\(\int \frac {1}{(d+e (F^{c (a+b x)})^n)^{4/3}} \, dx\) [15]

Optimal result

Integrand size = 19, antiderivative size = 161 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=-\frac {x}{2 d^{4/3}}+\frac {3}{b c d \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n} n \log (F)}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{d}+2 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {3} \sqrt [3]{d}}\right )}{b c d^{4/3} n \log (F)}+\frac {3 \log \left (\sqrt [3]{d}-\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}\right )}{2 b c d^{4/3} n \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\frac {\frac {6 \sqrt [3]{d}}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}+2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt [3]{d}}}{\sqrt {3}}\right )+2 \log \left (\sqrt [3]{d}-\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}\right )-\log \left (d^{2/3}+\sqrt [3]{d} \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}+\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{2/3}\right )}{2 b c d^{4/3} n \log (F)} \] Input:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^(-4/3),x]
 

Output:

((6*d^(1/3))/(d + e*(F^(c*(a + b*x)))^n)^(1/3) + 2*Sqrt[3]*ArcTan[(1 + (2* 
(d + e*(F^(c*(a + b*x)))^n)^(1/3))/d^(1/3))/Sqrt[3]] + 2*Log[d^(1/3) - (d 
+ e*(F^(c*(a + b*x)))^n)^(1/3)] - Log[d^(2/3) + d^(1/3)*(d + e*(F^(c*(a + 
b*x)))^n)^(1/3) + (d + e*(F^(c*(a + b*x)))^n)^(2/3)])/(2*b*c*d^(4/3)*n*Log 
[F])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2720, 798, 61, 67, 16, 1082, 217}

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.

\(\displaystyle \int \frac {1}{\left (e \left (F^{c (a+b x)}\right )^n+d\right )^{4/3}} \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int \frac {F^{-c (a+b x)}}{\left (e \left (F^{c (a+b x)}\right )^n+d\right )^{4/3}}dF^{c (a+b x)}}{b c \log (F)}\)

\(\Big \downarrow \) 798

\(\displaystyle \frac {\int \frac {F^{-c (a+b x)}}{\left (e \left (F^{c (a+b x)}\right )^n+d\right )^{4/3}}d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\)

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(d + e*(F^(c*(a + b*x)))^n)^(-4/3),x]
 

Output:

(3/(d*(d + e*(F^(c*(a + b*x)))^n)^(1/3)) + ((Sqrt[3]*ArcTan[(1 + (2*(d + e 
*(F^(c*(a + b*x)))^n)^(1/3))/d^(1/3))/Sqrt[3]])/d^(1/3) - Log[(F^(c*(a + b 
*x)))^n]/(2*d^(1/3)) + (3*Log[d^(1/3) - (d + e*(F^(c*(a + b*x)))^n)^(1/3)] 
)/(2*d^(1/3)))/d)/(b*c*n*Log[F])
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 61

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

rule 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

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 = 0.02 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}-d^{\frac {1}{3}}\right )}{d^{\frac {1}{3}}}-\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {2}{3}}+d^{\frac {1}{3}} {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+d^{\frac {2}{3}}\right )}{2 d^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}{d^{\frac {1}{3}}}+1\right )}{3}\right )}{d^{\frac {1}{3}}}}{d}+\frac {3}{d {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}}{\ln \left (F \right ) b c n}\)	\(159\)
default	\(\frac {\frac {\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}-d^{\frac {1}{3}}\right )}{d^{\frac {1}{3}}}-\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {2}{3}}+d^{\frac {1}{3}} {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+d^{\frac {2}{3}}\right )}{2 d^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}{d^{\frac {1}{3}}}+1\right )}{3}\right )}{d^{\frac {1}{3}}}}{d}+\frac {3}{d {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}}{\ln \left (F \right ) b c n}\)	\(159\)

Input:

