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

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

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 81, normalized size of antiderivative = 1.00, 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 12

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

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 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]

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*}

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Mathematica [A]  time = 0.42, size = 75, normalized size = 0.93 \[ \frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b (c+d x)}{d}\right )+b \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b (c+d x)}{d}\right )-\frac {d \sin ^2(a+b x)}{c+d x}}{d^2} \]

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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fricas [A]  time = 0.98, size = 130, normalized size = 1.60 \[ \frac {2 \, d \cos \left (b x + a\right )^{2} + 2 \, {\left (b d x + b c\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {2 \, {\left (b d x + b c\right )}}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {2 \, {\left (b d x + b c\right )}}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, d}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]

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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giac [B]  time = 1.37, size = 535, normalized size = 6.60 \[ \frac {{\left (2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, b^{3} c \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, a b^{2} d \operatorname {Ci}\left (\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) - 2 \, b^{3} c \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) + 2 \, a b^{2} d \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {Si}\left (-\frac {2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d\right )}}{d}\right ) + b^{2} d \cos \left (-\frac {2 \, {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )}}{d}\right ) - b^{2} d\right )} d^{2}}{2 \, {\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} d^{4} + b c d^{4} - a d^{5}\right )} b} \]

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*(2*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos_integral(2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d
*x + c)) + b*c - a*d)/d)*sin(-2*(b*c - a*d)/d) + 2*b^3*c*cos_integral(2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d
*x + c)) + b*c - a*d)/d)*sin(-2*(b*c - a*d)/d) - 2*a*b^2*d*cos_integral(2*((d*x + c)*(b - b*c/(d*x + c) + a*d/
(d*x + c)) + b*c - a*d)/d)*sin(-2*(b*c - a*d)/d) - 2*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos(-2*
(b*c - a*d)/d)*sin_integral(-2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) - 2*b^3*c*cos(-2
*(b*c - a*d)/d)*sin_integral(-2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + 2*a*b^2*d*cos
(-2*(b*c - a*d)/d)*sin_integral(-2*((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + b^2*d*cos(
-2*(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))/d) - b^2*d)*d^2/(((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c
))*d^4 + b*c*d^4 - a*d^5)*b)

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maple [A]  time = 0.02, size = 156, normalized size = 1.93 \[ \frac {-\frac {b^{2}}{2 \left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {b^{2} \left (-\frac {2 \cos \left (2 b x +2 a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}-\frac {2 \left (\frac {2 \Si \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \cos \left (\frac {-2 d a +2 c b}{d}\right )}{d}-\frac {2 \Ci \left (2 b x +2 a +\frac {-2 d a +2 c b}{d}\right ) \sin \left (\frac {-2 d a +2 c b}{d}\right )}{d}\right )}{d}\right )}{4}}{b} \]

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 = 0.61, size = 171, normalized size = 2.11 \[ \frac {16 \, b^{2} {\left (E_{2}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) + E_{2}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - b^{2} {\left (16 i \, E_{2}\left (\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right ) - 16 i \, E_{2}\left (-\frac {2 i \, b c + 2 i \, {\left (b x + a\right )} d - 2 i \, a d}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 32 \, b^{2}}{64 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2/(c + d*x)^2,x)

[Out]

int(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 \[ \int \frac {\sin ^{2}{\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \]

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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