\(\int \frac {x^2 (a+b \log (c (4 d+d g x^2)^p))}{\sqrt {4+g x^2}} \, dx\) [670]

Optimal result

Integrand size = 34, antiderivative size = 383 \[ \int \frac {x^2 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {2 b p}{g^{3/2} \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2}+\frac {a \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2}{8 g^{3/2}}-\frac {b p \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2}{8 g^{3/2}}+\frac {b \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2 \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )}{8 g^{3/2}}-\frac {2 \left (a+b \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )\right )}{g^{3/2} \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2}+\frac {2 b p \log \left (\sqrt {g} x+\sqrt {4+g x^2}\right )}{g^{3/2}}-\frac {2 \left (a+b \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )\right ) \log \left (\sqrt {g} x+\sqrt {4+g x^2}\right )}{g^{3/2}}-\frac {2 b p \log ^2\left (\sqrt {g} x+\sqrt {4+g x^2}\right )}{g^{3/2}}+\frac {4 b p \log \left (\sqrt {g} x+\sqrt {4+g x^2}\right ) \log \left (1+\frac {1}{4} \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2\right )}{g^{3/2}}+\frac {2 b p \operatorname {PolyLog}\left (2,-\frac {1}{4} \left (\sqrt {g} x+\sqrt {4+g x^2}\right )^2\right )}{g^{3/2}} \] Output:

2*b*p/g^(3/2)/(g^(1/2)*x+(g*x^2+4)^(1/2))^2+1/8*a*(g^(1/2)*x+(g*x^2+4)^(1/ 
2))^2/g^(3/2)-1/8*b*p*(g^(1/2)*x+(g*x^2+4)^(1/2))^2/g^(3/2)+1/8*b*(g^(1/2) 
*x+(g*x^2+4)^(1/2))^2*ln(c*(d*(g*x^2+4))^p)/g^(3/2)-2*(a+b*ln(c*(d*(g*x^2+ 
4))^p))/g^(3/2)/(g^(1/2)*x+(g*x^2+4)^(1/2))^2+2*b*p*ln(g^(1/2)*x+(g*x^2+4) 
^(1/2))/g^(3/2)-2*(a+b*ln(c*(d*(g*x^2+4))^p))*ln(g^(1/2)*x+(g*x^2+4)^(1/2) 
)/g^(3/2)-2*b*p*ln(g^(1/2)*x+(g*x^2+4)^(1/2))^2/g^(3/2)+4*b*p*ln(g^(1/2)*x 
+(g*x^2+4)^(1/2))*ln(1+1/4*(g^(1/2)*x+(g*x^2+4)^(1/2))^2)/g^(3/2)+2*b*p*po 
lylog(2,-1/4*(g^(1/2)*x+(g*x^2+4)^(1/2))^2)/g^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.75 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.43 \[ \int \frac {x^2 \left (a+b \log \left (c \left (4 d+d g x^2\right )^p\right )\right )}{\sqrt {4+g x^2}} \, dx=\frac {1}{2} \left (-\frac {4 b p x}{g \sqrt {4+g x^2}}-\frac {b p x^3}{\sqrt {4+g x^2}}+\frac {a x \sqrt {4+g x^2}}{g}-\frac {4 a \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}{g^{3/2}}+\frac {4 b p \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}{g^{3/2}}+\frac {8 i b p \pi \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}{g^{3/2}}+\frac {4 b p \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )^2}{g^{3/2}}-\frac {4 i b p \pi \log \left (1-i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )}{g^{3/2}}+\frac {8 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 )}{g^{3/2}}+\frac {4 i b p \pi \log \left (1+i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )}{g^{3/2}}+\frac {8 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 )}{g^{3/2}}-\frac {16 i b p \pi \log \left (1+e^{\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )}{g^{3/2}}+\frac {b x \sqrt {4+g x^2} \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )}{g}-\frac {4 b \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right ) \log \left (c \left (d \left (4+g x^2\right )\right )^p\right )}{g^{3/2}}-\frac {4 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 )}{g^{3/2}}+\frac {16 i b p \pi \log \left (\cosh \left (\frac {1}{2} \text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )\right )\right )}{g^{3/2}}+\frac {4 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 )}{g^{3/2}}-\frac {8 b p \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )}{g^{3/2}}-\frac {8 b p \operatorname {PolyLog}\left (2,i e^{-\text {arcsinh}\left (\frac {\sqrt {g} x}{2}\right )}\right )}{g^{3/2}}\right ) \] Input:

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

Output:

((-4*b*p*x)/(g*Sqrt[4 + g*x^2]) - (b*p*x^3)/Sqrt[4 + g*x^2] + (a*x*Sqrt[4 
+ g*x^2])/g - (4*a*ArcSinh[(Sqrt[g]*x)/2])/g^(3/2) + (4*b*p*ArcSinh[(Sqrt[ 
g]*x)/2])/g^(3/2) + ((8*I)*b*p*Pi*ArcSinh[(Sqrt[g]*x)/2])/g^(3/2) + (4*b*p 
*ArcSinh[(Sqrt[g]*x)/2]^2)/g^(3/2) - ((4*I)*b*p*Pi*Log[1 - I/E^ArcSinh[(Sq 
rt[g]*x)/2]])/g^(3/2) + (8*b*p*ArcSinh[(Sqrt[g]*x)/2]*Log[1 - I/E^ArcSinh[ 
(Sqrt[g]*x)/2]])/g^(3/2) + ((4*I)*b*p*Pi*Log[1 + I/E^ArcSinh[(Sqrt[g]*x)/2 
]])/g^(3/2) + (8*b*p*ArcSinh[(Sqrt[g]*x)/2]*Log[1 + I/E^ArcSinh[(Sqrt[g]*x 
)/2]])/g^(3/2) - ((16*I)*b*p*Pi*Log[1 + E^ArcSinh[(Sqrt[g]*x)/2]])/g^(3/2) 
 + (b*x*Sqrt[4 + g*x^2]*Log[c*(d*(4 + g*x^2))^p])/g - (4*b*ArcSinh[(Sqrt[g 
]*x)/2]*Log[c*(d*(4 + g*x^2))^p])/g^(3/2) - ((4*I)*b*p*Pi*Log[-Cos[(Pi + ( 
2*I)*ArcSinh[(Sqrt[g]*x)/2])/4]])/g^(3/2) + ((16*I)*b*p*Pi*Log[Cosh[ArcSin 
h[(Sqrt[g]*x)/2]/2]])/g^(3/2) + ((4*I)*b*p*Pi*Log[Sin[(Pi + (2*I)*ArcSinh[ 
(Sqrt[g]*x)/2])/4]])/g^(3/2) - (8*b*p*PolyLog[2, (-I)/E^ArcSinh[(Sqrt[g]*x 
)/2]])/g^(3/2) - (8*b*p*PolyLog[2, I/E^ArcSinh[(Sqrt[g]*x)/2]])/g^(3/2))/2
 

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.

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

\(\Big \downarrow \) 2929

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

Input:

Int[(x^2*(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 2929

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

1/2*a*(sqrt(g*x^2 + 4)*x/g - 4*arcsinh(1/2*sqrt(g)*x)/g^(3/2)) + b*integra 
te((p*x^2*log(g*x^2 + 4) + (p*log(d) + log(c))*x^2)/sqrt(g*x^2 + 4), x)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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