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^((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])

________________________________________________________________________________________

Rubi [A]  time = 0.067568, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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]

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 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]

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*}

Mathematica [A]  time = 0.0636177, 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])

________________________________________________________________________________________

Maple [B]  time = 0.062, size = 171, normalized size = 1.9 \begin{align*}{\frac{1}{d\ln \left ( f \right ) b}{f}^{bx+{\frac{a}{3}}} \left ({f}^{{\frac{e}{2}}} \right ) ^{-2} \left ({f}^{-{\frac{a}{3}}} \right ) ^{-2}}+{\frac{1}{2\,b{d}^{2}\ln \left ( f \right ) }\sqrt{-cd}\ln \left ({f}^{bx+{\frac{a}{3}}}-{\frac{1}{d}\sqrt{-cd} \left ({f}^{{\frac{e}{2}}} \right ) ^{-1} \left ({f}^{-{\frac{a}{3}}} \right ) ^{-1}} \right ) \left ({f}^{-{\frac{a}{3}}} \right ) ^{-3} \left ({f}^{{\frac{e}{2}}} \right ) ^{-3}}-{\frac{1}{2\,b{d}^{2}\ln \left ( f \right ) }\sqrt{-cd}\ln \left ({f}^{bx+{\frac{a}{3}}}+{\frac{1}{d}\sqrt{-cd} \left ({f}^{{\frac{e}{2}}} \right ) ^{-1} \left ({f}^{-{\frac{a}{3}}} \right ) ^{-1}} \right ) \left ({f}^{-{\frac{a}{3}}} \right ) ^{-3} \left ({f}^{{\frac{e}{2}}} \right ) ^{-3}} \end{align*}

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/d*(-c*d)^(1/2)/(f^(-1/3*a))/(f^(1/2*e)))-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/d*(-c*d)^(1/2)/(f^(-1/3*a))/(f^(1/2*e)))

________________________________________________________________________________________

Maxima [B]  time = 1.73656, size = 171, normalized size = 1.94 \begin{align*} -\frac{c f^{a - e} \log \left (\frac{d{\left (f^{3 \, b x + a}\right )}^{\frac{1}{3}} f^{e} - \sqrt{-c d f^{e}} f^{\frac{1}{3} \, a}}{d{\left (f^{3 \, b x + a}\right )}^{\frac{1}{3}} f^{e} + \sqrt{-c d f^{e}} f^{\frac{1}{3} \, a}}\right )}{2 \, \sqrt{-c d f^{e}} b d \log \left (f\right )} + \frac{{\left (f^{3 \, b x + a}\right )}^{\frac{1}{3}} f^{\frac{2}{3} \, a - e}}{b d \log \left (f\right )} \end{align*}

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]

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

________________________________________________________________________________________

Fricas [A]  time = 1.88714, size = 374, normalized size = 4.25 \begin{align*} \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 \left (f\right )}, -\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 \left (f\right )}\right ] \end{align*}

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))]

________________________________________________________________________________________

Sympy [B]  time = 1.6555, size = 253, normalized size = 2.88 \begin{align*} \begin{cases} \frac{e^{\frac{2 a \log{\left (f \right )}}{3}} e^{- e \log{\left (f \right )}} e^{\frac{\left (a + 3 b x\right ) \log{\left (f \right )}}{3}}}{b d \log{\left (f \right )}} & \text{for}\: b d e^{e \log{\left (f \right )}} \log{\left (f \right )} \neq 0 \\\frac{x \left (c^{2} e^{\frac{10 a \log{\left (f \right )}}{3}} + 2 c d e^{\frac{8 a \log{\left (f \right )}}{3}} e^{e \log{\left (f \right )}} + d^{2} e^{2 a \log{\left (f \right )}} e^{2 e \log{\left (f \right )}}\right )}{c^{2} d e^{\frac{8 a \log{\left (f \right )}}{3}} e^{e \log{\left (f \right )}} + 2 c d^{2} e^{2 a \log{\left (f \right )}} e^{2 e \log{\left (f \right )}} + d^{3} e^{\frac{4 a \log{\left (f \right )}}{3}} e^{3 e \log{\left (f \right )}}} & \text{otherwise} \end{cases} + \operatorname{RootSum}{\left (4 z^{2} b^{2} d^{3} e^{3 e \log{\left (f \right )}} \log{\left (f \right )}^{2} + c e^{2 a \log{\left (f \right )}}, \left ( i \mapsto i \log{\left (- 2 i b d e^{- \frac{2 a \log{\left (f \right )}}{3}} e^{e \log{\left (f \right )}} \log{\left (f \right )} + e^{\frac{\left (a + 3 b x\right ) \log{\left (f \right )}}{3}} \right )} \right )\right )} \end{align*}

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]

Piecewise((exp(2*a*log(f)/3)*exp(-e*log(f))*exp((a + 3*b*x)*log(f)/3)/(b*d*log(f)), Ne(b*d*exp(e*log(f))*log(f
), 0)), (x*(c**2*exp(10*a*log(f)/3) + 2*c*d*exp(8*a*log(f)/3)*exp(e*log(f)) + d**2*exp(2*a*log(f))*exp(2*e*log
(f)))/(c**2*d*exp(8*a*log(f)/3)*exp(e*log(f)) + 2*c*d**2*exp(2*a*log(f))*exp(2*e*log(f)) + d**3*exp(4*a*log(f)
/3)*exp(3*e*log(f))), True)) + 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))))

________________________________________________________________________________________

Giac [A]  time = 1.23517, size = 104, normalized size = 1.18 \begin{align*} -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 \left (f\right )} - \frac{f^{b x}}{b d f^{e} \log \left (f\right )}\right )} \end{align*}

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)))