Integrand size = 25, antiderivative size = 237 \[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\frac {i m \arcsin (c x)^2}{2 c}-\frac {m \arcsin (c x) \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {m \arcsin (c x) \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}+\frac {i m \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c} \] Output:
1/2*I*m*arcsin(c*x)^2/c-m*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/ (c*f-(c^2*f^2-g^2)^(1/2)))/c-m*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2 ))*g/(c*f+(c^2*f^2-g^2)^(1/2)))/c+arcsin(c*x)*ln(h*(g*x+f)^m)/c+I*m*polylo g(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*polylo g(2,I*(I*c*x+(-c^2*x^2+1)^(1/2))*g/(c*f+(c^2*f^2-g^2)^(1/2)))/c
Time = 0.00 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.04 \[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\frac {i m \arcsin (c x)^2}{2 c}-\frac {m \arcsin (c x) \log \left (1-\frac {i c e^{i \arcsin (c x)} g}{c^2 f-c \sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {m \arcsin (c x) \log \left (1-\frac {i c e^{i \arcsin (c x)} g}{c^2 f+c \sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}+\frac {i m \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {i m \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c} \] Input:
Integrate[Log[h*(f + g*x)^m]/Sqrt[1 - c^2*x^2],x]
Output:
((I/2)*m*ArcSin[c*x]^2)/c - (m*ArcSin[c*x]*Log[1 - (I*c*E^(I*ArcSin[c*x])* g)/(c^2*f - c*Sqrt[c^2*f^2 - g^2])])/c - (m*ArcSin[c*x]*Log[1 - (I*c*E^(I* ArcSin[c*x])*g)/(c^2*f + c*Sqrt[c^2*f^2 - g^2])])/c + (ArcSin[c*x]*Log[h*( f + g*x)^m])/c + (I*m*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f ^2 - g^2])])/c + (I*m*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f ^2 - g^2])])/c
Time = 0.75 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2851, 27, 5240, 5030, 2620, 2715, 2838}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx\) |
\(\Big \downarrow \) 2851 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-g m \int \frac {\arcsin (c x)}{c (f+g x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-\frac {g m \int \frac {\arcsin (c x)}{f+g x}dx}{c}\) |
\(\Big \downarrow \) 5240 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-\frac {g m \int \frac {\sqrt {1-c^2 x^2} \arcsin (c x)}{c f+c g x}d\arcsin (c x)}{c}\) |
\(\Big \downarrow \) 5030 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-\frac {g m \left (\int \frac {e^{i \arcsin (c x)} \arcsin (c x)}{c f-i e^{i \arcsin (c x)} g-\sqrt {c^2 f^2-g^2}}d\arcsin (c x)+\int \frac {e^{i \arcsin (c x)} \arcsin (c x)}{c f-i e^{i \arcsin (c x)} g+\sqrt {c^2 f^2-g^2}}d\arcsin (c x)-\frac {i \arcsin (c x)^2}{2 g}\right )}{c}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-\frac {g m \left (-\frac {\int \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )d\arcsin (c x)}{g}-\frac {\int \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )d\arcsin (c x)}{g}+\frac {\arcsin (c x) \log \left (1-\frac {i g e^{i \arcsin (c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{g}+\frac {\arcsin (c x) \log \left (1-\frac {i g e^{i \arcsin (c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{g}-\frac {i \arcsin (c x)^2}{2 g}\right )}{c}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-\frac {g m \left (\frac {i \int e^{-i \arcsin (c x)} \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )de^{i \arcsin (c x)}}{g}+\frac {i \int e^{-i \arcsin (c x)} \log \left (1-\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )de^{i \arcsin (c x)}}{g}+\frac {\arcsin (c x) \log \left (1-\frac {i g e^{i \arcsin (c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{g}+\frac {\arcsin (c x) \log \left (1-\frac {i g e^{i \arcsin (c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{g}-\frac {i \arcsin (c x)^2}{2 g}\right )}{c}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\arcsin (c x) \log \left (h (f+g x)^m\right )}{c}-\frac {g m \left (-\frac {i \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{g}-\frac {i \operatorname {PolyLog}\left (2,\frac {i e^{i \arcsin (c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{g}+\frac {\arcsin (c x) \log \left (1-\frac {i g e^{i \arcsin (c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{g}+\frac {\arcsin (c x) \log \left (1-\frac {i g e^{i \arcsin (c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{g}-\frac {i \arcsin (c x)^2}{2 g}\right )}{c}\) |
Input:
Int[Log[h*(f + g*x)^m]/Sqrt[1 - c^2*x^2],x]
Output:
(ArcSin[c*x]*Log[h*(f + g*x)^m])/c - (g*m*(((-1/2*I)*ArcSin[c*x]^2)/g + (A rcSin[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
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[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]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[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]
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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/Sqrt[(f_) + (g_.)* (x_)^2], x_Symbol] :> With[{u = IntHide[1/Sqrt[f + g*x^2], x]}, Simp[u*(a + b*Log[c*(d + e*x)^n]), x] - Simp[b*e*n Int[SimplifyIntegrand[u/(d + e*x) , x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && GtQ[f, 0]
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]
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]
\[\int \frac {\ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {-c^{2} x^{2}+1}}d x\]
Input:
int(ln(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x)
Output:
int(ln(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x)
\[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\int { \frac {\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {-c^{2} x^{2} + 1}} \,d x } \] Input:
integrate(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="fricas")
Output:
integral(-sqrt(-c^2*x^2 + 1)*log((g*x + f)^m*h)/(c^2*x^2 - 1), x)
\[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\int \frac {\log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \] Input:
integrate(ln(h*(g*x+f)**m)/(-c**2*x**2+1)**(1/2),x)
Output:
Integral(log(h*(f + g*x)**m)/sqrt(-(c*x - 1)*(c*x + 1)), x)
\[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\int { \frac {\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {-c^{2} x^{2} + 1}} \,d x } \] Input:
integrate(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate(log((g*x + f)^m*h)/sqrt(-c^2*x^2 + 1), x)
\[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\int { \frac {\log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {-c^{2} x^{2} + 1}} \,d x } \] Input:
integrate(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x, algorithm="giac")
Output:
integrate(log((g*x + f)^m*h)/sqrt(-c^2*x^2 + 1), x)
Timed out. \[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )}{\sqrt {1-c^2\,x^2}} \,d x \] Input:
int(log(h*(f + g*x)^m)/(1 - c^2*x^2)^(1/2),x)
Output:
int(log(h*(f + g*x)^m)/(1 - c^2*x^2)^(1/2), x)
\[ \int \frac {\log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx=\int \frac {\mathrm {log}\left (\left (g x +f \right )^{m} h \right )}{\sqrt {-c^{2} x^{2}+1}}d x \] Input:
int(log(h*(g*x+f)^m)/(-c^2*x^2+1)^(1/2),x)
Output:
int(log((f + g*x)**m*h)/sqrt( - c**2*x**2 + 1),x)