3.1.14 \(\int \frac {(a+b x)^2 \sin (c+d x)}{x^2} \, dx\) [14]

Optimal. Leaf size=72 \[ -\frac {b^2 \cos (c+d x)}{d}+a^2 d \cos (c) \text {Ci}(d x)+2 a b \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{x}+2 a b \cos (c) \text {Si}(d x)-a^2 d \sin (c) \text {Si}(d x) \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {6874, 2718, 3378, 3384, 3380, 3383} \begin {gather*} a^2 d \cos (c) \text {CosIntegral}(d x)-a^2 d \sin (c) \text {Si}(d x)-\frac {a^2 \sin (c+d x)}{x}+2 a b \sin (c) \text {CosIntegral}(d x)+2 a b \cos (c) \text {Si}(d x)-\frac {b^2 \cos (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^2*Sin[c + d*x])/x^2,x]

[Out]

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, 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^2} \, dx &=\int \left (b^2 \sin (c+d x)+\frac {a^2 \sin (c+d x)}{x^2}+\frac {2 a b \sin (c+d x)}{x}\right ) \, dx\\ &=a^2 \int \frac {\sin (c+d x)}{x^2} \, dx+(2 a b) \int \frac {\sin (c+d x)}{x} \, dx+b^2 \int \sin (c+d x) \, dx\\ &=-\frac {b^2 \cos (c+d x)}{d}-\frac {a^2 \sin (c+d x)}{x}+\left (a^2 d\right ) \int \frac {\cos (c+d x)}{x} \, dx+(2 a b \cos (c)) \int \frac {\sin (d x)}{x} \, dx+(2 a b \sin (c)) \int \frac {\cos (d x)}{x} \, dx\\ &=-\frac {b^2 \cos (c+d x)}{d}+2 a b \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{x}+2 a b \cos (c) \text {Si}(d x)+\left (a^2 d \cos (c)\right ) \int \frac {\cos (d x)}{x} \, dx-\left (a^2 d \sin (c)\right ) \int \frac {\sin (d x)}{x} \, dx\\ &=-\frac {b^2 \cos (c+d x)}{d}+a^2 d \cos (c) \text {Ci}(d x)+2 a b \text {Ci}(d x) \sin (c)-\frac {a^2 \sin (c+d x)}{x}+2 a b \cos (c) \text {Si}(d x)-a^2 d \sin (c) \text {Si}(d x)\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 64, normalized size = 0.89 \begin {gather*} -\frac {b^2 \cos (c+d x)}{d}+a \text {Ci}(d x) (a d \cos (c)+2 b \sin (c))-\frac {a^2 \sin (c+d x)}{x}-a (-2 b \cos (c)+a d \sin (c)) \text {Si}(d x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 74, normalized size = 1.03

