3.1.15 \(\int \frac {(a+b x)^2 \sin (c+d x)}{x^3} \, dx\) [15]

Optimal. Leaf size=121 \[ -\frac {a^2 d \cos (c+d x)}{2 x}+2 a b d \cos (c) \text {Ci}(d x)+b^2 \text {Ci}(d x) \sin (c)-\frac {1}{2} a^2 d^2 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{2 x^2}-\frac {2 a b \sin (c+d x)}{x}+b^2 \cos (c) \text {Si}(d x)-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x)-2 a b d \sin (c) \text {Si}(d x) \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6874, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {1}{2} a^2 d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} a^2 d^2 \cos (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{2 x^2}-\frac {a^2 d \cos (c+d x)}{2 x}+2 a b d \cos (c) \text {CosIntegral}(d x)-2 a b d \sin (c) \text {Si}(d x)-\frac {2 a b \sin (c+d x)}{x}+b^2 \sin (c) \text {CosIntegral}(d x)+b^2 \cos (c) \text {Si}(d x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Sin[c + d*x])/x^3,x]

[Out]

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

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Mathematica [A]
time = 0.26, size = 95, normalized size = 0.79 \begin {gather*} \frac {1}{2} \left (\text {Ci}(d x) \left (4 a b d \cos (c)+\left (2 b^2-a^2 d^2\right ) \sin (c)\right )-\frac {a (a d x \cos (c+d x)+(a+4 b x) \sin (c+d x))}{x^2}+\left (\left (2 b^2-a^2 d^2\right ) \cos (c)-4 a b d \sin (c)\right ) \text {Si}(d x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Sin[c + d*x])/x^3,x]

[Out]

(CosIntegral[d*x]*(4*a*b*d*Cos[c] + (2*b^2 - a^2*d^2)*Sin[c]) - (a*(a*d*x*Cos[c + d*x] + (a + 4*b*x)*Sin[c + d
*x]))/x^2 + ((2*b^2 - a^2*d^2)*Cos[c] - 4*a*b*d*Sin[c])*SinIntegral[d*x])/2

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Maple [A]
time = 0.10, size = 114, normalized size = 0.94

