3.21 \(\int \frac{(a+b \cos ^{-1}(c x)) \log (h (f+g x)^m)}{\sqrt{1-c^2 x^2}} \, dx\)

Optimal. Leaf size=374 \[ -\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{i \cos ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{i \cos ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{i \cos ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{i \cos ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{g e^{i \cos ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{2 b c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{g e^{i \cos ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{2 b c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c} \]

[Out]

((-I/6)*m*(a + b*ArcCos[c*x])^3)/(b^2*c) + (m*(a + b*ArcCos[c*x])^2*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f - Sqrt[
c^2*f^2 - g^2])])/(2*b*c) + (m*(a + b*ArcCos[c*x])^2*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])
])/(2*b*c) - ((a + b*ArcCos[c*x])^2*Log[h*(f + g*x)^m])/(2*b*c) - (I*m*(a + b*ArcCos[c*x])*PolyLog[2, -((E^(I*
ArcCos[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2]))])/c - (I*m*(a + b*ArcCos[c*x])*PolyLog[2, -((E^(I*ArcCos[c*x])*g)
/(c*f + Sqrt[c^2*f^2 - g^2]))])/c + (b*m*PolyLog[3, -((E^(I*ArcCos[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2]))])/c +
(b*m*PolyLog[3, -((E^(I*ArcCos[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2]))])/c

________________________________________________________________________________________

Rubi [A]  time = 0.611272, antiderivative size = 374, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4642, 4780, 4742, 4520, 2190, 2531, 2282, 6589} \[ -\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{i \cos ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{PolyLog}\left (2,-\frac{g e^{i \cos ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{i \cos ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{b m \text{PolyLog}\left (3,-\frac{g e^{i \cos ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{g e^{i \cos ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{2 b c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{g e^{i \cos ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{2 b c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcCos[c*x])*Log[h*(f + g*x)^m])/Sqrt[1 - c^2*x^2],x]

[Out]

((-I/6)*m*(a + b*ArcCos[c*x])^3)/(b^2*c) + (m*(a + b*ArcCos[c*x])^2*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f - Sqrt[
c^2*f^2 - g^2])])/(2*b*c) + (m*(a + b*ArcCos[c*x])^2*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])
])/(2*b*c) - ((a + b*ArcCos[c*x])^2*Log[h*(f + g*x)^m])/(2*b*c) - (I*m*(a + b*ArcCos[c*x])*PolyLog[2, -((E^(I*
ArcCos[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2]))])/c - (I*m*(a + b*ArcCos[c*x])*PolyLog[2, -((E^(I*ArcCos[c*x])*g)
/(c*f + Sqrt[c^2*f^2 - g^2]))])/c + (b*m*PolyLog[3, -((E^(I*ArcCos[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2]))])/c +
(b*m*PolyLog[3, -((E^(I*ArcCos[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2]))])/c

Rule 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[c*x])
^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] && NeQ[n,
-1]

Rule 4780

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

Rule 4742

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

Rule 4520

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :>
Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (-Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2,
2] + b*E^(I*(c + d*x))), x], x] - Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] + b*E^(I*(c
+ d*x))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^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 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 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 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{\left (a+b \cos ^{-1}(c x)\right ) \log \left (h (f+g x)^m\right )}{\sqrt{1-c^2 x^2}} \, dx &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac{(g m) \int \frac{\left (a+b \cos ^{-1}(c x)\right )^2}{f+g x} \, dx}{2 b c}\\ &=-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{(g m) \operatorname{Subst}\left (\int \frac{(a+b x)^2 \sin (x)}{c f+g \cos (x)} \, dx,x,\cos ^{-1}(c x)\right )}{2 b c}\\ &=-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac{(i g m) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^2}{c f+e^{i x} g-\sqrt{c^2 f^2-g^2}} \, dx,x,\cos ^{-1}(c x)\right )}{2 b c}+\frac{(i g m) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^2}{c f+e^{i x} g+\sqrt{c^2 f^2-g^2}} \, dx,x,\cos ^{-1}(c x)\right )}{2 b c}\\ &=-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{2 b c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{2 b c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{m \operatorname{Subst}\left (\int (a+b x) \log \left (1+\frac{e^{i x} g}{c f-\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}-\frac{m \operatorname{Subst}\left (\int (a+b x) \log \left (1+\frac{e^{i x} g}{c f+\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}\\ &=-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{2 b c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{2 b c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{(i b m) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{e^{i x} g}{c f-\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}+\frac{(i b m) \operatorname{Subst}\left (\int \text{Li}_2\left (-\frac{e^{i x} g}{c f+\sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}\\ &=-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{2 b c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{2 b c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{(b m) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{g x}{-c f+\sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{c}+\frac{(b m) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{g x}{c f+\sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{c}\\ &=-\frac{i m \left (a+b \cos ^{-1}(c x)\right )^3}{6 b^2 c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{2 b c}+\frac{m \left (a+b \cos ^{-1}(c x)\right )^2 \log \left (1+\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{2 b c}-\frac{\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}-\frac{i m \left (a+b \cos ^{-1}(c x)\right ) \text{Li}_2\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{b m \text{Li}_3\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{c}+\frac{b m \text{Li}_3\left (-\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{c}\\ \end{align*}

