Integrand size = 19, antiderivative size = 124 \[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=-\frac {x}{2 \sqrt [3]{d}}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{d}+2 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt {3} \sqrt [3]{d}}\right )}{b c \sqrt [3]{d} n \log (F)}+\frac {3 \log \left (\sqrt [3]{d}-\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}\right )}{2 b c \sqrt [3]{d} n \log (F)} \] Output:
-1/2*x/d^(1/3)+3^(1/2)*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/d^(1/3)/n/ln(F)+3/2*ln(d^(1/3)-(d+e*(F^(c*(b*x+a))) ^n)^(1/3))/b/c/d^(1/3)/n/ln(F)
Time = 0.49 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}}{\sqrt [3]{d}}}{\sqrt {3}}\right )+2 \log \left (\sqrt [3]{d}-\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}\right )-\log \left (d^{2/3}+\sqrt [3]{d} \sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}+\left (d+e \left (F^{c (a+b x)}\right )^n\right )^{2/3}\right )}{2 b c \sqrt [3]{d} n \log (F)} \] Input:
Integrate[(d + e*(F^(c*(a + b*x)))^n)^(-1/3),x]
Output:
(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^(1/3)*n*Log[F])
Time = 0.41 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2720, 798, 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.
\(\displaystyle \int \frac {1}{\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {F^{-c (a+b x)}}{\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}dF^{c (a+b x)}}{b c \log (F)}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\int \frac {F^{-c (a+b x)}}{\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}d\left (F^{c (a+b x)}\right )^n}{b c n \log (F)}\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {\frac {3}{2} \int \frac {1}{F^{2 c (a+b x)}+d^{2/3}+\sqrt [3]{d} \sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}d\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}-\frac {3 \int \frac {1}{\sqrt [3]{d}-\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}d\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}{2 \sqrt [3]{d}}-\frac {\log \left (\left (F^{c (a+b x)}\right )^n\right )}{2 \sqrt [3]{d}}}{b c n \log (F)}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\frac {3}{2} \int \frac {1}{F^{2 c (a+b x)}+d^{2/3}+\sqrt [3]{d} \sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}d\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}+\frac {3 \log \left (\sqrt [3]{d}-\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (\left (F^{c (a+b x)}\right )^n\right )}{2 \sqrt [3]{d}}}{b c n \log (F)}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {-\frac {3 \int \frac {1}{-F^{2 c (a+b x)}-3}d\left (\frac {2 \sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt [3]{d}}+1\right )}{\sqrt [3]{d}}+\frac {3 \log \left (\sqrt [3]{d}-\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (\left (F^{c (a+b x)}\right )^n\right )}{2 \sqrt [3]{d}}}{b c n \log (F)}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}}{\sqrt [3]{d}}+1}{\sqrt {3}}\right )}{\sqrt [3]{d}}+\frac {3 \log \left (\sqrt [3]{d}-\sqrt [3]{e \left (F^{c (a+b x)}\right )^n+d}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (\left (F^{c (a+b x)}\right )^n\right )}{2 \sqrt [3]{d}}}{b c n \log (F)}\) |
Input:
Int[(d + e*(F^(c*(a + b*x)))^n)^(-1/3),x]
Output:
((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)))/(b*c*n*Log[F])
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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]
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])
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]]
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]
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]]]
Time = 0.07 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}-d^{\frac {1}{3}}\right )}{d^{\frac {1}{3}}}-\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {2}{3}}+d^{\frac {1}{3}} {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+d^{\frac {2}{3}}\right )}{2 d^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}{d^{\frac {1}{3}}}+1\right )}{3}\right )}{d^{\frac {1}{3}}}}{\ln \left (F \right ) b c n}\) | \(129\) |
default | \(\frac {\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}-d^{\frac {1}{3}}\right )}{d^{\frac {1}{3}}}-\frac {\ln \left ({\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {2}{3}}+d^{\frac {1}{3}} {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}+d^{\frac {2}{3}}\right )}{2 d^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 {\left (d +e \left (F^{c \left (b x +a \right )}\right )^{n}\right )}^{\frac {1}{3}}}{d^{\frac {1}{3}}}+1\right )}{3}\right )}{d^{\frac {1}{3}}}}{\ln \left (F \right ) b c n}\) | \(129\) |