method	result	size
derivativedivides	\(d^{2} \left (a^{2} \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\sinIntegral \left (d x \right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )}{d}+\frac {b^{2} \left (\sinIntegral \left (d x \right ) \cos \left (c \right )+\cosineIntegral \left (d x \right ) \sin \left (c \right )\right )}{d^{2}}\right )\)	\(114\)
default	\(d^{2} \left (a^{2} \left (-\frac {\sin \left (d x +c \right )}{2 d^{2} x^{2}}-\frac {\cos \left (d x +c \right )}{2 d x}-\frac {\sinIntegral \left (d x \right ) \cos \left (c \right )}{2}-\frac {\cosineIntegral \left (d x \right ) \sin \left (c \right )}{2}\right )+\frac {2 a b \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )}{d}+\frac {b^{2} \left (\sinIntegral \left (d x \right ) \cos \left (c \right )+\cosineIntegral \left (d x \right ) \sin \left (c \right )\right )}{d^{2}}\right )\)	\(114\)
risch	\(-\cos \left (c \right ) \expIntegral \left (1, i d x \right ) a b d -\cos \left (c \right ) \expIntegral \left (1, -i d x \right ) a b d +\frac {i \cos \left (c \right ) \expIntegral \left (1, i d x \right ) a^{2} d^{2}}{4}-\frac {i \cos \left (c \right ) \expIntegral \left (1, -i d x \right ) a^{2} d^{2}}{4}-\frac {i \cos \left (c \right ) \expIntegral \left (1, i d x \right ) b^{2}}{2}+\frac {i \cos \left (c \right ) \expIntegral \left (1, -i d x \right ) b^{2}}{2}+i \sin \left (c \right ) \expIntegral \left (1, i d x \right ) a b d -i \sin \left (c \right ) \expIntegral \left (1, -i d x \right ) a b d +\frac {\sin \left (c \right ) \expIntegral \left (1, i d x \right ) a^{2} d^{2}}{4}+\frac {\sin \left (c \right ) \expIntegral \left (1, -i d x \right ) a^{2} d^{2}}{4}-\frac {\sin \left (c \right ) \expIntegral \left (1, i d x \right ) b^{2}}{2}-\frac {\sin \left (c \right ) \expIntegral \left (1, -i d x \right ) b^{2}}{2}-\frac {a^{2} d \cos \left (d x +c \right )}{2 x}+\frac {\left (-8 a b \,d^{4} x^{3}-2 a^{2} d^{4} x^{2}\right ) \sin \left (d x +c \right )}{4 d^{4} x^{4}}\)	\(239\)
meijerg	\(\frac {b^{2} \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {2 \gamma +2 \ln \left (x \right )+\ln \left (d^{2}\right )}{\sqrt {\pi }}-\frac {2 \gamma }{\sqrt {\pi }}-\frac {2 \ln \left (2\right )}{\sqrt {\pi }}-\frac {2 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}+\frac {2 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{2}+b^{2} \cos \left (c \right ) \sinIntegral \left (d x \right )+\frac {d^{2} a b \sqrt {\pi }\, \sin \left (c \right ) \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \sinIntegral \left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{2 \sqrt {d^{2}}}+\frac {d a b \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}+\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{2}+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+\ln \left (d^{2}\right )\right )}{\sqrt {\pi }}+\frac {-6 d^{2} x^{2}+4}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \gamma }{\sqrt {\pi }}+\frac {4 \ln \left (2\right )}{\sqrt {\pi }}+\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \cos \left (d x \right )}{\sqrt {\pi }\, d^{2} x^{2}}+\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}-\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a^{2} \sqrt {\pi }\, \cos \left (c \right ) d^{2} \left (-\frac {4 \cos \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{d^{2} x^{2} \sqrt {\pi }}-\frac {4 \sinIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{8}\)	\(386\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*sin(d*x+c)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 1.40, size = 189, normalized size = 1.56 \begin {gather*} -\frac {{\left ({\left (a^{2} {\left (-i \, \Gamma \left (-2, i \, d x\right ) + i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - a^{2} {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4} + 4 \, {\left (a b {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - a 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^{3} - 2 \, {\left (b^{2} {\left (-i \, \Gamma \left (-2, i \, d x\right ) + i \, \Gamma \left (-2, -i \, d x\right )\right )} \cos \left (c\right ) - b^{2} {\left (\Gamma \left (-2, i \, d x\right ) + \Gamma \left (-2, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2}\right )} x^{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} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^3,x, algorithm="maxima")

[Out]

-1/2*(((a^2*(-I*gamma(-2, I*d*x) + I*gamma(-2, -I*d*x))*cos(c) - a^2*(gamma(-2, I*d*x) + gamma(-2, -I*d*x))*si
n(c))*d^4 + 4*(a*b*(gamma(-2, I*d*x) + gamma(-2, -I*d*x))*cos(c) - a*b*(I*gamma(-2, I*d*x) - I*gamma(-2, -I*d*
x))*sin(c))*d^3 - 2*(b^2*(-I*gamma(-2, I*d*x) + I*gamma(-2, -I*d*x))*cos(c) - b^2*(gamma(-2, I*d*x) + gamma(-2
, -I*d*x))*sin(c))*d^2)*x^2 + 2*b^2*sin(d*x + c) + 2*(b^2*d*x + 2*a*b*d)*cos(d*x + c))/(d^2*x^2)

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Fricas [A]
time = 0.37, size = 147, normalized size = 1.21 \begin {gather*} -\frac {2 \, a^{2} d x \cos \left (d x + c\right ) - 2 \, {\left (2 \, a b d x^{2} \operatorname {Ci}\left (d x\right ) + 2 \, a b d x^{2} \operatorname {Ci}\left (-d x\right ) - {\left (a^{2} d^{2} - 2 \, b^{2}\right )} x^{2} \operatorname {Si}\left (d x\right )\right )} \cos \left (c\right ) + 2 \, {\left (4 \, a b x + a^{2}\right )} \sin \left (d x + c\right ) + {\left (8 \, a b d x^{2} \operatorname {Si}\left (d x\right ) + {\left (a^{2} d^{2} - 2 \, b^{2}\right )} x^{2} \operatorname {Ci}\left (d x\right ) + {\left (a^{2} d^{2} - 2 \, b^{2}\right )} x^{2} \operatorname {Ci}\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)^2*sin(d*x+c)/x^3,x, algorithm="fricas")

