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

Optimal. Leaf size=127 \[ \frac {c^{3/2} f^{a-\frac {5 e}{2}} \tan ^{-1}\left (\frac {\sqrt {d} f^{\frac {1}{2} (2 b x+e)}}{\sqrt {c}}\right )}{b d^{5/2} \log (f)}-\frac {c f^{\frac {1}{2} (2 a-5 e)+\frac {1}{2} (2 b x+e)}}{b d^2 \log (f)}+\frac {f^{\frac {1}{2} (2 a-5 e)+\frac {3}{2} (2 b x+e)}}{3 b d \log (f)} \]

[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)

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Rubi [A]  time = 0.08, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2248, 302, 205} \[ \frac {c^{3/2} f^{a-\frac {5 e}{2}} \tan ^{-1}\left (\frac {\sqrt {d} f^{\frac {1}{2} (2 b x+e)}}{\sqrt {c}}\right )}{b d^{5/2} \log (f)}-\frac {c f^{\frac {1}{2} (2 a-5 e)+\frac {1}{2} (2 b x+e)}}{b d^2 \log (f)}+\frac {f^{\frac {1}{2} (2 a-5 e)+\frac {3}{2} (2 b x+e)}}{3 b d \log (f)} \]

Antiderivative was successfully verified.

[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 205

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

Rule 302

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 2248

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*} \int \frac {f^{a+5 b x}}{c+d f^{e+2 b x}} \, dx &=\frac {f^{a-\frac {5 e}{2}} \operatorname {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}} \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.07, size = 86, normalized size = 0.68 \[ \frac {3 c^{3/2} f^{a-\frac {5 e}{2}} \tan ^{-1}\left (\frac {\sqrt {d} f^{b x+\frac {e}{2}}}{\sqrt {c}}\right )+\sqrt {d} f^{a+b x-2 e} \left (d f^{2 b x+e}-3 c\right )}{3 b d^{5/2} \log (f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 211, normalized size = 1.66 \[ \left [\frac {3 \, c f^{a - \frac {5}{2} \, e} \sqrt {-\frac {c}{d}} \log \left (\frac {2 \, d f^{b x + \frac {1}{2} \, e} \sqrt {-\frac {c}{d}} + d f^{2 \, b x + e} - c}{d f^{2 \, b x + e} + c}\right ) + 2 \, d f^{3 \, b x + \frac {3}{2} \, e} f^{a - \frac {5}{2} \, e} - 6 \, c f^{b x + \frac {1}{2} \, e} f^{a - \frac {5}{2} \, e}}{6 \, b d^{2} \log \relax (f)}, \frac {3 \, c f^{a - \frac {5}{2} \, e} \sqrt {\frac {c}{d}} \arctan \left (\frac {d f^{b x + \frac {1}{2} \, e} \sqrt {\frac {c}{d}}}{c}\right ) + d f^{3 \, b x + \frac {3}{2} \, e} f^{a - \frac {5}{2} \, e} - 3 \, c f^{b x + \frac {1}{2} \, e} f^{a - \frac {5}{2} \, e}}{3 \, b d^{2} \log \relax (f)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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giac [A]  time = 0.32, size = 122, normalized size = 0.96 \[ \frac {1}{3} \, f^{a} {\left (\frac {3 \, c^{2} \arctan \left (\frac {d f^{b x} f^{e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} b d^{2} f^{2 \, e} \log \relax (f)} + \frac {b^{2} d^{2} f^{3 \, b x} f^{2 \, e} \log \relax (f)^{2} - 3 \, b^{2} c d f^{b x} f^{e} \log \relax (f)^{2}}{b^{3} d^{3} f^{3 \, e} \log \relax (f)^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [B]  time = 0.10, size = 212, normalized size = 1.67 \[ -\frac {c \,f^{\frac {4 a}{5}} f^{-2 e} f^{b x +\frac {a}{5}}}{b \,d^{2} \ln \relax (f )}+\frac {f^{\frac {2 a}{5}} f^{-e} f^{3 b x +\frac {3 a}{5}}}{3 b d \ln \relax (f )}-\frac {\sqrt {-c d}\, c \,f^{a} f^{-\frac {5 e}{2}} \ln \left (-\frac {\sqrt {-c d}\, f^{\frac {a}{5}} f^{-\frac {e}{2}}}{d}+f^{b x +\frac {a}{5}}\right )}{2 b \,d^{3} \ln \relax (f )}+\frac {\sqrt {-c d}\, c \,f^{a} f^{-\frac {5 e}{2}} \ln \left (\frac {\sqrt {-c d}\, f^{\frac {a}{5}} f^{-\frac {e}{2}}}{d}+f^{b x +\frac {a}{5}}\right )}{2 b \,d^{3} \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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/(f^(1/2*e))/(f^(-1/5*a)
)/d*(-c*d)^(1/2))-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/(f^(1/2*e))/(
f^(-1/5*a))/d*(-c*d)^(1/2))

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maxima [A]  time = 0.97, size = 97, normalized size = 0.76 \[ \frac {c^{2} f^{a - 2 \, e} \arctan \left (\frac {d f^{b x + e}}{\sqrt {c d} f^{\frac {1}{2} \, e}}\right )}{\sqrt {c d} b d^{2} f^{\frac {1}{2} \, e} \log \relax (f)} + \frac {d f^{3 \, b x + a + e} - 3 \, c f^{b x + a}}{3 \, b d^{2} f^{2 \, e} \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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^(1/2*e)))/(sqrt(c*d)*b*d^2*f^(1/2*e)*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))

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mupad [B]  time = 3.56, size = 102, normalized size = 0.80 \[ \frac {f^a\,f^{3\,b\,x}}{3\,b\,d\,f^e\,\ln \relax (f)}-\frac {c\,f^a\,f^{b\,x}}{b\,d^2\,f^{2\,e}\,\ln \relax (f)}+\frac {c^2\,f^a\,{\mathrm {e}}^{-\frac {5\,e\,\ln \relax (f)}{2}}\,\mathrm {atan}\left (\frac {d\,f^{b\,x}\,{\mathrm {e}}^{\frac {e\,\ln \relax (f)}{2}}}{\sqrt {c\,d}}\right )}{b\,d^2\,\ln \relax (f)\,\sqrt {c\,d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 1.67, size = 185, normalized size = 1.46 \[ \operatorname {RootSum} {\left (4 z^{2} b^{2} d^{5} e^{5 e \log {\relax (f )}} \log {\relax (f )}^{2} + c^{3} e^{2 a \log {\relax (f )}}, \left (i \mapsto i \log {\left (\frac {2 i b d^{2} e^{- \frac {4 a \log {\relax (f )}}{5}} e^{2 e \log {\relax (f )}} \log {\relax (f )}}{c} + e^{\frac {\left (a + 5 b x\right ) \log {\relax (f )}}{5}} \right )} \right )\right )} + \frac {\left (\begin {cases} x \left (- c + d\right ) & \text {for}\: b = 0 \wedge f = 1 \\x \left (- c e^{a \log {\relax (f )}} + d e^{a \log {\relax (f )}} e^{e \log {\relax (f )}}\right ) & \text {for}\: b = 0 \\x \left (- c + d\right ) & \text {for}\: f = 1 \\- \frac {c e^{a \log {\relax (f )}} e^{b x \log {\relax (f )}}}{b \log {\relax (f )}} + \frac {d e^{a \log {\relax (f )}} e^{e \log {\relax (f )}} e^{3 b x \log {\relax (f )}}}{3 b \log {\relax (f )}} & \text {otherwise} \end {cases}\right ) e^{- 2 e \log {\relax (f )}}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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