Mathematica [B]  time = 5.56597, size = 1248, normalized size = 3.34 \[ \frac{-i b m \cos ^{-1}(c x)^3-3 i a m \cos ^{-1}(c x)^2+3 b m \log \left (\frac{e^{i \cos ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}+1\right ) \cos ^{-1}(c x)^2+3 b m \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f-\sqrt{c^2 f^2-g^2}\right )}{g}+1\right ) \cos ^{-1}(c x)^2+3 b m \log \left (\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}+1\right ) \cos ^{-1}(c x)^2+3 b m \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f+\sqrt{c^2 f^2-g^2}\right )}{g}+1\right ) \cos ^{-1}(c x)^2-3 b \log \left (h (f+g x)^m\right ) \cos ^{-1}(c x)^2-3 b m \log \left (\frac{\left (c f-\sqrt{c^2 f^2-g^2}\right ) \left (c x+i \sqrt{1-c^2 x^2}\right )}{g}+1\right ) \cos ^{-1}(c x)^2-3 b m \log \left (\frac{\left (c f+\sqrt{c^2 f^2-g^2}\right ) \left (c x+i \sqrt{1-c^2 x^2}\right )}{g}+1\right ) \cos ^{-1}(c x)^2+6 a m \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f-\sqrt{c^2 f^2-g^2}\right )}{g}+1\right ) \cos ^{-1}(c x)+12 b m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f-\sqrt{c^2 f^2-g^2}\right )}{g}+1\right ) \cos ^{-1}(c x)+6 a m \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f+\sqrt{c^2 f^2-g^2}\right )}{g}+1\right ) \cos ^{-1}(c x)-12 b m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f+\sqrt{c^2 f^2-g^2}\right )}{g}+1\right ) \cos ^{-1}(c x)-6 a m \log (f+g x) \cos ^{-1}(c x)-12 b m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \log \left (\frac{\left (c f-\sqrt{c^2 f^2-g^2}\right ) \left (c x+i \sqrt{1-c^2 x^2}\right )}{g}+1\right ) \cos ^{-1}(c x)+12 b m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \log \left (\frac{\left (c f+\sqrt{c^2 f^2-g^2}\right ) \left (c x+i \sqrt{1-c^2 x^2}\right )}{g}+1\right ) \cos ^{-1}(c x)-6 i b m \text{PolyLog}\left (2,\frac{e^{i \cos ^{-1}(c x)} g}{\sqrt{c^2 f^2-g^2}-c f}\right ) \cos ^{-1}(c x)-6 i b m \text{PolyLog}\left (2,-\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right ) \cos ^{-1}(c x)+24 i a m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(c f-g) \tan \left (\frac{1}{2} \cos ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2}}\right )+12 a m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f-\sqrt{c^2 f^2-g^2}\right )}{g}+1\right )-12 a m \sin ^{-1}\left (\frac{\sqrt{\frac{c f}{g}+1}}{\sqrt{2}}\right ) \log \left (\frac{e^{i \cos ^{-1}(c x)} \left (c f+\sqrt{c^2 f^2-g^2}\right )}{g}+1\right )-6 a m \sin ^{-1}(c x) \log (f+g x)+6 a \sin ^{-1}(c x) \log \left (h (f+g x)^m\right )-6 i a m \text{PolyLog}\left (2,\frac{e^{i \cos ^{-1}(c x)} \left (\sqrt{c^2 f^2-g^2}-c f\right )}{g}\right )-6 i a m \text{PolyLog}\left (2,-\frac{e^{i \cos ^{-1}(c x)} \left (c f+\sqrt{c^2 f^2-g^2}\right )}{g}\right )+6 b m \text{PolyLog}\left (3,\frac{e^{i \cos ^{-1}(c x)} g}{\sqrt{c^2 f^2-g^2}-c f}\right )+6 b m \text{PolyLog}\left (3,-\frac{e^{i \cos ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{6 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + b*ArcCos[c*x])*Log[h*(f + g*x)^m])/Sqrt[1 - c^2*x^2],x]

[Out]

