\(\int \log (c (d+\frac {e}{(f+g x)^2})^q) \, dx\) [633]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 59 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=\frac {2 \sqrt {e} q \arctan \left (\frac {\sqrt {d} (f+g x)}{\sqrt {e}}\right )}{\sqrt {d} g}+\frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right )}{g} \]

[Out]

(g*x+f)*ln(c*(d+e/(g*x+f)^2)^q)/g+2*q*arctan((g*x+f)*d^(1/2)/e^(1/2))*e^(1/2)/g/d^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2533, 2498, 269, 211} \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=\frac {2 \sqrt {e} q \arctan \left (\frac {\sqrt {d} (f+g x)}{\sqrt {e}}\right )}{\sqrt {d} g}+\frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right )}{g} \]

[In]

Int[Log[c*(d + e/(f + g*x)^2)^q],x]

[Out]

(2*Sqrt[e]*q*ArcTan[(Sqrt[d]*(f + g*x))/Sqrt[e]])/(Sqrt[d]*g) + ((f + g*x)*Log[c*(d + e/(f + g*x)^2)^q])/g

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 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2533

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*((f_.) + (g_.)*(x_))^(n_))^(p_.)]*(b_.))^(q_.), x_Symbol] :> Dist[1/g, Su
bst[Int[(a + b*Log[c*(d + e*x^n)^p])^q, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IGtQ[q
, 0] && (EqQ[q, 1] || IntegerQ[n])

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.34 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=\frac {\frac {2 \sqrt {e} q \arctan \left (\frac {\sqrt {d} (f+g x)}{\sqrt {e}}\right )}{\sqrt {d}}-2 f q \log (f+g x)+g x \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right )+f q \log \left (e+d (f+g x)^2\right )}{g} \]

[In]

Integrate[Log[c*(d + e/(f + g*x)^2)^q],x]

[Out]

((2*Sqrt[e]*q*ArcTan[(Sqrt[d]*(f + g*x))/Sqrt[e]])/Sqrt[d] - 2*f*q*Log[f + g*x] + g*x*Log[c*(d + e/(f + g*x)^2
)^q] + f*q*Log[e + d*(f + g*x)^2])/g

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(110\) vs. \(2(51)=102\).

Time = 0.93 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.88

method result size
parts \(\ln \left (c \left (d +\frac {e}{\left (g x +f \right )^{2}}\right )^{q}\right ) x +2 q e g \left (-\frac {f \ln \left (g x +f \right )}{g^{2} e}+\frac {\frac {f \ln \left (d \,g^{2} x^{2}+2 d f g x +d \,f^{2}+e \right )}{2 g}+\frac {e \arctan \left (\frac {2 d \,g^{2} x +2 d f g}{2 \sqrt {d e}\, g}\right )}{\sqrt {d e}\, g}}{e g}\right )\) \(111\)
default \(\ln \left (c \left (\frac {d \,g^{2} x^{2}+2 d f g x +d \,f^{2}+e}{\left (g x +f \right )^{2}}\right )^{q}\right ) x +2 q e g \left (-\frac {f \ln \left (g x +f \right )}{g^{2} e}+\frac {\frac {f \ln \left (d \,g^{2} x^{2}+2 d f g x +d \,f^{2}+e \right )}{2 g}+\frac {e \arctan \left (\frac {2 d \,g^{2} x +2 d f g}{2 \sqrt {d e}\, g}\right )}{\sqrt {d e}\, g}}{e g}\right )\) \(129\)

[In]

int(ln(c*(d+e/(g*x+f)^2)^q),x,method=_RETURNVERBOSE)

[Out]

ln(c*(d+e/(g*x+f)^2)^q)*x+2*q*e*g*(-f/g^2/e*ln(g*x+f)+1/e/g*(1/2*f/g*ln(d*g^2*x^2+2*d*f*g*x+d*f^2+e)+e/(d*e)^(
1/2)/g*arctan(1/2*(2*d*g^2*x+2*d*f*g)/(d*e)^(1/2)/g)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (51) = 102\).

Time = 0.34 (sec) , antiderivative size = 287, normalized size of antiderivative = 4.86 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=\left [\frac {g q x \log \left (\frac {d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e}{g^{2} x^{2} + 2 \, f g x + f^{2}}\right ) + f q \log \left (d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e\right ) - 2 \, f q \log \left (g x + f\right ) + g x \log \left (c\right ) + q \sqrt {-\frac {e}{d}} \log \left (\frac {d g^{2} x^{2} + 2 \, d f g x + d f^{2} + 2 \, {\left (d g x + d f\right )} \sqrt {-\frac {e}{d}} - e}{d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e}\right )}{g}, \frac {g q x \log \left (\frac {d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e}{g^{2} x^{2} + 2 \, f g x + f^{2}}\right ) + f q \log \left (d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e\right ) - 2 \, f q \log \left (g x + f\right ) + g x \log \left (c\right ) + 2 \, q \sqrt {\frac {e}{d}} \arctan \left (\frac {{\left (d g x + d f\right )} \sqrt {\frac {e}{d}}}{e}\right )}{g}\right ] \]

