\(\int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx\) [187]

Optimal result

Integrand size = 25, antiderivative size = 27 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {(a+b \log (c (e+f x)))^3}{3 b d f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2437, 12, 2339, 30} \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {(a+b \log (c (e+f x)))^3}{3 b d f} \]

[In]

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

[Out]

(a + b*Log[c*(e + f*x)])^3/(3*b*d*f)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2437

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a+b \log (c x))^2}{d x} \, dx,x,e+f x\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b \log (c x))^2}{x} \, dx,x,e+f x\right )}{d f} \\ & = \frac {\text {Subst}\left (\int x^2 \, dx,x,a+b \log (c (e+f x))\right )}{b d f} \\ & = \frac {(a+b \log (c (e+f x)))^3}{3 b d f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {(a+b \log (c (e+f x)))^3}{3 b d f} \]

[In]

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

[Out]

(a + b*Log[c*(e + f*x)])^3/(3*b*d*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(53\) vs. \(2(25)=50\).

Time = 0.50 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00

method	result	size
parallelrisch	\(\frac {\ln \left (c \left (f x +e \right )\right )^{3} b^{2} f +3 \ln \left (c \left (f x +e \right )\right )^{2} a b f +3 \ln \left (c \left (f x +e \right )\right ) a^{2} f}{3 d \,f^{2}}\)	\(54\)
risch	\(\frac {b^{2} \ln \left (c \left (f x +e \right )\right )^{3}}{3 d f}+\frac {a b \ln \left (c \left (f x +e \right )\right )^{2}}{d f}+\frac {a^{2} \ln \left (f x +e \right )}{d f}\)	\(58\)
parts	\(\frac {b^{2} \ln \left (c \left (f x +e \right )\right )^{3}}{3 d f}+\frac {a b \ln \left (c \left (f x +e \right )\right )^{2}}{d f}+\frac {a^{2} \ln \left (f x +e \right )}{d f}\)	\(58\)
norman	\(\frac {a^{2} \ln \left (c \left (f x +e \right )\right )}{d f}+\frac {a b \ln \left (c \left (f x +e \right )\right )^{2}}{d f}+\frac {b^{2} \ln \left (c \left (f x +e \right )\right )^{3}}{3 d f}\)	\(60\)
derivativedivides	\(\frac {\frac {c \,a^{2} \ln \left (c f x +c e \right )}{d}+\frac {c a b \ln \left (c f x +c e \right )^{2}}{d}+\frac {c \,b^{2} \ln \left (c f x +c e \right )^{3}}{3 d}}{c f}\)	\(64\)
default	\(\frac {\frac {c \,a^{2} \ln \left (c f x +c e \right )}{d}+\frac {c a b \ln \left (c f x +c e \right )^{2}}{d}+\frac {c \,b^{2} \ln \left (c f x +c e \right )^{3}}{3 d}}{c f}\)	\(64\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (25) = 50\).

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {b^{2} \log \left (c f x + c e\right )^{3} + 3 \, a b \log \left (c f x + c e\right )^{2} + 3 \, a^{2} \log \left (c f x + c e\right )}{3 \, d f} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (19) = 38\).

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {a^{2} \log {\left (d e + d f x \right )}}{d f} + \frac {a b \log {\left (c \left (e + f x\right ) \right )}^{2}}{d f} + \frac {b^{2} \log {\left (c \left (e + f x\right ) \right )}^{3}}{3 d f} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (25) = 50\).

Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 4.74 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=-a b {\left (\frac {2 \, \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} - \frac {\log \left (f x + e\right )^{2} + 2 \, \log \left (f x + e\right ) \log \left (c\right )}{d f}\right )} + \frac {b^{2} \log \left (c f x + c e\right )^{3}}{3 \, d f} + \frac {2 \, a b \log \left (c f x + c e\right ) \log \left (d f x + d e\right )}{d f} + \frac {a^{2} \log \left (d f x + d e\right )}{d f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (25) = 50\).

Time = 0.38 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.19 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {b^{2} \log \left (c f x + c e\right )^{3}}{3 \, d f} + \frac {a b \log \left (c f x + c e\right )^{2}}{d f} + \frac {a^{2} \log \left (f x + e\right )}{d f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {(a+b \log (c (e+f x)))^2}{d e+d f x} \, dx=\frac {3\,\ln \left (e+f\,x\right )\,a^2+3\,a\,b\,{\ln \left (c\,e+c\,f\,x\right )}^2+b^2\,{\ln \left (c\,e+c\,f\,x\right )}^3}{3\,d\,f} \]

[In]

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

[Out]

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