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

Optimal. Leaf size=93 \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^2 c \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.182426, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {4930, 4905, 4904, 3304, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sqrt{a^2 x^2+1} \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^2 c \sqrt{a^2 c x^2+c}}-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[ArcTan[a*x]]/(a^2*c*Sqrt[c + a^2*c*x^2])) + (Sqrt[Pi/2]*Sqrt[1 + a^2*x^2]*FresnelC[Sqrt[2/Pi]*Sqrt[ArcT
an[a*x]]])/(a^2*c*Sqrt[c + a^2*c*x^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 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 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 \sqrt{\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{c+a^2 c x^2}}+\frac{\int \frac{1}{\left (c+a^2 c x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{2 a}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{c+a^2 c x^2}}+\frac{\sqrt{1+a^2 x^2} \int \frac{1}{\left (1+a^2 x^2\right )^{3/2} \sqrt{\tan ^{-1}(a x)}} \, dx}{2 a c \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{c+a^2 c x^2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \frac{\cos (x)}{\sqrt{x}} \, dx,x,\tan ^{-1}(a x)\right )}{2 a^2 c \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{c+a^2 c x^2}}+\frac{\sqrt{1+a^2 x^2} \operatorname{Subst}\left (\int \cos \left (x^2\right ) \, dx,x,\sqrt{\tan ^{-1}(a x)}\right )}{a^2 c \sqrt{c+a^2 c x^2}}\\ &=-\frac{\sqrt{\tan ^{-1}(a x)}}{a^2 c \sqrt{c+a^2 c x^2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{1+a^2 x^2} C\left (\sqrt{\frac{2}{\pi }} \sqrt{\tan ^{-1}(a x)}\right )}{a^2 c \sqrt{c+a^2 c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.1514, size = 121, normalized size = 1.3 \[ \frac{-i \sqrt{a^2 x^2+1} \sqrt{-i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},-i \tan ^{-1}(a x)\right )+i \sqrt{a^2 x^2+1} \sqrt{i \tan ^{-1}(a x)} \text{Gamma}\left (\frac{1}{2},i \tan ^{-1}(a x)\right )-4 \tan ^{-1}(a x)}{4 a^2 c \sqrt{a^2 c x^2+c} \sqrt{\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.882, size = 0, normalized size = 0. \begin{align*} \int{x\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(x*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^(3/2),x)

[Out]

int(x*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(x*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(x*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{x \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(x*atan(a*x)**(1/2)/(a**2*c*x**2+c)**(3/2),x)

[Out]

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

[Out]

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