\(\int \frac {f+g x}{(d+e (F^{c (a+b x)})^n)^{3/2}} \, dx\) [49]

Optimal result

Integrand size = 25, antiderivative size = 335 \[ \int \frac {f+g x}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2}} \, dx=\frac {4 g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}+\frac {2 g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )^2}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}+\frac {2 (f+g x)}{b c d \sqrt {d+e \left (F^{c (a+b x)}\right )^n} n \log (F)}-\frac {2 (f+g x) \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b c d^{3/2} n \log (F)}-\frac {4 g \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}\right )}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}-\frac {2 g \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}\right )}{b^2 c^2 d^{3/2} n^2 \log ^2(F)} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.36 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2617, 2009}

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.

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

\(\Big \downarrow \) 2617

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

\(\Big \downarrow \) 2009

\(\displaystyle -g \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt {d}}\right )^2}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}+\frac {4 \text {arctanh}\left (\frac {\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}\right )}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}-\frac {4 \text {arctanh}\left (\frac {\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt {d}}\right )}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}+\frac {2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}\right )}{b^2 c^2 d^{3/2} n^2 \log ^2(F)}\right )-\frac {2 (f+g x) \text {arctanh}\left (\frac {\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt {d}}\right )}{b c d^{3/2} n \log (F)}+\frac {2 (f+g x)}{b c d n \log (F) \sqrt {e \left (F^{c (a+b x)}\right )^n+d}}\)

Input:

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

Output:

(2*(f + g*x))/(b*c*d*Sqrt[d + e*(F^(c*(a + b*x)))^n]*n*Log[F]) - (2*(f + g 
*x)*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]])/(b*c*d^(3/2)*n*Log[F 
]) - g*((-4*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]])/(b^2*c^2*d^( 
3/2)*n^2*Log[F]^2) - (2*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]]^2 
)/(b^2*c^2*d^(3/2)*n^2*Log[F]^2) + (4*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x))) 
^n]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*(F^(c*(a + b*x)))^n])]) 
/(b^2*c^2*d^(3/2)*n^2*Log[F]^2) + (2*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - 
 Sqrt[d + e*(F^(c*(a + b*x)))^n])])/(b^2*c^2*d^(3/2)*n^2*Log[F]^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2617

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + 
(d_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[(a + b*(F^(g*(e + f*x)))^ 
n)^p, x]}, Simp[(c + d*x)^m   u, x] - Simp[d*m   Int[(c + d*x)^(m - 1)*u, x 
], x]] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0] && LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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