Input:
int(1/(d+e*(F^(c*(b*x+a)))^n)^(1/3),x,method=_RETURNVERBOSE)
Output:
1/ln(F)/b/c/n*(1/d^(1/3)*ln((d+e*(F^(c*(b*x+a)))^n)^(1/3)-d^(1/3))-1/2/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))+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)))
Time = 0.09 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.01 \[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\left [\frac {\sqrt {3} d \sqrt {-\frac {1}{d^{\frac {2}{3}}}} \log \left (\frac {2 \, \sqrt {3} {\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} d^{\frac {2}{3}} \sqrt {-\frac {1}{d^{\frac {2}{3}}}} - \sqrt {3} d^{\frac {4}{3}} \sqrt {-\frac {1}{d^{\frac {2}{3}}}} + 2 \, F^{b c n x + a c n} e - {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} {\left (\sqrt {3} d \sqrt {-\frac {1}{d^{\frac {2}{3}}}} + 3 \, d^{\frac {2}{3}}\right )} + 3 \, d}{F^{b c n x + a c n}}\right ) - d^{\frac {2}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right ) + 2 \, d^{\frac {2}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right )}{2 \, b c d n \log \left (F\right )}, \frac {2 \, \sqrt {3} d^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} + \frac {2 \, \sqrt {3} {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}}}{3 \, d^{\frac {1}{3}}}\right ) - d^{\frac {2}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right ) + 2 \, d^{\frac {2}{3}} \log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right )}{2 \, b c d n \log \left (F\right )}\right ] \] Input:
integrate(1/(d+e*(F^((b*x+a)*c))^n)^(1/3),x, algorithm="fricas")
Output:
[1/2*(sqrt(3)*d*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)) - d^(2/3)*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^(2/3)*log((F^(b*c*n*x + a*c*n)*e + d)^(1/3) - d^(1/3)))/(b*c*d *n*log(F)), 1/2*(2*sqrt(3)*d^(2/3)*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^(2/3)*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^(2 /3)*log((F^(b*c*n*x + a*c*n)*e + d)^(1/3) - d^(1/3)))/(b*c*d*n*log(F))]
\[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\int \frac {1}{\sqrt [3]{d + e \left (F^{c \left (a + b x\right )}\right )^{n}}}\, dx \] Input:
integrate(1/(d+e*(F**((b*x+a)*c))**n)**(1/3),x)
Output:
Integral((d + e*(F**(c*(a + b*x)))**n)**(-1/3), x)
Time = 0.13 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} + d^{\frac {1}{3}}\right )}}{3 \, d^{\frac {1}{3}}}\right )}{b c d^{\frac {1}{3}} n \log \left (F\right )} - \frac {\log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right )}{2 \, b c d^{\frac {1}{3}} n \log \left (F\right )} + \frac {\log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}}\right )}{b c d^{\frac {1}{3}} n \log \left (F\right )} \] Input:
integrate(1/(d+e*(F^((b*x+a)*c))^n)^(1/3),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^(1/3)*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^(1/3)*n*l og(F)) + log((F^(b*c*n*x + a*c*n)*e + d)^(1/3) - d^(1/3))/(b*c*d^(1/3)*n*l og(F))
Time = 0.14 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\frac {\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} + d^{\frac {1}{3}}\right )}}{3 \, d^{\frac {1}{3}}}\right )}{d^{\frac {1}{3}}} - \frac {\log \left ({\left (F^{b c n x + a c n} e + d\right )}^{\frac {2}{3}} + {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} d^{\frac {1}{3}} + d^{\frac {2}{3}}\right )}{d^{\frac {1}{3}}} + \frac {2 \, \log \left ({\left | {\left (F^{b c n x + a c n} e + d\right )}^{\frac {1}{3}} - d^{\frac {1}{3}} \right |}\right )}{d^{\frac {1}{3}}}}{2 \, b c n \log \left (F\right )} \] Input:
integrate(1/(d+e*(F^((b*x+a)*c))^n)^(1/3),x, algorithm="giac")
Output:
1/2*(2*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))/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))/d^(1/3) + 2*log(abs((F^(b*c* n*x + a*c*n)*e + d)^(1/3) - d^(1/3)))/d^(1/3))/(b*c*n*log(F))
Timed out. \[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\int \frac {1}{{\left (d+e\,{\left (F^{c\,\left (a+b\,x\right )}\right )}^n\right )}^{1/3}} \,d x \] Input:
int(1/(d + e*(F^(c*(a + b*x)))^n)^(1/3),x)
Output:
int(1/(d + e*(F^(c*(a + b*x)))^n)^(1/3), x)
\[ \int \frac {1}{\sqrt [3]{d+e \left (F^{c (a+b x)}\right )^n}} \, dx=\int \frac {1}{\left (f^{b c n x +a c n} e +d \right )^{\frac {1}{3}}}d x \] Input:
int(1/(d+e*(F^((b*x+a)*c))^n)^(1/3),x)
Output:
int(1/(f**(a*c*n + b*c*n*x)*e + d)**(1/3),x)