3.16 \(\int \frac{(a+b \sin (c+d x^2))^2}{x^3} \, dx\)

Optimal. Leaf size=115 \[ -\frac{2 a^2+b^2}{4 x^2}+a b d \cos (c) \text{CosIntegral}\left (d x^2\right )-a b d \sin (c) \text{Si}\left (d x^2\right )-\frac{a b \sin \left (c+d x^2\right )}{x^2}+\frac{1}{2} b^2 d \sin (2 c) \text{CosIntegral}\left (2 d x^2\right )+\frac{1}{2} b^2 d \cos (2 c) \text{Si}\left (2 d x^2\right )+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.221334, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3403, 6, 3380, 3297, 3303, 3299, 3302, 3379} \[ -\frac{2 a^2+b^2}{4 x^2}+a b d \cos (c) \text{CosIntegral}\left (d x^2\right )-a b d \sin (c) \text{Si}\left (d x^2\right )-\frac{a b \sin \left (c+d x^2\right )}{x^2}+\frac{1}{2} b^2 d \sin (2 c) \text{CosIntegral}\left (2 d x^2\right )+\frac{1}{2} b^2 d \cos (2 c) \text{Si}\left (2 d x^2\right )+\frac{b^2 \cos \left (2 \left (c+d x^2\right )\right )}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])^2/x^3,x]

[Out]

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

Rule 3403

Int[((e_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3297

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 3303

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 3299

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 3302

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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

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

Mathematica [A]  time = 0.254218, size = 116, normalized size = 1.01 \[ \frac{-2 a^2+4 a b d x^2 \cos (c) \text{CosIntegral}\left (d x^2\right )-4 a b d x^2 \sin (c) \text{Si}\left (d x^2\right )-4 a b \sin \left (c+d x^2\right )+2 b^2 d x^2 \sin (2 c) \text{CosIntegral}\left (2 d x^2\right )+2 b^2 d x^2 \cos (2 c) \text{Si}\left (2 d x^2\right )+b^2 \cos \left (2 \left (c+d x^2\right )\right )-b^2}{4 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.247, size = 203, normalized size = 1.8 \begin{align*} -{\frac{{\it csgn} \left ( d{x}^{2} \right ){{\rm e}^{-2\,ic}}\pi \,{b}^{2}d}{4}}+{\frac{{\it Si} \left ( 2\,d{x}^{2} \right ){{\rm e}^{-2\,ic}}{b}^{2}d}{2}}-{\frac{i}{4}}{\it Ei} \left ( 1,-2\,id{x}^{2} \right ){{\rm e}^{-2\,ic}}{b}^{2}d+{\frac{i}{4}}{b}^{2}d{\it Ei} \left ( 1,-2\,id{x}^{2} \right ){{\rm e}^{2\,ic}}-{\frac{abd{\it Ei} \left ( 1,-id{x}^{2} \right ){{\rm e}^{ic}}}{2}}+{\frac{i}{2}}{\it csgn} \left ( d{x}^{2} \right ){{\rm e}^{-ic}}\pi \,abd-i{\it Si} \left ( d{x}^{2} \right ){{\rm e}^{-ic}}abd-{\frac{{\it Ei} \left ( 1,-id{x}^{2} \right ){{\rm e}^{-ic}}abd}{2}}-{\frac{{a}^{2}}{2\,{x}^{2}}}-{\frac{{b}^{2}}{4\,{x}^{2}}}-{\frac{ab\sin \left ( d{x}^{2}+c \right ) }{{x}^{2}}}+{\frac{{b}^{2}\cos \left ( 2\,d{x}^{2}+2\,c \right ) }{4\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*csgn(d*x^2)*exp(-2*I*c)*Pi*b^2*d+1/2*Si(2*d*x^2)*exp(-2*I*c)*b^2*d-1/4*I*Ei(1,-2*I*d*x^2)*exp(-2*I*c)*b^2
*d+1/4*I*b^2*d*Ei(1,-2*I*d*x^2)*exp(2*I*c)-1/2*a*b*d*Ei(1,-I*d*x^2)*exp(I*c)+1/2*I*csgn(d*x^2)*exp(-I*c)*Pi*a*
b*d-I*Si(d*x^2)*exp(-I*c)*a*b*d-1/2*Ei(1,-I*d*x^2)*exp(-I*c)*a*b*d-1/2/x^2*a^2-1/4*b^2/x^2-a*b*sin(d*x^2+c)/x^
2+1/4*b^2*cos(2*d*x^2+2*c)/x^2

