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

Optimal. Leaf size=514 \[ \frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {2 i b^2 m \text {Li}_4\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 i b^2 m \text {Li}_4\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c} \]

[Out]

1/12*I*m*(a+b*arcsin(c*x))^4/b^2/c+1/3*(a+b*arcsin(c*x))^3*ln(h*(g*x+f)^m)/b/c-1/3*m*(a+b*arcsin(c*x))^3*ln(1-
I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f^2-g^2)^(1/2)))/b/c-1/3*m*(a+b*arcsin(c*x))^3*ln(1-I*(I*c*x+(-c^2*x^
2+1)^(1/2))*g/(c*f+(c^2*f^2-g^2)^(1/2)))/b/c+I*m*(a+b*arcsin(c*x))^2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/
(c*f-(c^2*f^2-g^2)^(1/2)))/c+I*m*(a+b*arcsin(c*x))^2*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2-g^
2)^(1/2)))/c-2*b*m*(a+b*arcsin(c*x))*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f^2-g^2)^(1/2)))/c-2*b
*m*(a+b*arcsin(c*x))*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2-g^2)^(1/2)))/c-2*I*b^2*m*polylog(4
,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f-(c^2*f^2-g^2)^(1/2)))/c-2*I*b^2*m*polylog(4,I*(I*c*x+(-c^2*x^2+1)^(1/2))*
g/(c*f+(c^2*f^2-g^2)^(1/2)))/c

________________________________________________________________________________________

Rubi [A]  time = 0.75, antiderivative size = 514, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4641, 4779, 4741, 4519, 2190, 2531, 6609, 2282, 6589} \[ \frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}-\frac {2 i b^2 m \text {PolyLog}\left (4,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 i b^2 m \text {PolyLog}\left (4,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/12)*m*(a + b*ArcSin[c*x])^4)/(b^2*c) - (m*(a + b*ArcSin[c*x])^3*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqr
t[c^2*f^2 - g^2])])/(3*b*c) - (m*(a + b*ArcSin[c*x])^3*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g
^2])])/(3*b*c) + ((a + b*ArcSin[c*x])^3*Log[h*(f + g*x)^m])/(3*b*c) + (I*m*(a + b*ArcSin[c*x])^2*PolyLog[2, (I
*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c + (I*m*(a + b*ArcSin[c*x])^2*PolyLog[2, (I*E^(I*ArcSin[c
*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c - (2*b*m*(a + b*ArcSin[c*x])*PolyLog[3, (I*E^(I*ArcSin[c*x])*g)/(c*f -
Sqrt[c^2*f^2 - g^2])])/c - (2*b*m*(a + b*ArcSin[c*x])*PolyLog[3, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2
- g^2])])/c - ((2*I)*b^2*m*PolyLog[4, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c - ((2*I)*b^2*m*P
olyLog[4, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c

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 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 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 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4641

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSin[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 4741

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

Rule 4779

Int[(Log[(h_.)*((f_.) + (g_.)*(x_))^(m_.)]*((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.))/Sqrt[(d_) + (e_.)*(x_)^2]
, x_Symbol] :> Simp[(Log[h*(f + g*x)^m]*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(g*m)/(b
*c*Sqrt[d]*(n + 1)), Int[(a + b*ArcSin[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 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]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx &=\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^3}{f+g x} \, dx}{3 b c}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {(a+b x)^3 \cos (x)}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{3 b c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^3}{c f-i e^{i x} g-\sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^3}{c f-i e^{i x} g+\sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{3 b c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {m \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1-\frac {i e^{i x} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}+\frac {m \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1-\frac {i e^{i x} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {(2 i b m) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (\frac {i e^{i x} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}-\frac {(2 i b m) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (\frac {i e^{i x} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (\frac {i e^{i x} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}+\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (\frac {i e^{i x} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {\left (2 i b^2 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i g x}{c f-\sqrt {c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c}-\frac {\left (2 i b^2 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i g x}{c f+\sqrt {c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^3 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 b m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 i b^2 m \text {Li}_4\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {2 i b^2 m \text {Li}_4\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 78.46, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} x^{2} + 1} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 11.74, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2} \ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {-c^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\frac {1}{3} \, b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{3} \log \relax (h) + a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \relax (h) + b^{2} c \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} \log \left ({\left (g x + f\right )}^{m}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x} + 2 \, a b c \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) \log \left ({\left (g x + f\right )}^{m}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x} + a^{2} c \int \frac {\log \left ({\left (g x + f\right )}^{m}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x} + a^{2} \arctan \left (c x, \sqrt {-c^{2} x^{2} + 1}\right ) \log \relax (h)}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*c*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x)*log(h) + 2*a*
b*c*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x)*log(h) + b^2*c*int
egrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log((g*x + f)^m)/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + 2*a*
b*c*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log((g*x + f)^m)/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x) +
a^2*c*integrate(log((g*x + f)^m)/(sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + a^2*arctan2(c*x, sqrt(-c^2*x^2 + 1))*lo
g(h))/c

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2} \log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))**2*log(h*(f + g*x)**m)/sqrt(-(c*x - 1)*(c*x + 1)), x)

________________________________________________________________________________________