3.1.17 \(\int \frac {(a+b x)^2 \sin (c+d x)}{x^5} \, dx\) [17]

Optimal. Leaf size=248 \[ -\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{3} a b d^3 \cos (c) \text {Ci}(d x)-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)+\frac {1}{24} a^2 d^4 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x) \]

[Out]

-1/3*a*b*d^3*Ci(d*x)*cos(c)-1/12*a^2*d*cos(d*x+c)/x^3-1/3*a*b*d*cos(d*x+c)/x^2-1/2*b^2*d*cos(d*x+c)/x+1/24*a^2
*d^3*cos(d*x+c)/x-1/2*b^2*d^2*cos(c)*Si(d*x)+1/24*a^2*d^4*cos(c)*Si(d*x)-1/2*b^2*d^2*Ci(d*x)*sin(c)+1/24*a^2*d
^4*Ci(d*x)*sin(c)+1/3*a*b*d^3*Si(d*x)*sin(c)-1/4*a^2*sin(d*x+c)/x^4-2/3*a*b*sin(d*x+c)/x^3-1/2*b^2*sin(d*x+c)/
x^2+1/24*a^2*d^2*sin(d*x+c)/x^2+1/3*a*b*d^2*sin(d*x+c)/x

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Rubi [A]
time = 0.32, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 20, 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}{24} a^2 d^4 \sin (c) \text {CosIntegral}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {a^2 d^3 \cos (c+d x)}{24 x}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {1}{3} a b d^3 \cos (c) \text {CosIntegral}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {1}{2} b^2 d^2 \sin (c) \text {CosIntegral}(d x)-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)-\frac {b^2 \sin (c+d x)}{2 x^2}-\frac {b^2 d \cos (c+d x)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Sin[c + d*x])/x^5,x]

[Out]

-1/12*(a^2*d*Cos[c + d*x])/x^3 - (a*b*d*Cos[c + d*x])/(3*x^2) - (b^2*d*Cos[c + d*x])/(2*x) + (a^2*d^3*Cos[c +
d*x])/(24*x) - (a*b*d^3*Cos[c]*CosIntegral[d*x])/3 - (b^2*d^2*CosIntegral[d*x]*Sin[c])/2 + (a^2*d^4*CosIntegra
l[d*x]*Sin[c])/24 - (a^2*Sin[c + d*x])/(4*x^4) - (2*a*b*Sin[c + d*x])/(3*x^3) - (b^2*Sin[c + d*x])/(2*x^2) + (
a^2*d^2*Sin[c + d*x])/(24*x^2) + (a*b*d^2*Sin[c + d*x])/(3*x) - (b^2*d^2*Cos[c]*SinIntegral[d*x])/2 + (a^2*d^4
*Cos[c]*SinIntegral[d*x])/24 + (a*b*d^3*Sin[c]*SinIntegral[d*x])/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]

