3.1.30 \(\int \frac {\sin (c+d x)}{(a+b x)^2} \, dx\) [30]

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

[Out]

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

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Rubi [A]
time = 0.07, 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 = {3378, 3384, 3380, 3383} \begin {gather*} \frac {d \cos \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{b^2}-\frac {d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{b^2}-\frac {\sin (c+d x)}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(a + b*x)^2,x]

[Out]

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

Rule 3378

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 3380

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 3383

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 3384

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

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Mathematica [A]
time = 0.16, size = 66, normalized size = 0.92 \begin {gather*} \frac {d \cos \left (c-\frac {a d}{b}\right ) \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right )-\frac {b \sin (c+d x)}{a+b x}-d \sin \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(a + b*x)^2,x]

[Out]

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

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Maple [A]
time = 0.05, size = 107, normalized size = 1.49

method	result	size
derivativedivides	\(d \left (-\frac {\sin \left (d x +c \right )}{\left (d a -c b +b \left (d x +c \right )\right ) b}+\frac {\frac {\sinIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\cosineIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )\)	\(107\)
default	\(d \left (-\frac {\sin \left (d x +c \right )}{\left (d a -c b +b \left (d x +c \right )\right ) b}+\frac {\frac {\sinIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}+\frac {\cosineIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}}{b}\right )\)	\(107\)
risch	\(-\frac {d \,{\mathrm e}^{-\frac {i \left (d a -c b \right )}{b}} \expIntegral \left (1, -i d x -i c -\frac {i a d -i b c}{b}\right )}{2 b^{2}}-\frac {d \,{\mathrm e}^{\frac {i \left (d a -c b \right )}{b}} \expIntegral \left (1, i d x +i c +\frac {i \left (d a -c b \right )}{b}\right )}{2 b^{2}}-\frac {\left (-2 d x b -2 d a \right ) \sin \left (d x +c \right )}{2 b \left (b x +a \right ) \left (-d x b -d a \right )}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.37, size = 164, normalized size = 2.28 \begin {gather*} \frac {d^{2} {\left (-i \, E_{2}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + i \, E_{2}\left (-\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) + d^{2} {\left (E_{2}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + E_{2}\left (-\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{2 \, {\left ({\left (d x + c\right )} b^{2} - b^{2} c + a b d\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d^2*(-I*exp_integral_e(2, (I*(d*x + c)*b - I*b*c + I*a*d)/b) + I*exp_integral_e(2, -(I*(d*x + c)*b - I*b*
c + I*a*d)/b))*cos(-(b*c - a*d)/b) + d^2*(exp_integral_e(2, (I*(d*x + c)*b - I*b*c + I*a*d)/b) + exp_integral_
e(2, -(I*(d*x + c)*b - I*b*c + I*a*d)/b))*sin(-(b*c - a*d)/b))/(((d*x + c)*b^2 - b^2*c + a*b*d)*d)

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Fricas [A]
time = 0.34, size = 123, normalized size = 1.71 \begin {gather*} \frac {2 \, {\left (b d x + a d\right )} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {b d x + a d}{b}\right ) + {\left ({\left (b d x + a d\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (b d x + a d\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right )\right )} \cos \left (-\frac {b c - a d}{b}\right ) - 2 \, b \sin \left (d x + c\right )}{2 \, {\left (b^{3} x + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (73) = 146\).
time = 5.99, size = 518, normalized size = 7.19 \begin {gather*} \frac {{\left ({\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - b c d^{2} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a d^{3} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Ci}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + {\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} d^{2} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) - b c d^{2} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + a d^{3} \sin \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b c + a d}{b}\right ) + b d^{2} \sin \left (-\frac {{\left (b x + a\right )} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )}}{b}\right )\right )} b^{2}}{{\left ({\left (b x + a\right )} b^{4} {\left (\frac {b c}{b x + a} - \frac {a d}{b x + a} + d\right )} - b^{5} c + a b^{4} d\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(a + b*x)^2,x)

[Out]

int(sin(c + d*x)/(a + b*x)^2, x)

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