*d*x*sin_integral(2*(b*d*x + b*c)
/d)*tan(a)*tan(b*c/d) + 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2*tan(b*c/d) + 2*b*c*real_part(c
os_integral(-2*b*x - 2*b*c/d))*tan(a)^2*tan(b*c/d) - b*d*x*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d)
^2 + b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d)^2 - 2*b*d*x*sin_integral(2*(b*d*x + b*c)/d)*ta
n(b*c/d)^2 - 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)*tan(b*c/d)^2 - 2*b*c*real_part(cos_integral
(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d)^2 + b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*x)^2 - b*c*imag_p
art(cos_integral(-2*b*x - 2*b*c/d))*tan(b*x)^2 + 2*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*x)^2 + 2*b*d*x*re
al_part(cos_integral(2*b*x + 2*b*c/d))*tan(a) + 2*b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a) - b*c
*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(a)^2 + b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)^2 -
2*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(a)^2 - 2*b*d*x*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d) -
 2*b*d*x*real_part(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d) + 4*b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))
*tan(a)*tan(b*c/d) - 4*b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d))*tan(a)*tan(b*c/d) + 8*b*c*sin_integral(2*
(b*d*x + b*c)/d)*tan(a)*tan(b*c/d) - b*c*imag_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d)^2 + b*c*imag_part
(cos_integral(-2*b*x - 2*b*c/d))*tan(b*c/d)^2 - 2*b*c*sin_integral(2*(b*d*x + b*c)/d)*tan(b*c/d)^2 - 2*d*tan(b
*x)^2*tan(b*c/d)^2 - 4*d*tan(b*x)*tan(a)*tan(b*c/d)^2 - 2*d*tan(a)^2*tan(b*c/d)^2 + b*d*x*imag_part(cos_integr
al(2*b*x + 2*b*c/d)) - b*d*x*imag_part(cos_integral(-2*b*x - 2*b*c/d)) + 2*b*d*x*sin_integral(2*(b*d*x + b*c)/
d) + 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(a) + 2*b*c*real_part(cos_integral(-2*b*x - 2*b*c/d))*t
an(a) - 2*b*c*real_part(cos_integral(2*b*x + 2*b*c/d))*tan(b*c/d) - 2*b*c*real_part(cos_integral(-2*b*x - 2*b*
c/d))*tan(b*c/d) + b*c*imag_part(cos_integral(2*b*x + 2*b*c/d)) - b*c*imag_part(cos_integral(-2*b*x - 2*b*c/d)
) + 2*b*c*sin_integral(2*(b*d*x + b*c)/d) - 2*d*tan(b*x)^2 - 4*d*tan(b*x)*tan(a) - 2*d*tan(a)^2)/(d^3*x*tan(b*
x)^2*tan(a)^2*tan(b*c/d)^2 + c*d^2*tan(b*x)^2*tan(a)^2*tan(b*c/d)^2 + d^3*x*tan(b*x)^2*tan(a)^2 + d^3*x*tan(b*
x)^2*tan(b*c/d)^2 + d^3*x*tan(a)^2*tan(b*c/d)^2 + c*d^2*tan(b*x)^2*tan(a)^2 + c*d^2*tan(b*x)^2*tan(b*c/d)^2 +
c*d^2*tan(a)^2*tan(b*c/d)^2 + d^3*x*tan(b*x)^2 + d^3*x*tan(a)^2 + d^3*x*tan(b*c/d)^2 + c*d^2*tan(b*x)^2 + c*d^
2*tan(a)^2 + c*d^2*tan(b*c/d)^2 + d^3*x + c*d^2)