\(\int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx\) [213]

Optimal result

Integrand size = 25, antiderivative size = 31 \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {(a+b \log (c (e+f x)))^{1+p}}{b d f (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, 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} \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {(a+b \log (c (e+f x)))^{p+1}}{b d f (p+1)} \]

[In]

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

[Out]

(a + b*Log[c*(e + f*x)])^(1 + p)/(b*d*f*(1 + p))

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))^p}{d x} \, dx,x,e+f x\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b \log (c x))^p}{x} \, dx,x,e+f x\right )}{d f} \\ & = \frac {\text {Subst}\left (\int x^p \, dx,x,a+b \log (c (e+f x))\right )}{b d f} \\ & = \frac {(a+b \log (c (e+f x)))^{1+p}}{b d f (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {(a+b \log (c (e+f x)))^{1+p}}{b d f (1+p)} \]

[In]

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

[Out]

(a + b*Log[c*(e + f*x)])^(1 + p)/(b*d*f*(1 + p))

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {\left (a +b \ln \left (c \left (f x +e \right )\right )\right )^{p +1}}{b d f \left (p +1\right )}\)	\(32\)
parallelrisch	\(\frac {\ln \left (c \left (f x +e \right )\right ) \left (a +b \ln \left (c \left (f x +e \right )\right )\right )^{p} b f +\left (a +b \ln \left (c \left (f x +e \right )\right )\right )^{p} a f}{d \,f^{2} b \left (p +1\right )}\)	\(59\)
norman	\(\frac {\ln \left (c \left (f x +e \right )\right ) {\mathrm e}^{p \ln \left (a +b \ln \left (c \left (f x +e \right )\right )\right )}}{d f \left (p +1\right )}+\frac {a \,{\mathrm e}^{p \ln \left (a +b \ln \left (c \left (f x +e \right )\right )\right )}}{b d f \left (p +1\right )}\)	\(70\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {{\left (b \log \left (c f x + c e\right ) + a\right )} {\left (b \log \left (c f x + c e\right ) + a\right )}^{p}}{b d f p + b d f} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {\int \frac {\left (a + b \log {\left (c e + c f x \right )}\right )^{p}}{e + f x}\, dx}{d} \]

[In]

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

[Out]

Integral((a + b*log(c*e + c*f*x))**p/(e + f*x), x)/d

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {{\left (b \log \left (c f x + c e\right ) + a\right )}^{p + 1}}{b d f {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {{\left (b \log \left (c f x + c e\right ) + a\right )}^{p + 1}}{b d f {\left (p + 1\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx=\frac {{\left (a+b\,\ln \left (c\,\left (e+f\,x\right )\right )\right )}^{p+1}}{b\,d\,f\,\left (p+1\right )} \]

[In]

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

[Out]

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