\(\int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx\) [521]

Optimal result

Integrand size = 25, antiderivative size = 94 \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {x}{a}+\frac {b \text {arctanh}\left (\frac {b+2 c f^{c+d x}}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} d \log (f)}-\frac {\log \left (a+b f^{c+d x}+c f^{2 c+2 d x}\right )}{2 a d \log (f)} \]

[Out]

x/a-1/2*ln(a+b*f^(d*x+c)+c*f^(2*d*x+2*c))/a/d/ln(f)+b*arctanh((b+2*c*f^(d*x+c))/(-4*a*c+b^2)^(1/2))/a/d/ln(f)/
(-4*a*c+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2320, 719, 29, 648, 632, 212, 642} \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {b \text {arctanh}\left (\frac {b+2 c f^{c+d x}}{\sqrt {b^2-4 a c}}\right )}{a d \log (f) \sqrt {b^2-4 a c}}-\frac {\log \left (a+b f^{c+d x}+c f^{2 c+2 d x}\right )}{2 a d \log (f)}+\frac {x}{a} \]

[In]

Int[(a + b*f^(c + d*x) + c*f^(2*c + 2*d*x))^(-1),x]

[Out]

x/a + (b*ArcTanh[(b + 2*c*f^(c + d*x))/Sqrt[b^2 - 4*a*c]])/(a*Sqrt[b^2 - 4*a*c]*d*Log[f]) - Log[a + b*f^(c + d
*x) + c*f^(2*c + 2*d*x)]/(2*a*d*Log[f])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 719

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{x \left (a+b x+c x^2\right )} \, dx,x,f^{c+d x}\right )}{d \log (f)} \\ & = \frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,f^{c+d x}\right )}{a d \log (f)}+\frac {\text {Subst}\left (\int \frac {-b-c x}{a+b x+c x^2} \, dx,x,f^{c+d x}\right )}{a d \log (f)} \\ & = \frac {x}{a}-\frac {\text {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,f^{c+d x}\right )}{2 a d \log (f)}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,f^{c+d x}\right )}{2 a d \log (f)} \\ & = \frac {x}{a}-\frac {\log \left (a+b f^{c+d x}+c f^{2 c+2 d x}\right )}{2 a d \log (f)}+\frac {b \text {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c f^{c+d x}\right )}{a d \log (f)} \\ & = \frac {x}{a}+\frac {b \tanh ^{-1}\left (\frac {b+2 c f^{c+d x}}{\sqrt {b^2-4 a c}}\right )}{a \sqrt {b^2-4 a c} d \log (f)}-\frac {\log \left (a+b f^{c+d x}+c f^{2 c+2 d x}\right )}{2 a d \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.98 \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=-\frac {\frac {2 b \arctan \left (\frac {b+2 c f^{c+d x}}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-2 \log \left (f^{c+d x}\right )+\log \left (a+f^{c+d x} \left (b+c f^{c+d x}\right )\right )}{2 a d \log (f)} \]

[In]

Integrate[(a + b*f^(c + d*x) + c*f^(2*c + 2*d*x))^(-1),x]

[Out]

-1/2*((2*b*ArcTan[(b + 2*c*f^(c + d*x))/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] - 2*Log[f^(c + d*x)] + Log[a +
f^(c + d*x)*(b + c*f^(c + d*x))])/(a*d*Log[f])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(546\) vs. \(2(88)=176\).

Time = 0.08 (sec) , antiderivative size = 547, normalized size of antiderivative = 5.82

