3.1.13 \(\int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx\) [13]

3.1.13.1 Optimal result
3.1.13.2 Mathematica [A] (verified)
3.1.13.3 Rubi [A] (verified)
3.1.13.4 Maple [A] (verified)
3.1.13.5 Fricas [A] (verification not implemented)
3.1.13.6 Sympy [A] (verification not implemented)
3.1.13.7 Maxima [C] (verification not implemented)
3.1.13.8 Giac [C] (verification not implemented)
3.1.13.9 Mupad [F(-1)]

3.1.13.1 Optimal result

Integrand size = 17, antiderivative size = 62 \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=-\frac {2 a b \cos (c+d x)}{d}-\frac {b^2 x \cos (c+d x)}{d}+a^2 \operatorname {CosIntegral}(d x) \sin (c)+\frac {b^2 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x) \]

output
-2*a*b*cos(d*x+c)/d-b^2*x*cos(d*x+c)/d+a^2*cos(c)*Si(d*x)+a^2*Ci(d*x)*sin( 
c)+b^2*sin(d*x+c)/d^2
 
3.1.13.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=a^2 \operatorname {CosIntegral}(d x) \sin (c)+\frac {b (-d (2 a+b x) \cos (c+d x)+b \sin (c+d x))}{d^2}+a^2 \cos (c) \text {Si}(d x) \]

input
Integrate[((a + b*x)^2*Sin[c + d*x])/x,x]
 
output
a^2*CosIntegral[d*x]*Sin[c] + (b*(-(d*(2*a + b*x)*Cos[c + d*x]) + b*Sin[c 
+ d*x]))/d^2 + a^2*Cos[c]*SinIntegral[d*x]
 
3.1.13.3 Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {7293, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle a^2 \sin (c) \operatorname {CosIntegral}(d x)+a^2 \cos (c) \text {Si}(d x)-\frac {2 a b \cos (c+d x)}{d}+\frac {b^2 \sin (c+d x)}{d^2}-\frac {b^2 x \cos (c+d x)}{d}\)

input
Int[((a + b*x)^2*Sin[c + d*x])/x,x]
 
output
(-2*a*b*Cos[c + d*x])/d - (b^2*x*Cos[c + d*x])/d + a^2*CosIntegral[d*x]*Si 
n[c] + (b^2*Sin[c + d*x])/d^2 + a^2*Cos[c]*SinIntegral[d*x]
 

3.1.13.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.1.13.4 Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.27

method result size
derivativedivides \(a^{2} \left (\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )\right )-\frac {2 a b \cos \left (d x +c \right )}{d}+\frac {2 b^{2} c \cos \left (d x +c \right )}{d^{2}}+\frac {\left (c +1\right ) b^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}\) \(79\)
default \(a^{2} \left (\operatorname {Si}\left (d x \right ) \cos \left (c \right )+\operatorname {Ci}\left (d x \right ) \sin \left (c \right )\right )-\frac {2 a b \cos \left (d x +c \right )}{d}+\frac {2 b^{2} c \cos \left (d x +c \right )}{d^{2}}+\frac {\left (c +1\right ) b^{2} \left (\sin \left (d x +c \right )-\cos \left (d x +c \right ) \left (d x +c \right )\right )}{d^{2}}\) \(79\)
risch \(-\frac {{\mathrm e}^{-i c} \pi \,\operatorname {csgn}\left (d x \right ) a^{2}}{2}-\frac {i {\mathrm e}^{-i c} \operatorname {Ei}_{1}\left (-i d x \right ) a^{2}}{2}+\frac {i a^{2} {\mathrm e}^{i c} \operatorname {Ei}_{1}\left (-i d x \right )}{2}+{\mathrm e}^{-i c} \operatorname {Si}\left (d x \right ) a^{2}-\frac {b^{2} x \cos \left (d x +c \right )}{d}-\frac {2 a b \cos \left (d x +c \right )}{d}+\frac {b^{2} \sin \left (d x +c \right )}{d^{2}}\) \(107\)
meijerg \(\frac {2 b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {d x \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 a b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {2 a b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {2 \gamma +2 \ln \left (x \right )+\ln \left (d^{2}\right )}{\sqrt {\pi }}-\frac {2 \gamma }{\sqrt {\pi }}-\frac {2 \ln \left (2\right )}{\sqrt {\pi }}-\frac {2 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}+\frac {2 \,\operatorname {Ci}\left (d x \right )}{\sqrt {\pi }}\right )}{2}+a^{2} \cos \left (c \right ) \operatorname {Si}\left (d x \right )\) \(182\)

