\(\int \frac {a+b \log (c (4 d+d g x^2)^p)}{\sqrt {4+g x^2}} \, dx\) [671]

Optimal result

Integrand size = 31, antiderivative size = 174 \[ \int \frac {a+b \log \left (c \left (4 d+d g x^2\right )^p\right )}{\sqrt {4+g x^2}} \, dx=\frac {\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right ) \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {g}}+\frac {b p \log ^2\left (\frac {\sqrt {g} x}{2}+\sqrt {1+\frac {g x^2}{4}}\right )}{\sqrt {g}}-\frac {2 b p \log \left (\frac {\sqrt {g} x}{2}+\sqrt {1+\frac {g x^2}{4}}\right ) \log \left (1+\left (\frac {\sqrt {g} x}{2}+\sqrt {1+\frac {g x^2}{4}}\right )^2\right )}{\sqrt {g}}-\frac {b p \operatorname {PolyLog}\left (2,-\frac {1}{4} \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2\right )}{\sqrt {g}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.14 \[ \int \frac {a+b \log \left (c \left (4 d+d g x^2\right )^p\right )}{\sqrt {4+g x^2}} \, dx=\frac {a \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )-2 i b p \pi \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )-b p \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )^2+i b p \pi \log \left (1-i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )-2 b p \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right ) \log \left (1-i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )-i b p \pi \log \left (1+i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )-2 b p \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right ) \log \left (1+i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )+4 i b p \pi \log \left (1+e^{\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )+b \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right ) \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )+i b p \pi \log \left (-\cos \left (\frac {1}{4} \left (\pi +2 i \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )\right )\right )\right )-4 i b p \pi \log \left (\cosh \left (\frac {1}{2} \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )\right )\right )-i b p \pi \log \left (\sin \left (\frac {1}{4} \left (\pi +2 i \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )\right )\right )\right )+2 b p \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )+2 b p \operatorname {PolyLog}\left (2,i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )}{\sqrt {g}} \] Input:

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

Output:

(a*ArcSinh[(Sqrt[g]*x)/2] - (2*I)*b*p*Pi*ArcSinh[(Sqrt[g]*x)/2] - b*p*ArcS 
inh[(Sqrt[g]*x)/2]^2 + I*b*p*Pi*Log[1 - I/E^ArcSinh[(Sqrt[g]*x)/2]] - 2*b* 
p*ArcSinh[(Sqrt[g]*x)/2]*Log[1 - I/E^ArcSinh[(Sqrt[g]*x)/2]] - I*b*p*Pi*Lo 
g[1 + I/E^ArcSinh[(Sqrt[g]*x)/2]] - 2*b*p*ArcSinh[(Sqrt[g]*x)/2]*Log[1 + I 
/E^ArcSinh[(Sqrt[g]*x)/2]] + (4*I)*b*p*Pi*Log[1 + E^ArcSinh[(Sqrt[g]*x)/2] 
] + b*ArcSinh[(Sqrt[g]*x)/2]*Log[c*(d*(4 + g*x^2))^p] + I*b*p*Pi*Log[-Cos[ 
(Pi + (2*I)*ArcSinh[(Sqrt[g]*x)/2])/4]] - (4*I)*b*p*Pi*Log[Cosh[ArcSinh[(S 
qrt[g]*x)/2]/2]] - I*b*p*Pi*Log[Sin[(Pi + (2*I)*ArcSinh[(Sqrt[g]*x)/2])/4] 
] + 2*b*p*PolyLog[2, (-I)/E^ArcSinh[(Sqrt[g]*x)/2]] + 2*b*p*PolyLog[2, I/E 
^ArcSinh[(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.

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

\(\Big \downarrow \) 2923

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

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="frica 
s")
 

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="maxim 
a")
 

Output:

b*integrate((p*log(g*x^2 + 4) + p*log(d) + log(c))/sqrt(g*x^2 + 4), x) + a 
*arcsinh(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="giac" 
)
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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