3.1061 \(\int \frac{x}{(c+a^2 c x^2)^2 \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{8 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^2 c^2}-\frac{2 x}{3 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}} \]

[Out]

(-2*x)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (4*(1 - a^2*x^2))/(3*a^2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]
]) - (8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^2*c^2)

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Rubi [A]  time = 0.121547, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4932, 4970, 4406, 12, 3305, 3351} \[ -\frac{8 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^2 c^2}-\frac{2 x}{3 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

Int[x/((c + a^2*c*x^2)^2*ArcTan[a*x]^(5/2)),x]

[Out]

(-2*x)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (4*(1 - a^2*x^2))/(3*a^2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]
]) - (8*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*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 4970

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

Rule 4406

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(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 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[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)^{5/2}} \, dx &=-\frac{2 x}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{16}{3} \int \frac{x}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx\\ &=-\frac{2 x}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \frac{\cos (x) \sin (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^2 c^2}\\ &=-\frac{2 x}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{2 \sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^2 c^2}\\ &=-\frac{2 x}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{8 \operatorname{Subst}\left (\int \frac{\sin (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^2 c^2}\\ &=-\frac{2 x}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{16 \operatorname{Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{3 a^2 c^2}\\ &=-\frac{2 x}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{4 \left (1-a^2 x^2\right )}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{8 \sqrt{\pi } S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0779631, size = 88, normalized size = 0.87 \[ -\frac{2 \left (4 \sqrt{\pi } \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2} S\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )+\left (2-2 a^2 x^2\right ) \tan ^{-1}(a x)+a x\right )}{3 a^2 c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((c + a^2*c*x^2)^2*ArcTan[a*x]^(5/2)),x]

[Out]

(-2*(a*x + (2 - 2*a^2*x^2)*ArcTan[a*x] + 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)*FresnelS[(2*Sqrt[ArcTan[a*
x]])/Sqrt[Pi]]))/(3*a^2*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2))

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Maple [A]  time = 0.102, size = 59, normalized size = 0.6 \begin{align*} -{\frac{1}{3\,{a}^{2}{c}^{2}} \left ( 8\,\sqrt{\pi }{\it FresnelS} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \left ( \arctan \left ( ax \right ) \right ) ^{3/2}+4\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) +\sin \left ( 2\,\arctan \left ( ax \right ) \right ) \right ) \left ( \arctan \left ( ax \right ) \right ) ^{-{\frac{3}{2}}}} \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)^(5/2),x)

[Out]

-1/3/a^2/c^2*(8*Pi^(1/2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+4*cos(2*arctan(a*x))*arctan(
a*x)+sin(2*arctan(a*x)))/arctan(a*x)^(3/2)

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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)^(5/2),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)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate(x/((a^2*c*x^2 + c)^2*arctan(a*x)^(5/2)), x)