\(\int \frac {\cos ^2(a+b x^2)}{x^3} \, dx\) [14]

Optimal result

Integrand size = 14, antiderivative size = 57 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{4 x^2}-\frac {\cos \left (2 \left (a+b x^2\right )\right )}{4 x^2}-\frac {1}{2} b \operatorname {CosIntegral}\left (2 b x^2\right ) \sin (2 a)-\frac {1}{2} b \cos (2 a) \text {Si}\left (2 b x^2\right ) \]

[Out]

-1/4/x^2-1/4*cos(2*b*x^2+2*a)/x^2-1/2*b*cos(2*a)*Si(2*b*x^2)-1/2*b*Ci(2*b*x^2)*sin(2*a)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3485, 3461, 3378, 3384, 3380, 3383} \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {1}{2} b \sin (2 a) \operatorname {CosIntegral}\left (2 b x^2\right )-\frac {1}{2} b \cos (2 a) \text {Si}\left (2 b x^2\right )-\frac {\cos \left (2 \left (a+b x^2\right )\right )}{4 x^2}-\frac {1}{4 x^2} \]

[In]

Int[Cos[a + b*x^2]^2/x^3,x]

[Out]

-1/4*1/x^2 - Cos[2*(a + b*x^2)]/(4*x^2) - (b*CosIntegral[2*b*x^2]*Sin[2*a])/2 - (b*Cos[2*a]*SinIntegral[2*b*x^
2])/2

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 3461

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 3485

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {\cos ^2\left (a+b x^2\right )+b x^2 \operatorname {CosIntegral}\left (2 b x^2\right ) \sin (2 a)+b x^2 \cos (2 a) \text {Si}\left (2 b x^2\right )}{2 x^2} \]

[In]

Integrate[Cos[a + b*x^2]^2/x^3,x]

[Out]

-1/2*(Cos[a + b*x^2]^2 + b*x^2*CosIntegral[2*b*x^2]*Sin[2*a] + b*x^2*Cos[2*a]*SinIntegral[2*b*x^2])/x^2

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.64 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.72

method	result	size
risch	\(\frac {i {\mathrm e}^{-2 i a} \operatorname {Ei}_{1}\left (-2 i b \,x^{2}\right ) b \,x^{2}+{\mathrm e}^{-2 i a} \pi \,\operatorname {csgn}\left (b \,x^{2}\right ) b \,x^{2}-i {\mathrm e}^{2 i a} b \,\operatorname {Ei}_{1}\left (-2 i b \,x^{2}\right ) x^{2}-2 \,{\mathrm e}^{-2 i a} \operatorname {Si}\left (2 b \,x^{2}\right ) b \,x^{2}-\cos \left (2 b \,x^{2}+2 a \right )-1}{4 x^{2}}\)	\(98\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {b x^{2} \operatorname {Ci}\left (2 \, b x^{2}\right ) \sin \left (2 \, a\right ) + b x^{2} \cos \left (2 \, a\right ) \operatorname {Si}\left (2 \, b x^{2}\right ) + \cos \left (b x^{2} + a\right )^{2}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=\int \frac {\cos ^{2}{\left (a + b x^{2} \right )}}{x^{3}}\, dx \]

[In]

integrate(cos(b*x**2+a)**2/x**3,x)

[Out]

Integral(cos(a + b*x**2)**2/x**3, x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.07 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {{\left ({\left (i \, \Gamma \left (-1, 2 i \, b x^{2}\right ) - i \, \Gamma \left (-1, -2 i \, b x^{2}\right )\right )} \cos \left (2 \, a\right ) + {\left (\Gamma \left (-1, 2 i \, b x^{2}\right ) + \Gamma \left (-1, -2 i \, b x^{2}\right )\right )} \sin \left (2 \, a\right )\right )} b x^{2} + 1}{4 \, x^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (50) = 100\).

Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.88 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=-\frac {2 \, {\left (b x^{2} + a\right )} b^{2} \operatorname {Ci}\left (2 \, b x^{2}\right ) \sin \left (2 \, a\right ) - 2 \, a b^{2} \operatorname {Ci}\left (2 \, b x^{2}\right ) \sin \left (2 \, a\right ) - 2 \, {\left (b x^{2} + a\right )} b^{2} \cos \left (2 \, a\right ) \operatorname {Si}\left (-2 \, b x^{2}\right ) + 2 \, a b^{2} \cos \left (2 \, a\right ) \operatorname {Si}\left (-2 \, b x^{2}\right ) + b^{2} \cos \left (2 \, b x^{2} + 2 \, a\right ) + b^{2}}{4 \, b^{2} x^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^3} \, dx=\int \frac {{\cos \left (b\,x^2+a\right )}^2}{x^3} \,d x \]

[In]

int(cos(a + b*x^2)^2/x^3,x)

[Out]

int(cos(a + b*x^2)^2/x^3, x)