3.1.7 \(\int \frac {\sin (a+b x)}{(c+d x)^3} \, dx\) [7]

Optimal. Leaf size=104 \[ -\frac {b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {b^2 \text {Ci}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{2 d^3}-\frac {\sin (a+b x)}{2 d (c+d x)^2}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{2 d^3} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {b^2 \sin \left (a-\frac {b c}{d}\right ) \text {CosIntegral}\left (\frac {b c}{d}+b x\right )}{2 d^3}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{2 d^3}-\frac {b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin (a+b x)}{2 d (c+d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/(c + d*x)^3,x]

[Out]

-1/2*(b*Cos[a + b*x])/(d^2*(c + d*x)) - (b^2*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/(2*d^3) - Sin[a + b*
x]/(2*d*(c + d*x)^2) - (b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/(2*d^3)

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 (a+b x)}{(c+d x)^3} \, dx &=-\frac {\sin (a+b x)}{2 d (c+d x)^2}+\frac {b \int \frac {\cos (a+b x)}{(c+d x)^2} \, dx}{2 d}\\ &=-\frac {b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin (a+b x)}{2 d (c+d x)^2}-\frac {b^2 \int \frac {\sin (a+b x)}{c+d x} \, dx}{2 d^2}\\ &=-\frac {b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {\sin (a+b x)}{2 d (c+d x)^2}-\frac {\left (b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}-\frac {\left (b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}\\ &=-\frac {b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {b^2 \text {Ci}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{2 d^3}-\frac {\sin (a+b x)}{2 d (c+d x)^2}-\frac {b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.46, size = 87, normalized size = 0.84 \begin {gather*} -\frac {b^2 \text {Ci}\left (b \left (\frac {c}{d}+x\right )\right ) \sin \left (a-\frac {b c}{d}\right )+\frac {d (b (c+d x) \cos (a+b x)+d \sin (a+b x))}{(c+d x)^2}+b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (b \left (\frac {c}{d}+x\right )\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/(c + d*x)^3,x]

[Out]

-1/2*(b^2*CosIntegral[b*(c/d + x)]*Sin[a - (b*c)/d] + (d*(b*(c + d*x)*Cos[a + b*x] + d*Sin[a + b*x]))/(c + d*x
)^2 + b^2*Cos[a - (b*c)/d]*SinIntegral[b*(c/d + x)])/d^3

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Maple [A]
time = 0.10, size = 150, normalized size = 1.44

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\sin \left (b x +a \right )}{2 \left (-d a +c b +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-d a +c b +d \left (b x +a \right )\right ) d}-\frac {-\frac {\sinIntegral \left (-b x -a -\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}-\frac {\cosineIntegral \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}\right )\)	\(150\)
default	\(b^{2} \left (-\frac {\sin \left (b x +a \right )}{2 \left (-d a +c b +d \left (b x +a \right )\right )^{2} d}+\frac {-\frac {\cos \left (b x +a \right )}{\left (-d a +c b +d \left (b x +a \right )\right ) d}-\frac {-\frac {\sinIntegral \left (-b x -a -\frac {-d a +c b}{d}\right ) \cos \left (\frac {-d a +c b}{d}\right )}{d}-\frac {\cosineIntegral \left (b x +a +\frac {-d a +c b}{d}\right ) \sin \left (\frac {-d a +c b}{d}\right )}{d}}{d}}{2 d}\right )\)	\(150\)
risch	\(-\frac {i b^{2} {\mathrm e}^{\frac {i \left (d a -c b \right )}{d}} \expIntegral \left (1, -i b x -i a -\frac {-i a d +i b c}{d}\right )}{4 d^{3}}+\frac {i b^{2} {\mathrm e}^{-\frac {i \left (d a -c b \right )}{d}} \expIntegral \left (1, i b x +i a -\frac {i \left (d a -c b \right )}{d}\right )}{4 d^{3}}+\frac {i \left (2 i b^{3} d^{3} x^{3}+6 i b^{3} c \,d^{2} x^{2}+6 i b^{3} c^{2} d x +2 i b^{3} c^{3}\right ) \cos \left (b x +a \right )}{4 d^{2} \left (d x +c \right )^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {\left (2 d^{2} x^{2} b^{2}+4 b^{2} c d x +2 b^{2} c^{2}\right ) \sin \left (b x +a \right )}{4 d \left (d x +c \right )^{2} \left (d^{2} x^{2} b^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}\)	\(271\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)/(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

b^2*(-1/2*sin(b*x+a)/(-d*a+c*b+d*(b*x+a))^2/d+1/2*(-cos(b*x+a)/(-d*a+c*b+d*(b*x+a))/d-(-Si(-b*x-a-(-a*d+b*c)/d
)*cos((-a*d+b*c)/d)/d-Ci(b*x+a+(-a*d+b*c)/d)*sin((-a*d+b*c)/d)/d)/d)/d)