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^5} \, dx &=\int \left (\frac {a^2 \sin (c+d x)}{x^5}+\frac {2 a b \sin (c+d x)}{x^4}+\frac {b^2 \sin (c+d x)}{x^3}\right ) \, dx\\ &=a^2 \int \frac {\sin (c+d x)}{x^5} \, dx+(2 a b) \int \frac {\sin (c+d x)}{x^4} \, dx+b^2 \int \frac {\sin (c+d x)}{x^3} \, dx\\ &=-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {1}{4} \left (a^2 d\right ) \int \frac {\cos (c+d x)}{x^4} \, dx+\frac {1}{3} (2 a b d) \int \frac {\cos (c+d x)}{x^3} \, dx+\frac {1}{2} \left (b^2 d\right ) \int \frac {\cos (c+d x)}{x^2} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}-\frac {1}{12} \left (a^2 d^2\right ) \int \frac {\sin (c+d x)}{x^3} \, dx-\frac {1}{3} \left (a b d^2\right ) \int \frac {\sin (c+d x)}{x^2} \, dx-\frac {1}{2} \left (b^2 d^2\right ) \int \frac {\sin (c+d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{24} \left (a^2 d^3\right ) \int \frac {\cos (c+d x)}{x^2} \, dx-\frac {1}{3} \left (a b d^3\right ) \int \frac {\cos (c+d x)}{x} \, dx-\frac {1}{2} \left (b^2 d^2 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx-\frac {1}{2} \left (b^2 d^2 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4\right ) \int \frac {\sin (c+d x)}{x} \, dx-\frac {1}{3} \left (a b d^3 \cos (c)\right ) \int \frac {\cos (d x)}{x} \, dx+\frac {1}{3} \left (a b d^3 \sin (c)\right ) \int \frac {\sin (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{3} a b d^3 \cos (c) \text {Ci}(d x)-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)+\frac {1}{24} \left (a^2 d^4 \cos (c)\right ) \int \frac {\sin (d x)}{x} \, dx+\frac {1}{24} \left (a^2 d^4 \sin (c)\right ) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {a^2 d \cos (c+d x)}{12 x^3}-\frac {a b d \cos (c+d x)}{3 x^2}-\frac {b^2 d \cos (c+d x)}{2 x}+\frac {a^2 d^3 \cos (c+d x)}{24 x}-\frac {1}{3} a b d^3 \cos (c) \text {Ci}(d x)-\frac {1}{2} b^2 d^2 \text {Ci}(d x) \sin (c)+\frac {1}{24} a^2 d^4 \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{4 x^4}-\frac {2 a b \sin (c+d x)}{3 x^3}-\frac {b^2 \sin (c+d x)}{2 x^2}+\frac {a^2 d^2 \sin (c+d x)}{24 x^2}+\frac {a b d^2 \sin (c+d x)}{3 x}-\frac {1}{2} b^2 d^2 \cos (c) \text {Si}(d x)+\frac {1}{24} a^2 d^4 \cos (c) \text {Si}(d x)+\frac {1}{3} a b d^3 \sin (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 204, normalized size = 0.82 \begin {gather*} \frac {-2 a^2 d x \cos (c+d x)-8 a b d x^2 \cos (c+d x)-12 b^2 d x^3 \cos (c+d x)+a^2 d^3 x^3 \cos (c+d x)+d^2 x^4 \text {Ci}(d x) \left (-8 a b d \cos (c)+\left (-12 b^2+a^2 d^2\right ) \sin (c)\right )-6 a^2 \sin (c+d x)-16 a b x \sin (c+d x)-12 b^2 x^2 \sin (c+d x)+a^2 d^2 x^2 \sin (c+d x)+8 a b d^2 x^3 \sin (c+d x)+d^2 x^4 \left (-12 b^2 \cos (c)+a^2 d^2 \cos (c)+8 a b d \sin (c)\right ) \text {Si}(d x)}{24 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^2*Sin[c + d*x])/x^5,x]

[Out]

(-2*a^2*d*x*Cos[c + d*x] - 8*a*b*d*x^2*Cos[c + d*x] - 12*b^2*d*x^3*Cos[c + d*x] + a^2*d^3*x^3*Cos[c + d*x] + d
^2*x^4*CosIntegral[d*x]*(-8*a*b*d*Cos[c] + (-12*b^2 + a^2*d^2)*Sin[c]) - 6*a^2*Sin[c + d*x] - 16*a*b*x*Sin[c +
d*x] - 12*b^2*x^2*Sin[c + d*x] + a^2*d^2*x^2*Sin[c + d*x] + 8*a*b*d^2*x^3*Sin[c + d*x] + d^2*x^4*(-12*b^2*Cos
[c] + a^2*d^2*Cos[c] + 8*a*b*d*Sin[c])*SinIntegral[d*x])/(24*x^4)

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Maple [A]
time = 0.14, size = 201, normalized size = 0.81 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^2*sin(d*x+c)/x^5,x,method=_RETURNVERBOSE)

[Out]

d^4*(a^2*(-1/4*sin(d*x+c)/d^4/x^4-1/12*cos(d*x+c)/d^3/x^3+1/24*sin(d*x+c)/d^2/x^2+1/24*cos(d*x+c)/d/x+1/24*Si(
d*x)*cos(c)+1/24*Ci(d*x)*sin(c))+2/d*a*b*(-1/3*sin(d*x+c)/d^3/x^3-1/6*cos(d*x+c)/d^2/x^2+1/6*sin(d*x+c)/d/x+1/
6*Si(d*x)*sin(c)-1/6*Ci(d*x)*cos(c))+1/d^2*b^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)))

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Maxima [C] Result contains complex when optimal does not.
time = 1.58, size = 186, normalized size = 0.75 \begin {gather*} -\frac {{\left ({\left (a^{2} {\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) + a^{2} {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{6} - 8 \, {\left (a b {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) + a b {\left (-i \, \Gamma \left (-4, i \, d x\right ) + i \, \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{5} - 12 \, {\left (b^{2} {\left (i \, \Gamma \left (-4, i \, d x\right ) - i \, \Gamma \left (-4, -i \, d x\right )\right )} \cos \left (c\right ) + b^{2} {\left (\Gamma \left (-4, i \, d x\right ) + \Gamma \left (-4, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{4}\right )} x^{4} + 6 \, 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^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^5,x, algorithm="maxima")

[Out]

