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

Optimal. Leaf size=98 \[ -\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c}+\frac {i b \text {Li}_2\left (-\frac {i \sqrt {c x+1}}{\sqrt {1-c x}}\right )}{2 c}-\frac {i b \text {Li}_2\left (\frac {i \sqrt {c x+1}}{\sqrt {1-c x}}\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {206, 6681, 4849, 2391} \[ \frac {i b \text {PolyLog}\left (2,-\frac {i \sqrt {c x+1}}{\sqrt {1-c x}}\right )}{2 c}-\frac {i b \text {PolyLog}\left (2,\frac {i \sqrt {c x+1}}{\sqrt {1-c x}}\right )}{2 c}-\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Log[Sqrt[1 - c*x]/Sqrt[1 + c*x]])/c) + ((I/2)*b*PolyLog[2, ((-I)*Sqrt[1 + c*x])/Sqrt[1 - c*x]])/c - ((I/2
)*b*PolyLog[2, (I*Sqrt[1 + c*x])/Sqrt[1 - c*x]])/c

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4849

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I/(c*
x)]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I/(c*x)]/x, x], x]) /; FreeQ[{a, b, c}, 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 {a+b \cot ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {a+b \cot ^{-1}(x)}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}\\ &=-\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {i}{x}\right )}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}+\frac {(i b) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {i}{x}\right )}{x} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}\\ &=-\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{c}+\frac {i b \text {Li}_2\left (-\frac {i \sqrt {1+c x}}{\sqrt {1-c x}}\right )}{2 c}-\frac {i b \text {Li}_2\left (\frac {i \sqrt {1+c x}}{\sqrt {1-c x}}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 93, normalized size = 0.95 \[ -\frac {a \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )-\frac {1}{2} i b \text {Li}_2\left (-\frac {i \sqrt {c x+1}}{\sqrt {1-c x}}\right )+\frac {1}{2} i b \text {Li}_2\left (\frac {i \sqrt {c x+1}}{\sqrt {1-c x}}\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Log[Sqrt[1 - c*x]/Sqrt[1 + c*x]] - (I/2)*b*PolyLog[2, ((-I)*Sqrt[1 + c*x])/Sqrt[1 - c*x]] + (I/2)*b*PolyL
og[2, (I*Sqrt[1 + c*x])/Sqrt[1 - c*x]])/c)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \operatorname {arccot}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.97, size = 259, normalized size = 2.64 \[ -\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}+\frac {b \,\mathrm {arccot}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1-\frac {\left (i+\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{c}-\frac {b \,\mathrm {arccot}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (\frac {\left (i+\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}+1\right )}{c}+\frac {i b \dilog \left (\frac {\left (i+\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}+1\right )}{2 c}-\frac {i b \dilog \left (1-\frac {\left (i+\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{\frac {-c x +1}{c x +1}+1}\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/c*ln(c*x-1)+1/2*a/c*ln(c*x+1)+b/c*arccot((-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))-b/c*arccot((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln((I+(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/(
(-c*x+1)/(c*x+1)+1)+1)+1/2*I*b/c*dilog((I+(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1)+1)-1/2*I*b/c*di
log(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 \[ \frac {1}{2} \, a {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} + \frac {{\left ({\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt {c x + 1}, \sqrt {-c x + 1}\right ) + c \int \frac {e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (c x + 1\right ) - e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{{\left (c^{2} x^{2} - 1\right )} {\left (c x + 1\right )} - {\left (c^{2} x^{2} - 1\right )} {\left (c x - 1\right )}}\,{d x}\right )} b}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(log(c*x + 1)/c - log(c*x - 1)/c) + 1/2*((log(c*x + 1) - log(-c*x + 1))*arctan2(sqrt(c*x + 1), sqrt(-c*x
+ 1)) + 2*c*integrate(1/2*(e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(c*x + 1) - e^(1/2*log(c*x + 1) + 1/2*
log(-c*x + 1))*log(-c*x + 1))/((c^2*x^2 - 1)*(c*x + 1) - (c^2*x^2 - 1)*(c*x - 1)), x))*b/c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {a+b\,\mathrm {acot}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )}{c^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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