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

Optimal. Leaf size=256 \[ \frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {a^2 c x^2+c}} \]

[Out]

-2*I*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)^2*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+2*I*arctan(a*x)
*polylog(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-2*I*arctan(a*x)*polylog(2,I
*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-2*polylog(3,-I*(1+I*a*x)/(a^2*x^2+1)^(1/
2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+2*polylog(3,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^
2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4890, 4888, 4181, 2531, 2282, 6589} \[ \frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {2 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[1 + a^2*x^2]*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2)/(a*Sqrt[c + a^2*c*x^2]) + ((2*I)*Sqrt[1 + a
^2*x^2]*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - ((2*I)*Sqrt[1 + a^2*x^2]*Arc
Tan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - (2*Sqrt[1 + a^2*x^2]*PolyLog[3, (-I)*E^(I*
ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) + (2*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2
*c*x^2])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4888

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subst
[Int[(a + b*x)^p*Sec[x], x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0] &
& GtQ[d, 0]

Rule 4890

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*
d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {\tan ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}-\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (2 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (2 i \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {2 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 140, normalized size = 0.55 \[ \frac {2 \sqrt {c \left (a^2 x^2+1\right )} \left (i \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )-i \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )-\text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )+\text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )-i \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2\right )}{a c \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c*(1 + a^2*x^2)]*((-I)*ArcTan[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2 + I*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*Ar
cTan[a*x])] - I*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])] - PolyLog[3, (-I)*E^(I*ArcTan[a*x])] + PolyLog[3,
I*E^(I*ArcTan[a*x])]))/(a*c*Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{2}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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maple [F]  time = 0.58, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x \right )^{2}}{\sqrt {a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \left (a x\right )^{2}}{\sqrt {a^{2} c x^{2} + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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