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

Optimal. Leaf size=283 \[ \frac {i b \text {Li}_2\left (1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {i b \text {Li}_2\left (\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{2 c}-\frac {b^2 \text {Li}_3\left (\frac {2}{\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}+1}-1\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6681, 4850, 4988, 4884, 4994, 6610} \[ \frac {i b \text {PolyLog}\left (2,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {i b \text {PolyLog}\left (2,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right )}{2 c}-\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right )}{2 c}-\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4850

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTan[c*x])^(p - 1)*ArcTanh[1 - 2/(1 + I*c*x)])/(1 + c^2*x^2), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 4988

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (2-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}-\frac {i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1+i x}\right )}{1+x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {2 \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}-\frac {i b \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_2\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+\frac {i \sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 354, normalized size = 1.25 \[ -\frac {2 i b \text {Li}_2\left (-\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )-2 i b \text {Li}_2\left (\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )+4 \tanh ^{-1}\left (1-\frac {2 i}{-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+i}\right ) \left (a+b \tan ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2+b^2 \text {Li}_3\left (-\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right )-b^2 \text {Li}_3\left (\frac {\sqrt {1-c x}+i \sqrt {c x+1}}{\sqrt {1-c x}-i \sqrt {c x+1}}\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(4*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*ArcTanh[1 - (2*I)/(I - Sqrt[1 - c*x]/Sqrt[1 + c*x])] + (
2*I)*b*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, -((Sqrt[1 - c*x] + I*Sqrt[1 + c*x])/(Sqrt[1 - c*
x] - I*Sqrt[1 + c*x]))] - (2*I)*b*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, (Sqrt[1 - c*x] + I*Sq
rt[1 + c*x])/(Sqrt[1 - c*x] - I*Sqrt[1 + c*x])] + b^2*PolyLog[3, -((Sqrt[1 - c*x] + I*Sqrt[1 + c*x])/(Sqrt[1 -
c*x] - I*Sqrt[1 + c*x]))] - b^2*PolyLog[3, (Sqrt[1 - c*x] + I*Sqrt[1 + c*x])/(Sqrt[1 - c*x] - I*Sqrt[1 + c*x]
)])/c

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 2 \, a b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{2}}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))^2 + 2*a*b*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a^2)/(c^2
*x^2 - 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{2}}{c^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-(b*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^2/(c^2*x^2 - 1), x)

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maple [B]  time = 1.32, size = 837, normalized size = 2.96 \[ -\frac {a^{2} \ln \left (c x -1\right )}{2 c}+\frac {a^{2} \ln \left (c x +1\right )}{2 c}+\frac {b^{2} \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}+1\right )}{c}-\frac {i b^{2} \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{c}+\frac {b^{2} \polylog \left (3, -\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{2 c}-\frac {b^{2} \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1-\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}+\frac {2 i b^{2} \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, \frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {2 b^{2} \polylog \left (3, \frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {b^{2} \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1+\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}+\frac {2 i b^{2} \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {2 b^{2} \polylog \left (3, -\frac {1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}}{\sqrt {\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {2 a b \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{c}+\frac {2 a b \arctan \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}+1\right )}{c}-\frac {i a b \dilog \left (\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}+1\right )}{c}+\frac {i a b \dilog \left (1-\frac {\left (1+\frac {i \sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} - \frac {b^{2} \log \relax (2)^{2} \log \left (c x + 1\right ) - b^{2} \log \relax (2)^{2} \log \left (-c x + 1\right ) - 4 \, {\left (b^{2} \log \left (c x + 1\right ) - b^{2} \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right )^{2} - {\left (12 \, b^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + b^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} \log \relax (2)^{2} - 4 \, b^{2} \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) \log \left (c x + 1\right )}{c^{2} x^{2} - 1}\,{d x} + 4 \, b^{2} \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) \log \left (-c x + 1\right )}{c^{2} x^{2} - 1}\,{d x} + \frac {32 \, {\left ({\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {-c x + 1}, \sqrt {c x + 1}\right ) - c \int \frac {e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (c x + 1\right ) - e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{{\left (c^{2} x^{2} - 1\right )} {\left (c x + 1\right )} - {\left (c^{2} x^{2} - 1\right )} {\left (c x - 1\right )}}\,{d x}\right )} a b}{c}\right )} c}{32 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(log(c*x + 1)/c - log(c*x - 1)/c) - 1/32*(b^2*log(2)^2*log(c*x + 1) - b^2*log(2)^2*log(-c*x + 1) - 4*(
b^2*log(c*x + 1) - b^2*log(-c*x + 1))*arctan2(sqrt(-c*x + 1), sqrt(c*x + 1))^2 - (b^2*(log(c*x + 1)/c - log(c*
x - 1)/c)*log(2)^2 - 64*b^2*integrate(1/16*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))*l
og(c*x + 1)/(c^2*x^2 - 1), x) + 64*b^2*integrate(1/16*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan(sqrt(-c*x + 1)/sqrt(
c*x + 1))*log(-c*x + 1)/(c^2*x^2 - 1), x) - 384*b^2*integrate(1/16*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))^2/(c^2
*x^2 - 1), x) - 1024*a*b*integrate(1/16*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))/(c^2*x^2 - 1), x))*c)/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((a+b*atan((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**2/(-c**2*x**2+1),x)

[Out]

Timed out

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