((-3*I)*a*m*ArcCos[c*x]^2 - I*b*m*ArcCos[c*x]^3 + (24*I)*a*m*ArcSin[Sqrt[1 + (c*f)/g]/Sqrt[2]]*ArcTan[((c*f -
g)*Tan[ArcCos[c*x]/2])/Sqrt[c^2*f^2 - g^2]] + 3*b*m*ArcCos[c*x]^2*Log[1 + (E^(I*ArcCos[c*x])*g)/(c*f - Sqrt[c^
2*f^2 - g^2])] + 6*a*m*ArcCos[c*x]*Log[1 + (E^(I*ArcCos[c*x])*(c*f - Sqrt[c^2*f^2 - g^2]))/g] + 3*b*m*ArcCos[c
*x]^2*Log[1 + (E^(I*ArcCos[c*x])*(c*f - Sqrt[c^2*f^2 - g^2]))/g] + 12*a*m*ArcSin[Sqrt[1 + (c*f)/g]/Sqrt[2]]*Lo
g[1 + (E^(I*ArcCos[c*x])*(c*f - Sqrt[c^2*f^2 - g^2]))/g] + 12*b*m*ArcCos[c*x]*ArcSin[Sqrt[1 + (c*f)/g]/Sqrt[2]
]*Log[1 + (E^(I*ArcCos[c*x])*(c*f - Sqrt[c^2*f^2 - g^2]))/g] + 3*b*m*ArcCos[c*x]^2*Log[1 + (E^(I*ArcCos[c*x])*
g)/(c*f + Sqrt[c^2*f^2 - g^2])] + 6*a*m*ArcCos[c*x]*Log[1 + (E^(I*ArcCos[c*x])*(c*f + Sqrt[c^2*f^2 - g^2]))/g]
+ 3*b*m*ArcCos[c*x]^2*Log[1 + (E^(I*ArcCos[c*x])*(c*f + Sqrt[c^2*f^2 - g^2]))/g] - 12*a*m*ArcSin[Sqrt[1 + (c*
f)/g]/Sqrt[2]]*Log[1 + (E^(I*ArcCos[c*x])*(c*f + Sqrt[c^2*f^2 - g^2]))/g] - 12*b*m*ArcCos[c*x]*ArcSin[Sqrt[1 +
(c*f)/g]/Sqrt[2]]*Log[1 + (E^(I*ArcCos[c*x])*(c*f + Sqrt[c^2*f^2 - g^2]))/g] - 6*a*m*ArcCos[c*x]*Log[f + g*x]
- 6*a*m*ArcSin[c*x]*Log[f + g*x] - 3*b*ArcCos[c*x]^2*Log[h*(f + g*x)^m] + 6*a*ArcSin[c*x]*Log[h*(f + g*x)^m]
- 3*b*m*ArcCos[c*x]^2*Log[1 + ((c*f - Sqrt[c^2*f^2 - g^2])*(c*x + I*Sqrt[1 - c^2*x^2]))/g] - 12*b*m*ArcCos[c*x
]*ArcSin[Sqrt[1 + (c*f)/g]/Sqrt[2]]*Log[1 + ((c*f - Sqrt[c^2*f^2 - g^2])*(c*x + I*Sqrt[1 - c^2*x^2]))/g] - 3*b
*m*ArcCos[c*x]^2*Log[1 + ((c*f + Sqrt[c^2*f^2 - g^2])*(c*x + I*Sqrt[1 - c^2*x^2]))/g] + 12*b*m*ArcCos[c*x]*Arc
Sin[Sqrt[1 + (c*f)/g]/Sqrt[2]]*Log[1 + ((c*f + Sqrt[c^2*f^2 - g^2])*(c*x + I*Sqrt[1 - c^2*x^2]))/g] - (6*I)*b*
m*ArcCos[c*x]*PolyLog[2, (E^(I*ArcCos[c*x])*g)/(-(c*f) + Sqrt[c^2*f^2 - g^2])] - (6*I)*a*m*PolyLog[2, (E^(I*Ar
cCos[c*x])*(-(c*f) + Sqrt[c^2*f^2 - g^2]))/g] - (6*I)*b*m*ArcCos[c*x]*PolyLog[2, -((E^(I*ArcCos[c*x])*g)/(c*f
+ Sqrt[c^2*f^2 - g^2]))] - (6*I)*a*m*PolyLog[2, -((E^(I*ArcCos[c*x])*(c*f + Sqrt[c^2*f^2 - g^2]))/g)] + 6*b*m*
PolyLog[3, (E^(I*ArcCos[c*x])*g)/(-(c*f) + Sqrt[c^2*f^2 - g^2])] + 6*b*m*PolyLog[3, -((E^(I*ArcCos[c*x])*g)/(c
*f + Sqrt[c^2*f^2 - g^2]))])/(6*c)

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Maple [F]  time = 3.134, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arccos \left ( cx \right ) \right ) \ln \left ( h \left ( gx+f \right ) ^{m} \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c^{2} x^{2} + 1}{\left (b \arccos \left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c^2*x^2 + 1)*(b*arccos(c*x) + a)*log((g*x + f)^m*h)/(c^2*x^2 - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acos}{\left (c x \right )}\right ) \log{\left (h \left (f + g x\right )^{m} \right )}}{\sqrt{- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arccos \left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt{-c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccos(c*x) + a)*log((g*x + f)^m*h)/sqrt(-c^2*x^2 + 1), x)