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

Optimal. Leaf size=47 \[ \frac {\sqrt {\tan ^{-1}(a x)}}{a^3 c^2}-\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{2 a^3 c^2} \]

[Out]

-1/2*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^3/c^2+arctan(a*x)^(1/2)/a^3/c^2

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Rubi [A]  time = 0.11, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4970, 3312, 3304, 3352} \[ \frac {\sqrt {\tan ^{-1}(a x)}}{a^3 c^2}-\frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{2 a^3 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[ArcTan[a*x]]/(a^3*c^2) - (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(2*a^3*c^2)

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 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 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]

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])

Rubi steps

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

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Mathematica [C]  time = 0.22, size = 122, normalized size = 2.60 \[ \frac {-4 \sqrt {\pi } \sqrt {\tan ^{-1}(a x)} C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )+16 \tan ^{-1}(a x)+i \sqrt {2} \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 i \tan ^{-1}(a x)\right )-i \sqrt {2} \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},2 i \tan ^{-1}(a x)\right )}{16 a^3 c^2 \sqrt {\tan ^{-1}(a x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(16*ArcTan[a*x] - 4*Sqrt[Pi]*Sqrt[ArcTan[a*x]]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + I*Sqrt[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)*ArcTan[a*x]])/(16
*a^3*c^2*Sqrt[ArcTan[a*x]])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.43, size = 38, normalized size = 0.81 \[ -\frac {\FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {\pi }}{2 a^{3} c^{2}}+\frac {\sqrt {\arctan \left (a x \right )}}{a^{3} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^3/c^2+arctan(a*x)^(1/2)/a^3/c^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2}{\sqrt {\mathrm {atan}\left (a\,x\right )}\,{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(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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