\(\int \sqrt [3]{d+e (F^{c (a+b x)})^n} \, dx\) [12]

Optimal result

Integrand size = 19, antiderivative size = 159 \[ \int \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n} \, dx=-\frac {1}{2} \sqrt [3]{d} x+\frac {3 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{b c n \log (F)}-\frac {\sqrt {3} \sqrt [3]{d} \arctan \left (\frac {\sqrt [3]{d}+2 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {3} \sqrt [3]{d}}\right )}{b c n \log (F)}+\frac {3 \sqrt [3]{d} \log \left (\sqrt [3]{d}-\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}\right )}{2 b c n \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.13 \[ \int \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n} \, dx=\frac {6 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}-2 \sqrt {3} \sqrt [3]{d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt [3]{d}}}{\sqrt {3}}\right )+2 \sqrt [3]{d} \log \left (\sqrt [3]{d}-\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}\right )-\sqrt [3]{d} \log \left (d^{2/3}+\sqrt [3]{d} \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}+\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{2/3}\right )}{2 b c n \log (F)} \] Input:

Integrate[(d + e*(F^(c*(a + b*x)))^n)^(1/3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2720, 798, 60, 69, 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.

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[(d + e*(F^(c*(a + b*x)))^n)^(1/3),x]
 

Output:

(3*(d + e*(F^(c*(a + b*x)))^n)^(1/3) + d*(-((Sqrt[3]*ArcTan[(1 + (2*(d + e 
*(F^(c*(a + b*x)))^n)^(1/3))/d^(1/3))/Sqrt[3]])/d^(2/3)) - Log[(F^(c*(a + 
b*x)))^n]/(2*d^(2/3)) + (3*Log[d^(1/3) - (d + e*(F^(c*(a + b*x)))^n)^(1/3) 
])/(2*d^(2/3))))/(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 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 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   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 = 0.05 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {3 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+3 \left (\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}-d^{\frac {1}{3}}\right )}{3 d^{\frac {2}{3}}}-\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {2}{3}}+d^{\frac {1}{3}} {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+d^{\frac {2}{3}}\right )}{6 d^{\frac {2}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}{d^{\frac {1}{3}}}+1\right )}{3}\right )}{3 d^{\frac {2}{3}}}\right ) d}{\ln \left (F \right ) b c n}\)	\(154\)
default	\(\frac {3 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+3 \left (\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}-d^{\frac {1}{3}}\right )}{3 d^{\frac {2}{3}}}-\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {2}{3}}+d^{\frac {1}{3}} {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+d^{\frac {2}{3}}\right )}{6 d^{\frac {2}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}{d^{\frac {1}{3}}}+1\right )}{3}\right )}{3 d^{\frac {2}{3}}}\right ) d}{\ln \left (F \right ) b c n}\)	\(154\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.01 \[ \int \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n} \, dx=-\frac {2 \, \sqrt {3} d^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {2}{3}} + \sqrt {3} d}{3 \, d}\right ) + d^{\frac {1}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right ) - 2 \, d^{\frac {1}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right ) - 6 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}}}{2 \, b c n \log \left (F\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.21 \[ \int \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n} \, dx=-\frac {\sqrt {3} d^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} + d^{\frac {1}{3}}\right )}}{3 \, d^{\frac {1}{3}}}\right )}{b c n \log \left (F\right )} - \frac {d^{\frac {1}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right )}{2 \, b c n \log \left (F\right )} + \frac {d^{\frac {1}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right )}{b c n \log \left (F\right )} + \frac {3 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}}}{b c n \log \left (F\right )} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n} \, dx=-\frac {2 \, \sqrt {3} d^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} + d^{\frac {1}{3}}\right )}}{3 \, d^{\frac {1}{3}}}\right ) + d^{\frac {1}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right ) - 2 \, d^{\frac {1}{3}} \log \left ({\left | {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}} \right |}\right ) - 6 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}}}{2 \, b c n \log \left (F\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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