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

Optimal. Leaf size=53 \[ -\frac{a}{2 x^2}+\frac{1}{2} b d \cos (c) \text{CosIntegral}\left (d x^2\right )-\frac{1}{2} b d \sin (c) \text{Si}\left (d x^2\right )-\frac{b \sin \left (c+d x^2\right )}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0911441, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {14, 3379, 3297, 3303, 3299, 3302} \[ -\frac{a}{2 x^2}+\frac{1}{2} b d \cos (c) \text{CosIntegral}\left (d x^2\right )-\frac{1}{2} b d \sin (c) \text{Si}\left (d x^2\right )-\frac{b \sin \left (c+d x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x^2])/x^3,x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]))

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]

Rubi steps

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

Mathematica [A]  time = 0.0812176, size = 48, normalized size = 0.91 \[ -\frac{a-b d x^2 \cos (c) \text{CosIntegral}\left (d x^2\right )+b d x^2 \sin (c) \text{Si}\left (d x^2\right )+b \sin \left (c+d x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 47, normalized size = 0.9 \begin{align*} -{\frac{a}{2\,{x}^{2}}}+b \left ( -{\frac{\sin \left ( d{x}^{2}+c \right ) }{2\,{x}^{2}}}+d \left ({\frac{\cos \left ( c \right ){\it Ci} \left ( d{x}^{2} \right ) }{2}}-{\frac{\sin \left ( c \right ){\it Si} \left ( d{x}^{2} \right ) }{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.14813, size = 77, normalized size = 1.45 \begin{align*} \frac{1}{4} \,{\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 )} b d - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((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)
)*b*d - 1/2*a/x^2

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Fricas [A]  time = 1.9283, size = 197, normalized size = 3.72 \begin{align*} -\frac{2 \, b d x^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) -{\left (b d x^{2} \operatorname{Ci}\left (d x^{2}\right ) + b d x^{2} \operatorname{Ci}\left (-d x^{2}\right )\right )} \cos \left (c\right ) + 2 \, b \sin \left (d x^{2} + c\right ) + 2 \, a}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.14906, size = 134, normalized size = 2.53 \begin{align*} \frac{{\left (d x^{2} + c\right )} b d^{2} \cos \left (c\right ) \operatorname{Ci}\left (d x^{2}\right ) - b c d^{2} \cos \left (c\right ) \operatorname{Ci}\left (d x^{2}\right ) -{\left (d x^{2} + c\right )} b d^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) + b c d^{2} \sin \left (c\right ) \operatorname{Si}\left (d x^{2}\right ) - b d^{2} \sin \left (d x^{2} + c\right ) - a d^{2}}{2 \, 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))/x^3,x, algorithm="giac")

[Out]

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