\(\int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx\) [37]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=-\frac {c f^{\frac {1}{2} (2 a-5 e)+\frac {1}{2} (e+2 b x)}}{b d^2 \log (f)}+\frac {f^{\frac {1}{2} (2 a-5 e)+\frac {3}{2} (e+2 b x)}}{3 b d \log (f)}+\frac {c^{3/2} f^{a-\frac {5 e}{2}} \arctan \left (\frac {\sqrt {d} f^{\frac {1}{2} (e+2 b x)}}{\sqrt {c}}\right )}{b d^{5/2} \log (f)} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2280, 308, 211} \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {c^{3/2} f^{a-\frac {5 e}{2}} \arctan \left (\frac {\sqrt {d} f^{\frac {1}{2} (2 b x+e)}}{\sqrt {c}}\right )}{b d^{5/2} \log (f)}-\frac {c f^{\frac {1}{2} (2 a-5 e)+\frac {1}{2} (2 b x+e)}}{b d^2 \log (f)}+\frac {f^{\frac {1}{2} (2 a-5 e)+\frac {3}{2} (2 b x+e)}}{3 b d \log (f)} \]

[In]

Int[f^(a + 5*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

-((c*f^((2*a - 5*e)/2 + (e + 2*b*x)/2))/(b*d^2*Log[f])) + f^((2*a - 5*e)/2 + (3*(e + 2*b*x))/2)/(3*b*d*Log[f])
+ (c^(3/2)*f^(a - (5*e)/2)*ArcTan[(Sqrt[d]*f^((e + 2*b*x)/2))/Sqrt[c]])/(b*d^(5/2)*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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2280

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a-\frac {5 e}{2}} \text {Subst}\left (\int \frac {x^4}{c+d x^2} \, dx,x,f^{\frac {1}{2} (e+2 b x)}\right )}{b \log (f)} \\ & = \frac {f^{a-\frac {5 e}{2}} \text {Subst}\left (\int \left (-\frac {c}{d^2}+\frac {x^2}{d}+\frac {c^2}{d^2 \left (c+d x^2\right )}\right ) \, dx,x,f^{\frac {1}{2} (e+2 b x)}\right )}{b \log (f)} \\ & = -\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 {\left (c^2 f^{a-\frac {5 e}{2}}\right ) \text {Subst}\left (\int \frac {1}{c+d x^2} \, dx,x,f^{\frac {1}{2} (e+2 b x)}\right )}{b d^2 \log (f)} \\ & = -\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}} \tan ^{-1}\left (\frac {\sqrt {d} f^{\frac {1}{2} (e+2 b x)}}{\sqrt {c}}\right )}{b d^{5/2} \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.67 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^a \left (-\frac {c f^{-2 e+b x}}{d^2}+\frac {f^{-e+3 b x}}{3 d}+\frac {c^{3/2} f^{-5 e/2} \arctan \left (\frac {\sqrt {d} f^{\frac {e}{2}+b x}}{\sqrt {c}}\right )}{d^{5/2}}\right )}{b \log (f)} \]

[In]

Integrate[f^(a + 5*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

(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])

Maple [B] (verified)

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	\(\frac {f^{-e} f^{\frac {2 a}{5}} f^{3 b x +\frac {3 a}{5}}}{3 d \ln \left (f \right ) b}-\frac {c \,f^{-2 e} f^{\frac {4 a}{5}} f^{b x +\frac {a}{5}}}{d^{2} \ln \left (f \right ) b}+\frac {\sqrt {-c d}\, c \,f^{a} f^{-\frac {5 e}{2}} \ln \left (f^{b x +\frac {a}{5}}+\frac {\sqrt {-c d}\, f^{\frac {a}{5}} f^{-\frac {e}{2}}}{d}\right )}{2 d^{3} b \ln \left (f \right )}-\frac {\sqrt {-c d}\, c \,f^{a} f^{-\frac {5 e}{2}} \ln \left (f^{b x +\frac {a}{5}}-\frac {\sqrt {-c d}\, f^{\frac {a}{5}} f^{-\frac {e}{2}}}{d}\right )}{2 d^{3} b \ln \left (f \right )}\)	\(212\)

[In]

int(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x,method=_RETURNVERBOSE)

[Out]

