\(\int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx\) [33]

Optimal result

Integrand size = 22, antiderivative size = 50 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-\frac {e}{2}} \arctan \left (\frac {\sqrt {d} f^{\frac {e}{2}+b x}}{\sqrt {c}}\right )}{b \sqrt {c} \sqrt {d} \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2281, 211} \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-\frac {e}{2}} \arctan \left (\frac {\sqrt {d} f^{b x+\frac {e}{2}}}{\sqrt {c}}\right )}{b \sqrt {c} \sqrt {d} \log (f)} \]

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2281

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[d*e*(Log[F]/(g*h*Log[G]))]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Denominator[m]))], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-\frac {e}{2}} \arctan \left (\frac {\sqrt {d} f^{\frac {e}{2}+b x}}{\sqrt {c}}\right )}{b \sqrt {c} \sqrt {d} \log (f)} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(90\) vs. \(2(38)=76\).

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.82

method	result	size
risch	\(-\frac {f^{a} \ln \left (f^{b x +a}-\frac {f^{a} c}{\sqrt {-f^{e} c d}}\right )}{2 \sqrt {-f^{e} c d}\, b \ln \left (f \right )}+\frac {f^{a} \ln \left (f^{b x +a}+\frac {f^{a} c}{\sqrt {-f^{e} c d}}\right )}{2 \sqrt {-f^{e} c d}\, b \ln \left (f \right )}\)	\(91\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 179, normalized size of antiderivative = 3.58 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\left [-\frac {\sqrt {-c d f^{-2 \, a + e}} \log \left (\frac {d f^{2 \, b x + 2 \, a} f^{-2 \, a + e} - 2 \, \sqrt {-c d f^{-2 \, a + e}} f^{b x + a} - c}{d f^{2 \, b x + 2 \, a} f^{-2 \, a + e} + c}\right )}{2 \, b c d f^{-2 \, a + e} \log \left (f\right )}, -\frac {\sqrt {c d f^{-2 \, a + e}} \arctan \left (\frac {\sqrt {c d f^{-2 \, a + e}}}{d f^{b x + a} f^{-2 \, a + e}}\right )}{b c d f^{-2 \, a + e} \log \left (f\right )}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.02 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\operatorname {RootSum} {\left (4 z^{2} b^{2} c d e^{e \log {\left (f \right )}} \log {\left (f \right )}^{2} + e^{2 a \log {\left (f \right )}}, \left ( i \mapsto i \log {\left (2 i b c \log {\left (f \right )} + f^{a + b x} \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(4*_z**2*b**2*c*d*exp(e*log(f))*log(f)**2 + exp(2*a*log(f)), Lambda(_i, _i*log(2*_i*b*c*log(f) + f**(a
+ b*x))))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.74 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a} \arctan \left (\frac {d f^{b x + e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} b \log \left (f\right )} \]

[In]

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

[Out]

f^a*arctan(d*f^(b*x + e)/sqrt(c*d*f^e))/(sqrt(c*d*f^e)*b*log(f))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{2 \, a} \arctan \left (\frac {d f^{b x} f^{e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} b f^{a} \log \left (f\right )} \]

[In]

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

[Out]

f^(2*a)*arctan(d*f^(b*x)*f^e/sqrt(c*d*f^e))/(sqrt(c*d*f^e)*b*f^a*log(f))

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {f^{a+b x}}{c+d f^{e+2 b x}} \, dx=\frac {\mathrm {atan}\left (\frac {f^{a+b\,x}\,\sqrt {b^2\,c\,d\,f^e\,{\ln \left (f\right )}^2}}{b\,c\,\ln \left (f\right )\,\sqrt {f^{2\,a}}}\right )\,\sqrt {f^{2\,a}}}{\sqrt {b^2\,c\,d\,f^e\,{\ln \left (f\right )}^2}} \]

[In]

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

[Out]

(atan((f^(a + b*x)*(b^2*c*d*f^e*log(f)^2)^(1/2))/(b*c*log(f)*(f^(2*a))^(1/2)))*(f^(2*a))^(1/2))/(b^2*c*d*f^e*l
og(f)^2)^(1/2)