3.1028 \(\int \frac{1}{(c+a^2 c x^2)^{3/2} \tan ^{-1}(a x)^{3/2}} \, dx\)

Optimal. Leaf size=92 \[ -\frac{2 \sqrt{2 \pi } \sqrt{a^2 x^2+1} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a c \sqrt{a^2 c x^2+c}}-\frac{2}{a c \sqrt{a^2 c x^2+c} \sqrt{\tan ^{-1}(a x)}} \]

[Out]

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

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Rubi [A]  time = 0.212764, antiderivative size = 92, 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 = {4902, 4971, 4970, 3305, 3351} \[ -\frac{2 \sqrt{2 \pi } \sqrt{a^2 x^2+1} S\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a c \sqrt{a^2 c x^2+c}}-\frac{2}{a c \sqrt{a^2 c x^2+c} \sqrt{\tan ^{-1}(a x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(3/2)),x]

[Out]

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

Rule 4902

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

Rule 4971

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[(d^(q +
1/2)*Sqrt[1 + c^2*x^2])/Sqrt[d + e*x^2], Int[x^m*(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] && IGtQ[m, 0] && ILtQ[m + 2*q + 1, 0] &&  !(IntegerQ[q] || GtQ[d, 0])

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 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{1}{\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)^{3/2}} \, dx &=-\frac{2}{a c \sqrt{c+a^2 c x^2} \sqrt{\tan ^{-1}(a x)}}-(2 a) \int \frac{x}{\left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \, dx\\ &=-\frac{2}{a c \sqrt{c+a^2 c x^2} \sqrt{\tan ^{-1}(a x)}}-\frac{\left (2 a \sqrt{1+a^2 x^2}\right ) \int \frac{x}{\left (1+a^2 x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{c \sqrt{c+a^2 c x^2}}\\ &=-\frac{2}{a c \sqrt{c+a^2 c x^2} \sqrt{\tan ^{-1}(a x)}}-\frac{\left (2 \sqrt{1+a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\sin (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c \sqrt{c+a^2 c x^2}}\\ &=-\frac{2}{a c \sqrt{c+a^2 c x^2} \sqrt{\tan ^{-1}(a x)}}-\frac{\left (4 \sqrt{1+a^2 x^2}\right ) \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{2}{a c \sqrt{c+a^2 c x^2} \sqrt{\tan ^{-1}(a x)}}-\frac{2 \sqrt{2 \pi } \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 [C]  time = 0.127484, size = 107, normalized size = 1.16 \[ \frac{\sqrt{a^2 x^2+1} \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-i \tan ^{-1}(a x)\right )+\sqrt{a^2 x^2+1} \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},i \tan ^{-1}(a x)\right )-2}{a c \sqrt{a^2 c x^2+c} \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((c + a^2*c*x^2)^(3/2)*ArcTan[a*x]^(3/2)),x]

[Out]

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

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Maple [F]  time = 0.723, size = 0, normalized size = 0. \begin{align*} \int{ \left ({a}^{2}c{x}^{2}+c \right ) ^{-{\frac{3}{2}}} \left ( \arctan \left ( ax \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{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(1/(a^2*c*x^2+c)^(3/2)/arctan(a*x)^(3/2),x, algorithm="giac")

[Out]

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