input
int((b*x+a)^2*sin(d*x+c)/x,x,method=_RETURNVERBOSE)
 
output
a^2*(Si(d*x)*cos(c)+Ci(d*x)*sin(c))-2*a*b*cos(d*x+c)/d+2/d^2*b^2*c*cos(d*x 
+c)+(c+1)/d^2*b^2*(sin(d*x+c)-cos(d*x+c)*(d*x+c))
 
3.1.13.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.98 \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=\frac {a^{2} d^{2} \operatorname {Ci}\left (d x\right ) \sin \left (c\right ) + a^{2} d^{2} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + b^{2} \sin \left (d x + c\right ) - {\left (b^{2} d x + 2 \, a b d\right )} \cos \left (d x + c\right )}{d^{2}} \]

input
integrate((b*x+a)^2*sin(d*x+c)/x,x, algorithm="fricas")
 
output
(a^2*d^2*cos_integral(d*x)*sin(c) + a^2*d^2*cos(c)*sin_integral(d*x) + b^2 
*sin(d*x + c) - (b^2*d*x + 2*a*b*d)*cos(d*x + c))/d^2
 
3.1.13.6 Sympy [A] (verification not implemented)

Time = 1.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.48 \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=a^{2} \sin {\left (c \right )} \operatorname {Ci}{\left (d x \right )} + a^{2} \cos {\left (c \right )} \operatorname {Si}{\left (d x \right )} + 2 a b \left (\begin {cases} x \sin {\left (c \right )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) + b^{2} x \left (\begin {cases} x \sin {\left (c \right )} & \text {for}\: d = 0 \\- \frac {\cos {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases}\right ) - b^{2} \left (\begin {cases} \frac {x^{2} \sin {\left (c \right )}}{2} & \text {for}\: d = 0 \\- \frac {\begin {cases} \frac {\sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \cos {\left (c \right )} & \text {otherwise} \end {cases}}{d} & \text {otherwise} \end {cases}\right ) \]

input
integrate((b*x+a)**2*sin(d*x+c)/x,x)
 
output
a**2*sin(c)*Ci(d*x) + a**2*cos(c)*Si(d*x) + 2*a*b*Piecewise((x*sin(c), Eq( 
d, 0)), (-cos(c + d*x)/d, True)) + b**2*x*Piecewise((x*sin(c), Eq(d, 0)), 
(-cos(c + d*x)/d, True)) - b**2*Piecewise((x**2*sin(c)/2, Eq(d, 0)), (-Pie 
cewise((sin(c + d*x)/d, Ne(d, 0)), (x*cos(c), True))/d, True))
 
3.1.13.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=\frac {{\left (a^{2} {\left (-i \, {\rm Ei}\left (i \, d x\right ) + i \, {\rm Ei}\left (-i \, d x\right )\right )} \cos \left (c\right ) + a^{2} {\left ({\rm Ei}\left (i \, d x\right ) + {\rm Ei}\left (-i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2} + 2 \, b^{2} \sin \left (d x + c\right ) - 2 \, {\left (b^{2} d x + 2 \, a b d\right )} \cos \left (d x + c\right )}{2 \, d^{2}} \]

input
integrate((b*x+a)^2*sin(d*x+c)/x,x, algorithm="maxima")
 
output
1/2*((a^2*(-I*Ei(I*d*x) + I*Ei(-I*d*x))*cos(c) + a^2*(Ei(I*d*x) + Ei(-I*d* 
x))*sin(c))*d^2 + 2*b^2*sin(d*x + c) - 2*(b^2*d*x + 2*a*b*d)*cos(d*x + c)) 
/d^2
 
