3.710 \(\int \frac{x \sqrt{\tan ^{-1}(a x)}}{(c+a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=79 \[ \frac{\sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{8 a^2 c^2}-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{\sqrt{\tan ^{-1}(a x)}}{4 a^2 c^2} \]

[Out]

Sqrt[ArcTan[a*x]]/(4*a^2*c^2) - Sqrt[ArcTan[a*x]]/(2*a^2*c^2*(1 + a^2*x^2)) + (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTa
n[a*x]])/Sqrt[Pi]])/(8*a^2*c^2)

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Rubi [A]  time = 0.120755, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {4930, 4904, 3312, 3304, 3352} \[ \frac{\sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{8 a^2 c^2}-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac{\sqrt{\tan ^{-1}(a x)}}{4 a^2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[ArcTan[a*x]])/(c + a^2*c*x^2)^2,x]

[Out]

Sqrt[ArcTan[a*x]]/(4*a^2*c^2) - Sqrt[ArcTan[a*x]]/(2*a^2*c^2*(1 + a^2*x^2)) + (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTa
n[a*x]])/Sqrt[Pi]])/(8*a^2*c^2)

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 4904

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^q/c, Subst[Int[(a
+ b*x)^p/Cos[x]^(2*(q + 1)), x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && ILtQ
[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \frac{x \sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx}{4 a}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\cos ^2(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^2 c^2}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int \left (\frac{1}{2 \sqrt{x}}+\frac{\cos (2 x)}{2 \sqrt{x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{4 a^2 c^2}\\ &=\frac{\sqrt{\tan ^{-1}(a x)}}{4 a^2 c^2}-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c^2}\\ &=\frac{\sqrt{\tan ^{-1}(a x)}}{4 a^2 c^2}-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{4 a^2 c^2}\\ &=\frac{\sqrt{\tan ^{-1}(a x)}}{4 a^2 c^2}-\frac{\sqrt{\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac{\sqrt{\pi } C\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{8 a^2 c^2}\\ \end{align*}

Mathematica [C]  time = 0.280483, size = 136, normalized size = 1.72 \[ \frac{4 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )+\frac{-i \sqrt{2} \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-2 i \tan ^{-1}(a x)\right )+i \sqrt{2} \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )+\frac{16 \left (a^2 x^2-1\right ) \tan ^{-1}(a x)}{a^2 x^2+1}}{\sqrt{\tan ^{-1}(a x)}}}{64 a^2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*Sqrt[ArcTan[a*x]])/(c + a^2*c*x^2)^2,x]

[Out]

(4*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + ((16*(-1 + a^2*x^2)*ArcTan[a*x])/(1 + a^2*x^2) - I*Sqrt
[2]*Sqrt[(-I)*ArcTan[a*x]]*Gamma[1/2, (-2*I)*ArcTan[a*x]] + I*Sqrt[2]*Sqrt[I*ArcTan[a*x]]*Gamma[1/2, (2*I)*Arc
Tan[a*x]])/Sqrt[ArcTan[a*x]])/(64*a^2*c^2)

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Maple [A]  time = 0.112, size = 46, normalized size = 0.6 \begin{align*} -{\frac{\cos \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,{a}^{2}{c}^{2}}\sqrt{\arctan \left ( ax \right ) }}+{\frac{\sqrt{\pi }}{8\,{a}^{2}{c}^{2}}{\it FresnelC} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^2,x)

[Out]

-1/4/a^2/c^2*arctan(a*x)^(1/2)*cos(2*arctan(a*x))+1/8*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2/c^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x \sqrt{\operatorname{atan}{\left (a x \right )}}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*atan(a*x)**(1/2)/(a**2*c*x**2+c)**2,x)

[Out]

Integral(x*sqrt(atan(a*x))/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sqrt{\arctan \left (a x\right )}}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

integrate(x*sqrt(arctan(a*x))/(a^2*c*x^2 + c)^2, x)