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

Optimal. Leaf size=91 \[ \frac{x \sqrt{\tan ^{-1}(a x)}}{c \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a c \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.111199, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4905, 4904, 3296, 3305, 3351} \[ \frac{x \sqrt{\tan ^{-1}(a x)}}{c \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4905

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q + 1/2)*Sqrt[1
+ c^2*x^2])/Sqrt[d + e*x^2], Int[(1 + c^2*x^2)^q*(a + b*ArcTan[c*x])^p, 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 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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

Mathematica [A]  time = 0.0717767, size = 78, normalized size = 0.86 \[ \frac{2 a x \sqrt{\tan ^{-1}(a x)}-\sqrt{2 \pi } \sqrt{a^2 x^2+1} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{2 a c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.705, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{\arctan \left ( ax \right ) } \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(3/2),x)

[Out]

int(arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(3/2),x)

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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(arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(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(arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\operatorname{atan}{\left (a x \right )}}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(a*x)**(1/2)/(a**2*c*x**2+c)**(3/2),x)

[Out]

Integral(sqrt(atan(a*x))/(c*(a**2*x**2 + 1))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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