-1/2*(((a^2*(I*gamma(-4, I*d*x) - I*gamma(-4, -I*d*x))*cos(c) + a^2*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*sin
(c))*d^6 - 8*(a*b*(gamma(-4, I*d*x) + gamma(-4, -I*d*x))*cos(c) + a*b*(-I*gamma(-4, I*d*x) + I*gamma(-4, -I*d*
x))*sin(c))*d^5 - 12*(b^2*(I*gamma(-4, I*d*x) - I*gamma(-4, -I*d*x))*cos(c) + b^2*(gamma(-4, I*d*x) + gamma(-4
, -I*d*x))*sin(c))*d^4)*x^4 + 6*b^2*sin(d*x + c) + 2*(b^2*d*x + 2*a*b*d)*cos(d*x + c))/(d^2*x^4)

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Fricas [A]
time = 0.36, size = 222, normalized size = 0.90 \begin {gather*} -\frac {2 \, {\left (8 \, a b d x^{2} + 2 \, a^{2} d x - {\left (a^{2} d^{3} - 12 \, b^{2} d\right )} x^{3}\right )} \cos \left (d x + c\right ) + 2 \, {\left (4 \, a b d^{3} x^{4} \operatorname {Ci}\left (d x\right ) + 4 \, a b d^{3} x^{4} \operatorname {Ci}\left (-d x\right ) - {\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname {Si}\left (d x\right )\right )} \cos \left (c\right ) - 2 \, {\left (8 \, a b d^{2} x^{3} - 16 \, a b x + {\left (a^{2} d^{2} - 12 \, b^{2}\right )} x^{2} - 6 \, a^{2}\right )} \sin \left (d x + c\right ) - {\left (16 \, a b d^{3} x^{4} \operatorname {Si}\left (d x\right ) + {\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname {Ci}\left (d x\right ) + {\left (a^{2} d^{4} - 12 \, b^{2} d^{2}\right )} x^{4} \operatorname {Ci}\left (-d x\right )\right )} \sin \left (c\right )}{48 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^2*sin(d*x+c)/x^5,x, algorithm="fricas")

[Out]

-1/48*(2*(8*a*b*d*x^2 + 2*a^2*d*x - (a^2*d^3 - 12*b^2*d)*x^3)*cos(d*x + c) + 2*(4*a*b*d^3*x^4*cos_integral(d*x
) + 4*a*b*d^3*x^4*cos_integral(-d*x) - (a^2*d^4 - 12*b^2*d^2)*x^4*sin_integral(d*x))*cos(c) - 2*(8*a*b*d^2*x^3
- 16*a*b*x + (a^2*d^2 - 12*b^2)*x^2 - 6*a^2)*sin(d*x + c) - (16*a*b*d^3*x^4*sin_integral(d*x) + (a^2*d^4 - 12
*b^2*d^2)*x^4*cos_integral(d*x) + (a^2*d^4 - 12*b^2*d^2)*x^4*cos_integral(-d*x))*sin(c))/x^4

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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^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**2*sin(d*x+c)/x**5,x)

[Out]

Integral((a + b*x)**2*sin(c + d*x)/x**5, x)

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 4.04, size = 1712, normalized size = 6.90 \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^5,x, algorithm="giac")

[Out]

