Integrand size = 23, antiderivative size = 127 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=-\frac {c f^{\frac {1}{2} (2 a-5 e)+\frac {1}{2} (e+2 b x)}}{b d^2 \log (f)}+\frac {f^{\frac {1}{2} (2 a-5 e)+\frac {3}{2} (e+2 b x)}}{3 b d \log (f)}+\frac {c^{3/2} f^{a-\frac {5 e}{2}} \arctan \left (\frac {\sqrt {d} f^{\frac {1}{2} (e+2 b x)}}{\sqrt {c}}\right )}{b d^{5/2} \log (f)} \]
-c*f^(b*x+a-2*e)/b/d^2/ln(f)+1/3*f^(3*b*x+a-e)/b/d/ln(f)+c^(3/2)*f^(a-5/2* e)*arctan(f^(1/2*e+b*x)*d^(1/2)/c^(1/2))/b/d^(5/2)/ln(f)
Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.67 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^a \left (-\frac {c f^{-2 e+b x}}{d^2}+\frac {f^{-e+3 b x}}{3 d}+\frac {c^{3/2} f^{-5 e/2} \arctan \left (\frac {\sqrt {d} f^{\frac {e}{2}+b x}}{\sqrt {c}}\right )}{d^{5/2}}\right )}{b \log (f)} \]
(f^a*(-((c*f^(-2*e + b*x))/d^2) + f^(-e + 3*b*x)/(3*d) + (c^(3/2)*ArcTan[( Sqrt[d]*f^(e/2 + b*x))/Sqrt[c]])/(d^(5/2)*f^((5*e)/2))))/(b*Log[f])
Time = 0.25 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2678, 254, 2009}
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.
\(\displaystyle \int \frac {f^{a+5 b x}}{d f^{2 b x+e}+c} \, dx\) |
\(\Big \downarrow \) 2678 |
\(\displaystyle \frac {f^{a-\frac {5 e}{2}} \int \frac {f^{2 (e+2 b x)}}{d f^{e+2 b x}+c}df^{\frac {1}{2} (e+2 b x)}}{b \log (f)}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {f^{a-\frac {5 e}{2}} \int \left (\frac {f^{e+2 b x}}{d}+\frac {c^2}{d^2 \left (d f^{e+2 b x}+c\right )}-\frac {c}{d^2}\right )df^{\frac {1}{2} (e+2 b x)}}{b \log (f)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f^{a-\frac {5 e}{2}} \left (\frac {c^{3/2} \arctan \left (\frac {\sqrt {d} f^{\frac {1}{2} (2 b x+e)}}{\sqrt {c}}\right )}{d^{5/2}}-\frac {c f^{\frac {1}{2} (2 b x+e)}}{d^2}+\frac {f^{\frac {3}{2} (2 b x+e)}}{3 d}\right )}{b \log (f)}\) |
(f^(a - (5*e)/2)*(-((c*f^((e + 2*b*x)/2))/d^2) + f^((3*(e + 2*b*x))/2)/(3* d) + (c^(3/2)*ArcTan[(Sqrt[d]*f^((e + 2*b*x)/2))/Sqrt[c]])/d^(5/2)))/(b*Lo g[f])
3.1.37.3.1 Defintions of rubi rules used
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_ .) + (g_.)*(x_))), x_Symbol] :> With[{m = FullSimplify[g*h*(Log[G]/(d*e*Log [F]))]}, Simp[Denominator[m]*(G^(f*h - c*g*(h/d))/(d*e*Log[F])) Subst[Int [x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^(e*((c + d*x)/De nominator[m]))], x] /; LeQ[m, -1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(211\) vs. \(2(85)=170\).
