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

Optimal. Leaf size=370 \[ \frac {i \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right ) \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right ) \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {b \sqrt {1-c^2 x^2} \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {b \sqrt {1-c^2 x^2} \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]

[Out]

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

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Rubi [A]  time = 0.61, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4778, 4774, 3321, 2264, 2190, 2279, 2391} \[ \frac {b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right ) \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right ) \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/(Sqrt[c^2
*f^2 - g^2]*Sqrt[d - c^2*d*x^2]) - (I*Sqrt[1 - c^2*x^2]*(a + b*ArcCos[c*x])*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f
 + Sqrt[c^2*f^2 - g^2])])/(Sqrt[c^2*f^2 - g^2]*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*PolyLog[2, -((E^(I*
ArcCos[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2]))])/(Sqrt[c^2*f^2 - g^2]*Sqrt[d - c^2*d*x^2]) - (b*Sqrt[1 - c^2*x^2
]*PolyLog[2, -((E^(I*ArcCos[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2]))])/(Sqrt[c^2*f^2 - g^2]*Sqrt[d - c^2*d*x^2])

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
 Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 3321

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

Rule 4774

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> -Dist[(c^(m + 1)*Sqrt[d])^(-1), Subst[Int[(a + b*x)^n*(c*f + g*Cos[x])^m, x], x, ArcCos[c*x]], x] /; FreeQ[
{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rule 4778

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a +
b*ArcCos[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ
[p - 1/2] &&  !GtQ[d, 0]

