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

Optimal. Leaf size=62 \[ -\frac {2 a b \cos (c+d x)}{d}-\frac {b^2 x \cos (c+d x)}{d}+a^2 \text {Ci}(d x) \sin (c)+\frac {b^2 \sin (c+d x)}{d^2}+a^2 \cos (c) \text {Si}(d x) \]

[Out]

-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

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Rubi [A]
time = 0.13, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6874, 2718, 3384, 3380, 3383, 3377, 2717} \begin {gather*} a^2 \sin (c) \text {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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Sin[c + d*x])/x,x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 51, normalized size = 0.82 \begin {gather*} a^2 \text {Ci}(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) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Sin[c + d*x])/x,x]

[Out]

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]*SinIntegra
l[d*x]

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Maple [A]
time = 0.10, size = 79, normalized size = 1.27

method	result	size
derivativedivides	\(a^{2} \left (\sinIntegral \left (d x \right ) \cos \left (c \right )+\cosineIntegral \left (d x \right ) \sin \left (c \right )\right )-\frac {2 a b \cos \left (d x +c \right )}{d}+\frac {2 b^{2} c \cos \left (d x +c \right )}{d^{2}}+\frac {\left (c +1\right ) b^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}\)	\(79\)
default	\(a^{2} \left (\sinIntegral \left (d x \right ) \cos \left (c \right )+\cosineIntegral \left (d x \right ) \sin \left (c \right )\right )-\frac {2 a b \cos \left (d x +c \right )}{d}+\frac {2 b^{2} c \cos \left (d x +c \right )}{d^{2}}+\frac {\left (c +1\right ) b^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )}{d^{2}}\)	\(79\)
meijerg	\(\frac {2 b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {d x \sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (-\frac {d x \cos \left (d x \right )}{2 \sqrt {\pi }}+\frac {\sin \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 a b \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {2 a b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {2 \gamma +2 \ln \left (x \right )+\ln \left (d^{2}\right )}{\sqrt {\pi }}-\frac {2 \gamma }{\sqrt {\pi }}-\frac {2 \ln \left (2\right )}{\sqrt {\pi }}-\frac {2 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}+\frac {2 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{2}+a^{2} \cos \left (c \right ) \sinIntegral \left (d x \right )\)	\(182\)
risch	\(-\frac {{\mathrm e}^{-i c} \pi \,\mathrm {csgn}\left (d x \right ) a^{2}}{2}+{\mathrm e}^{-i c} \sinIntegral \left (d x \right ) a^{2}-\frac {i {\mathrm e}^{-i c} \expIntegral \left (1, -i d x \right ) a^{2}}{2}+\frac {i a^{2} {\mathrm e}^{i c} \expIntegral \left (1, -i d x \right )}{2}+\frac {b^{2} \cos \left (d x +c \right ) x^{2}}{-d x -2 c}+\frac {2 b \cos \left (d x +c \right ) a x}{-d x -2 c}+\frac {2 b^{2} \cos \left (d x +c \right ) c x}{d \left (-d x -2 c \right )}+\frac {4 b \cos \left (d x +c \right ) a c}{d \left (-d x -2 c \right )}-\frac {b^{2} \sin \left (d x +c \right ) x}{d \left (-d x -2 c \right )}-\frac {2 b^{2} \sin \left (d x +c \right ) c}{d^{2} \left (-d x -2 c \right )}\)	\(210\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)-(d*x+c
)*cos(d*x+c))

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Maxima [C] Result contains complex when optimal does not.
time = 0.54, size = 80, normalized size = 1.29 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 78, normalized size = 1.26 \begin {gather*} \frac {2 \, a^{2} d^{2} \cos \left (c\right ) \operatorname {Si}\left (d x\right ) + 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 ) + {\left (a^{2} d^{2} \operatorname {Ci}\left (d x\right ) + a^{2} d^{2} \operatorname {Ci}\left (-d x\right )\right )} \sin \left (c\right )}{2 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 2.24, size = 92, normalized size = 1.48 \begin {gather*} 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)), (-P
iecewise((sin(c + d*x)/d, Ne(d, 0)), (x*cos(c), True))/d, True))

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 5.82, size = 551, normalized size = 8.89 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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_integra
l(-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_part(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*tan(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x)^2)/x,x)

[Out]

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

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