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Maxima [C]  time = 1.2421, size = 167, normalized size = 1.45 \begin{align*} \frac{1}{2} \,{\left ({\left (\Gamma \left (-1, i \, d x^{2}\right ) + \Gamma \left (-1, -i \, d x^{2}\right )\right )} \cos \left (c\right ) -{\left (i \, \Gamma \left (-1, i \, d x^{2}\right ) - i \, \Gamma \left (-1, -i \, d x^{2}\right )\right )} \sin \left (c\right )\right )} a b d + \frac{{\left ({\left ({\left (i \, \Gamma \left (-1, 2 i \, d x^{2}\right ) - i \, \Gamma \left (-1, -2 i \, d x^{2}\right )\right )} \cos \left (2 \, c\right ) +{\left (\Gamma \left (-1, 2 i \, d x^{2}\right ) + \Gamma \left (-1, -2 i \, d x^{2}\right )\right )} \sin \left (2 \, c\right )\right )} d x^{2} - 1\right )} b^{2}}{4 \, x^{2}} - \frac{a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.06748, size = 425, normalized size = 3.7 \begin{align*} \frac{2 \, b^{2} d x^{2} \cos \left (2 \, c\right ) \operatorname{Si}\left (2 \, d x^{2}\right ) - 4 \, a b d x^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) + 2 \, b^{2} \cos \left (d x^{2} + c\right )^{2} - 4 \, a b \sin \left (d x^{2} + c\right ) - 2 \, a^{2} - 2 \, b^{2} + 2 \,{\left (a b d x^{2} \operatorname{Ci}\left (d x^{2}\right ) + a b d x^{2} \operatorname{Ci}\left (-d x^{2}\right )\right )} \cos \left (c\right ) +{\left (b^{2} d x^{2} \operatorname{Ci}\left (2 \, d x^{2}\right ) + b^{2} d x^{2} \operatorname{Ci}\left (-2 \, d x^{2}\right )\right )} \sin \left (2 \, c\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \sin{\left (c + d x^{2} \right )}\right )^{2}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.12347, size = 305, normalized size = 2.65 \begin{align*} \frac{4 \,{\left (d x^{2} + c\right )} a b d^{2} \cos \left (c\right ) \operatorname{Ci}\left (d x^{2}\right ) - 4 \, a b c d^{2} \cos \left (c\right ) \operatorname{Ci}\left (d x^{2}\right ) + 2 \,{\left (d x^{2} + c\right )} b^{2} d^{2} \operatorname{Ci}\left (2 \, d x^{2}\right ) \sin \left (2 \, c\right ) - 2 \, b^{2} c d^{2} \operatorname{Ci}\left (2 \, d x^{2}\right ) \sin \left (2 \, c\right ) - 4 \,{\left (d x^{2} + c\right )} a b d^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) + 4 \, a b c d^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) - 2 \,{\left (d x^{2} + c\right )} b^{2} d^{2} \cos \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{2}\right ) + 2 \, b^{2} c d^{2} \cos \left (2 \, c\right ) \operatorname{Si}\left (-2 \, d x^{2}\right ) + b^{2} d^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) - 4 \, a b d^{2} \sin \left (d x^{2} + c\right ) - 2 \, a^{2} d^{2} - b^{2} d^{2}}{4 \, d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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