Rubi steps

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

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Mathematica [B]  time = 2.10, size = 930, normalized size = 2.51 \[ \frac {\frac {a \log (f+g x)}{\sqrt {d}}-\frac {a \log \left (d \left (f x c^2+g\right )+\sqrt {d} \sqrt {g^2-c^2 f^2} \sqrt {d-c^2 d x^2}\right )}{\sqrt {d}}-\frac {b \sqrt {1-c^2 x^2} \left (2 \cos ^{-1}(c x) \tanh ^{-1}\left (\frac {(c f+g) \cot \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )-2 \cos ^{-1}\left (-\frac {c f}{g}\right ) \tanh ^{-1}\left (\frac {(g-c f) \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )+\left (\cos ^{-1}\left (-\frac {c f}{g}\right )-2 i \tanh ^{-1}\left (\frac {(c f+g) \cot \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )+2 i \tanh ^{-1}\left (\frac {(g-c f) \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )\right ) \log \left (\frac {e^{-\frac {1}{2} i \cos ^{-1}(c x)} \sqrt {g^2-c^2 f^2}}{\sqrt {2} \sqrt {g} \sqrt {c (f+g x)}}\right )+\left (\cos ^{-1}\left (-\frac {c f}{g}\right )+2 i \left (\tanh ^{-1}\left (\frac {(c f+g) \cot \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )-\tanh ^{-1}\left (\frac {(g-c f) \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )\right )\right ) \log \left (\frac {e^{\frac {1}{2} i \cos ^{-1}(c x)} \sqrt {g^2-c^2 f^2}}{\sqrt {2} \sqrt {g} \sqrt {c (f+g x)}}\right )-\left (\cos ^{-1}\left (-\frac {c f}{g}\right )-2 i \tanh ^{-1}\left (\frac {(g-c f) \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )\right ) \log \left (\frac {(c f+g) \left (-i c f+i g+\sqrt {g^2-c^2 f^2}\right ) \left (\tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )-i\right )}{g \left (c f+g+\sqrt {g^2-c^2 f^2} \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )\right )}\right )-\left (\cos ^{-1}\left (-\frac {c f}{g}\right )+2 i \tanh ^{-1}\left (\frac {(g-c f) \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )}{\sqrt {g^2-c^2 f^2}}\right )\right ) \log \left (\frac {(c f+g) \left (i c f-i g+\sqrt {g^2-c^2 f^2}\right ) \left (\tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )+i\right )}{g \left (c f+g+\sqrt {g^2-c^2 f^2} \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )\right )}\right )+i \left (\text {Li}_2\left (\frac {\left (c f-i \sqrt {g^2-c^2 f^2}\right ) \left (c f+g-\sqrt {g^2-c^2 f^2} \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )\right )}{g \left (c f+g+\sqrt {g^2-c^2 f^2} \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )\right )}\right )-\text {Li}_2\left (\frac {\left (c f+i \sqrt {g^2-c^2 f^2}\right ) \left (c f+g-\sqrt {g^2-c^2 f^2} \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )\right )}{g \left (c f+g+\sqrt {g^2-c^2 f^2} \tan \left (\frac {1}{2} \cos ^{-1}(c x)\right )\right )}\right )\right )\right )}{\sqrt {d-c^2 d x^2}}}{\sqrt {g^2-c^2 f^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a*Log[f + g*x])/Sqrt[d] - (a*Log[d*(g + c^2*f*x) + Sqrt[d]*Sqrt[-(c^2*f^2) + g^2]*Sqrt[d - c^2*d*x^2]])/Sqrt
[d] - (b*Sqrt[1 - c^2*x^2]*(2*ArcCos[c*x]*ArcTanh[((c*f + g)*Cot[ArcCos[c*x]/2])/Sqrt[-(c^2*f^2) + g^2]] - 2*A
rcCos[-((c*f)/g)]*ArcTanh[((-(c*f) + g)*Tan[ArcCos[c*x]/2])/Sqrt[-(c^2*f^2) + g^2]] + (ArcCos[-((c*f)/g)] - (2
*I)*ArcTanh[((c*f + g)*Cot[ArcCos[c*x]/2])/Sqrt[-(c^2*f^2) + g^2]] + (2*I)*ArcTanh[((-(c*f) + g)*Tan[ArcCos[c*
x]/2])/Sqrt[-(c^2*f^2) + g^2]])*Log[Sqrt[-(c^2*f^2) + g^2]/(Sqrt[2]*E^((I/2)*ArcCos[c*x])*Sqrt[g]*Sqrt[c*(f +
g*x)])] + (ArcCos[-((c*f)/g)] + (2*I)*(ArcTanh[((c*f + g)*Cot[ArcCos[c*x]/2])/Sqrt[-(c^2*f^2) + g^2]] - ArcTan
h[((-(c*f) + g)*Tan[ArcCos[c*x]/2])/Sqrt[-(c^2*f^2) + g^2]]))*Log[(E^((I/2)*ArcCos[c*x])*Sqrt[-(c^2*f^2) + g^2
])/(Sqrt[2]*Sqrt[g]*Sqrt[c*(f + g*x)])] - (ArcCos[-((c*f)/g)] - (2*I)*ArcTanh[((-(c*f) + g)*Tan[ArcCos[c*x]/2]
)/Sqrt[-(c^2*f^2) + g^2]])*Log[((c*f + g)*((-I)*c*f + I*g + Sqrt[-(c^2*f^2) + g^2])*(-I + Tan[ArcCos[c*x]/2]))
/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Tan[ArcCos[c*x]/2]))] - (ArcCos[-((c*f)/g)] + (2*I)*ArcTanh[((-(c*f) + g
)*Tan[ArcCos[c*x]/2])/Sqrt[-(c^2*f^2) + g^2]])*Log[((c*f + g)*(I*c*f - I*g + Sqrt[-(c^2*f^2) + g^2])*(I + Tan[
ArcCos[c*x]/2]))/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Tan[ArcCos[c*x]/2]))] + I*(PolyLog[2, ((c*f - I*Sqrt[-(c
^2*f^2) + g^2])*(c*f + g - Sqrt[-(c^2*f^2) + g^2]*Tan[ArcCos[c*x]/2]))/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Ta
n[ArcCos[c*x]/2]))] - PolyLog[2, ((c*f + I*Sqrt[-(c^2*f^2) + g^2])*(c*f + g - Sqrt[-(c^2*f^2) + g^2]*Tan[ArcCo
s[c*x]/2]))/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Tan[ArcCos[c*x]/2]))])))/Sqrt[d - c^2*d*x^2])/Sqrt[-(c^2*f^2)
 + g^2]