3.1.13.8 Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.28 (sec) , antiderivative size = 551, normalized size of antiderivative = 8.89 \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=-\frac {a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a^{2} d^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a^{2} d^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, a^{2} d^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, b^{2} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, a^{2} d^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a^{2} d^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 4 \, a b d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, b^{2} d x \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, a^{2} d^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a^{2} d^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) + 2 \, b^{2} d x \tan \left (\frac {1}{2} \, c\right )^{2} - a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) + a^{2} d^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a^{2} d^{2} \operatorname {Si}\left (d x\right ) - 4 \, a b d \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4 \, a b d \tan \left (\frac {1}{2} \, c\right )^{2} - 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b^{2} d x + 4 \, a b d - 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, {\left (d^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + d^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d^{2}\right )}} \]

input
integrate((b*x+a)^2*sin(d*x+c)/x,x, algorithm="giac")
 
output
-1/2*(a^2*d^2*imag_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2* 
c)^2 - a^2*d^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/ 
2*c)^2 + 2*a^2*d^2*sin_integral(d*x)*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 
 2*a^2*d^2*real_part(cos_integral(d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c) 
- 2*a^2*d^2*real_part(cos_integral(-d*x))*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c 
) - 2*b^2*d*x*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - a^2*d^2*imag_part(cos_ 
integral(d*x))*tan(1/2*d*x + 1/2*c)^2 + a^2*d^2*imag_part(cos_integral(-d* 
x))*tan(1/2*d*x + 1/2*c)^2 - 2*a^2*d^2*sin_integral(d*x)*tan(1/2*d*x + 1/2 
*c)^2 + a^2*d^2*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - a^2*d^2*imag_p 
art(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^2*sin_integral(d*x)*tan(1/2 
*c)^2 - 4*a*b*d*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 - 2*b^2*d*x*tan(1/2*d* 
x + 1/2*c)^2 - 2*a^2*d^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d 
^2*real_part(cos_integral(-d*x))*tan(1/2*c) + 2*b^2*d*x*tan(1/2*c)^2 - a^2 
*d^2*imag_part(cos_integral(d*x)) + a^2*d^2*imag_part(cos_integral(-d*x)) 
- 2*a^2*d^2*sin_integral(d*x) - 4*a*b*d*tan(1/2*d*x + 1/2*c)^2 + 4*a*b*d*t 
an(1/2*c)^2 - 4*b^2*tan(1/2*d*x + 1/2*c)*tan(1/2*c)^2 + 2*b^2*d*x + 4*a*b* 
d - 4*b^2*tan(1/2*d*x + 1/2*c))/(d^2*tan(1/2*d*x + 1/2*c)^2*tan(1/2*c)^2 + 
 d^2*tan(1/2*d*x + 1/2*c)^2 + d^2*tan(1/2*c)^2 + d^2)
 
3.1.13.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x)^2 \sin (c+d x)}{x} \, dx=b^2\,\cos \left (c\right )\,\left (\frac {\sin \left (d\,x\right )}{d^2}-\frac {x\,\cos \left (d\,x\right )}{d}\right )+b^2\,\sin \left (c\right )\,\left (\frac {\cos \left (d\,x\right )}{d^2}+\frac {x\,\sin \left (d\,x\right )}{d}\right )+a^2\,\mathrm {cosint}\left (d\,x\right )\,\sin \left (c\right )+a^2\,\mathrm {sinint}\left (d\,x\right )\,\cos \left (c\right )-\frac {2\,a\,b\,\cos \left (d\,x\right )\,\cos \left (c\right )}{d}+\frac {2\,a\,b\,\sin \left (d\,x\right )\,\sin \left (c\right )}{d} \]

input
int((sin(c + d*x)*(a + b*x)^2)/x,x)
 
output
b^2*cos(c)*(sin(d*x)/d^2 - (x*cos(d*x))/d) + b^2*sin(c)*(cos(d*x)/d^2 + (x 
*sin(d*x))/d) + a^2*cosint(d*x)*sin(c) + a^2*sinint(d*x)*cos(c) - (2*a*b*c 
os(d*x)*cos(c))/d + (2*a*b*sin(d*x)*sin(c))/d