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

Optimal. Leaf size=127 \[ \frac{3 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{32 a^3 c^2}-\frac{x \tan ^{-1}(a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}-\frac{3 \sqrt{\tan ^{-1}(a x)}}{8 a^3 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{5 a^3 c^2}+\frac{3 \sqrt{\tan ^{-1}(a x)}}{16 a^3 c^2} \]

[Out]

(3*Sqrt[ArcTan[a*x]])/(16*a^3*c^2) - (3*Sqrt[ArcTan[a*x]])/(8*a^3*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x]^(3/2))/(
2*a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(5*a^3*c^2) + (3*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]
])/(32*a^3*c^2)

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Rubi [A]  time = 0.188436, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4936, 4930, 4904, 3312, 3304, 3352} \[ \frac{3 \sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )}{32 a^3 c^2}-\frac{x \tan ^{-1}(a x)^{3/2}}{2 a^2 c^2 \left (a^2 x^2+1\right )}-\frac{3 \sqrt{\tan ^{-1}(a x)}}{8 a^3 c^2 \left (a^2 x^2+1\right )}+\frac{\tan ^{-1}(a x)^{5/2}}{5 a^3 c^2}+\frac{3 \sqrt{\tan ^{-1}(a x)}}{16 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[ArcTan[a*x]])/(16*a^3*c^2) - (3*Sqrt[ArcTan[a*x]])/(8*a^3*c^2*(1 + a^2*x^2)) - (x*ArcTan[a*x]^(3/2))/(
2*a^2*c^2*(1 + a^2*x^2)) + ArcTan[a*x]^(5/2)/(5*a^3*c^2) + (3*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]
])/(32*a^3*c^2)

Rule 4936

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^2)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[(a + b*ArcTan
[c*x])^(p + 1)/(2*b*c^3*d^2*(p + 1)), x] + (Dist[(b*p)/(2*c), Int[(x*(a + b*ArcTan[c*x])^(p - 1))/(d + e*x^2)^
2, x], x] - Simp[(x*(a + b*ArcTan[c*x])^p)/(2*c^2*d*(d + e*x^2)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c
^2*d] && GtQ[p, 0]

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

Mathematica [C]  time = 0.353764, size = 187, normalized size = 1.47 \[ \frac{\frac{15 \left (-i \sqrt{2} \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-2 i \tan ^{-1}(a x)\right )+i \sqrt{2} \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},2 i \tan ^{-1}(a x)\right )+8 \tan ^{-1}(a x)\right )}{\sqrt{\tan ^{-1}(a x)}}+\frac{16 \sqrt{\tan ^{-1}(a x)} \left (15 \left (a^2 x^2-1\right )+16 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2-40 a x \tan ^{-1}(a x)\right )}{a^2 x^2+1}+60 \left (\sqrt{\pi } \text{FresnelC}\left (\frac{2 \sqrt{\tan ^{-1}(a x)}}{\sqrt{\pi }}\right )-2 \sqrt{\tan ^{-1}(a x)}\right )}{1280 a^3 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((16*Sqrt[ArcTan[a*x]]*(15*(-1 + a^2*x^2) - 40*a*x*ArcTan[a*x] + 16*(1 + a^2*x^2)*ArcTan[a*x]^2))/(1 + a^2*x^2
) + 60*(-2*Sqrt[ArcTan[a*x]] + Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]]) + (15*(8*ArcTan[a*x] - I*Sqr
t[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)*Ar
cTan[a*x]]))/Sqrt[ArcTan[a*x]])/(1280*a^3*c^2)

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Maple [A]  time = 0.102, size = 81, normalized size = 0.6 \begin{align*}{\frac{1}{5\,{a}^{3}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{5}{2}}}}-{\frac{\sin \left ( 2\,\arctan \left ( ax \right ) \right ) }{4\,{a}^{3}{c}^{2}} \left ( \arctan \left ( ax \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{3\,\cos \left ( 2\,\arctan \left ( ax \right ) \right ) }{16\,{a}^{3}{c}^{2}}\sqrt{\arctan \left ( ax \right ) }}+{\frac{3\,\sqrt{\pi }}{32\,{a}^{3}{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^2*arctan(a*x)^(3/2)/(a^2*c*x^2+c)^2,x)

[Out]

1/5*arctan(a*x)^(5/2)/a^3/c^2-1/4/a^3/c^2*arctan(a*x)^(3/2)*sin(2*arctan(a*x))-3/16/a^3/c^2*arctan(a*x)^(1/2)*
cos(2*arctan(a*x))+3/32*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^3/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^2*arctan(a*x)^(3/2)/(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^2*arctan(a*x)^(3/2)/(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^{2} \operatorname{atan}^{\frac{3}{2}}{\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**2*atan(a*x)**(3/2)/(a**2*c*x**2+c)**2,x)

[Out]

Integral(x**2*atan(a*x)**(3/2)/(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^{2} \arctan \left (a x\right )^{\frac{3}{2}}}{{\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^2*arctan(a*x)^(3/2)/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

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