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Maxima [C] Result contains complex when optimal does not.
time = 0.42, size = 199, normalized size = 1.91 \begin {gather*} -\frac {b^{3} {\left (i \, E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) - i \, E_{3}\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^{3} {\left (E_{3}\left (\frac {i \, b c + i \, {\left (b x + a\right )} d - i \, a d}{d}\right ) + E_{3}\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^{2} c^{2} d - 2 \, a b c d^{2} + {\left (b x + a\right )}^{2} d^{3} + a^{2} d^{3} + 2 \, {\left (b c d^{2} - a d^{3}\right )} {\left (b x + a\right )}\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/2*(b^3*(I*exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) - I*exp_integral_e(3, -(I*b*c + I*(b*x + a)*
d - I*a*d)/d))*cos(-(b*c - a*d)/d) + b^3*(exp_integral_e(3, (I*b*c + I*(b*x + a)*d - I*a*d)/d) + exp_integral_
e(3, -(I*b*c + I*(b*x + a)*d - I*a*d)/d))*sin(-(b*c - a*d)/d))/((b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + a
^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (96) = 192\).
time = 0.43, size = 209, normalized size = 2.01 \begin {gather*} -\frac {2 \, d^{2} \sin \left (b x + a\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {Si}\left (\frac {b d x + b c}{d}\right ) + 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) + {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + b c}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*(2*d^2*sin(b*x + a) + 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(-(b*c - a*d)/d)*sin_integral((b*d*x + b
*c)/d) + 2*(b*d^2*x + b*c*d)*cos(b*x + a) + ((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral((b*d*x + b*c)/
d) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos_integral(-(b*d*x + b*c)/d))*sin(-(b*c - a*d)/d))/(d^5*x^2 + 2*c
*d^4*x + c^2*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + b x \right )}}{\left (c + d x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*x+c)**3,x)

[Out]

Integral(sin(a + b*x)/(c + d*x)**3, x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 3.78, size = 5727, normalized size = 55.07 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)/(d*x+c)^3,x, algorithm="giac")

[Out]

-1/4*(b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b^2*d^2*
x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^2*x^2*sin_int
egral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^2*x^2*real_part(cos_integral(b*x
+ b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) + 2*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*ta
n(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) - 2*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*t
an(1/2*a)*tan(1/2*b*c/d)^2 - 2*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan
(1/2*b*c/d)^2 + 2*b^2*c*d*x*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2
- 2*b^2*c*d*x*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 4*b^2*c*d*x
*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b^2*d^2*x^2*imag_part(cos_integr
al(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2 + b^2*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^
2*tan(1/2*a)^2 - 2*b^2*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2 + 4*b^2*d^2*x^2*imag_
part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) - 4*b^2*d^2*x^2*imag_part(cos_integra
l(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 8*b^2*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/
2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 4*b^2*c*d*x*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a
)^2*tan(1/2*b*c/d) + 4*b^2*c*d*x*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c
/d) - b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 + b^2*d^2*x^2*imag_part
(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 2*b^2*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan
(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 4*b^2*c*d*x*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(
1/2*b*c/d)^2 - 4*b^2*c*d*x*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 +
b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - b^2*d^2*x^2*imag_part(cos_int
egral(-b*x - b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^2*x^2*sin_integral((b*d*x + b*c)/d)*tan(1/2*a)^2*
tan(1/2*b*c/d)^2 + b^2*c^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 -
b^2*c^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*c^2*sin_in
tegral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b^2*d^2*x^2*real_part(cos_integral(b*
x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a) + 2*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan
(1/2*a) - 2*b^2*c*d*x*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2 + 2*b^2*c*d*x*imag_part
(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2 - 4*b^2*c*d*x*sin_integral((b*d*x + b*c)/d)*tan(1/2*b
*x)^2*tan(1/2*a)^2 - 2*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d) - 2*b^2*
d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d) + 8*b^2*c*d*x*imag_part(cos_integr
al(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) - 8*b^2*c*d*x*imag_part(cos_integral(-b*x - b*c/d))*
tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d) + 16*b^2*c*d*x*sin_integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*a
)*tan(1/2*b*c/d) + 2*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d) + 2*b^2*d^2*
x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d) + 2*b^2*c^2*real_part(cos_integral(b*x +
b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) + 2*b^2*c^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*
b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d) - 2*b^2*c*d*x*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*b
*c/d)^2 + 2*b^2*c*d*x*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 4*b^2*c*d*x*sin_
integral((b*d*x + b*c)/d)*tan(1/2*b*x)^2*tan(1/2*b*c/d)^2 - 2*b^2*d^2*x^2*real_part(cos_integral(b*x + b*c/d))
*tan(1/2*a)*tan(1/2*b*c/d)^2 - 2*b^2*d^2*x^2*real_part(cos_integral(-b*x - b*c/d))*tan(1/2*a)*tan(1/2*b*c/d)^2
- 2*b^2*c^2*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 - 2*b^2*c^2*real_
part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2*tan(1/2*a)*tan(1/2*b*c/d)^2 + 2*b^2*c*d*x*imag_part(cos_integr
al(b*x + b*c/d))*tan(1/2*a)^2*tan(1/2*b*c/d)^2 - 2*b^2*c*d*x*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*a)^
2*tan(1/2*b*c/d)^2 + 4*b^2*c*d*x*sin_integral((b*d*x + b*c)/d)*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + 2*b*d^2*x*tan(1
/2*b*x)^2*tan(1/2*a)^2*tan(1/2*b*c/d)^2 + b^2*d^2*x^2*imag_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2 - b^
2*d^2*x^2*imag_part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2 + 2*b^2*d^2*x^2*sin_integral((b*d*x + b*c)/d)*t
an(1/2*b*x)^2 + 4*b^2*c*d*x*real_part(cos_integral(b*x + b*c/d))*tan(1/2*b*x)^2*tan(1/2*a) + 4*b^2*c*d*x*real_
part(cos_integral(-b*x - b*c/d))*tan(1/2*b*x)^2...

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (a+b\,x\right )}{{\left (c+d\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)/(c + d*x)^3,x)

[Out]

int(sin(a + b*x)/(c + d*x)^3, x)

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