3.278 \(\int \frac {a+b \log (c (d+e x)^n)}{\sqrt {f-g x} \sqrt {f+g x}} \, dx\)

Optimal. Leaf size=510 \[ \frac {f \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {f-g x} \sqrt {f+g x}}+\frac {i b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \text {Li}_2\left (-\frac {e e^{i \sin ^{-1}\left (\frac {g x}{f}\right )} f}{i d g-\sqrt {e^2 f^2-d^2 g^2}}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}+\frac {i b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \text {Li}_2\left (-\frac {e e^{i \sin ^{-1}\left (\frac {g x}{f}\right )} f}{i d g+\sqrt {e^2 f^2-d^2 g^2}}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}-\frac {b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right ) \log \left (1+\frac {e f e^{i \sin ^{-1}\left (\frac {g x}{f}\right )}}{-\sqrt {e^2 f^2-d^2 g^2}+i d g}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}-\frac {b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right ) \log \left (1+\frac {e f e^{i \sin ^{-1}\left (\frac {g x}{f}\right )}}{\sqrt {e^2 f^2-d^2 g^2}+i d g}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}+\frac {i b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right )^2}{2 g \sqrt {f-g x} \sqrt {f+g x}} \]

[Out]

1/2*I*b*f*n*arcsin(g*x/f)^2*(1-g^2*x^2/f^2)^(1/2)/g/(-g*x+f)^(1/2)/(g*x+f)^(1/2)+f*arcsin(g*x/f)*(a+b*ln(c*(e*
x+d)^n))*(1-g^2*x^2/f^2)^(1/2)/g/(-g*x+f)^(1/2)/(g*x+f)^(1/2)-b*f*n*arcsin(g*x/f)*ln(1+e*(I*g*x/f+(1-g^2*x^2/f
^2)^(1/2))*f/(I*d*g-(-d^2*g^2+e^2*f^2)^(1/2)))*(1-g^2*x^2/f^2)^(1/2)/g/(-g*x+f)^(1/2)/(g*x+f)^(1/2)-b*f*n*arcs
in(g*x/f)*ln(1+e*(I*g*x/f+(1-g^2*x^2/f^2)^(1/2))*f/(I*d*g+(-d^2*g^2+e^2*f^2)^(1/2)))*(1-g^2*x^2/f^2)^(1/2)/g/(
-g*x+f)^(1/2)/(g*x+f)^(1/2)+I*b*f*n*polylog(2,-e*(I*g*x/f+(1-g^2*x^2/f^2)^(1/2))*f/(I*d*g-(-d^2*g^2+e^2*f^2)^(
1/2)))*(1-g^2*x^2/f^2)^(1/2)/g/(-g*x+f)^(1/2)/(g*x+f)^(1/2)+I*b*f*n*polylog(2,-e*(I*g*x/f+(1-g^2*x^2/f^2)^(1/2
))*f/(I*d*g+(-d^2*g^2+e^2*f^2)^(1/2)))*(1-g^2*x^2/f^2)^(1/2)/g/(-g*x+f)^(1/2)/(g*x+f)^(1/2)

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Rubi [A]  time = 0.64, antiderivative size = 510, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {2407, 216, 2404, 12, 4741, 4521, 2190, 2279, 2391} \[ \frac {i b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \text {PolyLog}\left (2,-\frac {e f e^{i \sin ^{-1}\left (\frac {g x}{f}\right )}}{-\sqrt {e^2 f^2-d^2 g^2}+i d g}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}+\frac {i b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \text {PolyLog}\left (2,-\frac {e f e^{i \sin ^{-1}\left (\frac {g x}{f}\right )}}{\sqrt {e^2 f^2-d^2 g^2}+i d g}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}+\frac {f \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g \sqrt {f-g x} \sqrt {f+g x}}-\frac {b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right ) \log \left (1+\frac {e f e^{i \sin ^{-1}\left (\frac {g x}{f}\right )}}{-\sqrt {e^2 f^2-d^2 g^2}+i d g}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}-\frac {b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right ) \log \left (1+\frac {e f e^{i \sin ^{-1}\left (\frac {g x}{f}\right )}}{\sqrt {e^2 f^2-d^2 g^2}+i d g}\right )}{g \sqrt {f-g x} \sqrt {f+g x}}+\frac {i b f n \sqrt {1-\frac {g^2 x^2}{f^2}} \sin ^{-1}\left (\frac {g x}{f}\right )^2}{2 g \sqrt {f-g x} \sqrt {f+g x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(Sqrt[f - g*x]*Sqrt[f + g*x]),x]

