\(\int \sqrt {d+e (F^{c (a+b x)})^n} (f+g x) \, dx\) [47]

Optimal result

Integrand size = 25, antiderivative size = 367 \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=-\frac {4 \sqrt {d+e \left (F^{c (a+b x)}\right )^n} g}{b^2 c^2 n^2 \log ^2(F)}+\frac {4 \sqrt {d} g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b^2 c^2 n^2 \log ^2(F)}+\frac {2 \sqrt {d} g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )^2}{b^2 c^2 n^2 \log ^2(F)}+\frac {2 \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x)}{b c n \log (F)}-\frac {2 \sqrt {d} (f+g x) \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b c n \log (F)}-\frac {4 \sqrt {d} g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}\right )}{b^2 c^2 n^2 \log ^2(F)}-\frac {2 \sqrt {d} g \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}\right )}{b^2 c^2 n^2 \log ^2(F)} \] Output:

-4*(d+e*(F^(c*(b*x+a)))^n)^(1/2)*g/b^2/c^2/n^2/ln(F)^2+4*d^(1/2)*g*arctanh 
((d+e*(F^(c*(b*x+a)))^n)^(1/2)/d^(1/2))/b^2/c^2/n^2/ln(F)^2+2*d^(1/2)*g*ar 
ctanh((d+e*(F^(c*(b*x+a)))^n)^(1/2)/d^(1/2))^2/b^2/c^2/n^2/ln(F)^2+2*(d+e* 
(F^(c*(b*x+a)))^n)^(1/2)*(g*x+f)/b/c/n/ln(F)-2*d^(1/2)*(g*x+f)*arctanh((d+ 
e*(F^(c*(b*x+a)))^n)^(1/2)/d^(1/2))/b/c/n/ln(F)-4*d^(1/2)*g*arctanh((d+e*( 
F^(c*(b*x+a)))^n)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(d+e*(F^(c*(b*x+a)) 
)^n)^(1/2)))/b^2/c^2/n^2/ln(F)^2-2*d^(1/2)*g*polylog(2,1-2*d^(1/2)/(d^(1/2 
)-(d+e*(F^(c*(b*x+a)))^n)^(1/2)))/b^2/c^2/n^2/ln(F)^2
 

Mathematica [F]

\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx \] Input:

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

Output:

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

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.

\(\displaystyle \int (f+g x) \sqrt {e \left (F^{c (a+b x)}\right )^n+d} \, dx\)

\(\Big \downarrow \) 2618

\(\displaystyle \int (f+g x) \sqrt {e \left (F^{a c+b c x}\right )^n+d}dx\)

\(\Big \downarrow \) 2619

\(\displaystyle \int (f+g x) \sqrt {e \left (F^{a c+b c x}\right )^n+d}dx\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2618

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m 
_.), x_Symbol] :> Int[(c + d*x)^m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] 
 /; FreeQ[{F, a, b, c, d, g, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v 
, x] && IntegerQ[m]
 

rule 2619

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + 
 (d_.)*(x_))^(m_.), x_Symbol] :> Unintegrable[(a + b*(F^(g*(e + f*x)))^n)^p 
*(c + d*x)^m, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x]
 

Maple [F]

\[\int \sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}\, \left (g x +f \right )d x\]

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\int \sqrt {d + e \left (F^{a c + b c x}\right )^{n}} \left (f + g x\right )\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\int { \sqrt {{\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d} {\left (g x + f\right )} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\int { \sqrt {{\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d} {\left (g x + f\right )} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\int \left (f+g\,x\right )\,\sqrt {d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x) \, dx=\left (\int \sqrt {f^{b c n x +a c n} e +d}d x \right ) f +\left (\int \sqrt {f^{b c n x +a c n} e +d}\, x d x \right ) g \] Input:

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

Output:

int(sqrt(f**(a*c*n + b*c*n*x)*e + d),x)*f + int(sqrt(f**(a*c*n + b*c*n*x)* 
e + d)*x,x)*g