int(1/(d+e*(F^(c*(b*x+a)))^n)^(4/3),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 560, normalized size of antiderivative = 3.48 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\left [\frac {\sqrt {3} {\left (F^{b c n x + a c n} d e + d^{2}\right )} \sqrt {-\frac {1}{d^{\frac {2}{3}}}} \log \left (\frac {2 \, \sqrt {3} {\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} d^{\frac {2}{3}} \sqrt {-\frac {1}{d^{\frac {2}{3}}}} - \sqrt {3} d^{\frac {4}{3}} \sqrt {-\frac {1}{d^{\frac {2}{3}}}} + 2 \, F^{b c n x + a c n} e - {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} {\left (\sqrt {3} d \sqrt {-\frac {1}{d^{\frac {2}{3}}}} + 3 \, d^{\frac {2}{3}}\right )} + 3 \, d}{F^{b c n x + a c n}}\right ) - {\left (F^{b c n x + a c n} d^{\frac {2}{3}} e + d^{\frac {5}{3}}\right )} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right ) + 2 \, {\left (F^{b c n x + a c n} d^{\frac {2}{3}} e + d^{\frac {5}{3}}\right )} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right ) + 6 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} d}{2 \, {\left (F^{b c n x + a c n} b c d^{2} e n \log \left (F\right ) + b c d^{3} n \log \left (F\right )\right )}}, \frac {\frac {2 \, \sqrt {3} {\left (F^{b c n x + a c n} d e + d^{2}\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} + \frac {2 \, \sqrt {3} {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}}}{3 \, d^{\frac {1}{3}}}\right )}{d^{\frac {1}{3}}} - {\left (F^{b c n x + a c n} d^{\frac {2}{3}} e + d^{\frac {5}{3}}\right )} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right ) + 2 \, {\left (F^{b c n x + a c n} d^{\frac {2}{3}} e + d^{\frac {5}{3}}\right )} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right ) + 6 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} d}{2 \, {\left (F^{b c n x + a c n} b c d^{2} e n \log \left (F\right ) + b c d^{3} n \log \left (F\right )\right )}}\right ] \] Input:

integrate(1/(d+e*(F^((b*x+a)*c))^n)^(4/3),x, algorithm="fricas")
 

Output:

[1/2*(sqrt(3)*(F^(b*c*n*x + a*c*n)*d*e + d^2)*sqrt(-1/d^(2/3))*log((2*sqrt 
(3)*(F^(b*c*n*x + a*c*n)*e + d)^(2/3)*d^(2/3)*sqrt(-1/d^(2/3)) - sqrt(3)*d 
^(4/3)*sqrt(-1/d^(2/3)) + 2*F^(b*c*n*x + a*c*n)*e - (F^(b*c*n*x + a*c*n)*e 
 + d)^(1/3)*(sqrt(3)*d*sqrt(-1/d^(2/3)) + 3*d^(2/3)) + 3*d)/F^(b*c*n*x + a 
*c*n)) - (F^(b*c*n*x + a*c*n)*d^(2/3)*e + d^(5/3))*log((F^(b*c*n*x + a*c*n 
)*e + d)^(2/3) + (F^(b*c*n*x + a*c*n)*e + d)^(1/3)*d^(1/3) + d^(2/3)) + 2* 
(F^(b*c*n*x + a*c*n)*d^(2/3)*e + d^(5/3))*log((F^(b*c*n*x + a*c*n)*e + d)^ 
(1/3) - d^(1/3)) + 6*(F^(b*c*n*x + a*c*n)*e + d)^(2/3)*d)/(F^(b*c*n*x + a* 
c*n)*b*c*d^2*e*n*log(F) + b*c*d^3*n*log(F)), 1/2*(2*sqrt(3)*(F^(b*c*n*x + 
a*c*n)*d*e + d^2)*arctan(1/3*sqrt(3) + 2/3*sqrt(3)*(F^(b*c*n*x + a*c*n)*e 
+ d)^(1/3)/d^(1/3))/d^(1/3) - (F^(b*c*n*x + a*c*n)*d^(2/3)*e + d^(5/3))*lo 
g((F^(b*c*n*x + a*c*n)*e + d)^(2/3) + (F^(b*c*n*x + a*c*n)*e + d)^(1/3)*d^ 
(1/3) + d^(2/3)) + 2*(F^(b*c*n*x + a*c*n)*d^(2/3)*e + d^(5/3))*log((F^(b*c 
*n*x + a*c*n)*e + d)^(1/3) - d^(1/3)) + 6*(F^(b*c*n*x + a*c*n)*e + d)^(2/3 
)*d)/(F^(b*c*n*x + a*c*n)*b*c*d^2*e*n*log(F) + b*c*d^3*n*log(F))]
 

Sympy [F]

\[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\int \frac {1}{\left (d + e \left (F^{c \left (a + b x\right )}\right )^{n}\right )^{\frac {4}{3}}}\, dx \] Input:

integrate(1/(d+e*(F**((b*x+a)*c))**n)**(4/3),x)
 

Output:

Integral((d + e*(F**(c*(a + b*x)))**n)**(-4/3), x)
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} + d^{\frac {1}{3}}\right )}}{3 \, d^{\frac {1}{3}}}\right )}{b c d^{\frac {4}{3}} n \log \left (F\right )} - \frac {\log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right )}{2 \, b c d^{\frac {4}{3}} n \log \left (F\right )} + \frac {\log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right )}{b c d^{\frac {4}{3}} n \log \left (F\right )} + \frac {3}{{\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} b c d n \log \left (F\right )} \] Input:

integrate(1/(d+e*(F^((b*x+a)*c))^n)^(4/3),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (136) = 272\).