method	result	size
derivativedivides	\(d \left (a^{2} \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )+\frac {2 a b \left (\sinIntegral \left (d x \right ) \cos \left (c \right )+\cosineIntegral \left (d x \right ) \sin \left (c \right )\right )}{d}-\frac {b^{2} \cos \left (d x +c \right )}{d^{2}}\right )\)	\(74\)
default	\(d \left (a^{2} \left (-\frac {\sin \left (d x +c \right )}{d x}-\sinIntegral \left (d x \right ) \sin \left (c \right )+\cosineIntegral \left (d x \right ) \cos \left (c \right )\right )+\frac {2 a b \left (\sinIntegral \left (d x \right ) \cos \left (c \right )+\cosineIntegral \left (d x \right ) \sin \left (c \right )\right )}{d}-\frac {b^{2} \cos \left (d x +c \right )}{d^{2}}\right )\)	\(74\)
risch	\(-i \cos \left (c \right ) \expIntegral \left (1, i d x \right ) a b -\frac {d \cos \left (c \right ) a^{2} \expIntegral \left (1, i d x \right )}{2}+i \cos \left (c \right ) \expIntegral \left (1, -i d x \right ) a b -\frac {d \cos \left (c \right ) a^{2} \expIntegral \left (1, -i d x \right )}{2}-\sin \left (c \right ) \expIntegral \left (1, i d x \right ) a b +\frac {i d \sin \left (c \right ) a^{2} \expIntegral \left (1, i d x \right )}{2}-\sin \left (c \right ) \expIntegral \left (1, -i d x \right ) a b -\frac {i d \sin \left (c \right ) a^{2} \expIntegral \left (1, -i d x \right )}{2}-\frac {b^{2} \cos \left (d x +c \right )}{d}-\frac {a^{2} \sin \left (d x +c \right )}{x}\)	\(146\)
meijerg	\(\frac {b^{2} \sin \left (c \right ) \sin \left (d x \right )}{d}+\frac {b^{2} \sqrt {\pi }\, \cos \left (c \right ) \left (\frac {1}{\sqrt {\pi }}-\frac {\cos \left (d x \right )}{\sqrt {\pi }}\right )}{d}+a b \sqrt {\pi }\, \sin \left (c \right ) \left (\frac {2 \gamma +2 \ln \left (x \right )+\ln \left (d^{2}\right )}{\sqrt {\pi }}-\frac {2 \gamma }{\sqrt {\pi }}-\frac {2 \ln \left (2\right )}{\sqrt {\pi }}-\frac {2 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}+\frac {2 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )+2 a b \cos \left (c \right ) \sinIntegral \left (d x \right )+\frac {a^{2} \sqrt {\pi }\, \sin \left (c \right ) d^{2} \left (-\frac {4 d^{2} \cos \left (x \sqrt {d^{2}}\right )}{x \left (d^{2}\right )^{\frac {3}{2}} \sqrt {\pi }}-\frac {4 \sinIntegral \left (x \sqrt {d^{2}}\right )}{\sqrt {\pi }}\right )}{4 \sqrt {d^{2}}}+\frac {a^{2} \sqrt {\pi }\, \cos \left (c \right ) d \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \gamma }{\sqrt {\pi }}-\frac {4 \ln \left (2\right )}{\sqrt {\pi }}-\frac {4 \ln \left (\frac {d x}{2}\right )}{\sqrt {\pi }}-\frac {4 \sin \left (d x \right )}{\sqrt {\pi }\, d x}+\frac {4 \cosineIntegral \left (d x \right )}{\sqrt {\pi }}\right )}{4}\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.90, size = 122, normalized size = 1.69 \begin {gather*} \frac {{\left ({\left (a^{2} {\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \cos \left (c\right ) - a^{2} {\left (i \, \Gamma \left (-1, i \, d x\right ) - i \, \Gamma \left (-1, -i \, d x\right )\right )} \sin \left (c\right )\right )} d^{2} + 2 \, {\left (a b {\left (i \, \Gamma \left (-1, i \, d x\right ) - i \, \Gamma \left (-1, -i \, d x\right )\right )} \cos \left (c\right ) + a b {\left (\Gamma \left (-1, i \, d x\right ) + \Gamma \left (-1, -i \, d x\right )\right )} \sin \left (c\right )\right )} d\right )} x - 2 \, {\left (b^{2} x + 2 \, a b\right )} \cos \left (d x + c\right )}{2 \, d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 111, normalized size = 1.54 \begin {gather*} -\frac {2 \, b^{2} x \cos \left (d x + c\right ) + 2 \, a^{2} d \sin \left (d x + c\right ) - {\left (a^{2} d^{2} x \operatorname {Ci}\left (d x\right ) + a^{2} d^{2} x \operatorname {Ci}\left (-d x\right ) + 4 \, a b d x \operatorname {Si}\left (d x\right )\right )} \cos \left (c\right ) + 2 \, {\left (a^{2} d^{2} x \operatorname {Si}\left (d x\right ) - a b d x \operatorname {Ci}\left (d x\right ) - a b d x \operatorname {Ci}\left (-d x\right )\right )} \sin \left (c\right )}{2 \, d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 5.91, size = 743, normalized size = 10.32 \begin {gather*} -\frac {a^{2} d^{2} x \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^{2} d^{2} x \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 \, a^{2} d^{2} x \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^{2} d^{2} x \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 \, a^{2} d^{2} x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) + 2 \, a b d x \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 b d x \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} + 4 \, a b d x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a^{2} d^{2} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - a^{2} d^{2} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 