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

Optimal. Leaf size=79 \[ \frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{8 a^2 c^2}-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {\sqrt {\tan ^{-1}(a x)}}{4 a^2 c^2} \]

[Out]

1/8*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2/c^2+1/4*arctan(a*x)^(1/2)/a^2/c^2-1/2*arctan(a*x)^(1/2
)/a^2/c^2/(a^2*x^2+1)

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Rubi [A]  time = 0.12, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4930, 4904, 3312, 3304, 3352} \[ \frac {\sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{8 a^2 c^2}-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (a^2 x^2+1\right )}+\frac {\sqrt {\tan ^{-1}(a x)}}{4 a^2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[ArcTan[a*x]]/(4*a^2*c^2) - Sqrt[ArcTan[a*x]]/(2*a^2*c^2*(1 + a^2*x^2)) + (Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTa
n[a*x]])/Sqrt[Pi]])/(8*a^2*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 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 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]

Rubi steps

\begin {align*} \int \frac {x \sqrt {\tan ^{-1}(a x)}}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2 \sqrt {\tan ^{-1}(a x)}} \, dx}{4 a}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\cos ^2(x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{4 a^2 c^2}\\ &=-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\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 )}{4 a^2 c^2}\\ &=\frac {\sqrt {\tan ^{-1}(a x)}}{4 a^2 c^2}-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\cos (2 x)}{\sqrt {x}} \, dx,x,\tan ^{-1}(a x)\right )}{8 a^2 c^2}\\ &=\frac {\sqrt {\tan ^{-1}(a x)}}{4 a^2 c^2}-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\operatorname {Subst}\left (\int \cos \left (2 x^2\right ) \, dx,x,\sqrt {\tan ^{-1}(a x)}\right )}{4 a^2 c^2}\\ &=\frac {\sqrt {\tan ^{-1}(a x)}}{4 a^2 c^2}-\frac {\sqrt {\tan ^{-1}(a x)}}{2 a^2 c^2 \left (1+a^2 x^2\right )}+\frac {\sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )}{8 a^2 c^2}\\ \end {align*}

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Mathematica [C]  time = 0.28, size = 136, normalized size = 1.72 \[ \frac {4 \sqrt {\pi } C\left (\frac {2 \sqrt {\tan ^{-1}(a x)}}{\sqrt {\pi }}\right )+\frac {\frac {16 \left (a^2 x^2-1\right ) \tan ^{-1}(a x)}{a^2 x^2+1}-i \sqrt {2} \sqrt {-i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},-2 i \tan ^{-1}(a x)\right )+i \sqrt {2} \sqrt {i \tan ^{-1}(a x)} \Gamma \left (\frac {1}{2},2 i \tan ^{-1}(a x)\right )}{\sqrt {\tan ^{-1}(a x)}}}{64 a^2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*Sqrt[Pi]*FresnelC[(2*Sqrt[ArcTan[a*x]])/Sqrt[Pi]] + ((16*(-1 + a^2*x^2)*ArcTan[a*x])/(1 + a^2*x^2) - 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)*Arc
Tan[a*x]])/Sqrt[ArcTan[a*x]])/(64*a^2*c^2)

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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*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^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]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.39, size = 46, normalized size = 0.58 \[ -\frac {\sqrt {\arctan \left (a x \right )}\, \cos \left (2 \arctan \left (a x \right )\right )}{4 a^{2} c^{2}}+\frac {\FresnelC \left (\frac {2 \sqrt {\arctan \left (a x \right )}}{\sqrt {\pi }}\right ) \sqrt {\pi }}{8 a^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/a^2/c^2*arctan(a*x)^(1/2)*cos(2*arctan(a*x))+1/8*FresnelC(2*arctan(a*x)^(1/2)/Pi^(1/2))*Pi^(1/2)/a^2/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*arctan(a*x)^(1/2)/(a^2*c*x^2+c)^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 = -0.01 \[ \int \frac {x\,\sqrt {\mathrm {atan}\left (a\,x\right )}}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x*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 \[ \frac {\int \frac {x \sqrt {\operatorname {atan}{\left (a x \right )}}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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