Integrand size = 24, antiderivative size = 94 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=\frac {x}{a}+\frac {b \text {arctanh}\left (\frac {b+2 c f^{g+h x}}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} h \log (f)}-\frac {\log \left (a+b f^{g+h x}+c f^{2 g+2 h x}\right )}{2 a h \log (f)} \] Output:
x/a+b*arctanh((b+2*c*f^(h*x+g))/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)^(1/2)/h /ln(f)-1/2*ln(a+b*f^(h*x+g)+c*f^(2*h*x+2*g))/a/h/ln(f)
Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=-\frac {\frac {2 b \arctan \left (\frac {b+2 c f^{g+h x}}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 \log \left (f^{g+h x}\right )+\log \left (a+f^{g+h x} \left (b+c f^{g+h x}\right )\right )}{2 a h \log (f)} \] Input:
Integrate[(a + b*f^(g + h*x) + c*f^(2*(g + h*x)))^(-1),x]
Output:
-1/2*((2*b*ArcTan[(b + 2*c*f^(g + h*x))/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4 *a*c] - 2*Log[f^(g + h*x)] + Log[a + f^(g + h*x)*(b + c*f^(g + h*x))])/(a* h*Log[f])
Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2720, 1144, 25, 1142, 1083, 219, 1103}
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.
\(\displaystyle \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {f^{-g-h x}}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}}{h \log (f)}\) |
\(\Big \downarrow \) 1144 |
\(\displaystyle \frac {\frac {\int -\frac {c f^{g+h x}+b}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}}{a}+\frac {\log \left (f^{g+h x}\right )}{a}}{h \log (f)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\log \left (f^{g+h x}\right )}{a}-\frac {\int \frac {c f^{g+h x}+b}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}}{a}}{h \log (f)}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\frac {\log \left (f^{g+h x}\right )}{a}-\frac {\frac {1}{2} b \int \frac {1}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}+\frac {1}{2} \int \frac {2 c f^{g+h x}+b}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}}{a}}{h \log (f)}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\frac {\log \left (f^{g+h x}\right )}{a}-\frac {\frac {1}{2} \int \frac {2 c f^{g+h x}+b}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}-b \int \frac {1}{-f^{2 g+2 h x}+b^2-4 a c}d\left (2 c f^{g+h x}+b\right )}{a}}{h \log (f)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\log \left (f^{g+h x}\right )}{a}-\frac {\frac {1}{2} \int \frac {2 c f^{g+h x}+b}{b f^{g+h x}+c f^{2 g+2 h x}+a}df^{g+h x}-\frac {b \text {arctanh}\left (\frac {b+2 c f^{g+h x}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a}}{h \log (f)}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\frac {\log \left (f^{g+h x}\right )}{a}-\frac {\frac {1}{2} \log \left (a+b f^{g+h x}+c f^{2 g+2 h x}\right )-\frac {b \text {arctanh}\left (\frac {b+2 c f^{g+h x}}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}}}{a}}{h \log (f)}\) |
Input:
Int[(a + b*f^(g + h*x) + c*f^(2*(g + h*x)))^(-1),x]
Output:
(Log[f^(g + h*x)]/a - (-((b*ArcTanh[(b + 2*c*f^(g + h*x))/Sqrt[b^2 - 4*a*c ]])/Sqrt[b^2 - 4*a*c]) + Log[a + b*f^(g + h*x) + c*f^(2*g + 2*h*x)]/2)/a)/ (h*Log[f])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Simp[e*(Log[RemoveContent[d + e*x, x]]/(c*d^2 - b*d*e + a*e^2)), x] + S imp[1/(c*d^2 - b*d*e + a*e^2) Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, 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]]]
Leaf count of result is larger than twice the leaf count of optimal. \(545\) vs. \(2(88)=176\).