[Out]

((I/2)*b*f*n*Sqrt[1 - (g^2*x^2)/f^2]*ArcSin[(g*x)/f]^2)/(g*Sqrt[f - g*x]*Sqrt[f + g*x]) - (b*f*n*Sqrt[1 - (g^2
*x^2)/f^2]*ArcSin[(g*x)/f]*Log[1 + (e*E^(I*ArcSin[(g*x)/f])*f)/(I*d*g - Sqrt[e^2*f^2 - d^2*g^2])])/(g*Sqrt[f -
g*x]*Sqrt[f + g*x]) - (b*f*n*Sqrt[1 - (g^2*x^2)/f^2]*ArcSin[(g*x)/f]*Log[1 + (e*E^(I*ArcSin[(g*x)/f])*f)/(I*d
*g + Sqrt[e^2*f^2 - d^2*g^2])])/(g*Sqrt[f - g*x]*Sqrt[f + g*x]) + (f*Sqrt[1 - (g^2*x^2)/f^2]*ArcSin[(g*x)/f]*(
a + b*Log[c*(d + e*x)^n]))/(g*Sqrt[f - g*x]*Sqrt[f + g*x]) + (I*b*f*n*Sqrt[1 - (g^2*x^2)/f^2]*PolyLog[2, -((e*
E^(I*ArcSin[(g*x)/f])*f)/(I*d*g - Sqrt[e^2*f^2 - d^2*g^2]))])/(g*Sqrt[f - g*x]*Sqrt[f + g*x]) + (I*b*f*n*Sqrt[
1 - (g^2*x^2)/f^2]*PolyLog[2, -((e*E^(I*ArcSin[(g*x)/f])*f)/(I*d*g + Sqrt[e^2*f^2 - d^2*g^2]))])/(g*Sqrt[f - g
*x]*Sqrt[f + g*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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]

Rule 2391

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]

Rule 2404

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/Sqrt[(f_) + (g_.)*(x_)^2], x_Symbol] :> With[{u = Int
Hide[1/Sqrt[f + g*x^2], x]}, Simp[u*(a + b*Log[c*(d + e*x)^n]), x] - Dist[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]

Rule 2407

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/(Sqrt[(f1_) + (g1_.)*(x_)]*Sqrt[(f2_) + (g2_.)*(x_)])
, x_Symbol] :> Dist[Sqrt[1 + (g1*g2*x^2)/(f1*f2)]/(Sqrt[f1 + g1*x]*Sqrt[f2 + g2*x]), Int[(a + b*Log[c*(d + e*x
)^n])/Sqrt[1 + (g1*g2*x^2)/(f1*f2)], x], x] /; FreeQ[{a, b, c, d, e, f1, g1, f2, g2, n}, x] && EqQ[f2*g1 + f1*
g2, 0]

Rule 4521

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] + (Dist[I, Int[((e + f*x)^m*E^(I*(c + d*x)))/(I*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)))/(I*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] && NegQ[a^2 - b^2]

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]

