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

Optimal. Leaf size=89 \[ -\frac {a d \cos (c+d x)}{2 x}+b d \cos (c) \text {Ci}(d x)-\frac {1}{2} a d^2 \text {Ci}(d x) \sin (c)-\frac {a \sin (c+d x)}{2 x^2}-\frac {b \sin (c+d x)}{x}-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-b d \sin (c) \text {Si}(d x) \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6874, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {1}{2} a d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} a d^2 \cos (c) \text {Si}(d x)-\frac {a \sin (c+d x)}{2 x^2}-\frac {a d \cos (c+d x)}{2 x}+b d \cos (c) \text {CosIntegral}(d x)-b d \sin (c) \text {Si}(d x)-\frac {b \sin (c+d x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Sin[c + d*x])/x^3,x]

[Out]

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

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]

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

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Mathematica [A]
time = 0.18, size = 76, normalized size = 0.85 \begin {gather*} -\frac {a d x \cos (c+d x)+d x^2 \text {Ci}(d x) (-2 b \cos (c)+a d \sin (c))+a \sin (c+d x)+2 b x \sin (c+d x)+d x^2 (a d \cos (c)+2 b \sin (c)) \text {Si}(d x)}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Sin[c + d*x])/x^3,x]

[Out]

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

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Maple [A]
time = 0.09, size = 88, normalized size = 0.99

