3.83 \(\int \frac {(a+b \sin ^{-1}(c x))^3 \log (h (f+g x)^m)}{\sqrt {1-c^2 x^2}} \, dx\)

Optimal. Leaf size=634 \[ -\frac {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_4\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_4\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^5}{20 b^2 c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {6 b^3 m \text {Li}_5\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {6 b^3 m \text {Li}_5\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c} \]

[Out]

1/20*I*m*(a+b*arcsin(c*x))^5/b^2/c+1/4*(a+b*arcsin(c*x))^4*ln(h*(g*x+f)^m)/b/c-1/4*m*(a+b*arcsin(c*x))^4*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-1/4*m*(a+b*arcsin(c*x))^4*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))^3*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))^3*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-3*b*m*(a+b*arcsin(c*x))^2*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-3
*b*m*(a+b*arcsin(c*x))^2*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-6*I*b^2*m*(a+b*
arcsin(c*x))*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-6*I*b^2*m*(a+b*arcsin(c*x))
*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+6*b^3*m*polylog(5,I*(I*c*x+(-c^2*x^2+1)
^(1/2))*g/(c*f-(c^2*f^2-g^2)^(1/2)))/c+6*b^3*m*polylog(5,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 = 0.87, antiderivative size = 634, normalized size of antiderivative = 1.00, number of steps used = 15, 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} \[ -\frac {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (4,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (4,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}-\frac {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {6 b^3 m \text {PolyLog}\left (5,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {6 b^3 m \text {PolyLog}\left (5,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcSin[c*x])^3*Log[h*(f + g*x)^m])/Sqrt[1 - c^2*x^2],x]

[Out]

((I/20)*m*(a + b*ArcSin[c*x])^5)/(b^2*c) - (m*(a + b*ArcSin[c*x])^4*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqr
t[c^2*f^2 - g^2])])/(4*b*c) - (m*(a + b*ArcSin[c*x])^4*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g
^2])])/(4*b*c) + ((a + b*ArcSin[c*x])^4*Log[h*(f + g*x)^m])/(4*b*c) + (I*m*(a + b*ArcSin[c*x])^3*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])^3*PolyLog[2, (I*E^(I*ArcSin[c
*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c - (3*b*m*(a + b*ArcSin[c*x])^2*PolyLog[3, (I*E^(I*ArcSin[c*x])*g)/(c*f
- Sqrt[c^2*f^2 - g^2])])/c - (3*b*m*(a + b*ArcSin[c*x])^2*PolyLog[3, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*
f^2 - g^2])])/c - ((6*I)*b^2*m*(a + b*ArcSin[c*x])*PolyLog[4, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^
2])])/c - ((6*I)*b^2*m*(a + b*ArcSin[c*x])*PolyLog[4, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/c
+ (6*b^3*m*PolyLog[5, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/c + (6*b^3*m*PolyLog[5, (I*E^(I*Ar
cSin[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 )^3 \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx &=\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}-\frac {(g m) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^4}{f+g x} \, dx}{4 b c}\\ &=\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {(a+b x)^4 \cos (x)}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^5}{20 b^2 c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^4}{c f-i e^{i x} g-\sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^4}{c f-i e^{i x} g+\sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {m \operatorname {Subst}\left (\int (a+b x)^3 \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)^3 \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 )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \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 )^3 \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 {(3 i b m) \operatorname {Subst}\left (\int (a+b x)^2 \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 {(3 i b m) \operatorname {Subst}\left (\int (a+b x)^2 \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 )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \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 )^3 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 (6 b^2 m\right ) \operatorname {Subst}\left (\int (a+b x) \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 (6 b^2 m\right ) \operatorname {Subst}\left (\int (a+b x) \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 )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \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 )^3 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \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 {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \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 {\left (6 i b^3 m\right ) \operatorname {Subst}\left (\int \text {Li}_4\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 (6 i b^3 m\right ) \operatorname {Subst}\left (\int \text {Li}_4\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 )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \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 )^3 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \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 {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \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 {\left (6 b^3 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4\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 (6 b^3 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_4\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 )^5}{20 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{4 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^4 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{4 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^4 \log \left (h (f+g x)^m\right )}{4 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3 \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 )^3 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {3 b m \left (a+b \sin ^{-1}(c x)\right )^2 \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 {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \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 {6 i b^2 m \left (a+b \sin ^{-1}(c x)\right ) \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 {6 b^3 m \text {Li}_5\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {6 b^3 m \text {Li}_5\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}\\ \end {align*}

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Mathematica [F]  time = 157.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \sin ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((a + b*ArcSin[c*x])^3*Log[h*(f + g*x)^m])/Sqrt[1 - c^2*x^2],x]

[Out]

Integrate[((a + b*ArcSin[c*x])^3*Log[h*(f + g*x)^m])/Sqrt[1 - c^2*x^2], x]

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b^{3} \arcsin \left (c x\right )^{3} + 3 \, a b^{2} \arcsin \left (c x\right )^{2} + 3 \, a^{2} b \arcsin \left (c x\right ) + a^{3}\right )} \sqrt {-c^{2} x^{2} + 1} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^3*log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-(b^3*arcsin(c*x)^3 + 3*a*b^2*arcsin(c*x)^2 + 3*a^2*b*arcsin(c*x) + a^3)*sqrt(-c^2*x^2 + 1)*log((g*x
+ f)^m*h)/(c^2*x^2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^3*log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)^3*log((g*x + f)^m*h)/sqrt(-c^2*x^2 + 1), x)

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maple [F]  time = 13.88, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{3} \ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {-c^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [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*arcsin(c*x))^3*log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^3}{\sqrt {1-c^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(h*(f + g*x)^m)*(a + b*asin(c*x))^3)/(1 - c^2*x^2)^(1/2),x)

[Out]

int((log(h*(f + g*x)^m)*(a + b*asin(c*x))^3)/(1 - c^2*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{3} \log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))**3*ln(h*(g*x+f)**m)/(-c**2*x**2+1)**(1/2),x)

[Out]

Integral((a + b*asin(c*x))**3*log(h*(f + g*x)**m)/sqrt(-(c*x - 1)*(c*x + 1)), x)

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