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

Optimal. Leaf size=390 \[ \frac {i m \left (a+b \sin ^{-1}(c x)\right )^3}{6 b^2 c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{2 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{2 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {b m \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {b m \text {Li}_3\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c} \]

[Out]

1/6*I*m*(a+b*arcsin(c*x))^3/b^2/c+1/2*(a+b*arcsin(c*x))^2*ln(h*(g*x+f)^m)/b/c-1/2*m*(a+b*arcsin(c*x))^2*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-1/2*m*(a+b*arcsin(c*x))^2*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))*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))*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-b*m*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-b*m*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

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Rubi [A]  time = 0.62, antiderivative size = 390, normalized size of antiderivative = 1.00, 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 = {4641, 4779, 4741, 4519, 2190, 2531, 2282, 6589} \[ \frac {i m \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}-\frac {b m \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {b m \text {PolyLog}\left (3,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{2 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{2 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right ) \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx &=\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {(g m) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{f+g x} \, dx}{2 b c}\\ &=\frac {\left (a+b \sin ^{-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 \cos (x)}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3}{6 b^2 c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{c f-i e^{i x} g-\sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^2}{c f-i e^{i x} g+\sqrt {c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{2 b c}\\ &=\frac {i m \left (a+b \sin ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{2 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{2 b c}+\frac {\left (a+b \sin ^{-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 {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) \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 )^3}{6 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{2 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{2 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right ) \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 ) \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 b m) \operatorname {Subst}\left (\int \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 b m) \operatorname {Subst}\left (\int \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 )^3}{6 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{2 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{2 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right ) \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 ) \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 {(b m) \operatorname {Subst}\left (\int \frac {\text {Li}_2\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 {(b m) \operatorname {Subst}\left (\int \frac {\text {Li}_2\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 )^3}{6 b^2 c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{2 b c}-\frac {m \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{2 b c}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac {i m \left (a+b \sin ^{-1}(c x)\right ) \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 ) \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 {b m \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 {b m \text {Li}_3\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 [B]  time = 9.79, size = 2724, normalized size = 6.98 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(m*ArcSin[c*x]*(2*a + b*ArcSin[c*x])*Log[f + g*x])/(2*c) + (a*ArcSin[c*x]*(-(m*Log[f + g*x]) + Log[h*(f + g*x)
^m]))/c + (b*f*(-(m*Log[f + g*x]) + Log[h*(f + g*x)^m])*((-I)*ArcSin[c*x]*(Log[1 + (I*E^(I*ArcSin[c*x])*g)/(-(
c*f) + Sqrt[c^2*f^2 - g^2])] - Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])]) - PolyLog[2, ((-I
)*E^(I*ArcSin[c*x])*g)/(-(c*f) + Sqrt[c^2*f^2 - g^2])] + PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^
2 - g^2])]))/Sqrt[c^2*f^2 - g^2] + (a*g*m*(-1/2*(((3*I)/2)*Pi*ArcSin[c*x] - (I/2)*ArcSin[c*x]^2 + 2*Pi*Log[1 +
E^((-I)*ArcSin[c*x])] - Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 2*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] - 2*Pi*L
og[Cos[ArcSin[c*x]/2]] + Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - (2*I)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c*(