method	result	size
derivativedivides	\(d^{2} \left (a \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\sinIntegral \left (d x \right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )}{d}\right )\)	\(88\)
default	\(d^{2} \left (a \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\sinIntegral \left (d x \right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {b \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )}{d}\right )\)	\(88\)
risch	\(-\frac {\cos \left (c \right ) \expIntegral \left (1, i d x \right ) b d}{2}-\frac {\cos \left (c \right ) \expIntegral \left (1, -i d x \right ) b d}{2}+\frac {i \cos \left (c \right ) \expIntegral \left (1, i d x \right ) a \,d^{2}}{4}-\frac {i \cos \left (c \right ) \expIntegral \left (1, -i d x \right ) a \,d^{2}}{4}+\frac {i \sin \left (c \right ) \expIntegral \left (1, i d x \right ) b d}{2}-\frac {i \sin \left (c \right ) \expIntegral \left (1, -i d x \right ) b d}{2}+\frac {\sin \left (c \right ) \expIntegral \left (1, i d x \right ) a \,d^{2}}{4}+\frac {\sin \left (c \right ) \expIntegral \left (1, -i d x \right ) a \,d^{2}}{4}-\frac {a d \cos \left (d x +c \right )}{2 x}+\frac {\left (-4 d^{3} x^{3} b -2 a \,d^{3} x^{2}\right ) \sin \left (d x +c \right )}{4 d^{3} x^{4}}\)	\(164\)
meijerg	\(\frac {d^{2} b \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \sinIntegral \left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {d b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}+\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {a \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 d^{2} x^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, d^{2} x^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}-\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a \sqrt {\pi }\, \cos \left (c \right ) d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{d^{2} x^{2} \sqrt {\pi }}-\frac {4 \sinIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{8}\)	\(311\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.56, size = 112, normalized size = 1.26 \begin {gather*} -\frac {{\left ({\left (a {\left (-i \, \Gamma \left (-2, i \, d x\right ) + i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - a {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{3} + 2 \, {\left (b {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - b {\left (i \, \Gamma \left (-2, i \, d x\right ) - i \, \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2}\right )} x^{2} + 2 \, b \cos \left (d x + c\right )}{2 \, d x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 111, normalized size = 1.25 \begin {gather*} -\frac {2 \, a d x \cos \left (d x + c\right ) + 2 \, {\left (a d^{2} x^{2} \operatorname {Si}\left (d x\right ) - b d x^{2} \operatorname {Ci}\left (d x\right ) - b d x^{2} \operatorname {Ci}\left (-d x\right )\right )} \cos \left (c\right ) + 2 \, {\left (2 \, b x + a\right )} \sin \left (d x + c\right ) + {\left (a d^{2} x^{2} \operatorname {Ci}\left (d x\right ) + a d^{2} x^{2} \operatorname {Ci}\left (-d x\right ) + 4 \, b d x^{2} \operatorname {Si}\left (d x\right )\right )} \sin \left (c\right )}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 4.47, size = 796, normalized size = 8.94 \begin {gather*} \frac {a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{2} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d^{2} x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{2} x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{2} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 4 \, b d x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, b d x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 8 \, b d x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, a d^{2} x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 2 \, a d^{2} x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a d^{2} x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) + a d^{2} x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) - 2 \, a d^{2} x^{2} \operatorname {Si}\left (d x\right ) - 4 \, b d x^{2} \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, b d x^{2} \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 8 \, b d x^{2} \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) + 2 \, b d x^{2} \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) + 2 \, a d x \tan \left (\frac {1}{2} \, d x\right )^{2} + 8 \, a d x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 8 \, b x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 2 \, a d x \tan \left (\frac {1}{2} \, c\right )^{2} + 8 \, b x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, a \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 4 \, a \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a d x - 8 \, b x \tan \left (\frac {1}{2} \, d x\right ) - 8 \, b x \tan \left (\frac {1}{2} \, c\right ) - 4 \, a \tan \left (\frac {1}{2} \, d x\right ) - 4 \, a \tan \left (\frac {1}{2} \, c\right )}{4 \, {\left (x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + x^{2} \tan \left (\frac {1}{2} \, d x\right )^{2} + x^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*x^2*imag_part(cos_integral(-d*
x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*d^2*x^2*real
_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a*d^2*x^2*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*
tan(1/2*c) - 2*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*b*d*x^2*real_part(cos_inte
gral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + a*d^2*x^2*im
ag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2 - 4*b*d*x^2*imag_par
t(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*b*d*x^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/
2*c) - 8*b*d*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) + a*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*
c)^2 - a*d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a*d^2*x^2*sin_integral(d*x)*tan(1/2*c)^2 + 2*b
*d*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 2*b*d*x^2*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 -
2*a*d^2*x^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a*d^2*x^2*real_part(cos_integral(-d*x))*tan(1/2*c) - 2
*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*c)^2 - 2*b*d*x^2*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*
a*d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a*d^2*x^2*imag_part(cos_integral(d*x)) + a*d^2*x^2*imag_part(cos_integral(
-d*x)) - 2*a*d^2*x^2*sin_integral(d*x) - 4*b*d*x^2*imag_part(cos_integral(d*x))*tan(1/2*c) + 4*b*d*x^2*imag_pa
rt(cos_integral(-d*x))*tan(1/2*c) - 8*b*d*x^2*sin_integral(d*x)*tan(1/2*c) + 2*b*d*x^2*real_part(cos_integral(
d*x)) + 2*b*d*x^2*real_part(cos_integral(-d*x)) + 2*a*d*x*tan(1/2*d*x)^2 + 8*a*d*x*tan(1/2*d*x)*tan(1/2*c) + 8
*b*x*tan(1/2*d*x)^2*tan(1/2*c) + 2*a*d*x*tan(1/2*c)^2 + 8*b*x*tan(1/2*d*x)*tan(1/2*c)^2 + 4*a*tan(1/2*d*x)^2*t
an(1/2*c) + 4*a*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a*d*x - 8*b*x*tan(1/2*d*x) - 8*b*x*tan(1/2*c) - 4*a*tan(1/2*d*x)
- 4*a*tan(1/2*c))/(x^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + x^2*tan(1/2*d*x)^2 + x^2*tan(1/2*c)^2 + 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 )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)*(a + b*x))/x^3,x)

[Out]

int((sin(c + d*x)*(a + b*x))/x^3, x)

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