3.10.91 \(\int \frac {x^2}{(c+a^2 c x^2)^2 \text {ArcTan}(a x)^{3/2}} \, dx\) [991]

Optimal. Leaf size=60 \[ -\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\text {ArcTan}(a x)}}+\frac {2 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{a^3 c^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5062, 5090, 4491, 12, 3386, 3432} \begin {gather*} \frac {2 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{a^3 c^2}-\frac {2 x^2}{a c^2 \left (a^2 x^2+1\right ) \sqrt {\text {ArcTan}(a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2)/(a*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) + (2*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(a^3*
c^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3386

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 3432

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

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 5062

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[
(f*x)^m*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^(p + 1)/(b*c*d*(p + 1))), x] - Dist[f*(m/(b*c*(p + 1))), Int[
(f*x)^(m - 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e
, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[p, -1]

Rule 5090

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 \tan ^{-1}(a x)^{3/2}} \, dx &=-\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {4 \int \frac {x}{\left (c+a^2 c x^2\right )^2 \sqrt {\tan ^{-1}(a x)}} \, dx}{a}\\ &=-\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {4 \text {Subst}\left (\int \frac {\cos (x) \sin (x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^2}\\ &=-\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {4 \text {Subst}\left (\int \frac {\sin (2 x)}{2 \sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^2}\\ &=-\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {2 \text {Subst}\left (\int \frac {\sin (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{a^3 c^2}\\ &=-\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {4 \text {Subst}\left (\int \sin \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{a^3 c^2}\\ &=-\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\tan ^{-1}(a x)}}+\frac {2 \sqrt {\pi } S\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{a^3 c^2}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 60, normalized size = 1.00 \begin {gather*} -\frac {2 x^2}{a c^2 \left (1+a^2 x^2\right ) \sqrt {\text {ArcTan}(a x)}}+\frac {2 \sqrt {\pi } S\left (\frac {2 \sqrt {\text {ArcTan}(a x)}}{\sqrt {\pi }}\right )}{a^3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^2)/(a*c^2*(1 + a^2*x^2)*Sqrt[ArcTan[a*x]]) + (2*Sqrt[Pi]*FresnelS[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]])/(a^3*
c^2)

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Maple [A]
time = 0.30, size = 46, normalized size = 0.77

method	result	size
default	\(\frac {2 \sqrt {\arctan \left (a x \right )}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right )+\cos \left (2 \arctan \left (a x \right )\right )-1}{c^{2} a^{3} \sqrt {\arctan \left (a x \right )}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/c^2/a^3*(2*arctan(a*x)^(1/2)*Pi^(1/2)*FresnelS(2*arctan(a*x)^(1/2)/Pi^(1/2))+cos(2*arctan(a*x))-1)/arctan(a*
x)^(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(x^2/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),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(x^2/(a^2*c*x^2+c)^2/arctan(a*x)^(3/2),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 {x^{2}}{a^{4} x^{4} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + 2 a^{2} x^{2} \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )} + \operatorname {atan}^{\frac {3}{2}}{\left (a x \right )}}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a**4*x**4*atan(a*x)**(3/2) + 2*a**2*x**2*atan(a*x)**(3/2) + atan(a*x)**(3/2)), x)/c**2

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{{\mathrm {atan}\left (a\,x\right )}^{3/2}\,{\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(x^2/(atan(a*x)^(3/2)*(c + a^2*c*x^2)^2),x)

[Out]

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

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