3.35 \(\int \frac {f^{a+3 b x}}{c+d f^{e+2 b x}} \, dx\)

Optimal. Leaf size=88 \[ \frac {f^{\frac {1}{2} (2 a-3 e)+\frac {1}{2} (2 b x+e)}}{b d \log (f)}-\frac {\sqrt {c} f^{a-\frac {3 e}{2}} \tan ^{-1}\left (\frac {\sqrt {d} f^{\frac {1}{2} (2 b x+e)}}{\sqrt {c}}\right )}{b d^{3/2} \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2248, 321, 205} \[ \frac {f^{\frac {1}{2} (2 a-3 e)+\frac {1}{2} (2 b x+e)}}{b d \log (f)}-\frac {\sqrt {c} f^{a-\frac {3 e}{2}} \tan ^{-1}\left (\frac {\sqrt {d} f^{\frac {1}{2} (2 b x+e)}}{\sqrt {c}}\right )}{b d^{3/2} \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + 3*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

f^((2*a - 3*e)/2 + (e + 2*b*x)/2)/(b*d*Log[f]) - (Sqrt[c]*f^(a - (3*e)/2)*ArcTan[(Sqrt[d]*f^((e + 2*b*x)/2))/S
qrt[c]])/(b*d^(3/2)*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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+3 b x}}{c+d f^{e+2 b x}} \, dx &=\frac {f^{a-\frac {3 e}{2}} \operatorname {Subst}\left (\int \frac {x^2}{c+d x^2} \, dx,x,f^{\frac {1}{2} (e+2 b x)}\right )}{b \log (f)}\\ &=\frac {f^{\frac {1}{2} (2 a-3 e)+\frac {1}{2} (e+2 b x)}}{b d \log (f)}-\frac {\left (c f^{a-\frac {3 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 \log (f)}\\ &=\frac {f^{\frac {1}{2} (2 a-3 e)+\frac {1}{2} (e+2 b x)}}{b d \log (f)}-\frac {\sqrt {c} f^{a-\frac {3 e}{2}} \tan ^{-1}\left (\frac {\sqrt {d} f^{\frac {1}{2} (e+2 b x)}}{\sqrt {c}}\right )}{b d^{3/2} \log (f)}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 67, normalized size = 0.76 \[ \frac {f^a \left (\frac {f^{b x-e}}{d}-\frac {\sqrt {c} f^{-3 e/2} \tan ^{-1}\left (\frac {\sqrt {d} f^{b x+\frac {e}{2}}}{\sqrt {c}}\right )}{d^{3/2}}\right )}{b \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + 3*b*x)/(c + d*f^(e + 2*b*x)),x]

[Out]

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

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fricas [A]  time = 0.44, size = 166, normalized size = 1.89 \[ \left [\frac {f^{a - \frac {3}{2} \, e} \sqrt {-\frac {c}{d}} \log \left (-\frac {2 \, d f^{b x + \frac {1}{2} \, e} \sqrt {-\frac {c}{d}} - d f^{2 \, b x + e} + c}{d f^{2 \, b x + e} + c}\right ) + 2 \, f^{b x + \frac {1}{2} \, e} f^{a - \frac {3}{2} \, e}}{2 \, b d \log \relax (f)}, -\frac {f^{a - \frac {3}{2} \, e} \sqrt {\frac {c}{d}} \arctan \left (\frac {d f^{b x + \frac {1}{2} \, e} \sqrt {\frac {c}{d}}}{c}\right ) - f^{b x + \frac {1}{2} \, e} f^{a - \frac {3}{2} \, e}}{b d \log \relax (f)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(3*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="fricas")

[Out]

[1/2*(f^(a - 3/2*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*f^(b*x + 1/2*e)*f^(a - 3/2*e))/(b*d*log(f)), -(f^(a - 3/2*e)*sqrt(c/d)*arctan(d*f^(b*x + 1/2*e)*sqrt(c
/d)/c) - f^(b*x + 1/2*e)*f^(a - 3/2*e))/(b*d*log(f))]

