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

Optimal. Leaf size=141 \[ \frac {i \left (a+b \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b c}-\frac {\log \left (1+e^{2 i \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {i b \text {Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {206, 6681, 4626, 3719, 2190, 2279, 2391} \[ \frac {i b \text {PolyLog}\left (2,-e^{2 i \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right )}{2 c}+\frac {i \left (a+b \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{2 b c}-\frac {\log \left (1+e^{2 i \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )}\right ) \left (a+b \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2)/(b*c) - ((a + b*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*Log
[1 + E^((2*I)*ArcCos[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c + ((I/2)*b*PolyLog[2, -E^((2*I)*ArcCos[Sqrt[1 - c*x]/Sq
rt[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 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 3719

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

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

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

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Mathematica [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {a+b \cos ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{1-c^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \arccos \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arccos(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 \[ \int -\frac {b \arccos \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a}{c^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 171, normalized size = 1.21 \[ -\frac {a \ln \left (c x -1\right )}{2 c}+\frac {a \ln \left (c x +1\right )}{2 c}+\frac {i b \arccos \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2}}{2 c}-\frac {b \arccos \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \ln \left (1+\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+i \sqrt {1-\frac {-c x +1}{c x +1}}\right )^{2}\right )}{c}+\frac {i b \polylog \left (2, -\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+i \sqrt {1-\frac {-c x +1}{c x +1}}\right )^{2}\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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