[Out]

-1/4*(2*a^2*d*x*cos(d*x + c) - 2*(2*a*b*d*x^2*cos_integral(d*x) + 2*a*b*d*x^2*cos_integral(-d*x) - (a^2*d^2 -
2*b^2)*x^2*sin_integral(d*x))*cos(c) + 2*(4*a*b*x + a^2)*sin(d*x + c) + (8*a*b*d*x^2*sin_integral(d*x) + (a^2*
d^2 - 2*b^2)*x^2*cos_integral(d*x) + (a^2*d^2 - 2*b^2)*x^2*cos_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 )^{2} \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)**2*sin(d*x+c)/x**3,x)

[Out]

Integral((a + b*x)**2*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 = 4.33, size = 1182, normalized size = 9.77 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^3,x, algorithm="giac")

[Out]

1/4*(a^2*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^2*x^2*imag_part(cos_integral
(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^2*
x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^2*x^2*real_part(cos_integral(-d*x))*tan(1
/2*d*x)^2*tan(1/2*c) - 4*a*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 4*a*b*d*x^2*real
_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^
2 + a^2*d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2
- 8*a*b*d*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 8*a*b*d*x^2*imag_part(cos_integral(-d*x
))*tan(1/2*d*x)^2*tan(1/2*c) - 16*a*b*d*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) + a^2*d^2*x^2*imag_par
t(cos_integral(d*x))*tan(1/2*c)^2 - a^2*d^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a^2*d^2*x^2*sin
_integral(d*x)*tan(1/2*c)^2 - 2*b^2*x^2*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*b^2*x^2*i
mag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 4*b^2*x^2*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*
c)^2 + 4*a*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 4*a*b*d*x^2*real_part(cos_integral(-d*x))*tan
(1/2*d*x)^2 - 2*a^2*d^2*x^2*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d^2*x^2*real_part(cos_integral(-d*
x))*tan(1/2*c) + 4*b^2*x^2*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*b^2*x^2*real_part(cos_in
tegral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 4*a*b*d*x^2*real_part(cos_integral(d*x))*tan(1/2*c)^2 - 4*a*b*d*x^2*
real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*a^2*d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^2*x^2*imag_part(cos
_integral(d*x)) + a^2*d^2*x^2*imag_part(cos_integral(-d*x)) - 2*a^2*d^2*x^2*sin_integral(d*x) + 2*b^2*x^2*imag
_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 2*b^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 4*b^2*x^2*s
in_integral(d*x)*tan(1/2*d*x)^2 - 8*a*b*d*x^2*imag_part(cos_integral(d*x))*tan(1/2*c) + 8*a*b*d*x^2*imag_part(
cos_integral(-d*x))*tan(1/2*c) - 16*a*b*d*x^2*sin_integral(d*x)*tan(1/2*c) - 2*b^2*x^2*imag_part(cos_integral(
d*x))*tan(1/2*c)^2 + 2*b^2*x^2*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 4*b^2*x^2*sin_integral(d*x)*tan(1/
2*c)^2 + 4*a*b*d*x^2*real_part(cos_integral(d*x)) + 4*a*b*d*x^2*real_part(cos_integral(-d*x)) + 2*a^2*d*x*tan(
1/2*d*x)^2 + 4*b^2*x^2*real_part(cos_integral(d*x))*tan(1/2*c) + 4*b^2*x^2*real_part(cos_integral(-d*x))*tan(1
/2*c) + 8*a^2*d*x*tan(1/2*d*x)*tan(1/2*c) + 16*a*b*x*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2*d*x*tan(1/2*c)^2 + 16*a
*b*x*tan(1/2*d*x)*tan(1/2*c)^2 + 2*b^2*x^2*imag_part(cos_integral(d*x)) - 2*b^2*x^2*imag_part(cos_integral(-d*
x)) + 4*b^2*x^2*sin_integral(d*x) + 4*a^2*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*tan(1/2*d*x)*tan(1/2*c)^2 - 2*a^2*
d*x - 16*a*b*x*tan(1/2*d*x) - 16*a*b*x*tan(1/2*c) - 4*a^2*tan(1/2*d*x) - 4*a^2*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 )}^2}{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)^2)/x^3,x)

[Out]

int((sin(c + d*x)*(a + b*x)^2)/x^3, x)

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