[In]

integrate(log(c*(d+e/(g*x+f)^2)^q),x, algorithm="fricas")

[Out]

[(g*q*x*log((d*g^2*x^2 + 2*d*f*g*x + d*f^2 + e)/(g^2*x^2 + 2*f*g*x + f^2)) + f*q*log(d*g^2*x^2 + 2*d*f*g*x + d
*f^2 + e) - 2*f*q*log(g*x + f) + g*x*log(c) + q*sqrt(-e/d)*log((d*g^2*x^2 + 2*d*f*g*x + d*f^2 + 2*(d*g*x + d*f
)*sqrt(-e/d) - e)/(d*g^2*x^2 + 2*d*f*g*x + d*f^2 + e)))/g, (g*q*x*log((d*g^2*x^2 + 2*d*f*g*x + d*f^2 + e)/(g^2
*x^2 + 2*f*g*x + f^2)) + f*q*log(d*g^2*x^2 + 2*d*f*g*x + d*f^2 + e) - 2*f*q*log(g*x + f) + g*x*log(c) + 2*q*sq
rt(e/d)*arctan((d*g*x + d*f)*sqrt(e/d)/e))/g]

Sympy [F(-1)]

Timed out. \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=\text {Timed out} \]

[In]

integrate(ln(c*(d+e/(g*x+f)**2)**q),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(log(c*(d+e/(g*x+f)^2)^q),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (51) = 102\).

Time = 0.53 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.36 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=d e g^{4} q {\left (\frac {f \log \left (d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e\right )}{d e g^{5}} - \frac {2 \, f \log \left ({\left | g x + f \right |}\right )}{d e g^{5}} + \frac {2 \, \arctan \left (\frac {d g x + d f}{\sqrt {d e}}\right )}{\sqrt {d e} d g^{5}}\right )} + q x \log \left (d g^{2} x^{2} + 2 \, d f g x + d f^{2} + e\right ) - q x \log \left (g^{2} x^{2} + 2 \, f g x + f^{2}\right ) + x \log \left (c\right ) \]

[In]

integrate(log(c*(d+e/(g*x+f)^2)^q),x, algorithm="giac")

[Out]

d*e*g^4*q*(f*log(d*g^2*x^2 + 2*d*f*g*x + d*f^2 + e)/(d*e*g^5) - 2*f*log(abs(g*x + f))/(d*e*g^5) + 2*arctan((d*
g*x + d*f)/sqrt(d*e))/(sqrt(d*e)*d*g^5)) + q*x*log(d*g^2*x^2 + 2*d*f*g*x + d*f^2 + e) - q*x*log(g^2*x^2 + 2*f*
g*x + f^2) + x*log(c)

Mupad [B] (verification not implemented)

Time = 1.85 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.76 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^2}\right )^q\right ) \, dx=x\,\ln \left (c\,{\left (d+\frac {e}{{\left (f+g\,x\right )}^2}\right )}^q\right )-\frac {2\,f\,q\,\ln \left (f+g\,x\right )}{g}+\frac {\ln \left (e\,\sqrt {-d\,e}-3\,d\,f^2\,\sqrt {-d\,e}+4\,d\,e\,f+d\,e\,g\,x-3\,d\,f\,g\,x\,\sqrt {-d\,e}\right )\,\left (q\,\sqrt {-d\,e}+d\,f\,q\right )}{d\,g}-\frac {\ln \left (3\,d\,f^2\,\sqrt {-d\,e}-e\,\sqrt {-d\,e}+4\,d\,e\,f+d\,e\,g\,x+3\,d\,f\,g\,x\,\sqrt {-d\,e}\right )\,\left (q\,\sqrt {-d\,e}-d\,f\,q\right )}{d\,g} \]

[In]

int(log(c*(d + e/(f + g*x)^2)^q),x)

[Out]

x*log(c*(d + e/(f + g*x)^2)^q) - (2*f*q*log(f + g*x))/g + (log(e*(-d*e)^(1/2) - 3*d*f^2*(-d*e)^(1/2) + 4*d*e*f
 + d*e*g*x - 3*d*f*g*x*(-d*e)^(1/2))*(q*(-d*e)^(1/2) + d*f*q))/(d*g) - (log(3*d*f^2*(-d*e)^(1/2) - e*(-d*e)^(1
/2) + 4*d*e*f + d*e*g*x + 3*d*f*g*x*(-d*e)^(1/2))*(q*(-d*e)^(1/2) - d*f*q))/(d*g)