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)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.66 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\left [\frac {3 \, c f^{a - \frac {5}{2} \, e} \sqrt {-\frac {c}{d}} \log \left (\frac {2 \, d f^{b x + \frac {1}{2} \, e} \sqrt {-\frac {c}{d}} + d f^{2 \, b x + e} - c}{d f^{2 \, b x + e} + c}\right ) + 2 \, d f^{3 \, b x + \frac {3}{2} \, e} f^{a - \frac {5}{2} \, e} - 6 \, c f^{b x + \frac {1}{2} \, e} f^{a - \frac {5}{2} \, e}}{6 \, b d^{2} \log \left (f\right )}, \frac {3 \, c f^{a - \frac {5}{2} \, e} \sqrt {\frac {c}{d}} \arctan \left (\frac {d f^{b x + \frac {1}{2} \, e} \sqrt {\frac {c}{d}}}{c}\right ) + d f^{3 \, b x + \frac {3}{2} \, e} f^{a - \frac {5}{2} \, e} - 3 \, c f^{b x + \frac {1}{2} \, e} f^{a - \frac {5}{2} \, e}}{3 \, b d^{2} \log \left (f\right )}\right ] \]

[In]

integrate(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="fricas")

[Out]

[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))]

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.46 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\operatorname {RootSum} {\left (4 z^{2} b^{2} d^{5} e^{5 e \log {\left (f \right )}} \log {\left (f \right )}^{2} + c^{3} e^{2 a \log {\left (f \right )}}, \left ( i \mapsto i \log {\left (\frac {2 i b d^{2} e^{- \frac {4 a \log {\left (f \right )}}{5}} e^{2 e \log {\left (f \right )}} \log {\left (f \right )}}{c} + e^{\frac {\left (a + 5 b x\right ) \log {\left (f \right )}}{5}} \right )} \right )\right )} + \frac {\left (\begin {cases} x \left (- c + d\right ) & \text {for}\: b = 0 \wedge f = 1 \\x \left (- c e^{a \log {\left (f \right )}} + d e^{a \log {\left (f \right )}} e^{e \log {\left (f \right )}}\right ) & \text {for}\: b = 0 \\x \left (- c + d\right ) & \text {for}\: f = 1 \\- \frac {c e^{a \log {\left (f \right )}} e^{b x \log {\left (f \right )}}}{b \log {\left (f \right )}} + \frac {d e^{a \log {\left (f \right )}} e^{e \log {\left (f \right )}} e^{3 b x \log {\left (f \right )}}}{3 b \log {\left (f \right )}} & \text {otherwise} \end {cases}\right ) e^{- 2 e \log {\left (f \right )}}}{d^{2}} \]

[In]

integrate(f**(5*b*x+a)/(c+d*f**(2*b*x+e)),x)

[Out]

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*lo
g(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*log(f))/d**2

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {c^{2} f^{a - 2 \, e} \arctan \left (\frac {d f^{b x + e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} b d^{2} \log \left (f\right )} + \frac {d f^{3 \, b x + a + e} - 3 \, c f^{b x + a}}{3 \, b d^{2} f^{2 \, e} \log \left (f\right )} \]

[In]

integrate(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="maxima")

[Out]

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*log(f)) + 1/3*(d*f^(3*b*x + a + e) -
3*c*f^(b*x + a))/(b*d^2*f^(2*e)*log(f))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a} {\left (\frac {3 \, c^{2} \arctan \left (\frac {d f^{b x} f^{e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} d^{2} f^{2 \, e} \log \left (f\right )} + \frac {d^{2} f^{3 \, b x} f^{2 \, e} \log \left (f\right )^{2} - 3 \, c d f^{b x} f^{e} \log \left (f\right )^{2}}{d^{3} f^{3 \, e} \log \left (f\right )^{3}}\right )}}{3 \, b} \]

[In]

integrate(f^(5*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.80 \[ \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^a\,f^{3\,b\,x}}{3\,b\,d\,f^e\,\ln \left (f\right )}-\frac {c\,f^a\,f^{b\,x}}{b\,d^2\,f^{2\,e}\,\ln \left (f\right )}+\frac {c^2\,f^a\,{\mathrm {e}}^{-\frac {5\,e\,\ln \left (f\right )}{2}}\,\mathrm {atan}\left (\frac {d\,f^{b\,x}\,{\mathrm {e}}^{\frac {e\,\ln \left (f\right )}{2}}}{\sqrt {c\,d}}\right )}{b\,d^2\,\ln \left (f\right )\,\sqrt {c\,d}} \]

[In]

int(f^(a + 5*b*x)/(c + d*f^(e + 2*b*x)),x)

[Out]

(f^a*f^(3*b*x))/(3*b*d*f^e*log(f)) - (c*f^a*f^(b*x))/(b*d^2*f^(2*e)*log(f)) + (c^2*f^a*exp(-(5*e*log(f))/2)*at
an((d*f^(b*x)*exp((e*log(f))/2))/(c*d)^(1/2)))/(b*d^2*log(f)*(c*d)^(1/2))