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

Optimal. Leaf size=72 \[ \frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (\frac {b c}{d}+b x\right )}{d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^2}-\frac {\sin (a+b x)}{d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3297, 3303, 3299, 3302} \[ \frac {b \cos \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{d^2}-\frac {b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^2}-\frac {\sin (a+b x)}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(c + d*x)^2,x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 66, normalized size = 0.92 \[ \frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right )-b \sin \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )-\frac {d \sin (a+b x)}{c+d x}}{d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/(c + d*x)^2,x]

[Out]

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

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fricas [A]  time = 0.63, size = 124, normalized size = 1.72 \[ -\frac {2 \, {\left (b d x + b c\right )} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) - {\left ({\left (b d x + b c\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b d x + b c\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + 2 \, d \sin \left (b x + a\right )}{2 \, {\left (d^{3} x + c d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.96, size = 521, normalized size = 7.24 \[ \frac {{\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + b^{3} c \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) - a b^{2} d \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Ci}\left (\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + {\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} b^{2} \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + b^{3} c \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) - a b^{2} d \sin \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} + b c - a d}{d}\right ) + b^{2} d \sin \left (-\frac {{\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )}}{d}\right )\right )} d^{2}}{{\left ({\left (d x + c\right )} {\left (b - \frac {b c}{d x + c} + \frac {a d}{d x + c}\right )} d^{4} + b c d^{4} - a d^{5}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*cos(-(b*c - a*d)/d)*cos_integral(((d*x + c)*(b - b*c/(d*x +
c) + a*d/(d*x + c)) + b*c - a*d)/d) + b^3*c*cos(-(b*c - a*d)/d)*cos_integral(((d*x + c)*(b - b*c/(d*x + c) +
a*d/(d*x + c)) + b*c - a*d)/d) - a*b^2*d*cos(-(b*c - a*d)/d)*cos_integral(((d*x + c)*(b - b*c/(d*x + c) + a*d/
(d*x + c)) + b*c - a*d)/d) + (d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c))*b^2*sin(-(b*c - a*d)/d)*sin_integra
l(-((d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + b^3*c*sin(-(b*c - a*d)/d)*sin_integral(-((
d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) - a*b^2*d*sin(-(b*c - a*d)/d)*sin_integral(-((d*x
+ c)*(b - b*c/(d*x + c) + a*d/(d*x + c)) + b*c - a*d)/d) + b^2*d*sin(-(d*x + c)*(b - b*c/(d*x + c) + a*d/(d*x
+ c))/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 = 107, normalized size = 1.49 \[ b \left (-\frac {\sin \left (b x +a \right )}{\left (\left (b x +a \right ) d -d a +c b \right ) d}+\frac {\frac {\Si \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}+\frac {\Ci \left (b x +a +\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}}{d}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-sin(b*x+a)/((b*x+a)*d-d*a+c*b)/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 = 0.52, size = 164, normalized size = 2.28 \[ -\frac {b^{2} {\left (i \, E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + b^{2} {\left (E_{2}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{2}\left (-\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )}{2 \, {\left (b c d + {\left (b x + a\right )} d^{2} - a d^{2}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(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*(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*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 \[ \int \frac {\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/(c + d*x)^2,x)

[Out]

int(sin(a + b*x)/(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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