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

Optimal. Leaf size=50 \[ a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {b e p \log (d (f+g x)+e)}{d g} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2483, 2448, 263, 31} \[ a x+\frac {b (f+g x) \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )}{g}+\frac {b e p \log (d (f+g x)+e)}{d g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 263

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 2448

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 2483

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

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Mathematica [A]  time = 0.04, size = 70, normalized size = 1.40 \[ a x+b x \log \left (c \left (d+\frac {e}{f+g x}\right )^p\right )-b e g p \left (\frac {f \log (f+g x)}{e g^2}-\frac {(d f+e) \log (d f+d g x+e)}{d e g^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*x - b*e*g*p*((f*Log[f + g*x])/(e*g^2) - ((e + d*f)*Log[e + d*f + d*g*x])/(d*e*g^2)) + b*x*Log[c*(d + e/(f +
g*x))^p]

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fricas [A]  time = 0.44, size = 76, normalized size = 1.52 \[ \frac {b d g p x \log \left (\frac {d g x + d f + e}{g x + f}\right ) - b d f p \log \left (g x + f\right ) + b d g x \log \relax (c) + a d g x + {\left (b d f + b e\right )} p \log \left (d g x + d f + e\right )}{d g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.21, size = 177, normalized size = 3.54 \[ \frac {{\left (d f g e^{\left (-2\right )} - {\left (d f + e\right )} g e^{\left (-2\right )}\right )} {\left (d p e^{2} \log \left (-d + \frac {d g x + d f + e}{g x + f}\right ) + d e^{2} \log \relax (c) - \frac {{\left (d g x + d f + e\right )} p e^{2} \log \left (-d + \frac {d g x + d f + e}{g x + f}\right )}{g x + f} + \frac {{\left (d g x + d f + e\right )} p e^{2} \log \left (\frac {d g x + d f + e}{g x + f}\right )}{g x + f}\right )} b}{d^{2} g^{2} - \frac {{\left (d g x + d f + e\right )} d g^{2}}{g x + f}} + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 81, normalized size = 1.62 \[ -\frac {b f p \ln \left (g x +f \right )}{g}+\frac {b f p \ln \left (d g x +d f +e \right )}{g}+b x \ln \left (c \left (\frac {d g x +d f +e}{g x +f}\right )^{p}\right )+a x +\frac {b e p \ln \left (d g x +d f +e \right )}{d g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.67, size = 70, normalized size = 1.40 \[ -b e g p {\left (\frac {f \log \left (g x + f\right )}{e g^{2}} - \frac {{\left (d f + e\right )} \log \left (d g x + d f + e\right )}{d e g^{2}}\right )} + b x \log \left (c {\left (d + \frac {e}{g x + f}\right )}^{p}\right ) + a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.42, size = 61, normalized size = 1.22 \[ a\,x+b\,x\,\ln \left (c\,{\left (d+\frac {e}{f+g\,x}\right )}^p\right )-\frac {b\,f\,p\,\ln \left (f+g\,x\right )}{g}+\frac {b\,p\,\ln \left (e+d\,f+d\,g\,x\right )\,\left (e+d\,f\right )}{d\,g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.05, size = 114, normalized size = 2.28 \[ a x + b \left (\begin {cases} x \log {\left (c \left (\frac {e}{f}\right )^{p} \right )} & \text {for}\: d = 0 \wedge g = 0 \\x \log {\left (c \left (d + \frac {e}{f}\right )^{p} \right )} & \text {for}\: g = 0 \\- \frac {f p \log {\left (f + g x \right )}}{g} + p x \log {\relax (e )} - p x \log {\left (f + g x \right )} + p x + x \log {\relax (c )} & \text {for}\: d = 0 \\\frac {f p \log {\left (d + \frac {e}{f + g x} \right )}}{g} + p x \log {\left (d + \frac {e}{f + g x} \right )} + x \log {\relax (c )} + \frac {e p \log {\left (d f + d g x + e \right )}}{d g} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*Piecewise((x*log(c*(e/f)**p), Eq(d, 0) & Eq(g, 0)), (x*log(c*(d + e/f)**p), Eq(g, 0)), (-f*p*log(f + g
*x)/g + p*x*log(e) - p*x*log(f + g*x) + p*x + x*log(c), Eq(d, 0)), (f*p*log(d + e/(f + g*x))/g + p*x*log(d + e
/(f + g*x)) + x*log(c) + e*p*log(d*f + d*g*x + e)/(d*g), True))

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