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

Optimal. Leaf size=31 \[ \frac{(a+b \log (c (e+f x)))^{p+1}}{b d f (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0751913, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2390, 12, 2302, 30} \[ \frac{(a+b \log (c (e+f x)))^{p+1}}{b d f (p+1)} \]

Antiderivative was successfully verified.

[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 2390

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]

Rule 12

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

Rule 2302

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 30

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

Rubi steps

\begin{align*} \int \frac{(a+b \log (c (e+f x)))^p}{d e+d f x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b \log (c x))^p}{d x} \, dx,x,e+f x\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b \log (c x))^p}{x} \, dx,x,e+f x\right )}{d f}\\ &=\frac{\operatorname{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]  time = 0.0119069, size = 31, normalized size = 1. \[ \frac{(a+b \log (c (e+f x)))^{p+1}}{b d f (p+1)} \]

Antiderivative was successfully verified.

[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))

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Maple [A]  time = 0.064, size = 32, normalized size = 1. \begin{align*}{\frac{ \left ( a+b\ln \left ( c \left ( fx+e \right ) \right ) \right ) ^{1+p}}{bdf \left ( 1+p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80185, size = 96, normalized size = 3.1 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.24523, size = 43, normalized size = 1.39 \begin{align*} \frac{{\left (b \log \left ({\left (f x + e\right )} c\right ) + a\right )}^{p + 1}}{b d f{\left (p + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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