\(\int \frac {a+b \log (c (4 d-d g x^2)^p)}{\sqrt {4-g x^2}} \, dx\) [684]

Optimal result

Integrand size = 33, antiderivative size = 192 \[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\frac {\arcsin \left (\frac {\sqrt {g} x}{2}\right ) \left (a+b \log \left (c \left (4 d-d g x^2\right )^p\right )\right )}{\sqrt {g}}-\frac {i b p \log ^2\left (\frac {1}{2} i \sqrt {g} x+\sqrt {1-\frac {g x^2}{4}}\right )}{\sqrt {g}}+\frac {2 i b p \log \left (\frac {1}{2} i \sqrt {g} x+\sqrt {1-\frac {g x^2}{4}}\right ) \log \left (1+\left (\frac {1}{2} i \sqrt {g} x+\sqrt {1-\frac {g x^2}{4}}\right )^2\right )}{\sqrt {g}}+\frac {i b p \operatorname {PolyLog}\left (2,-\frac {1}{4} \left (i \sqrt {g} x+\sqrt {4-g x^2}\right )^2\right )}{\sqrt {g}} \] Output:

arcsin(1/2*g^(1/2)*x)*(a+b*ln(c*(-d*g*x^2+4*d)^p))/g^(1/2)-I*b*p*ln(1/2*I* 
g^(1/2)*x+1/2*(-g*x^2+4)^(1/2))^2/g^(1/2)+2*I*b*p*ln(1/2*I*g^(1/2)*x+1/2*( 
-g*x^2+4)^(1/2))*ln(1+(1/2*I*g^(1/2)*x+1/2*(-g*x^2+4)^(1/2))^2)/g^(1/2)+I* 
b*p*polylog(2,-1/4*(I*g^(1/2)*x+(-g*x^2+4)^(1/2))^2)/g^(1/2)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.97 \[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\frac {a \arcsin \left (\frac {\sqrt {g} x}{2}\right )-2 i b p \pi \arcsin \left (\frac {\sqrt {g} x}{2}\right )+i b p \arcsin \left (\frac {\sqrt {g} x}{2}\right )^2-4 b p \pi \log \left (1+e^{-i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )-b p \pi \log \left (1-i e^{i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )-2 b p \arcsin \left (\frac {\sqrt {g} x}{2}\right ) \log \left (1-i e^{i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )+b p \pi \log \left (1+i e^{i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )-2 b p \arcsin \left (\frac {\sqrt {g} x}{2}\right ) \log \left (1+i e^{i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )+b \arcsin \left (\frac {\sqrt {g} x}{2}\right ) \log \left (c \left (d \left (4-g x^2\right )\right )^p\right )+4 b p \pi \log \left (\cos \left (\frac {1}{2} \arcsin \left (\frac {\sqrt {g} x}{2}\right )\right )\right )-b p \pi \log \left (-\cos \left (\frac {1}{4} \left (\pi +2 \arcsin \left (\frac {\sqrt {g} x}{2}\right )\right )\right )\right )+b p \pi \log \left (\sin \left (\frac {1}{4} \left (\pi +2 \arcsin \left (\frac {\sqrt {g} x}{2}\right )\right )\right )\right )+2 i b p \operatorname {PolyLog}\left (2,-i e^{i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )+2 i b p \operatorname {PolyLog}\left (2,i e^{i \arcsin \left (\frac {\sqrt {g} x}{2}\right )}\right )}{\sqrt {g}} \] Input:

Integrate[(a + b*Log[c*(4*d - d*g*x^2)^p])/Sqrt[4 - g*x^2],x]
 

Output:

(a*ArcSin[(Sqrt[g]*x)/2] - (2*I)*b*p*Pi*ArcSin[(Sqrt[g]*x)/2] + I*b*p*ArcS 
in[(Sqrt[g]*x)/2]^2 - 4*b*p*Pi*Log[1 + E^((-I)*ArcSin[(Sqrt[g]*x)/2])] - b 
*p*Pi*Log[1 - I*E^(I*ArcSin[(Sqrt[g]*x)/2])] - 2*b*p*ArcSin[(Sqrt[g]*x)/2] 
*Log[1 - I*E^(I*ArcSin[(Sqrt[g]*x)/2])] + b*p*Pi*Log[1 + I*E^(I*ArcSin[(Sq 
rt[g]*x)/2])] - 2*b*p*ArcSin[(Sqrt[g]*x)/2]*Log[1 + I*E^(I*ArcSin[(Sqrt[g] 
*x)/2])] + b*ArcSin[(Sqrt[g]*x)/2]*Log[c*(d*(4 - g*x^2))^p] + 4*b*p*Pi*Log 
[Cos[ArcSin[(Sqrt[g]*x)/2]/2]] - b*p*Pi*Log[-Cos[(Pi + 2*ArcSin[(Sqrt[g]*x 
)/2])/4]] + b*p*Pi*Log[Sin[(Pi + 2*ArcSin[(Sqrt[g]*x)/2])/4]] + (2*I)*b*p* 
PolyLog[2, (-I)*E^(I*ArcSin[(Sqrt[g]*x)/2])] + (2*I)*b*p*PolyLog[2, I*E^(I 
*ArcSin[(Sqrt[g]*x)/2])])/Sqrt[g]
 

Rubi [F]

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.