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giac [A]  time = 0.36, size = 77, normalized size = 0.88 \[ -f^{a} {\left (\frac {c \arctan \left (\frac {d f^{b x} f^{e}}{\sqrt {c d f^{e}}}\right )}{\sqrt {c d f^{e}} b d f^{e} \log \relax (f)} - \frac {f^{b x}}{b d f^{e} \log \relax (f)}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(3*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.10, size = 171, normalized size = 1.94 \[ \frac {f^{\frac {2 a}{3}} f^{-e} f^{b x +\frac {a}{3}}}{b d \ln \relax (f )}+\frac {\sqrt {-c d}\, f^{a} f^{-\frac {3 e}{2}} \ln \left (-\frac {\sqrt {-c d}\, f^{\frac {a}{3}} f^{-\frac {e}{2}}}{d}+f^{b x +\frac {a}{3}}\right )}{2 b \,d^{2} \ln \relax (f )}-\frac {\sqrt {-c d}\, f^{a} f^{-\frac {3 e}{2}} \ln \left (\frac {\sqrt {-c d}\, f^{\frac {a}{3}} f^{-\frac {e}{2}}}{d}+f^{b x +\frac {a}{3}}\right )}{2 b \,d^{2} \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(3*b*x+a)/(c+d*f^(2*b*x+e)),x)

[Out]

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

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maxima [A]  time = 1.00, size = 76, normalized size = 0.86 \[ -\frac {c f^{a - e} \arctan \left (\frac {d f^{b x + e}}{\sqrt {c d} f^{\frac {1}{2} \, e}}\right )}{\sqrt {c d} b d f^{\frac {1}{2} \, e} \log \relax (f)} + \frac {f^{b x + a - e}}{b d \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(3*b*x+a)/(c+d*f^(2*b*x+e)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.55, size = 66, normalized size = 0.75 \[ -\frac {f^a\,{\mathrm {e}}^{-\frac {3\,e\,\ln \relax (f)}{2}}\,\left (c\,\mathrm {atan}\left (\frac {d\,f^{b\,x}\,{\mathrm {e}}^{\frac {e\,\ln \relax (f)}{2}}}{\sqrt {c\,d}}\right )-f^{b\,x}\,{\mathrm {e}}^{\frac {e\,\ln \relax (f)}{2}}\,\sqrt {c\,d}\right )}{b\,d\,\ln \relax (f)\,\sqrt {c\,d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + 3*b*x)/(c + d*f^(e + 2*b*x)),x)

[Out]

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

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sympy [A]  time = 1.35, size = 110, normalized size = 1.25 \[ \operatorname {RootSum} {\left (4 z^{2} b^{2} d^{3} e^{3 e \log {\relax (f )}} \log {\relax (f )}^{2} + c e^{2 a \log {\relax (f )}}, \left (i \mapsto i \log {\left (- 2 i b d e^{- \frac {2 a \log {\relax (f )}}{3}} e^{e \log {\relax (f )}} \log {\relax (f )} + e^{\frac {\left (a + 3 b x\right ) \log {\relax (f )}}{3}} \right )} \right )\right )} + \frac {\left (\begin {cases} x & \text {for}\: b = 0 \vee f = 1 \\\frac {e^{b x \log {\relax (f )}}}{b \log {\relax (f )}} & \text {otherwise} \end {cases}\right ) e^{a \log {\relax (f )}} e^{- e \log {\relax (f )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(3*b*x+a)/(c+d*f**(2*b*x+e)),x)

[Out]

RootSum(4*_z**2*b**2*d**3*exp(3*e*log(f))*log(f)**2 + c*exp(2*a*log(f)), Lambda(_i, _i*log(-2*_i*b*d*exp(-2*a*
log(f)/3)*exp(e*log(f))*log(f) + exp((a + 3*b*x)*log(f)/3)))) + Piecewise((x, Eq(b, 0) | Eq(f, 1)), (exp(b*x*l
og(f))/(b*log(f)), True))*exp(a*log(f))*exp(-e*log(f))/d

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