Time = 0.16 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.73 \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\frac {{\left (\pi - \pi \mathrm {sgn}\left (F\right ) - 2 \, \sqrt {3} \log \left ({\left | F \right |}\right )\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (F^{b c n x} F^{a c n} e + d\right )}^{\frac {1}{3}} + d^{\frac {1}{3}}\right )}}{3 \, d^{\frac {1}{3}}}\right )}{\pi ^{2} b c d^{\frac {4}{3}} n \mathrm {sgn}\left (F\right ) - \pi ^{2} b c d^{\frac {4}{3}} n - 2 \, b c d^{\frac {4}{3}} n \log \left ({\left | F \right |}\right )^{2}} - \frac {{\left (\sqrt {3} \pi \mathrm {sgn}\left (F\right ) - \sqrt {3} \pi - 2 \, \log \left ({\left | F \right |}\right )\right )} \log \left ({\left (F^{b c n x} F^{a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x} F^{a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right )}{2 \, {\left (\pi ^{2} b c d^{\frac {4}{3}} n \mathrm {sgn}\left (F\right ) - \pi ^{2} b c d^{\frac {4}{3}} n - 2 \, b c d^{\frac {4}{3}} n \log \left ({\left | F \right |}\right )^{2}\right )}} + \frac {\log \left ({\left | {\left (F^{b c n x} F^{a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}} \right |}\right )}{b c d^{\frac {4}{3}} n \log \left (F\right )} + \frac {3}{{\left (F^{b c n x} F^{a c n} e + d\right )}^{\frac {1}{3}} b c d n \log \left (F\right )} \] Input:

integrate(1/(d+e*(F^((b*x+a)*c))^n)^(4/3),x, algorithm="giac")
 

Output:

(pi - pi*sgn(F) - 2*sqrt(3)*log(abs(F)))*arctan(1/3*sqrt(3)*(2*(F^(b*c*n*x 
)*F^(a*c*n)*e + d)^(1/3) + d^(1/3))/d^(1/3))/(pi^2*b*c*d^(4/3)*n*sgn(F) - 
pi^2*b*c*d^(4/3)*n - 2*b*c*d^(4/3)*n*log(abs(F))^2) - 1/2*(sqrt(3)*pi*sgn( 
F) - sqrt(3)*pi - 2*log(abs(F)))*log((F^(b*c*n*x)*F^(a*c*n)*e + d)^(2/3) + 
 (F^(b*c*n*x)*F^(a*c*n)*e + d)^(1/3)*d^(1/3) + d^(2/3))/(pi^2*b*c*d^(4/3)* 
n*sgn(F) - pi^2*b*c*d^(4/3)*n - 2*b*c*d^(4/3)*n*log(abs(F))^2) + log(abs(( 
F^(b*c*n*x)*F^(a*c*n)*e + d)^(1/3) - d^(1/3)))/(b*c*d^(4/3)*n*log(F)) + 3/ 
((F^(b*c*n*x)*F^(a*c*n)*e + d)^(1/3)*b*c*d*n*log(F))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\int \frac {1}{{\left (d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n\right )}^{4/3}} \,d x \] Input:

int(1/(d + e*(F^(c*(a + b*x)))^n)^(4/3),x)
                                                                                    
                                                                                    
 

Output:

int(1/(d + e*(F^(c*(a + b*x)))^n)^(4/3), x)
 

Reduce [F]

\[ \int \frac {1}{\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{4/3}} \, dx=\int \frac {1}{f^{b c n x +a c n} \left (f^{b c n x +a c n} e +d \right )^{\frac {1}{3}} e +\left (f^{b c n x +a c n} e +d \right )^{\frac {1}{3}} d}d x \] Input:

int(1/(d+e*(F^((b*x+a)*c))^n)^(4/3),x)
 

Output:

int(1/(f**(a*c*n + b*c*n*x)*(f**(a*c*n + b*c*n*x)*e + d)**(1/3)*e + (f**(a 
*c*n + b*c*n*x)*e + d)**(1/3)*d),x)