method	result	size
risch	\(\frac {4 \ln \left (f \right )^{2} a c \,d^{2} x}{4 \ln \left (f \right )^{2} a^{2} c \,d^{2}-\ln \left (f \right )^{2} a \,b^{2} d^{2}}-\frac {\ln \left (f \right )^{2} b^{2} d^{2} x}{4 \ln \left (f \right )^{2} a^{2} c \,d^{2}-\ln \left (f \right )^{2} a \,b^{2} d^{2}}+\frac {4 \ln \left (f \right )^{2} a \,c^{2} d}{4 \ln \left (f \right )^{2} a^{2} c \,d^{2}-\ln \left (f \right )^{2} a \,b^{2} d^{2}}-\frac {\ln \left (f \right )^{2} b^{2} c d}{4 \ln \left (f \right )^{2} a^{2} c \,d^{2}-\ln \left (f \right )^{2} a \,b^{2} d^{2}}-\frac {2 \ln \left (f^{d x +c}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 c b}\right ) c}{\left (4 c a -b^{2}\right ) d \ln \left (f \right )}+\frac {\ln \left (f^{d x +c}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 c b}\right ) b^{2}}{2 a \left (4 c a -b^{2}\right ) d \ln \left (f \right )}+\frac {\ln \left (f^{d x +c}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 c b}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 a \left (4 c a -b^{2}\right ) d \ln \left (f \right )}-\frac {2 \ln \left (f^{d x +c}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 c b}\right ) c}{\left (4 c a -b^{2}\right ) d \ln \left (f \right )}+\frac {\ln \left (f^{d x +c}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 c b}\right ) b^{2}}{2 a \left (4 c a -b^{2}\right ) d \ln \left (f \right )}-\frac {\ln \left (f^{d x +c}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 c b}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 a \left (4 c a -b^{2}\right ) d \ln \left (f \right )}\)	\(547\)

[In]

int(1/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x,method=_RETURNVERBOSE)

[Out]

4/(4*ln(f)^2*a^2*c*d^2-ln(f)^2*a*b^2*d^2)*ln(f)^2*a*c*d^2*x-1/(4*ln(f)^2*a^2*c*d^2-ln(f)^2*a*b^2*d^2)*ln(f)^2*
b^2*d^2*x+4/(4*ln(f)^2*a^2*c*d^2-ln(f)^2*a*b^2*d^2)*ln(f)^2*a*c^2*d-1/(4*ln(f)^2*a^2*c*d^2-ln(f)^2*a*b^2*d^2)*
ln(f)^2*b^2*c*d-2/(4*a*c-b^2)/d/ln(f)*ln(f^(d*x+c)-1/2*(-b^2+(-4*a*b^2*c+b^4)^(1/2))/c/b)*c+1/2/a/(4*a*c-b^2)/
d/ln(f)*ln(f^(d*x+c)-1/2*(-b^2+(-4*a*b^2*c+b^4)^(1/2))/c/b)*b^2+1/2/a/(4*a*c-b^2)/d/ln(f)*ln(f^(d*x+c)-1/2*(-b
^2+(-4*a*b^2*c+b^4)^(1/2))/c/b)*(-4*a*b^2*c+b^4)^(1/2)-2/(4*a*c-b^2)/d/ln(f)*ln(f^(d*x+c)+1/2*(b^2+(-4*a*b^2*c
+b^4)^(1/2))/c/b)*c+1/2/a/(4*a*c-b^2)/d/ln(f)*ln(f^(d*x+c)+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/c/b)*b^2-1/2/a/(4*
a*c-b^2)/d/ln(f)*ln(f^(d*x+c)+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/c/b)*(-4*a*b^2*c+b^4)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.29 \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\left [\frac {2 \, {\left (b^{2} - 4 \, a c\right )} d x \log \left (f\right ) + \sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} f^{2 \, d x + 2 \, c} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} f^{d x + c} + \sqrt {b^{2} - 4 \, a c} b}{c f^{2 \, d x + 2 \, c} + b f^{d x + c} + a}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c f^{2 \, d x + 2 \, c} + b f^{d x + c} + a\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d \log \left (f\right )}, \frac {2 \, {\left (b^{2} - 4 \, a c\right )} d x \log \left (f\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c f^{d x + c} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c f^{2 \, d x + 2 \, c} + b f^{d x + c} + a\right )}{2 \, {\left (a b^{2} - 4 \, a^{2} c\right )} d \log \left (f\right )}\right ] \]

[In]

integrate(1/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x, algorithm="fricas")

[Out]