-1/48*(a^2*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integr
al(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^4*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a^2*d^
4*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*d^4*x^4*real_part(cos_integral(-d*x))*tan
(1/2*d*x)^2*tan(1/2*c) - 8*a*b*d^3*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 8*a*b*d^3*x^
4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integral(d*x))*tan(1/2
*d*x)^2 + a^2*d^4*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^4*x^4*sin_integral(d*x)*tan(1/2*d
*x)^2 - 16*a*b*d^3*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 16*a*b*d^3*x^4*imag_part(cos_i
ntegral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 32*a*b*d^3*x^4*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c) + a^2*d^
4*x^4*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 2*a
^2*d^4*x^4*sin_integral(d*x)*tan(1/2*c)^2 - 12*b^2*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2
*c)^2 + 12*b^2*d^2*x^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 24*b^2*d^2*x^4*sin_integral
(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 + 8*a*b*d^3*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 8*a*b*d^3*x^4*
real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 2*a^2*d^4*x^4*real_part(cos_integral(d*x))*tan(1/2*c) - 2*a^2*d
^4*x^4*real_part(cos_integral(-d*x))*tan(1/2*c) + 24*b^2*d^2*x^4*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*t
an(1/2*c) + 24*b^2*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 8*a*b*d^3*x^4*real_part(c
os_integral(d*x))*tan(1/2*c)^2 - 8*a*b*d^3*x^4*real_part(cos_integral(-d*x))*tan(1/2*c)^2 - 2*a^2*d^3*x^3*tan(
1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^4*x^4*imag_part(cos_integral(d*x)) + a^2*d^4*x^4*imag_part(cos_integral(-d*x))
- 2*a^2*d^4*x^4*sin_integral(d*x) + 12*b^2*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 - 12*b^2*d^2*x
^4*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2 + 24*b^2*d^2*x^4*sin_integral(d*x)*tan(1/2*d*x)^2 - 16*a*b*d^3
*x^4*imag_part(cos_integral(d*x))*tan(1/2*c) + 16*a*b*d^3*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c) - 32*a*
b*d^3*x^4*sin_integral(d*x)*tan(1/2*c) - 12*b^2*d^2*x^4*imag_part(cos_integral(d*x))*tan(1/2*c)^2 + 12*b^2*d^2
*x^4*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 - 24*b^2*d^2*x^4*sin_integral(d*x)*tan(1/2*c)^2 + 8*a*b*d^3*x^
4*real_part(cos_integral(d*x)) + 8*a*b*d^3*x^4*real_part(cos_integral(-d*x)) + 2*a^2*d^3*x^3*tan(1/2*d*x)^2 +
24*b^2*d^2*x^4*real_part(cos_integral(d*x))*tan(1/2*c) + 24*b^2*d^2*x^4*real_part(cos_integral(-d*x))*tan(1/2*
c) + 8*a^2*d^3*x^3*tan(1/2*d*x)*tan(1/2*c) + 32*a*b*d^2*x^3*tan(1/2*d*x)^2*tan(1/2*c) + 2*a^2*d^3*x^3*tan(1/2*
c)^2 + 32*a*b*d^2*x^3*tan(1/2*d*x)*tan(1/2*c)^2 + 24*b^2*d*x^3*tan(1/2*d*x)^2*tan(1/2*c)^2 + 12*b^2*d^2*x^4*im
ag_part(cos_integral(d*x)) - 12*b^2*d^2*x^4*imag_part(cos_integral(-d*x)) + 24*b^2*d^2*x^4*sin_integral(d*x) +
4*a^2*d^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^2*x^2*tan(1/2*d*x)*tan(1/2*c)^2 + 16*a*b*d*x^2*tan(1/2*d*x)
^2*tan(1/2*c)^2 - 2*a^2*d^3*x^3 - 32*a*b*d^2*x^3*tan(1/2*d*x) - 24*b^2*d*x^3*tan(1/2*d*x)^2 - 32*a*b*d^2*x^3*t
an(1/2*c) - 96*b^2*d*x^3*tan(1/2*d*x)*tan(1/2*c) - 24*b^2*d*x^3*tan(1/2*c)^2 + 4*a^2*d*x*tan(1/2*d*x)^2*tan(1/
2*c)^2 - 4*a^2*d^2*x^2*tan(1/2*d*x) - 16*a*b*d*x^2*tan(1/2*d*x)^2 - 4*a^2*d^2*x^2*tan(1/2*c) - 64*a*b*d*x^2*ta
n(1/2*d*x)*tan(1/2*c) - 48*b^2*x^2*tan(1/2*d*x)^2*tan(1/2*c) - 16*a*b*d*x^2*tan(1/2*c)^2 - 48*b^2*x^2*tan(1/2*
d*x)*tan(1/2*c)^2 + 24*b^2*d*x^3 - 4*a^2*d*x*tan(1/2*d*x)^2 - 16*a^2*d*x*tan(1/2*d*x)*tan(1/2*c) - 64*a*b*x*ta
n(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d*x*tan(1/2*c)^2 - 64*a*b*x*tan(1/2*d*x)*tan(1/2*c)^2 + 16*a*b*d*x^2 + 48*b^2*
x^2*tan(1/2*d*x) + 48*b^2*x^2*tan(1/2*c) - 24*a^2*tan(1/2*d*x)^2*tan(1/2*c) - 24*a^2*tan(1/2*d*x)*tan(1/2*c)^2
+ 4*a^2*d*x + 64*a*b*x*tan(1/2*d*x) + 64*a*b*x*tan(1/2*c) + 24*a^2*tan(1/2*d*x) + 24*a^2*tan(1/2*c))/(x^4*tan
(1/2*d*x)^2*tan(1/2*c)^2 + x^4*tan(1/2*d*x)^2 + x^4*tan(1/2*c)^2 + x^4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sin \left (c+d\,x\right )\,{\left (a+b\,x\right )}^2}{x^5} \,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^5,x)

[Out]

int((sin(c + d*x)*(a + b*x)^2)/x^5, x)

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