3.10.34 \(\int \frac {1}{(c+a^2 c x^2)^2 \sqrt {\text {ArcTan}(a x)}} \, dx\) [934]

Optimal. Leaf size=47 \[ \frac {\sqrt {\text {ArcTan}(a x)}}{a c^2}+\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{2 a c^2} \]

[Out]

1/2*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a/c^2+arctan(a*x)^(1/2)/a/c^2

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Rubi [A]
time = 0.05, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5024, 3393, 3385, 3433} \begin {gather*} \frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{2 a c^2}+\frac {\sqrt {\text {ArcTan}(a x)}}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3385

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 3393

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 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 5024

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])

Rubi steps

\begin {align*} \int \frac {1}{\left (c+a^2 c x^2\right )^2 \sqrt {\tan ^{-1}(a x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a c^2}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{2 \sqrt {x}}+\frac {\cos (2 x)}{2 \sqrt {x}}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a c^2}\\ &=\frac {\sqrt {\tan ^{-1}(a x)}}{a c^2}+\frac {\text {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{2 a c^2}\\ &=\frac {\sqrt {\tan ^{-1}(a x)}}{a c^2}+\frac {\text {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a c^2}\\ &=\frac {\sqrt {\tan ^{-1}(a x)}}{a c^2}+\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{2 a c^2}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 43, normalized size = 0.91 \begin {gather*} \frac {2 \sqrt {\text {ArcTan}(a x)}+\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{2 a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.20, size = 38, normalized size = 0.81

method result size
default \(\frac {\pi \FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+2 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }}{2 c^{2} a \sqrt {\pi }}\) \(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2*c*x^2+c)^2/arctan(a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/c^2/a/Pi^(1/2)*(Pi*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))+2*arctan(a*x)^(1/2)*Pi^(1/2))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{a^{4} x^{4} \sqrt {\operatorname {atan}{\left (a x \right )}} + 2 a^{2} x^{2} \sqrt {\operatorname {atan}{\left (a x \right )}} + \sqrt {\operatorname {atan}{\left (a x \right )}}}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(a**4*x**4*sqrt(atan(a*x)) + 2*a**2*x**2*sqrt(atan(a*x)) + sqrt(atan(a*x))), x)/c**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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