4 \, a b d x \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 ) - 4 \, a b d x \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 ) + a^{2} d^{2} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + a^{2} d^{2} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a b d x \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + 2 \, a b d x \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, d x\right )^{2} - 4 \, a b d x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, d x\right )^{2} + 2 \, a^{2} d^{2} x \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, a^{2} d^{2} x \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, a^{2} d^{2} x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right ) + 2 \, a b d x \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a b d x \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, a b d x \operatorname {Si}\left (d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, b^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} - a^{2} d^{2} x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) - a^{2} d^{2} x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) - 4 \, a b d x \Re \left ( \operatorname {Ci}\left (d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 4 \, a b d x \Re \left ( \operatorname {Ci}\left (-d x\right ) \right ) \tan \left (\frac {1}{2} \, c\right ) - 4 \, a^{2} d \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 4 \, a^{2} d \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right )^{2} - 2 \, a b d x \Im \left ( \operatorname {Ci}\left (d x\right ) \right ) + 2 \, a b d x \Im \left ( \operatorname {Ci}\left (-d x\right ) \right ) - 4 \, a b d x \operatorname {Si}\left (d x\right ) - 2 \, b^{2} x \tan \left (\frac {1}{2} \, d x\right )^{2} - 8 \, b^{2} x \tan \left (\frac {1}{2} \, d x\right ) \tan \left (\frac {1}{2} \, c\right ) - 2 \, b^{2} x \tan \left (\frac {1}{2} \, c\right )^{2} + 4 \, a^{2} d \tan \left (\frac {1}{2} \, d x\right ) + 4 \, a^{2} d \tan \left (\frac {1}{2} \, c\right ) + 2 \, b^{2} x}{2 \, {\left (d x \tan \left (\frac {1}{2} \, d x\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + d x \tan \left (\frac {1}{2} \, d x\right )^{2} + d x \tan \left (\frac {1}{2} \, c\right )^{2} + d x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2*d^2*x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + a^2*d^2*x*real_part(cos_integral(-d
*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*a^2*d^2*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 2*a^2*
d^2*x*imag_part(cos_integral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c) + 4*a^2*d^2*x*sin_integral(d*x)*tan(1/2*d*x)^2*t
an(1/2*c) + 2*a*b*d*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*a*b*d*x*imag_part(cos_integ
ral(-d*x))*tan(1/2*d*x)^2*tan(1/2*c)^2 + 4*a*b*d*x*sin_integral(d*x)*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^2*x*r
eal_part(cos_integral(d*x))*tan(1/2*d*x)^2 - a^2*d^2*x*real_part(cos_integral(-d*x))*tan(1/2*d*x)^2 - 4*a*b*d*
x*real_part(cos_integral(d*x))*tan(1/2*d*x)^2*tan(1/2*c) - 4*a*b*d*x*real_part(cos_integral(-d*x))*tan(1/2*d*x
)^2*tan(1/2*c) + a^2*d^2*x*real_part(cos_integral(d*x))*tan(1/2*c)^2 + a^2*d^2*x*real_part(cos_integral(-d*x))
*tan(1/2*c)^2 - 2*a*b*d*x*imag_part(cos_integral(d*x))*tan(1/2*d*x)^2 + 2*a*b*d*x*imag_part(cos_integral(-d*x)
)*tan(1/2*d*x)^2 - 4*a*b*d*x*sin_integral(d*x)*tan(1/2*d*x)^2 + 2*a^2*d^2*x*imag_part(cos_integral(d*x))*tan(1
/2*c) - 2*a^2*d^2*x*imag_part(cos_integral(-d*x))*tan(1/2*c) + 4*a^2*d^2*x*sin_integral(d*x)*tan(1/2*c) + 2*a*
b*d*x*imag_part(cos_integral(d*x))*tan(1/2*c)^2 - 2*a*b*d*x*imag_part(cos_integral(-d*x))*tan(1/2*c)^2 + 4*a*b
*d*x*sin_integral(d*x)*tan(1/2*c)^2 + 2*b^2*x*tan(1/2*d*x)^2*tan(1/2*c)^2 - a^2*d^2*x*real_part(cos_integral(d
*x)) - a^2*d^2*x*real_part(cos_integral(-d*x)) - 4*a*b*d*x*real_part(cos_integral(d*x))*tan(1/2*c) - 4*a*b*d*x
*real_part(cos_integral(-d*x))*tan(1/2*c) - 4*a^2*d*tan(1/2*d*x)^2*tan(1/2*c) - 4*a^2*d*tan(1/2*d*x)*tan(1/2*c
)^2 - 2*a*b*d*x*imag_part(cos_integral(d*x)) + 2*a*b*d*x*imag_part(cos_integral(-d*x)) - 4*a*b*d*x*sin_integra
l(d*x) - 2*b^2*x*tan(1/2*d*x)^2 - 8*b^2*x*tan(1/2*d*x)*tan(1/2*c) - 2*b^2*x*tan(1/2*c)^2 + 4*a^2*d*tan(1/2*d*x
) + 4*a^2*d*tan(1/2*c) + 2*b^2*x)/(d*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + d*x*tan(1/2*d*x)^2 + d*x*tan(1/2*c)^2 + d
*x)

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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^2} \,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^2,x)

[Out]

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

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