3.1.36 \(\int \frac {\sin (c+d x)}{(a+b x)^3} \, dx\) [36]

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

[Out]

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

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Rubi [A]
time = 0.09, 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 {d^2 \sin \left (c-\frac {a d}{b}\right ) \text {CosIntegral}\left (\frac {a d}{b}+d x\right )}{2 b^3}-\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (x d+\frac {a d}{b}\right )}{2 b^3}-\frac {d \cos (c+d x)}{2 b^2 (a+b x)}-\frac {\sin (c+d x)}{2 b (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(a + b*x)^3,x]

[Out]

-1/2*(d*Cos[c + d*x])/(b^2*(a + b*x)) - (d^2*CosIntegral[(a*d)/b + d*x]*Sin[c - (a*d)/b])/(2*b^3) - Sin[c + d*
x]/(2*b*(a + b*x)^2) - (d^2*Cos[c - (a*d)/b]*SinIntegral[(a*d)/b + d*x])/(2*b^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 (c+d x)}{(a+b x)^3} \, dx &=-\frac {\sin (c+d x)}{2 b (a+b x)^2}+\frac {d \int \frac {\cos (c+d x)}{(a+b x)^2} \, dx}{2 b}\\ &=-\frac {d \cos (c+d x)}{2 b^2 (a+b x)}-\frac {\sin (c+d x)}{2 b (a+b x)^2}-\frac {d^2 \int \frac {\sin (c+d x)}{a+b x} \, dx}{2 b^2}\\ &=-\frac {d \cos (c+d x)}{2 b^2 (a+b x)}-\frac {\sin (c+d x)}{2 b (a+b x)^2}-\frac {\left (d^2 \cos \left (c-\frac {a d}{b}\right )\right ) \int \frac {\sin \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{2 b^2}-\frac {\left (d^2 \sin \left (c-\frac {a d}{b}\right )\right ) \int \frac {\cos \left (\frac {a d}{b}+d x\right )}{a+b x} \, dx}{2 b^2}\\ &=-\frac {d \cos (c+d x)}{2 b^2 (a+b x)}-\frac {d^2 \text {Ci}\left (\frac {a d}{b}+d x\right ) \sin \left (c-\frac {a d}{b}\right )}{2 b^3}-\frac {\sin (c+d x)}{2 b (a+b x)^2}-\frac {d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (\frac {a d}{b}+d x\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 87, normalized size = 0.84 \begin {gather*} -\frac {d^2 \text {Ci}\left (d \left (\frac {a}{b}+x\right )\right ) \sin \left (c-\frac {a d}{b}\right )+\frac {b (d (a+b x) \cos (c+d x)+b \sin (c+d x))}{(a+b x)^2}+d^2 \cos \left (c-\frac {a d}{b}\right ) \text {Si}\left (d \left (\frac {a}{b}+x\right )\right )}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(a + b*x)^3,x]

[Out]

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

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Maple [A]
time = 0.05, size = 145, normalized size = 1.39

method	result	size
derivativedivides	\(d^{2} \left (-\frac {\sin \left (d x +c \right )}{2 \left (d a -c b +b \left (d x +c \right )\right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (d a -c b +b \left (d x +c \right )\right ) b}-\frac {\frac {\sinIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\cosineIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}}{b}}{2 b}\right )\)	\(145\)
default	\(d^{2} \left (-\frac {\sin \left (d x +c \right )}{2 \left (d a -c b +b \left (d x +c \right )\right )^{2} b}+\frac {-\frac {\cos \left (d x +c \right )}{\left (d a -c b +b \left (d x +c \right )\right ) b}-\frac {\frac {\sinIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \cos \left (\frac {d a -c b}{b}\right )}{b}-\frac {\cosineIntegral \left (d x +c +\frac {d a -c b}{b}\right ) \sin \left (\frac {d a -c b}{b}\right )}{b}}{b}}{2 b}\right )\)	\(145\)
risch	\(-\frac {i d^{2} {\mathrm e}^{-\frac {i \left (d a -c b \right )}{b}} \expIntegral \left (1, -i d x -i c -\frac {i a d -i b c}{b}\right )}{4 b^{3}}+\frac {i d^{2} {\mathrm e}^{\frac {i \left (d a -c b \right )}{b}} \expIntegral \left (1, i d x +i c +\frac {i \left (d a -c b \right )}{b}\right )}{4 b^{3}}+\frac {i \left (-2 i b^{3} d^{3} x^{3}-6 i a \,b^{2} d^{3} x^{2}-6 i a^{2} b \,d^{3} x -2 i a^{3} d^{3}\right ) \cos \left (d x +c \right )}{4 b^{2} \left (b x +a \right )^{2} \left (-d^{2} x^{2} b^{2}-2 a b \,d^{2} x -d^{2} a^{2}\right )}-\frac {\left (-2 d^{2} x^{2} b^{2}-4 a b \,d^{2} x -2 d^{2} a^{2}\right ) \sin \left (d x +c \right )}{4 b \left (b x +a \right )^{2} \left (-d^{2} x^{2} b^{2}-2 a b \,d^{2} x -d^{2} a^{2}\right )}\)	\(275\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.39, size = 199, normalized size = 1.91 \begin {gather*} \frac {d^{3} {\left (-i \, E_{3}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + i \, E_{3}\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^{3} {\left (E_{3}\left (\frac {i \, {\left (d x + c\right )} b - i \, b c + i \, a d}{b}\right ) + E_{3}\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 )}^{2} b^{3} + b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2} - 2 \, {\left (b^{3} c - a b^{2} d\right )} {\left (d x + c\right )}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (98) = 196\).
time = 0.36, size = 210, normalized size = 2.02 \begin {gather*} -\frac {2 \, b^{2} \sin \left (d x + c\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \cos \left (-\frac {b c - a d}{b}\right ) \operatorname {Si}\left (\frac {b d x + a d}{b}\right ) + 2 \, {\left (b^{2} d x + a b d\right )} \cos \left (d x + c\right ) - {\left ({\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \operatorname {Ci}\left (\frac {b d x + a d}{b}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} \operatorname {Ci}\left (-\frac {b d x + a d}{b}\right )\right )} \sin \left (-\frac {b c - a d}{b}\right )}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 3.04, 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(d*x+c)/(b*x+a)^3,x, algorithm="giac")

[Out]

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

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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 )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(a + b*x)^3,x)

[Out]

int(sin(c + d*x)/(a + b*x)^3, x)

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