Integrand size = 15, antiderivative size = 87 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=-\frac {f^x}{4 a \left (b+a f^{2 x}\right )^2 \log (f)}+\frac {f^x}{8 a b \left (b+a f^{2 x}\right ) \log (f)}+\frac {\arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)} \] Output:
-1/4*f^x/a/(b+a*f^(2*x))^2/ln(f)+1/8*f^x/a/b/(b+a*f^(2*x))/ln(f)+1/8*arcta n(a^(1/2)*f^x/b^(1/2))/a^(3/2)/b^(3/2)/ln(f)
Time = 0.65 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {\frac {\sqrt {a} \sqrt {b} f^x \left (-b+a f^{2 x}\right )}{\left (b+a f^{2 x}\right )^2}+\arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{8 a^{3/2} b^{3/2} \log (f)} \] Input:
Integrate[(b/f^x + a*f^x)^(-3),x]
Output:
((Sqrt[a]*Sqrt[b]*f^x*(-b + a*f^(2*x)))/(b + a*f^(2*x))^2 + ArcTan[(Sqrt[a ]*f^x)/Sqrt[b]])/(8*a^(3/2)*b^(3/2)*Log[f])
Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2720, 252, 215, 218}
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.
\(\displaystyle \int \frac {1}{\left (a f^x+b f^{-x}\right )^3} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {f^{2 x}}{\left (a f^{2 x}+b\right )^3}df^x}{\log (f)}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\int \frac {1}{\left (a f^{2 x}+b\right )^2}df^x}{4 a}-\frac {f^x}{4 a \left (a f^{2 x}+b\right )^2}}{\log (f)}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{a f^{2 x}+b}df^x}{2 b}+\frac {f^x}{2 b \left (a f^{2 x}+b\right )}}{4 a}-\frac {f^x}{4 a \left (a f^{2 x}+b\right )^2}}{\log (f)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\frac {\arctan \left (\frac {\sqrt {a} f^x}{\sqrt {b}}\right )}{2 \sqrt {a} b^{3/2}}+\frac {f^x}{2 b \left (a f^{2 x}+b\right )}}{4 a}-\frac {f^x}{4 a \left (a f^{2 x}+b\right )^2}}{\log (f)}\) |
Input:
Int[(b/f^x + a*f^x)^(-3),x]
Output:
(-1/4*f^x/(a*(b + a*f^(2*x))^2) + (f^x/(2*b*(b + a*f^(2*x))) + ArcTan[(Sqr t[a]*f^x)/Sqrt[b]]/(2*Sqrt[a]*b^(3/2)))/(4*a))/Log[f]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, 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 = 0.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\frac {\frac {f^{3 x}}{8 b}-\frac {f^{x}}{8 a}}{\left (b +a \,f^{2 x}\right )^{2}}+\frac {\arctan \left (\frac {a \,f^{x}}{\sqrt {a b}}\right )}{8 a b \sqrt {a b}}}{\ln \left (f \right )}\) | \(62\) |
default | \(\frac {\frac {\frac {f^{3 x}}{8 b}-\frac {f^{x}}{8 a}}{\left (b +a \,f^{2 x}\right )^{2}}+\frac {\arctan \left (\frac {a \,f^{x}}{\sqrt {a b}}\right )}{8 a b \sqrt {a b}}}{\ln \left (f \right )}\) | \(62\) |
