\(\int (d+e (F^{c (a+b x)})^n)^{3/2} (f+g x) \, dx\) [46]

Optimal result

Integrand size = 25, antiderivative size = 451 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=-\frac {16 d \sqrt {d+e \left (F^{c (a+b x)}\right )^n} g}{3 b^2 c^2 n^2 \log ^2(F)}-\frac {4 \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} g}{9 b^2 c^2 n^2 \log ^2(F)}+\frac {16 d^{3/2} g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{3 b^2 c^2 n^2 \log ^2(F)}+\frac {2 d^{3/2} g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )^2}{b^2 c^2 n^2 \log ^2(F)}+\frac {2 d \sqrt {d+e \left (F^{c (a+b x)}\right )^n} (f+g x)}{b c n \log (F)}+\frac {2 \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x)}{3 b c n \log (F)}-\frac {2 d^{3/2} (f+g x) \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b c n \log (F)}-\frac {4 d^{3/2} g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}\right )}{b^2 c^2 n^2 \log ^2(F)}-\frac {2 d^{3/2} g \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}\right )}{b^2 c^2 n^2 \log ^2(F)} \] Output:

-16/3*d*(d+e*(F^(c*(b*x+a)))^n)^(1/2)*g/b^2/c^2/n^2/ln(F)^2-4/9*(d+e*(F^(c 
*(b*x+a)))^n)^(3/2)*g/b^2/c^2/n^2/ln(F)^2+16/3*d^(3/2)*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^(3/2)*g*arctanh((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*(d+e*(F^(c*(b 
*x+a)))^n)^(1/2)*(g*x+f)/b/c/n/ln(F)+2/3*(d+e*(F^(c*(b*x+a)))^n)^(3/2)*(g* 
x+f)/b/c/n/ln(F)-2*d^(3/2)*(g*x+f)*arctanh((d+e*(F^(c*(b*x+a)))^n)^(1/2)/d 
^(1/2))/b/c/n/ln(F)-4*d^(3/2)*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^(3/2)*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]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx \] Input:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^(3/2)*(f + g*x),x]
 

Output:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^(3/2)*(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.

\(\displaystyle \int (f+g x) \left (e \left (F^{c (a+b x)}\right )^n+d\right )^{3/2} \, dx\)

\(\Big \downarrow \) 2618

\(\displaystyle \int (f+g x) \left (e \left (F^{a c+b c x}\right )^n+d\right )^{3/2}dx\)

\(\Big \downarrow \) 2619

\(\displaystyle \int (f+g x) \left (e \left (F^{a c+b c x}\right )^n+d\right )^{3/2}dx\)

Input:

Int[(d + e*(F^(c*(a + b*x)))^n)^(3/2)*(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]

\[\int {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {3}{2}} \left (g x +f \right )d x\]

Input:

int((d+e*(F^(c*(b*x+a)))^n)^(3/2)*(g*x+f),x)
 

Output:

int((d+e*(F^(c*(b*x+a)))^n)^(3/2)*(g*x+f),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^(3/2)*(g*x+f),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\text {Timed out} \] Input:

integrate((d+e*(F**((b*x+a)*c))**n)**(3/2)*(g*x+f),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\int { {\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )}^{\frac {3}{2}} {\left (g x + f\right )} \,d x } \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^(3/2)*(g*x+f),x, algorithm="maxima")
 

Output:

1/3*f*(3*d^(3/2)*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*((F^(b*c*n*x + a*c* 
n)*e + d)^(3/2) + 3*sqrt(F^(b*c*n*x + a*c*n)*e + d)*d)/(b*c*n*log(F))) + g 
*integrate((F^(b*c*n*x)*F^(a*c*n)*e*x + d*x)*sqrt(F^(b*c*n*x)*F^(a*c*n)*e 
+ d), x)
 