Time = 0.06 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.67
method | result | size |
risch | \(\frac {f^{-e} f^{\frac {2 a}{5}} f^{3 b x +\frac {3 a}{5}}}{3 d \ln \left (f \right ) b}-\frac {c \,f^{-2 e} f^{\frac {4 a}{5}} f^{b x +\frac {a}{5}}}{d^{2} \ln \left (f \right ) b}+\frac {\sqrt {-c d}\, c \,f^{a} f^{-\frac {5 e}{2}} \ln \left (f^{b x +\frac {a}{5}}+\frac {\sqrt {-c d}\, f^{\frac {a}{5}} f^{-\frac {e}{2}}}{d}\right )}{2 d^{3} b \ln \left (f \right )}-\frac {\sqrt {-c d}\, c \,f^{a} f^{-\frac {5 e}{2}} \ln \left (f^{b x +\frac {a}{5}}-\frac {\sqrt {-c d}\, f^{\frac {a}{5}} f^{-\frac {e}{2}}}{d}\right )}{2 d^{3} b \ln \left (f \right )}\) | \(212\) |
1/3/(f^(1/2*e))^2/(f^(-1/5*a))^2/d/ln(f)/b*(f^(b*x+1/5*a))^3-c/(f^(1/2*e)) ^4/(f^(-1/5*a))^4/d^2/ln(f)/b*f^(b*x+1/5*a)+1/2/d^3*(-c*d)^(1/2)*c/b/(f^(- 1/5*a))^5/(f^(1/2*e))^5/ln(f)*ln(f^(b*x+1/5*a)+1/d*(-c*d)^(1/2)/(f^(-1/5*a ))/(f^(1/2*e)))-1/2/d^3*(-c*d)^(1/2)*c/b/(f^(-1/5*a))^5/(f^(1/2*e))^5/ln(f )*ln(f^(b*x+1/5*a)-1/d*(-c*d)^(1/2)/(f^(-1/5*a))/(f^(1/2*e)))
Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.66 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\left [\frac {3 \, c f^{a - \frac {5}{2} \, e} \sqrt {-\frac {c}{d}} \log \left (\frac {2 \, d f^{b x + \frac {1}{2} \, e} \sqrt {-\frac {c}{d}} + d f^{2 \, b x + e} - c}{d f^{2 \, b x + e} + c}\right ) + 2 \, d f^{3 \, b x + \frac {3}{2} \, e} f^{a - \frac {5}{2} \, e} - 6 \, c f^{b x + \frac {1}{2} \, e} f^{a - \frac {5}{2} \, e}}{6 \, b d^{2} \log \left (f\right )}, \frac {3 \, c f^{a - \frac {5}{2} \, e} \sqrt {\frac {c}{d}} \arctan \left (\frac {d f^{b x + \frac {1}{2} \, e} \sqrt {\frac {c}{d}}}{c}\right ) + d f^{3 \, b x + \frac {3}{2} \, e} f^{a - \frac {5}{2} \, e} - 3 \, c f^{b x + \frac {1}{2} \, e} f^{a - \frac {5}{2} \, e}}{3 \, b d^{2} \log \left (f\right )}\right ] \]
[1/6*(3*c*f^(a - 5/2*e)*sqrt(-c/d)*log((2*d*f^(b*x + 1/2*e)*sqrt(-c/d) + d *f^(2*b*x + e) - c)/(d*f^(2*b*x + e) + c)) + 2*d*f^(3*b*x + 3/2*e)*f^(a - 5/2*e) - 6*c*f^(b*x + 1/2*e)*f^(a - 5/2*e))/(b*d^2*log(f)), 1/3*(3*c*f^(a - 5/2*e)*sqrt(c/d)*arctan(d*f^(b*x + 1/2*e)*sqrt(c/d)/c) + d*f^(3*b*x + 3/ 2*e)*f^(a - 5/2*e) - 3*c*f^(b*x + 1/2*e)*f^(a - 5/2*e))/(b*d^2*log(f))]
Time = 0.75 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.46 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\operatorname {RootSum} {\left (4 z^{2} b^{2} d^{5} e^{5 e \log {\left (f \right )}} \log {\left (f \right )}^{2} + c^{3} e^{2 a \log {\left (f \right )}}, \left ( i \mapsto i \log {\left (\frac {2 i b d^{2} e^{- \frac {4 a \log {\left (f \right )}}{5}} e^{2 e \log {\left (f \right )}} \log {\left (f \right )}}{c} + e^{\frac {\left (a + 5 b x\right ) \log {\left (f \right )}}{5}} \right )} \right )\right )} + \frac {\left (\begin {cases} x \left (- c + d\right ) & \text {for}\: b = 0 \wedge f = 1 \\x \left (- c e^{a \log {\left (f \right )}} + d e^{a \log {\left (f \right )}} e^{e \log {\left (f \right )}}\right ) & \text {for}\: b = 0 \\x \left (- c + d\right ) & \text {for}\: f = 1 \\- \frac {c e^{a \log {\left (f \right )}} e^{b x \log {\left (f \right )}}}{b \log {\left (f \right )}} + \frac {d e^{a \log {\left (f \right )}} e^{e \log {\left (f \right )}} e^{3 b x \log {\left (f \right )}}}{3 b \log {\left (f \right )}} & \text {otherwise} \end {cases}\right ) e^{- 2 e \log {\left (f \right )}}}{d^{2}} \]
RootSum(4*_z**2*b**2*d**5*exp(5*e*log(f))*log(f)**2 + c**3*exp(2*a*log(f)) , Lambda(_i, _i*log(2*_i*b*d**2*exp(-4*a*log(f)/5)*exp(2*e*log(f))*log(f)/ c + exp((a + 5*b*x)*log(f)/5)))) + Piecewise((x*(-c + d), Eq(b, 0) & Eq(f, 1)), (x*(-c*exp(a*log(f)) + d*exp(a*log(f))*exp(e*log(f))), Eq(b, 0)), (x *(-c + d), Eq(f, 1)), (-c*exp(a*log(f))*exp(b*x*log(f))/(b*log(f)) + d*exp (a*log(f))*exp(e*log(f))*exp(3*b*x*log(f))/(3*b*log(f)), True))*exp(-2*e*l og(f))/d**2
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {c^{2} f^{a - 2 \, e} \arctan \left (\frac {d f^{b x + e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} b d^{2} \log \left (f\right )} + \frac {d f^{3 \, b x + a + e} - 3 \, c f^{b x + a}}{3 \, b d^{2} f^{2 \, e} \log \left (f\right )} \]
c^2*f^(a - 2*e)*arctan(d*f^(b*x + e)/sqrt(c*d*f^e))/(sqrt(c*d*f^e)*b*d^2*l og(f)) + 1/3*(d*f^(3*b*x + a + e) - 3*c*f^(b*x + a))/(b*d^2*f^(2*e)*log(f) )
Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a} {\left (\frac {3 \, c^{2} \arctan \left (\frac {d f^{b x} f^{e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} d^{2} f^{2 \, e} \log \left (f\right )} + \frac {d^{2} f^{3 \, b x} f^{2 \, e} \log \left (f\right )^{2} - 3 \, c d f^{b x} f^{e} \log \left (f\right )^{2}}{d^{3} f^{3 \, e} \log \left (f\right )^{3}}\right )}}{3 \, b} \]
1/3*f^a*(3*c^2*arctan(d*f^(b*x)*f^e/sqrt(c*d*f^e))/(sqrt(c*d*f^e)*d^2*f^(2 *e)*log(f)) + (d^2*f^(3*b*x)*f^(2*e)*log(f)^2 - 3*c*d*f^(b*x)*f^e*log(f)^2 )/(d^3*f^(3*e)*log(f)^3))/b
Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^a\,f^{3\,b\,x}}{3\,b\,d\,f^e\,\ln \left (f\right )}-\frac {c\,f^a\,f^{b\,x}}{b\,d^2\,f^{2\,e}\,\ln \left (f\right )}+\frac {c^2\,f^a\,{\mathrm {e}}^{-\frac {5\,e\,\ln \left (f\right )}{2}}\,\mathrm {atan}\left (\frac {d\,f^{b\,x}\,{\mathrm {e}}^{\frac {e\,\ln \left (f\right )}{2}}}{\sqrt {c\,d}}\right )}{b\,d^2\,\ln \left (f\right )\,\sqrt {c\,d}} \]