3.992 \(\int \frac{x}{(c+a^2 c x^2)^2 \tan ^{-1}(a x)^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ \frac{2 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{a^2 c^2}-\frac{2 x}{a c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (a^2 x^2+1\right )}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (a^2 x^2+1\right )}+\frac{4 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.169586, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4932, 4930, 4904, 3312, 3304, 3352} \[ \frac{2 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{a^2 c^2}-\frac{2 x}{a c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (a^2 x^2+1\right )}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (a^2 x^2+1\right )}+\frac{4 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x/((c + a^2*c*x^2)^2*ArcTan[a*x]^(3/2)),x]

[Out]

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

Rule 4932

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*(x_))/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(x*(a + b*ArcTan
[c*x])^(p + 1))/(b*c*d*(p + 1)*(d + e*x^2)), x] + (-Dist[4/(b^2*(p + 1)*(p + 2)), Int[(x*(a + b*ArcTan[c*x])^(
p + 2))/(d + e*x^2)^2, x], x] - Simp[((1 - c^2*x^2)*(a + b*ArcTan[c*x])^(p + 2))/(b^2*e*(p + 1)*(p + 2)*(d + e
*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[p, -1] && NeQ[p, -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}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{3/2}} \, dx &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+16 \int \frac{x \sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx\\ &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \int \frac{1}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx}{a}\\ &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \operatorname{Subst}\left (\int \frac{\cos ^2(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^2}\\ &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \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 )}{a^2 c^2}\\ &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{4 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^2 c^2}\\ &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{4 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \operatorname{Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{a^2 c^2}\\ &=-\frac{2 x}{a c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{4 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2}-\frac{8 \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{4 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{a^2 c^2 \left (1+a^2 x^2\right )}+\frac{2 \sqrt{\pi } C\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{a^2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0917111, size = 48, normalized size = 0.35 \[ \frac{2 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )-\frac{\sin \left (2 \tan ^{-1}(a x)\right )}{\sqrt{\tan ^{-1}(a x)}}}{a^2 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((c + a^2*c*x^2)^2*ArcTan[a*x]^(3/2)),x]

[Out]

(2*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] - Sin[2*ArcTan[a*x]]/Sqrt[ArcTan[a*x]])/(a^2*c^2)

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Maple [A]  time = 0.098, size = 47, normalized size = 0.3 \begin{align*}{\frac{1}{{a}^{2}{c}^{2}} \left ( 2\,\sqrt{\arctan \left ( ax \right ) }\sqrt{\pi }{\it FresnelC} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) -\sin \left ( 2\,\arctan \left ( ax \right ) \right ) \right ){\frac{1}{\sqrt{\arctan \left ( ax \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x)

[Out]

1/a^2/c^2*(2*arctan(a*x)^(1/2)*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))-sin(2*arctan(a*x)))/arctan(a*x)
^(1/2)

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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/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),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/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),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}{a^{4} x^{4} \operatorname{atan}^{\frac{3}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname{atan}^{\frac{3}{2}}{\left (a x \right )} + \operatorname{atan}^{\frac{3}{2}}{\left (a x \right )}}\, dx}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a**2*c*x**2+c)**2/atan(a*x)**(3/2),x)

[Out]

Integral(x/(a**4*x**4*atan(a*x)**(3/2) + 2*a**2*x**2*atan(a*x)**(3/2) + atan(a*x)**(3/2)), x)/c**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/((a^2*c*x^2 + c)^2*arctan(a*x)^(3/2)), x)