\(\displaystyle \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx\)

\(\Big \downarrow \) 2923

\(\displaystyle \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}}dx\)

Input:

Int[(a + b*Log[c*(4*d - d*g*x^2)^p])/Sqrt[4 - g*x^2],x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2923

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + 
 (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Unintegrable[(f + g*x^s)^r*(a + b*Log 
[c*(d + e*x^n)^p])^q, x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q, r, s}, x]
 

Maple [F]

\[\int \frac {a +b \ln \left (c \left (-d g \,x^{2}+4 d \right )^{p}\right )}{\sqrt {-g \,x^{2}+4}}d x\]

Input:

int((a+b*ln(c*(-d*g*x^2+4*d)^p))/(-g*x^2+4)^(1/2),x)
 

Output:

int((a+b*ln(c*(-d*g*x^2+4*d)^p))/(-g*x^2+4)^(1/2),x)
 

Fricas [F]

\[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\int { \frac {b \log \left ({\left (-d g x^{2} + 4 \, d\right )}^{p} c\right ) + a}{\sqrt {-g x^{2} + 4}} \,d x } \] Input:

integrate((a+b*log(c*(-d*g*x^2+4*d)^p))/(-g*x^2+4)^(1/2),x, algorithm="fri 
cas")
 

Output:

integral(-(sqrt(-g*x^2 + 4)*b*log((-d*g*x^2 + 4*d)^p*c) + sqrt(-g*x^2 + 4) 
*a)/(g*x^2 - 4), x)
 

Sympy [F]

\[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\int \frac {a + b \log {\left (c \left (- d g x^{2} + 4 d\right )^{p} \right )}}{\sqrt {- g x^{2} + 4}}\, dx \] Input:

integrate((a+b*ln(c*(-d*g*x**2+4*d)**p))/(-g*x**2+4)**(1/2),x)
 

Output:

Integral((a + b*log(c*(-d*g*x**2 + 4*d)**p))/sqrt(-g*x**2 + 4), x)
 

Maxima [F]

\[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\int { \frac {b \log \left ({\left (-d g x^{2} + 4 \, d\right )}^{p} c\right ) + a}{\sqrt {-g x^{2} + 4}} \,d x } \] Input:

integrate((a+b*log(c*(-d*g*x^2+4*d)^p))/(-g*x^2+4)^(1/2),x, algorithm="max 
ima")
 

Output:

b*integrate((p*log(d) + log((-g*x^2 + 4)^p) + log(c))/sqrt(-g*x^2 + 4), x) 
 + a*arcsin(1/2*sqrt(g)*x)/sqrt(g)
 

Giac [F]

\[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\int { \frac {b \log \left ({\left (-d g x^{2} + 4 \, d\right )}^{p} c\right ) + a}{\sqrt {-g x^{2} + 4}} \,d x } \] Input:

integrate((a+b*log(c*(-d*g*x^2+4*d)^p))/(-g*x^2+4)^(1/2),x, algorithm="gia 
c")
 

Output:

integrate((b*log((-d*g*x^2 + 4*d)^p*c) + a)/sqrt(-g*x^2 + 4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (4\,d-d\,g\,x^2\right )}^p\right )}{\sqrt {4-g\,x^2}} \,d x \] Input:

int((a + b*log(c*(4*d - d*g*x^2)^p))/(4 - g*x^2)^(1/2),x)
 

Output:

int((a + b*log(c*(4*d - d*g*x^2)^p))/(4 - g*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \frac {a+b \log \left (c \left (4 d-d g x^2\right )^p\right )}{\sqrt {4-g x^2}} \, dx=\frac {\sqrt {g}\, \mathit {asin} \left (\frac {\sqrt {g}\, x}{2}\right ) a +\left (\int \frac {\mathrm {log}\left (\left (-d g \,x^{2}+4 d \right )^{p} c \right )}{\sqrt {-g \,x^{2}+4}}d x \right ) b g}{g} \] Input:

int((a+b*log(c*(-d*g*x^2+4*d)^p))/(-g*x^2+4)^(1/2),x)
 

Output:

(sqrt(g)*asin((sqrt(g)*x)/2)*a + int(log(( - d*g*x**2 + 4*d)**p*c)/sqrt( - 
 g*x**2 + 4),x)*b*g)/g