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

Optimal. Leaf size=94 \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c f^{c+d x}}{\sqrt{b^2-4 a c}}\right )}{a d \log (f) \sqrt{b^2-4 a c}}-\frac{\log \left (a+b f^{c+d x}+c f^{2 c+2 d x}\right )}{2 a d \log (f)}+\frac{x}{a} \]

[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])

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Rubi [A]  time = 0.108036, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2282, 705, 29, 634, 618, 206, 628} \[ \frac{b \tanh ^{-1}\left (\frac{b+2 c f^{c+d x}}{\sqrt{b^2-4 a c}}\right )}{a d \log (f) \sqrt{b^2-4 a c}}-\frac{\log \left (a+b f^{c+d x}+c f^{2 c+2 d x}\right )}{2 a d \log (f)}+\frac{x}{a} \]

Antiderivative was successfully verified.

[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 2282

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]]]

Rule 705

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 29

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

Rule 634

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 618

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 628

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]

Rubi steps

\begin{align*} \int \frac{1}{a+b f^{c+d x}+c f^{2 c+2 d x}} \, dx &=\frac{\operatorname{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{\operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,f^{c+d x}\right )}{a d \log (f)}+\frac{\operatorname{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{\operatorname{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 \operatorname{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 \operatorname{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]  time = 0.137525, size = 93, normalized size = 0.99 \[ -\frac{\frac{2 b \tan ^{-1}\left (\frac{b+2 c f^{c+d x}}{\sqrt{4 a c-b^2}}\right )}{d \log (f) \sqrt{4 a c-b^2}}+\frac{\log \left (a+f^{c+d x} \left (b+c f^{c+d x}\right )\right )}{d \log (f)}-2 x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.085, size = 547, normalized size = 5.8 \begin{align*} 4\,{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}ac{d}^{2}x}{4\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}c{d}^{2}- \left ( \ln \left ( f \right ) \right ) ^{2}a{b}^{2}{d}^{2}}}-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{d}^{2}x}{4\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}c{d}^{2}- \left ( \ln \left ( f \right ) \right ) ^{2}a{b}^{2}{d}^{2}}}+4\,{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}a{c}^{2}d}{4\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}c{d}^{2}- \left ( \ln \left ( f \right ) \right ) ^{2}a{b}^{2}{d}^{2}}}-{\frac{ \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}cd}{4\, \left ( \ln \left ( f \right ) \right ) ^{2}{a}^{2}c{d}^{2}- \left ( \ln \left ( f \right ) \right ) ^{2}a{b}^{2}{d}^{2}}}-2\,{\frac{c}{d \left ( 4\,ac-{b}^{2} \right ) \ln \left ( f \right ) }\ln \left ({f}^{dx+c}-1/2\,{\frac{-{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}}{bc}} \right ) }+{\frac{{b}^{2}}{2\,a \left ( 4\,ac-{b}^{2} \right ) d\ln \left ( f \right ) }\ln \left ({f}^{dx+c}-{\frac{1}{2\,bc} \left ( -{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) }+{\frac{1}{2\,a \left ( 4\,ac-{b}^{2} \right ) d\ln \left ( f \right ) }\ln \left ({f}^{dx+c}-{\frac{1}{2\,bc} \left ( -{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) \sqrt{-4\,a{b}^{2}c+{b}^{4}}}-2\,{\frac{c}{d \left ( 4\,ac-{b}^{2} \right ) \ln \left ( f \right ) }\ln \left ({f}^{dx+c}+1/2\,{\frac{{b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}}}{bc}} \right ) }+{\frac{{b}^{2}}{2\,a \left ( 4\,ac-{b}^{2} \right ) d\ln \left ( f \right ) }\ln \left ({f}^{dx+c}+{\frac{1}{2\,bc} \left ({b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) }-{\frac{1}{2\,a \left ( 4\,ac-{b}^{2} \right ) d\ln \left ( f \right ) }\ln \left ({f}^{dx+c}+{\frac{1}{2\,bc} \left ({b}^{2}+\sqrt{-4\,a{b}^{2}c+{b}^{4}} \right ) } \right ) \sqrt{-4\,a{b}^{2}c+{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))/b/c)*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))/b/c)*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))/b/c)*(-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))/b/c)*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))/b/c)*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))/b/c)*(-4*a*b^2*c+b^4)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68393, size = 711, normalized size = 7.56 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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Sympy [A]  time = 0.420756, size = 104, normalized size = 1.11 \begin{align*} \operatorname{RootSum}{\left (z^{2} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{c f^{2 \, d x + 2 \, c} + b f^{d x + c} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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