\(\int (d+e (F^{c (a+b x)})^n)^{3/2} \, dx\) [5]

Optimal result

Integrand size = 19, antiderivative size = 118 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} \, dx=\frac {2 d \sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{b c n \log (F)}+\frac {2 \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2}}{3 b c n \log (F)}-\frac {2 d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{b c n \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} \, dx=\frac {2 \sqrt {d+e \left (F^{c (a+b x)}\right )^n} \left (4 d+e \left (F^{c (a+b x)}\right )^n\right )-6 d^{3/2} \text {arctanh}\left (\frac {\sqrt {d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {d}}\right )}{3 b c n \log (F)} \] Input:

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

Output:

(2*Sqrt[d + e*(F^(c*(a + b*x)))^n]*(4*d + e*(F^(c*(a + b*x)))^n) - 6*d^(3/ 
2)*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]])/(3*b*c*n*Log[F])
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2720, 798, 60, 60, 73, 221}

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.

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

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int F^{-c (a+b x)} \left (e \left (F^{c (a+b x)}\right )^n+d\right )^{3/2}dF^{c (a+b x)}}{b c \log (F)}\)

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {d \left (\frac {2 d \int \frac {1}{\frac {F^{2 c (a+b x)}}{e}-\frac {d}{e}}d\sqrt {e \left (F^{c (a+b x)}\right )^n+d}}{e}+2 \sqrt {e \left (F^{c (a+b x)}\right )^n+d}\right )+\frac {2}{3} \left (e \left (F^{c (a+b x)}\right )^n+d\right )^{3/2}}{b c n \log (F)}\)

\(\Big \downarrow \) 221

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

Input:

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

Output:

((2*(d + e*(F^(c*(a + b*x)))^n)^(3/2))/3 + d*(2*Sqrt[d + e*(F^(c*(a + b*x) 
))^n] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*(F^(c*(a + b*x)))^n]/Sqrt[d]]))/(b*c* 
n*Log[F])
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69

method	result	size
derivativedivides	\(\frac {\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {3}{2}}}{3}+2 d \sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}-2 d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}}{\sqrt {d}}\right )}{\ln \left (F \right ) b c n}\)	\(82\)
default	\(\frac {\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {3}{2}}}{3}+2 d \sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}-2 d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d +e \left (F^{c \left (b x +a \right )}\right )^{n}}}{\sqrt {d}}\right )}{\ln \left (F \right ) b c n}\)	\(82\)
risch	\(\frac {2 \left (e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}+4 d \right ) \sqrt {d +e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}}{3 n c b \ln \left (F \right )}-\frac {2 d^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d +e \,{\mathrm e}^{n \ln \left ({\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}\right )}}}{\sqrt {d}}\right )}{\ln \left (F \right ) b c n}\)	\(100\)

Input:

int((d+e*(F^(c*(b*x+a)))^n)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.72 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} \, dx=\left [\frac {3 \, d^{\frac {3}{2}} \log \left (\frac {F^{b c n x + a c n} e - 2 \, \sqrt {F^{b c n x + a c n} e + d} \sqrt {d} + 2 \, d}{F^{b c n x + a c n}}\right ) + 2 \, {\left (F^{b c n x + a c n} e + 4 \, d\right )} \sqrt {F^{b c n x + a c n} e + d}}{3 \, b c n \log \left (F\right )}, \frac {2 \, {\left (3 \, \sqrt {-d} d \arctan \left (\frac {\sqrt {-d}}{\sqrt {F^{b c n x + a c n} e + d}}\right ) + {\left (F^{b c n x + a c n} e + 4 \, d\right )} \sqrt {F^{b c n x + a c n} e + d}\right )}}{3 \, b c n \log \left (F\right )}\right ] \] Input:

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

Output:

[1/3*(3*d^(3/2)*log((F^(b*c*n*x + a*c*n)*e - 2*sqrt(F^(b*c*n*x + a*c*n)*e 
+ d)*sqrt(d) + 2*d)/F^(b*c*n*x + a*c*n)) + 2*(F^(b*c*n*x + a*c*n)*e + 4*d) 
*sqrt(F^(b*c*n*x + a*c*n)*e + d))/(b*c*n*log(F)), 2/3*(3*sqrt(-d)*d*arctan 
(sqrt(-d)/sqrt(F^(b*c*n*x + a*c*n)*e + d)) + (F^(b*c*n*x + a*c*n)*e + 4*d) 
*sqrt(F^(b*c*n*x + a*c*n)*e + d))/(b*c*n*log(F))]
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} \, dx=\frac {d^{\frac {3}{2}} \log \left (\frac {\sqrt {F^{b c n x + a c n} e + d} - \sqrt {d}}{\sqrt {F^{b c n x + a c n} e + d} + \sqrt {d}}\right )}{b c n \log \left (F\right )} + \frac {2 \, {\left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {3}{2}} + 3 \, \sqrt {F^{b c n x + a c n} e + d} d\right )}}{3 \, b c n \log \left (F\right )} \] Input:

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

Output:

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)) + 2/3*((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))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.76 \[ \int \left (d+e \left (F^{c (a+b x)}\right )^n\right )^{3/2} \, dx=\frac {2 \, {\left (\frac {3 \, d^{2} \arctan \left (\frac {\sqrt {F^{b c n x + a c n} e + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {3}{2}} + 3 \, \sqrt {F^{b c n x + a c n} e + d} d\right )}}{3 \, b c n \log \left (F\right )} \] Input:

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

Output:

2/3*(3*d^2*arctan(sqrt(F^(b*c*n*x + a*c*n)*e + d)/sqrt(-d))/sqrt(-d) + (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))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*f**(a*c*n + b*c*n*x)*sqrt(f**(a*c*n + b*c*n*x)*e + d)*e + 2*sqrt(f**(a* 
c*n + b*c*n*x)*e + d)*d + 3*int(sqrt(f**(a*c*n + b*c*n*x)*e + d),x)*log(f) 
*b*c*d*n)/(3*log(f)*b*c*n)