Rubi steps

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

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Mathematica [B]  time = 4.78, size = 1077, normalized size = 2.11 \[ \frac {\tan ^{-1}\left (\frac {g x}{\sqrt {f-g x} \sqrt {f+g x}}\right ) \left (a-b n \log (d+e x)+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {i b n \sqrt {f-g x} \sqrt {\frac {f+g x}{f-g x}} \left (\log ^2\left (i-\sqrt {\frac {f+g x}{f-g x}}\right )+2 \log (d+e x) \log \left (i-\sqrt {\frac {f+g x}{f-g x}}\right )+2 \log \left (\frac {1}{2} \left (1-i \sqrt {\frac {f+g x}{f-g x}}\right )\right ) \log \left (i-\sqrt {\frac {f+g x}{f-g x}}\right )-2 \log \left (\frac {\sqrt {e f-d g}-\sqrt {e f+d g} \sqrt {\frac {f+g x}{f-g x}}}{\sqrt {e f-d g}-i \sqrt {e f+d g}}\right ) \log \left (i-\sqrt {\frac {f+g x}{f-g x}}\right )-2 \log \left (\frac {\sqrt {e f-d g}+\sqrt {e f+d g} \sqrt {\frac {f+g x}{f-g x}}}{\sqrt {e f-d g}+i \sqrt {e f+d g}}\right ) \log \left (i-\sqrt {\frac {f+g x}{f-g x}}\right )-\log ^2\left (\sqrt {\frac {f+g x}{f-g x}}+i\right )-2 \log (d+e x) \log \left (\sqrt {\frac {f+g x}{f-g x}}+i\right )-2 \log \left (\frac {1}{2} \left (i \sqrt {\frac {f+g x}{f-g x}}+1\right )\right ) \log \left (\sqrt {\frac {f+g x}{f-g x}}+i\right )+2 \log \left (\sqrt {\frac {f+g x}{f-g x}}+i\right ) \log \left (\frac {\sqrt {e f-d g}-\sqrt {e f+d g} \sqrt {\frac {f+g x}{f-g x}}}{\sqrt {e f-d g}+i \sqrt {e f+d g}}\right )+2 \log \left (\sqrt {\frac {f+g x}{f-g x}}+i\right ) \log \left (\frac {\sqrt {e f-d g}+\sqrt {e f+d g} \sqrt {\frac {f+g x}{f-g x}}}{\sqrt {e f-d g}-i \sqrt {e f+d g}}\right )-2 \text {Li}_2\left (\frac {1}{2}-\frac {1}{2} i \sqrt {\frac {f+g x}{f-g x}}\right )+2 \text {Li}_2\left (\frac {1}{2} i \sqrt {\frac {f+g x}{f-g x}}+\frac {1}{2}\right )+2 \text {Li}_2\left (\frac {\sqrt {e f+d g} \left (1-i \sqrt {\frac {f+g x}{f-g x}}\right )}{i \sqrt {e f-d g}+\sqrt {e f+d g}}\right )-2 \text {Li}_2\left (\frac {\sqrt {e f+d g} \left (i \sqrt {\frac {f+g x}{f-g x}}+1\right )}{\sqrt {e f+d g}-i \sqrt {e f-d g}}\right )-2 \text {Li}_2\left (\frac {\sqrt {e f+d g} \left (i \sqrt {\frac {f+g x}{f-g x}}+1\right )}{i \sqrt {e f-d g}+\sqrt {e f+d g}}\right )+2 \text {Li}_2\left (\frac {\sqrt {e f+d g} \left (\sqrt {\frac {f+g x}{f-g x}}+i\right )}{\sqrt {e f-d g}+i \sqrt {e f+d g}}\right )\right )}{2 g \sqrt {f+g x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(Sqrt[f - g*x]*Sqrt[f + g*x]),x]

[Out]

