3.1060 \(\int \frac{x^2}{(c+a^2 c x^2)^2 \tan ^{-1}(a x)^{5/2}} \, dx\)

Optimal. Leaf size=180 \[ \frac{8 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^3 c^2}-\frac{2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (a^2 x^2+1\right )}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (a^2 x^2+1\right )}+\frac{16 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2} \]

[Out]

(-2*x^2)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (8*x)/(3*a^2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) + (16*S
qrt[ArcTan[a*x]])/(3*a^3*c^2) - (32*Sqrt[ArcTan[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)) + (16*(1 - a^2*x^2)*Sqrt[ArcT
an[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)) + (8*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^3*c^2)

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Rubi [A]  time = 0.2631, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {4942, 4932, 4930, 4904, 3312, 3304, 3352} \[ \frac{8 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^3 c^2}-\frac{2 x^2}{3 a c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (a^2 x^2+1\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (a^2 x^2+1\right )}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (a^2 x^2+1\right )}+\frac{16 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2)/(3*a*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)) - (8*x)/(3*a^2*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) + (16*S
qrt[ArcTan[a*x]])/(3*a^3*c^2) - (32*Sqrt[ArcTan[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)) + (16*(1 - a^2*x^2)*Sqrt[ArcT
an[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)) + (8*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(3*a^3*c^2)

Rule 4942

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

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^2}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{5/2}} \, dx &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}+\frac{4 \int \frac{x}{\left (c+a^2 c x^2\right )^2 \tan ^{-1}(a x)^{3/2}} \, dx}{3 a}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{64 \int \frac{x \sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx}{3 a}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \int \frac{1}{\left (c+a^2 c x^2\right )^2 \sqrt{\tan ^{-1}(a x)}} \, dx}{3 a^2}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \operatorname{Subst}\left (\int \frac{\cos ^2(x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^3 c^2}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \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 )}{3 a^3 c^2}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{16 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{8 \operatorname{Subst}\left (\int \frac{\cos (2 x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{3 a^3 c^2}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{16 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \operatorname{Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{3 a^3 c^2}\\ &=-\frac{2 x^2}{3 a c^2 \left (1+a^2 x^2\right ) \tan ^{-1}(a x)^{3/2}}-\frac{8 x}{3 a^2 c^2 \left (1+a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}+\frac{16 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2}-\frac{32 \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{16 \left (1-a^2 x^2\right ) \sqrt{\tan ^{-1}(a x)}}{3 a^3 c^2 \left (1+a^2 x^2\right )}+\frac{8 \sqrt{\pi } C\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{3 a^3 c^2}\\ \end{align*}

Mathematica [C]  time = 0.341508, size = 162, normalized size = 0.9 \[ \frac{\sqrt{2} \left (a^2 x^2+1\right ) \left (-i \tan ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-2 i \tan ^{-1}(a x)\right )+\sqrt{2} \left (a^2 x^2+1\right ) \left (i \tan ^{-1}(a x)\right )^{3/2} \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )+4 \sqrt{\pi } \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2} \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )-2 a x \left (a x+4 \tan ^{-1}(a x)\right )}{3 a^3 c^2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*a*x*(a*x + 4*ArcTan[a*x]) + 4*Sqrt[Pi]*(1 + a^2*x^2)*ArcTan[a*x]^(3/2)*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt
[Pi]] + Sqrt[2]*(1 + a^2*x^2)*((-I)*ArcTan[a*x])^(3/2)*Gamma[1/2, (-2*I)*ArcTan[a*x]] + Sqrt[2]*(1 + a^2*x^2)*
(I*ArcTan[a*x])^(3/2)*Gamma[1/2, (2*I)*ArcTan[a*x]])/(3*a^3*c^2*(1 + a^2*x^2)*ArcTan[a*x]^(3/2))

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Maple [A]  time = 0.11, size = 62, normalized size = 0.3 \begin{align*} -{\frac{1}{3\,{a}^{3}{c}^{2}} \left ( -8\,\sqrt{\pi }{\it FresnelC} \left ( 2\,{\frac{\sqrt{\arctan \left ( ax \right ) }}{\sqrt{\pi }}} \right ) \left ( \arctan \left ( ax \right ) \right ) ^{3/2}+4\,\sin \left ( 2\,\arctan \left ( ax \right ) \right ) \arctan \left ( ax \right ) -\cos \left ( 2\,\arctan \left ( ax \right ) \right ) +1 \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^2/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x)

[Out]

-1/3/a^3/c^2*(-8*Pi^(1/2)*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*arctan(a*x)^(3/2)+4*sin(2*arctan(a*x))*arctan
(a*x)-cos(2*arctan(a*x))+1)/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^2/(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^2/(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**2/(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^{2}}{{\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^2/(a^2*c*x^2+c)^2/arctan(a*x)^(5/2),x, algorithm="giac")

[Out]

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