Time = 0.12 (sec) , antiderivative size = 546, normalized size of antiderivative = 5.81
method | result | size |
risch | \(\frac {4 \ln \left (f \right )^{2} a c \,h^{2} x}{4 \ln \left (f \right )^{2} a^{2} c \,h^{2}-\ln \left (f \right )^{2} a \,b^{2} h^{2}}-\frac {\ln \left (f \right )^{2} b^{2} h^{2} x}{4 \ln \left (f \right )^{2} a^{2} c \,h^{2}-\ln \left (f \right )^{2} a \,b^{2} h^{2}}+\frac {4 \ln \left (f \right )^{2} a c g h}{4 \ln \left (f \right )^{2} a^{2} c \,h^{2}-\ln \left (f \right )^{2} a \,b^{2} h^{2}}-\frac {\ln \left (f \right )^{2} b^{2} g h}{4 \ln \left (f \right )^{2} a^{2} c \,h^{2}-\ln \left (f \right )^{2} a \,b^{2} h^{2}}-\frac {2 \ln \left (f^{h x +g}-\frac {-b^{2}+\sqrt {-4 b^{2} a c +b^{4}}}{2 b c}\right ) c}{\left (4 a c -b^{2}\right ) h \ln \left (f \right )}+\frac {\ln \left (f^{h x +g}-\frac {-b^{2}+\sqrt {-4 b^{2} a c +b^{4}}}{2 b c}\right ) b^{2}}{2 a \left (4 a c -b^{2}\right ) h \ln \left (f \right )}+\frac {\ln \left (f^{h x +g}-\frac {-b^{2}+\sqrt {-4 b^{2} a c +b^{4}}}{2 b c}\right ) \sqrt {-4 b^{2} a c +b^{4}}}{2 a \left (4 a c -b^{2}\right ) h \ln \left (f \right )}-\frac {2 \ln \left (f^{h x +g}+\frac {b^{2}+\sqrt {-4 b^{2} a c +b^{4}}}{2 b c}\right ) c}{\left (4 a c -b^{2}\right ) h \ln \left (f \right )}+\frac {\ln \left (f^{h x +g}+\frac {b^{2}+\sqrt {-4 b^{2} a c +b^{4}}}{2 b c}\right ) b^{2}}{2 a \left (4 a c -b^{2}\right ) h \ln \left (f \right )}-\frac {\ln \left (f^{h x +g}+\frac {b^{2}+\sqrt {-4 b^{2} a c +b^{4}}}{2 b c}\right ) \sqrt {-4 b^{2} a c +b^{4}}}{2 a \left (4 a c -b^{2}\right ) h \ln \left (f \right )}\) | \(546\) |
Input:
int(1/(a+b*f^(h*x+g)+c*f^(2*h*x+2*g)),x,method=_RETURNVERBOSE)
Output:
4/(4*ln(f)^2*a^2*c*h^2-ln(f)^2*a*b^2*h^2)*ln(f)^2*a*c*h^2*x-1/(4*ln(f)^2*a ^2*c*h^2-ln(f)^2*a*b^2*h^2)*ln(f)^2*b^2*h^2*x+4/(4*ln(f)^2*a^2*c*h^2-ln(f) ^2*a*b^2*h^2)*ln(f)^2*a*c*g*h-1/(4*ln(f)^2*a^2*c*h^2-ln(f)^2*a*b^2*h^2)*ln (f)^2*b^2*g*h-2/(4*a*c-b^2)/h/ln(f)*ln(f^(h*x+g)-1/2*(-b^2+(-4*a*b^2*c+b^4 )^(1/2))/b/c)*c+1/2/a/(4*a*c-b^2)/h/ln(f)*ln(f^(h*x+g)-1/2*(-b^2+(-4*a*b^2 *c+b^4)^(1/2))/b/c)*b^2+1/2/a/(4*a*c-b^2)/h/ln(f)*ln(f^(h*x+g)-1/2*(-b^2+( -4*a*b^2*c+b^4)^(1/2))/b/c)*(-4*a*b^2*c+b^4)^(1/2)-2/(4*a*c-b^2)/h/ln(f)*l n(f^(h*x+g)+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*c+1/2/a/(4*a*c-b^2)/h/ln (f)*ln(f^(h*x+g)+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*b^2-1/2/a/(4*a*c-b^ 2)/h/ln(f)*ln(f^(h*x+g)+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*(-4*a*b^2*c+ b^4)^(1/2)
Time = 0.09 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.29 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=\left [\frac {2 \, {\left (b^{2} - 4 \, a c\right )} h x \log \left (f\right ) + \sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} f^{2 \, h x + 2 \, g} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} f^{h x + g} + \sqrt {b^{2} - 4 \, a c} b}{c f^{2 \, h x + 2 \, g} + b f^{h x + g} + a}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c f^{2 \, h x + 2 \, g} + b f^{h x + g} + a\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} h \log \left (f\right )}, \frac {2 \, {\left (b^{2} - 4 \, a c\right )} h x \log \left (f\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c f^{h x + g} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c f^{2 \, h x + 2 \, g} + b f^{h x + g} + a\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} h \log \left (f\right )}\right ] \] Input:
integrate(1/(a+b*f^(h*x+g)+c*f^(2*h*x+2*g)),x, algorithm="fricas")
Output:
[1/2*(2*(b^2 - 4*a*c)*h*x*log(f) + sqrt(b^2 - 4*a*c)*b*log((2*c^2*f^(2*h*x + 2*g) + b^2 - 2*a*c + 2*(b*c + sqrt(b^2 - 4*a*c)*c)*f^(h*x + g) + sqrt(b ^2 - 4*a*c)*b)/(c*f^(2*h*x + 2*g) + b*f^(h*x + g) + a)) - (b^2 - 4*a*c)*lo g(c*f^(2*h*x + 2*g) + b*f^(h*x + g) + a))/((a*b^2 - 4*a^2*c)*h*log(f)), 1/ 2*(2*(b^2 - 4*a*c)*h*x*log(f) + 2*sqrt(-b^2 + 4*a*c)*b*arctan(-(2*sqrt(-b^ 2 + 4*a*c)*c*f^(h*x + g) + sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) - (b^2 - 4 *a*c)*log(c*f^(2*h*x + 2*g) + b*f^(h*x + g) + a))/((a*b^2 - 4*a^2*c)*h*log (f))]
Time = 0.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=\operatorname {RootSum} {\left (z^{2} \cdot \left (4 a^{2} c h^{2} \log {\left (f \right )}^{2} - a b^{2} h^{2} \log {\left (f \right )}^{2}\right ) + z \left (4 a c h \log {\left (f \right )} - b^{2} h \log {\left (f \right )}\right ) + c, \left ( i \mapsto i \log {\left (f^{g + h x} + \frac {- 4 i a^{2} c h \log {\left (f \right )} + i a b^{2} h \log {\left (f \right )} - 2 a c + b^{2}}{b c} \right )} \right )\right )} + \frac {x}{a} \] Input:
integrate(1/(a+b*f**(h*x+g)+c*f**(2*h*x+2*g)),x)
Output:
RootSum(_z**2*(4*a**2*c*h**2*log(f)**2 - a*b**2*h**2*log(f)**2) + _z*(4*a* c*h*log(f) - b**2*h*log(f)) + c, Lambda(_i, _i*log(f**(g + h*x) + (-4*_i*a **2*c*h*log(f) + _i*a*b**2*h*log(f) - 2*a*c + b**2)/(b*c)))) + x/a
Exception generated. \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*f^(h*x+g)+c*f^(2*h*x+2*g)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.12 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.17 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=-\frac {\frac {2 \, b \arctan \left (\frac {2 \, c f^{h x} f^{g} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a \log \left (f\right )} + \frac {\log \left (c f^{2 \, h x} f^{2 \, g} + b f^{h x} f^{g} + a\right )}{a \log \left (f\right )} - \frac {2 \, \log \left ({\left | f \right |}^{h x} {\left | f \right |}^{g}\right )}{a \log \left (f\right )}}{2 \, h} \] Input:
integrate(1/(a+b*f^(h*x+g)+c*f^(2*h*x+2*g)),x, algorithm="giac")
Output:
-1/2*(2*b*arctan((2*c*f^(h*x)*f^g + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4* a*c)*a*log(f)) + log(c*f^(2*h*x)*f^(2*g) + b*f^(h*x)*f^g + a)/(a*log(f)) - 2*log(abs(f)^(h*x)*abs(f)^g)/(a*log(f)))/h
Time = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=\frac {x}{a}-\frac {\ln \left (a+c\,f^{2\,h\,x}\,f^{2\,g}+b\,f^{h\,x}\,f^g\right )}{2\,a\,h\,\ln \left (f\right )}-\frac {b\,\mathrm {atan}\left (\frac {b+2\,c\,f^{h\,x}\,f^g}{\sqrt {4\,a\,c-b^2}}\right )}{a\,h\,\ln \left (f\right )\,\sqrt {4\,a\,c-b^2}} \] Input:
int(1/(a + b*f^(g + h*x) + c*f^(2*g + 2*h*x)),x)
Output:
x/a - log(a + c*f^(2*h*x)*f^(2*g) + b*f^(h*x)*f^g)/(2*a*h*log(f)) - (b*ata n((b + 2*c*f^(h*x)*f^g)/(4*a*c - b^2)^(1/2)))/(a*h*log(f)*(4*a*c - b^2)^(1 /2))
Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.48 \[ \int \frac {1}{a+b f^{g+h x}+c f^{2 (g+h x)}} \, dx=\frac {-2 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 f^{h x +g} c +b}{\sqrt {4 a c -b^{2}}}\right ) b -4 \,\mathrm {log}\left (f^{2 h x +2 g} c +f^{h x +g} b +a \right ) a c +\mathrm {log}\left (f^{2 h x +2 g} c +f^{h x +g} b +a \right ) b^{2}+8 \,\mathrm {log}\left (f \right ) a c h x -2 \,\mathrm {log}\left (f \right ) b^{2} h x}{2 \,\mathrm {log}\left (f \right ) a h \left (4 a c -b^{2}\right )} \] Input:
int(1/(a+b*f^(h*x+g)+c*f^(2*h*x+2*g)),x)
Output:
( - 2*sqrt(4*a*c - b**2)*atan((2*f**(g + h*x)*c + b)/sqrt(4*a*c - b**2))*b - 4*log(f**(2*g + 2*h*x)*c + f**(g + h*x)*b + a)*a*c + log(f**(2*g + 2*h* x)*c + f**(g + h*x)*b + a)*b**2 + 8*log(f)*a*c*h*x - 2*log(f)*b**2*h*x)/(2 *log(f)*a*h*(4*a*c - b**2))