(ArcTan[(g*x)/(Sqrt[f - g*x]*Sqrt[f + g*x])]*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))/g - ((I/2)*b*n*Sqr
t[f - g*x]*Sqrt[(f + g*x)/(f - g*x)]*(2*Log[d + e*x]*Log[I - Sqrt[(f + g*x)/(f - g*x)]] + Log[I - Sqrt[(f + g*
x)/(f - g*x)]]^2 + 2*Log[I - Sqrt[(f + g*x)/(f - g*x)]]*Log[(1 - I*Sqrt[(f + g*x)/(f - g*x)])/2] - 2*Log[d + e
*x]*Log[I + Sqrt[(f + g*x)/(f - g*x)]] - 2*Log[(1 + I*Sqrt[(f + g*x)/(f - g*x)])/2]*Log[I + Sqrt[(f + g*x)/(f
- g*x)]] - Log[I + Sqrt[(f + g*x)/(f - g*x)]]^2 - 2*Log[I - Sqrt[(f + g*x)/(f - g*x)]]*Log[(Sqrt[e*f - d*g] -
Sqrt[e*f + d*g]*Sqrt[(f + g*x)/(f - g*x)])/(Sqrt[e*f - d*g] - I*Sqrt[e*f + d*g])] + 2*Log[I + Sqrt[(f + g*x)/(
f - g*x)]]*Log[(Sqrt[e*f - d*g] - Sqrt[e*f + d*g]*Sqrt[(f + g*x)/(f - g*x)])/(Sqrt[e*f - d*g] + I*Sqrt[e*f + d
*g])] + 2*Log[I + Sqrt[(f + g*x)/(f - g*x)]]*Log[(Sqrt[e*f - d*g] + Sqrt[e*f + d*g]*Sqrt[(f + g*x)/(f - g*x)])
/(Sqrt[e*f - d*g] - I*Sqrt[e*f + d*g])] - 2*Log[I - Sqrt[(f + g*x)/(f - g*x)]]*Log[(Sqrt[e*f - d*g] + Sqrt[e*f
+ d*g]*Sqrt[(f + g*x)/(f - g*x)])/(Sqrt[e*f - d*g] + I*Sqrt[e*f + d*g])] - 2*PolyLog[2, 1/2 - (I/2)*Sqrt[(f +
g*x)/(f - g*x)]] + 2*PolyLog[2, 1/2 + (I/2)*Sqrt[(f + g*x)/(f - g*x)]] + 2*PolyLog[2, (Sqrt[e*f + d*g]*(1 - I
*Sqrt[(f + g*x)/(f - g*x)]))/(I*Sqrt[e*f - d*g] + Sqrt[e*f + d*g])] - 2*PolyLog[2, (Sqrt[e*f + d*g]*(1 + I*Sqr
t[(f + g*x)/(f - g*x)]))/((-I)*Sqrt[e*f - d*g] + Sqrt[e*f + d*g])] - 2*PolyLog[2, (Sqrt[e*f + d*g]*(1 + I*Sqrt
[(f + g*x)/(f - g*x)]))/(I*Sqrt[e*f - d*g] + Sqrt[e*f + d*g])] + 2*PolyLog[2, (Sqrt[e*f + d*g]*(I + Sqrt[(f +
g*x)/(f - g*x)]))/(Sqrt[e*f - d*g] + I*Sqrt[e*f + d*g])]))/(g*Sqrt[f + g*x])

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {g x + f} \sqrt {-g x + f} b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt {g x + f} \sqrt {-g x + f} a}{g^{2} x^{2} - f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(-g*x+f)^(1/2)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

integral(-(sqrt(g*x + f)*sqrt(-g*x + f)*b*log((e*x + d)^n*c) + sqrt(g*x + f)*sqrt(-g*x + f)*a)/(g^2*x^2 - f^2)
, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(-g*x+f)^(1/2)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/(sqrt(g*x + f)*sqrt(-g*x + f)), x)

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maple [F]  time = 0.65, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \left (e x +d \right )^{n}\right )+a}{\sqrt {-g x +f}\, \sqrt {g x +f}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*ln(c*(e*x+d)^n)+a)/(-g*x+f)^(1/2)/(g*x+f)^(1/2),x)

[Out]

int((b*ln(c*(e*x+d)^n)+a)/(-g*x+f)^(1/2)/(g*x+f)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(-g*x+f)^(1/2)/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

b*integrate((log((e*x + d)^n) + log(c))/(sqrt(g*x + f)*sqrt(-g*x + f)), x) + a*arcsin(g*x/f)/g

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(d + e*x)^n))/((f + g*x)^(1/2)*(f - g*x)^(1/2)),x)

[Out]

int((a + b*log(c*(d + e*x)^n))/((f + g*x)^(1/2)*(f - g*x)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))/(-g*x+f)**(1/2)/(g*x+f)**(1/2),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))/(sqrt(f - g*x)*sqrt(f + g*x)), x)

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