-c^(-1) - f/g)*g) + ((I/2)*Pi*ArcSin[c*x] - (I/2)*ArcSin[c*x]^2 + 2*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + Pi*Log[
1 - I*E^(I*ArcSin[c*x])] + 2*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 2*Pi*Log[Cos[ArcSin[c*x]/2]] - Pi*Log[
Sin[(Pi + 2*ArcSin[c*x])/4]] - (2*I)*PolyLog[2, I*E^(I*ArcSin[c*x])])/(2*c*(c^(-1) - f/g)*g) + (((-1/2*I)*ArcS
in[c*x]^2)/g + (ArcSin[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/g + (ArcSin[c*x]*Log
[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/g - (I*PolyLog[2, ((-I)*E^(I*ArcSin[c*x])*g)/(-(c*f
) + Sqrt[c^2*f^2 - g^2])])/g - (I*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/g)/(c^2*(-c
^(-1) - f/g)*(c^(-1) - f/g))))/c - a*c*g*m*(-1/2*(((3*I)/2)*Pi*ArcSin[c*x] - (I/2)*ArcSin[c*x]^2 + 2*Pi*Log[1
+ E^((-I)*ArcSin[c*x])] - Pi*Log[1 + I*E^(I*ArcSin[c*x])] + 2*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] - 2*Pi*
Log[Cos[ArcSin[c*x]/2]] + Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - (2*I)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^
3*(-c^(-1) - f/g)*g) + ((I/2)*Pi*ArcSin[c*x] - (I/2)*ArcSin[c*x]^2 + 2*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + Pi*L
og[1 - I*E^(I*ArcSin[c*x])] + 2*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 2*Pi*Log[Cos[ArcSin[c*x]/2]] - Pi*L
og[Sin[(Pi + 2*ArcSin[c*x])/4]] - (2*I)*PolyLog[2, I*E^(I*ArcSin[c*x])])/(2*c^3*(c^(-1) - f/g)*g) + (f^2*(((-1
/2*I)*ArcSin[c*x]^2)/g + (ArcSin[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/g + (ArcSi
n[c*x]*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/g - (I*PolyLog[2, ((-I)*E^(I*ArcSin[c*x])
*g)/(-(c*f) + Sqrt[c^2*f^2 - g^2])])/g - (I*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/g
))/(c^2*(-c^(-1) - f/g)*(c^(-1) - f/g)*g^2)) + (b*(-(m*Log[f + g*x]) + Log[h*(f + g*x)^m])*(ArcSin[c*x]^2 - 2*
c*f*((Pi*ArcTan[(g + c*f*Tan[ArcSin[c*x]/2])/Sqrt[c^2*f^2 - g^2]])/Sqrt[c^2*f^2 - g^2] + (2*ArcCos[-((c*f)/g)]
*ArcTanh[((c*f - g)*Cot[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]] + (Pi - 2*ArcSin[c*x])*ArcTanh[((c*f
+ g)*Tan[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]] + (ArcCos[-((c*f)/g)] + (2*I)*(ArcTanh[((c*f - g)*Co
t[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]] + ArcTanh[((c*f + g)*Tan[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^
2*f^2) + g^2]]))*Log[((1/2 + I/2)*Sqrt[-(c^2*f^2) + g^2])/(E^((I/2)*ArcSin[c*x])*Sqrt[g]*Sqrt[c*f + c*g*x])] +
(ArcCos[-((c*f)/g)] - (2*I)*ArcTanh[((c*f - g)*Cot[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]] - (2*I)*A
rcTanh[((c*f + g)*Tan[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]])*Log[((1/2 - I/2)*E^((I/2)*ArcSin[c*x])
*Sqrt[-(c^2*f^2) + g^2])/(Sqrt[g]*Sqrt[c*f + c*g*x])] - (ArcCos[-((c*f)/g)] + (2*I)*ArcTanh[((c*f - g)*Cot[(Pi
+ 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]])*Log[((c*f + g)*(-(c*f) + g - I*Sqrt[-(c^2*f^2) + g^2])*(1 + I*C
ot[(Pi + 2*ArcSin[c*x])/4]))/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Cot[(Pi + 2*ArcSin[c*x])/4]))] - (ArcCos[-((
c*f)/g)] - (2*I)*ArcTanh[((c*f - g)*Cot[(Pi + 2*ArcSin[c*x])/4])/Sqrt[-(c^2*f^2) + g^2]])*Log[((c*f + g)*(I*c*
f - I*g + Sqrt[-(c^2*f^2) + g^2])*(I + Cot[(Pi + 2*ArcSin[c*x])/4]))/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Cot[
(Pi + 2*ArcSin[c*x])/4]))] + I*(PolyLog[2, ((c*f - I*Sqrt[-(c^2*f^2) + g^2])*(c*f + g - Sqrt[-(c^2*f^2) + g^2]
*Cot[(Pi + 2*ArcSin[c*x])/4]))/(g*(c*f + g + Sqrt[-(c^2*f^2) + g^2]*Cot[(Pi + 2*ArcSin[c*x])/4]))] - PolyLog[2
, ((c*f + I*Sqrt[-(c^2*f^2) + g^2])*(c*f + g - Sqrt[-(c^2*f^2) + g^2]*Cot[(Pi + 2*ArcSin[c*x])/4]))/(g*(c*f +
g + Sqrt[-(c^2*f^2) + g^2]*Cot[(Pi + 2*ArcSin[c*x])/4]))]))/Sqrt[-(c^2*f^2) + g^2])))/(2*c) - (b*g*m*(((-1/3*I
)*ArcSin[c*x]^3)/g + (ArcSin[c*x]^2*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/g + (ArcSin[
c*x]^2*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/g - ((2*I)*ArcSin[c*x]*PolyLog[2, (I*E^(I
*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/g - ((2*I)*ArcSin[c*x]*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f
+ Sqrt[c^2*f^2 - g^2])])/g + (2*PolyLog[3, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/g + (2*PolyLo
g[3, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/g))/(2*c)

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fricas [F]  time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} x^{2} + 1} {\left (b \arcsin \left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))*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*arcsin(c*x) + a)*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 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {-c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 9.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right ) \ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {-c^{2} x^{2}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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