risch | \(\frac {f^{x} \left (-b +a \,f^{2 x}\right )}{8 \ln \left (f \right ) a \left (b +a \,f^{2 x}\right )^{2} b}-\frac {\ln \left (f^{x}-\frac {b}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, b a \ln \left (f \right )}+\frac {\ln \left (f^{x}+\frac {b}{\sqrt {-a b}}\right )}{16 \sqrt {-a b}\, b a \ln \left (f \right )}\) | \(102\) |
Input:
int(1/(b/(f^x)+a*f^x)^3,x,method=_RETURNVERBOSE)
Output:
1/ln(f)*((1/8/b*(f^x)^3-1/8/a*f^x)/((f^x)^2*a+b)^2+1/8/a/b/(a*b)^(1/2)*arc tan(a*f^x/(a*b)^(1/2)))
Time = 0.08 (sec) , antiderivative size = 261, normalized size of antiderivative = 3.00 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=\left [\frac {2 \, a^{2} b f^{3 \, x} - 2 \, a b^{2} f^{x} - {\left (\sqrt {-a b} a^{2} f^{4 \, x} + 2 \, \sqrt {-a b} a b f^{2 \, x} + \sqrt {-a b} b^{2}\right )} \log \left (\frac {a f^{2 \, x} - 2 \, \sqrt {-a b} f^{x} - b}{a f^{2 \, x} + b}\right )}{16 \, {\left (a^{4} b^{2} f^{4 \, x} \log \left (f\right ) + 2 \, a^{3} b^{3} f^{2 \, x} \log \left (f\right ) + a^{2} b^{4} \log \left (f\right )\right )}}, \frac {a^{2} b f^{3 \, x} - a b^{2} f^{x} - {\left (\sqrt {a b} a^{2} f^{4 \, x} + 2 \, \sqrt {a b} a b f^{2 \, x} + \sqrt {a b} b^{2}\right )} \arctan \left (\frac {\sqrt {a b}}{a f^{x}}\right )}{8 \, {\left (a^{4} b^{2} f^{4 \, x} \log \left (f\right ) + 2 \, a^{3} b^{3} f^{2 \, x} \log \left (f\right ) + a^{2} b^{4} \log \left (f\right )\right )}}\right ] \] Input:
integrate(1/(b/(f^x)+a*f^x)^3,x, algorithm="fricas")
Output:
[1/16*(2*a^2*b*f^(3*x) - 2*a*b^2*f^x - (sqrt(-a*b)*a^2*f^(4*x) + 2*sqrt(-a *b)*a*b*f^(2*x) + sqrt(-a*b)*b^2)*log((a*f^(2*x) - 2*sqrt(-a*b)*f^x - b)/( a*f^(2*x) + b)))/(a^4*b^2*f^(4*x)*log(f) + 2*a^3*b^3*f^(2*x)*log(f) + a^2* b^4*log(f)), 1/8*(a^2*b*f^(3*x) - a*b^2*f^x - (sqrt(a*b)*a^2*f^(4*x) + 2*s qrt(a*b)*a*b*f^(2*x) + sqrt(a*b)*b^2)*arctan(sqrt(a*b)/(a*f^x)))/(a^4*b^2* f^(4*x)*log(f) + 2*a^3*b^3*f^(2*x)*log(f) + a^2*b^4*log(f))]
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {a f^{- x} - b f^{- 3 x}}{8 a^{3} b \log {\left (f \right )} + 16 a^{2} b^{2} f^{- 2 x} \log {\left (f \right )} + 8 a b^{3} f^{- 4 x} \log {\left (f \right )}} + \frac {\operatorname {RootSum} {\left (256 z^{2} a^{3} b^{3} + 1, \left ( i \mapsto i \log {\left (- 16 i a^{2} b + f^{- x} \right )} \right )\right )}}{\log {\left (f \right )}} \] Input:
integrate(1/(b/(f**x)+a*f**x)**3,x)
Output:
(a/f**x - b/f**(3*x))/(8*a**3*b*log(f) + 16*a**2*b**2*log(f)/f**(2*x) + 8* a*b**3*log(f)/f**(4*x)) + RootSum(256*_z**2*a**3*b**3 + 1, Lambda(_i, _i*l og(-16*_i*a**2*b + f**(-x))))/log(f)
Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=-\frac {\frac {b}{f^{3 \, x}} - \frac {a}{f^{x}}}{8 \, {\left (a^{3} b + \frac {a b^{3}}{f^{4 \, x}} + \frac {2 \, a^{2} b^{2}}{f^{2 \, x}}\right )} \log \left (f\right )} - \frac {\arctan \left (\frac {b}{\sqrt {a b} f^{x}}\right )}{8 \, \sqrt {a b} a b \log \left (f\right )} \] Input:
integrate(1/(b/(f^x)+a*f^x)^3,x, algorithm="maxima")
Output:
-1/8*(b/f^(3*x) - a/f^x)/((a^3*b + a*b^3/f^(4*x) + 2*a^2*b^2/f^(2*x))*log( f)) - 1/8*arctan(b/(sqrt(a*b)*f^x))/(sqrt(a*b)*a*b*log(f))
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {\arctan \left (\frac {a f^{x}}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a b \log \left (f\right )} + \frac {a f^{3 \, x} - b f^{x}}{8 \, {\left (a f^{2 \, x} + b\right )}^{2} a b \log \left (f\right )} \] Input:
integrate(1/(b/(f^x)+a*f^x)^3,x, algorithm="giac")
Output:
1/8*arctan(a*f^x/sqrt(a*b))/(sqrt(a*b)*a*b*log(f)) + 1/8*(a*f^(3*x) - b*f^ x)/((a*f^(2*x) + b)^2*a*b*log(f))
Time = 23.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {f^x}{8\,\left (a\,b^2\,\ln \left (f\right )+a^2\,b\,f^{2\,x}\,\ln \left (f\right )\right )}-\frac {f^x}{4\,\left (a\,b^2\,\ln \left (f\right )+a^3\,f^{4\,x}\,\ln \left (f\right )+2\,a^2\,b\,f^{2\,x}\,\ln \left (f\right )\right )}+\frac {\mathrm {atan}\left (\frac {f^x\,\sqrt {a^3\,b^3\,{\ln \left (f\right )}^2}}{a\,b^2\,\ln \left (f\right )}\right )}{8\,\sqrt {a^3\,b^3\,{\ln \left (f\right )}^2}} \] Input:
int(1/(b/f^x + a*f^x)^3,x)
Output:
f^x/(8*(a*b^2*log(f) + a^2*b*f^(2*x)*log(f))) - f^x/(4*(a*b^2*log(f) + a^3 *f^(4*x)*log(f) + 2*a^2*b*f^(2*x)*log(f))) + atan((f^x*(a^3*b^3*log(f)^2)^ (1/2))/(a*b^2*log(f)))/(8*(a^3*b^3*log(f)^2)^(1/2))
Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\left (b f^{-x}+a f^x\right )^3} \, dx=\frac {f^{4 x} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {f^{x} a}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+2 f^{2 x} \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {f^{x} a}{\sqrt {b}\, \sqrt {a}}\right ) a b +\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {f^{x} a}{\sqrt {b}\, \sqrt {a}}\right ) b^{2}+f^{3 x} a^{2} b -f^{x} a \,b^{2}}{8 \,\mathrm {log}\left (f \right ) a^{2} b^{2} \left (f^{4 x} a^{2}+2 f^{2 x} a b +b^{2}\right )} \] Input:
int(1/(b/(f^x)+a*f^x)^3,x)
Output:
(f**(4*x)*sqrt(b)*sqrt(a)*atan((f**x*a)/(sqrt(b)*sqrt(a)))*a**2 + 2*f**(2* x)*sqrt(b)*sqrt(a)*atan((f**x*a)/(sqrt(b)*sqrt(a)))*a*b + sqrt(b)*sqrt(a)* atan((f**x*a)/(sqrt(b)*sqrt(a)))*b**2 + f**(3*x)*a**2*b - f**x*a*b**2)/(8* log(f)*a**2*b**2*(f**(4*x)*a**2 + 2*f**(2*x)*a*b + b**2))