[1/2*(2*(b^2 - 4*a*c)*d*x*log(f) + sqrt(b^2 - 4*a*c)*b*log((2*c^2*f^(2*d*x + 2*c) + b^2 - 2*a*c + 2*(b*c + sqr
t(b^2 - 4*a*c)*c)*f^(d*x + c) + sqrt(b^2 - 4*a*c)*b)/(c*f^(2*d*x + 2*c) + b*f^(d*x + c) + a)) - (b^2 - 4*a*c)*
log(c*f^(2*d*x + 2*c) + b*f^(d*x + c) + a))/((a*b^2 - 4*a^2*c)*d*log(f)), 1/2*(2*(b^2 - 4*a*c)*d*x*log(f) + 2*
sqrt(-b^2 + 4*a*c)*b*arctan(-(2*sqrt(-b^2 + 4*a*c)*c*f^(d*x + c) + sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) - (b^2
- 4*a*c)*log(c*f^(2*d*x + 2*c) + b*f^(d*x + c) + a))/((a*b^2 - 4*a^2*c)*d*log(f))]

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.11 \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\operatorname {RootSum} {\left (z^{2} \cdot \left (4 a^{2} c d^{2} \log {\left (f \right )}^{2} - a b^{2} d^{2} \log {\left (f \right )}^{2}\right ) + z \left (4 a c d \log {\left (f \right )} - b^{2} d \log {\left (f \right )}\right ) + c, \left ( i \mapsto i \log {\left (f^{c + d x} + \frac {- 4 i a^{2} c d \log {\left (f \right )} + i a b^{2} d \log {\left (f \right )} - 2 a c + b^{2}}{b c} \right )} \right )\right )} + \frac {x}{a} \]

[In]

integrate(1/(a+b*f**(d*x+c)+c*f**(2*d*x+2*c)),x)

[Out]

RootSum(_z**2*(4*a**2*c*d**2*log(f)**2 - a*b**2*d**2*log(f)**2) + _z*(4*a*c*d*log(f) - b**2*d*log(f)) + c, Lam
bda(_i, _i*log(f**(c + d*x) + (-4*_i*a**2*c*d*log(f) + _i*a*b**2*d*log(f) - 2*a*c + b**2)/(b*c)))) + x/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.17 \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=-\frac {\frac {2 \, b \arctan \left (\frac {2 \, c f^{d x} f^{c} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a \log \left (f\right )} + \frac {\log \left (c f^{2 \, d x} f^{2 \, c} + b f^{d x} f^{c} + a\right )}{a \log \left (f\right )} - \frac {2 \, \log \left ({\left | f \right |}^{d x} {\left | f \right |}^{c}\right )}{a \log \left (f\right )}}{2 \, d} \]

[In]

integrate(1/(a+b*f^(d*x+c)+c*f^(2*d*x+2*c)),x, algorithm="giac")

[Out]

-1/2*(2*b*arctan((2*c*f^(d*x)*f^c + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*a*log(f)) + log(c*f^(2*d*x)*f^(
2*c) + b*f^(d*x)*f^c + a)/(a*log(f)) - 2*log(abs(f)^(d*x)*abs(f)^c)/(a*log(f)))/d

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx=\frac {x}{a}-\frac {\ln \left (a+c\,f^{2\,d\,x}\,f^{2\,c}+b\,f^{d\,x}\,f^c\right )}{2\,a\,d\,\ln \left (f\right )}-\frac {b\,\mathrm {atan}\left (\frac {b+2\,c\,f^{d\,x}\,f^c}{\sqrt {4\,a\,c-b^2}}\right )}{a\,d\,\ln \left (f\right )\,\sqrt {4\,a\,c-b^2}} \]

[In]

int(1/(a + b*f^(c + d*x) + c*f^(2*c + 2*d*x)),x)

[Out]

x/a - log(a + c*f^(2*d*x)*f^(2*c) + b*f^(d*x)*f^c)/(2*a*d*log(f)) - (b*atan((b + 2*c*f^(d*x)*f^c)/(4*a*c - b^2
)^(1/2)))/(a*d*log(f)*(4*a*c - b^2)^(1/2))