Giac [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\int { {\left ({\left (F^{{\left (b x + a\right )} c}\right )}^{n} e + d\right )}^{\frac {3}{2}} {\left (g x + f\right )} \,d x } \] Input:

integrate((d+e*(F^((b*x+a)*c))^n)^(3/2)*(g*x+f),x, algorithm="giac")
 

Output:

integrate(((F^((b*x + a)*c))^n*e + d)^(3/2)*(g*x + f), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\int \left (f+g\,x\right )\,{\left (d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n\right )}^{3/2} \,d x \] Input:

int((f + g*x)*(d + e*(F^(c*(a + b*x)))^n)^(3/2),x)
 

Output:

int((f + g*x)*(d + e*(F^(c*(a + b*x)))^n)^(3/2), x)
 

Reduce [F]

\[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} (f+g x) \, dx=\frac {6 f^{b c n x +a c n} \sqrt {f^{b c n x +a c n} e +d}\, \mathrm {log}\left (f \right ) b c e f n +6 f^{b c n x +a c n} \sqrt {f^{b c n x +a c n} e +d}\, \mathrm {log}\left (f \right ) b c e g n x -4 f^{b c n x +a c n} \sqrt {f^{b c n x +a c n} e +d}\, e g +24 \sqrt {f^{b c n x +a c n} e +d}\, \mathrm {log}\left (f \right ) b c d f n +24 \sqrt {f^{b c n x +a c n} e +d}\, \mathrm {log}\left (f \right ) b c d g n x -52 \sqrt {f^{b c n x +a c n} e +d}\, d g +9 \left (\int \frac {\sqrt {f^{b c n x +a c n} e +d}}{f^{b c n x +a c n} e +d}d x \right ) \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{2} f \,n^{2}-24 \left (\int \frac {\sqrt {f^{b c n x +a c n} e +d}}{f^{b c n x +a c n} e +d}d x \right ) \mathrm {log}\left (f \right ) b c \,d^{2} g n +9 \left (\int \frac {\sqrt {f^{b c n x +a c n} e +d}\, x}{f^{b c n x +a c n} e +d}d x \right ) \mathrm {log}\left (f \right )^{2} b^{2} c^{2} d^{2} g \,n^{2}}{9 \mathrm {log}\left (f \right )^{2} b^{2} c^{2} n^{2}} \] Input:

int((d+e*(F^((b*x+a)*c))^n)^(3/2)*(g*x+f),x)
 

Output:

(6*f**(a*c*n + b*c*n*x)*sqrt(f**(a*c*n + b*c*n*x)*e + d)*log(f)*b*c*e*f*n 
+ 6*f**(a*c*n + b*c*n*x)*sqrt(f**(a*c*n + b*c*n*x)*e + d)*log(f)*b*c*e*g*n 
*x - 4*f**(a*c*n + b*c*n*x)*sqrt(f**(a*c*n + b*c*n*x)*e + d)*e*g + 24*sqrt 
(f**(a*c*n + b*c*n*x)*e + d)*log(f)*b*c*d*f*n + 24*sqrt(f**(a*c*n + b*c*n* 
x)*e + d)*log(f)*b*c*d*g*n*x - 52*sqrt(f**(a*c*n + b*c*n*x)*e + d)*d*g + 9 
*int(sqrt(f**(a*c*n + b*c*n*x)*e + d)/(f**(a*c*n + b*c*n*x)*e + d),x)*log( 
f)**2*b**2*c**2*d**2*f*n**2 - 24*int(sqrt(f**(a*c*n + b*c*n*x)*e + d)/(f** 
(a*c*n + b*c*n*x)*e + d),x)*log(f)*b*c*d**2*g*n + 9*int((sqrt(f**(a*c*n + 
b*c*n*x)*e + d)*x)/(f**(a*c*n + b*c*n*x)*e + d),x)*log(f)**2*b**2*c**2*d** 
2*g*n**2)/(9*log(f)**2*b**2*c**2*n**2)