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arccos \left (c x\right ) + a\right )}}{c^{2} d g x^{3} + c^{2} d f x^{2} - d g x - d f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*arccos(c*x) + a)/(c^2*d*g*x^3 + c^2*d*f*x^2 - d*g*x - d*f), x)

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giac [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((a+b*arccos(c*x))/(g*x+f)/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^3+t_nostep)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [A]  time = 0.65, size = 487, normalized size = 1.32 \[ -\frac {a \ln \left (\frac {-\frac {2 d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}+\frac {2 c^{2} d f \left (x +\frac {f}{g}\right )}{g}+2 \sqrt {-\frac {d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}}\, \sqrt {-c^{2} d \left (x +\frac {f}{g}\right )^{2}+\frac {2 c^{2} d f \left (x +\frac {f}{g}\right )}{g}-\frac {d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}}}{x +\frac {f}{g}}\right )}{g \sqrt {-\frac {d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}}}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \ln \left (\frac {-\left (c x +i \sqrt {-c^{2} x^{2}+1}\right ) g -c f +\sqrt {c^{2} f^{2}-g^{2}}}{-c f +\sqrt {c^{2} f^{2}-g^{2}}}\right ) \arccos \left (c x \right )-i \ln \left (\frac {\left (c x +i \sqrt {-c^{2} x^{2}+1}\right ) g +c f +\sqrt {c^{2} f^{2}-g^{2}}}{c f +\sqrt {c^{2} f^{2}-g^{2}}}\right ) \arccos \left (c x \right )+\dilog \left (\frac {-\left (c x +i \sqrt {-c^{2} x^{2}+1}\right ) g -c f +\sqrt {c^{2} f^{2}-g^{2}}}{-c f +\sqrt {c^{2} f^{2}-g^{2}}}\right )-\dilog \left (\frac {\left (c x +i \sqrt {-c^{2} x^{2}+1}\right ) g +c f +\sqrt {c^{2} f^{2}-g^{2}}}{c f +\sqrt {c^{2} f^{2}-g^{2}}}\right )\right )}{\sqrt {c^{2} f^{2}-g^{2}}\, d \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(c*x))/(g*x+f)/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-a/g/(-d*(c^2*f^2-g^2)/g^2)^(1/2)*ln((-2*d*(c^2*f^2-g^2)/g^2+2*c^2*d*f/g*(x+f/g)+2*(-d*(c^2*f^2-g^2)/g^2)^(1/2
)*(-c^2*d*(x+f/g)^2+2*c^2*d*f/g*(x+f/g)-d*(c^2*f^2-g^2)/g^2)^(1/2))/(x+f/g))-b*(-d*(c^2*x^2-1))^(1/2)/(c^2*f^2
-g^2)^(1/2)*(-c^2*x^2+1)^(1/2)*(I*ln((-(c*x+I*(-c^2*x^2+1)^(1/2))*g-c*f+(c^2*f^2-g^2)^(1/2))/(-c*f+(c^2*f^2-g^
2)^(1/2)))*arccos(c*x)-I*ln(((c*x+I*(-c^2*x^2+1)^(1/2))*g+c*f+(c^2*f^2-g^2)^(1/2))/(c*f+(c^2*f^2-g^2)^(1/2)))*
arccos(c*x)+dilog((-(c*x+I*(-c^2*x^2+1)^(1/2))*g-c*f+(c^2*f^2-g^2)^(1/2))/(-c*f+(c^2*f^2-g^2)^(1/2)))-dilog(((
c*x+I*(-c^2*x^2+1)^(1/2))*g+c*f+(c^2*f^2-g^2)^(1/2))/(c*f+(c^2*f^2-g^2)^(1/2))))/d/(c^2*x^2-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))/(g*x+f)/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {acos}{\left (c x \right )}}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (f + g x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acos(c*x))/(g*x+f)/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*acos(c*x))/(sqrt(-d*(c*x - 1)*(c*x + 1))*(f + g*x)), x)

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