3.5.23 \(\int (a+b \log (c (d (e+f x)^p)^q)) \, dx\) [423]

Optimal. Leaf size=34 \[ a x-b p q x+\frac {b (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2436, 2332, 2495} \begin {gather*} a x+\frac {b (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f}-b p q x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[a + b*Log[c*(d*(e + f*x)^p)^q],x]

[Out]

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

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

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

Rule 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 34, normalized size = 1.00 \begin {gather*} a x-b p q x+\frac {b (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 42, normalized size = 1.24

method result size
default \(a x +b \ln \left (c \left (d \left (f x +e \right )^{p}\right )^{q}\right ) x -b p q x +\frac {b q p e \ln \left (f x +e \right )}{f}\) \(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 48, normalized size = 1.41 \begin {gather*} -b f p q {\left (\frac {x}{f} - \frac {e \log \left (f x + e\right )}{f^{2}}\right )} + b x \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 52, normalized size = 1.53 \begin {gather*} \frac {b f q x \log \left (d\right ) + b f x \log \left (c\right ) - {\left (b f p q - a f\right )} x + {\left (b f p q x + b p q e\right )} \log \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.21, size = 53, normalized size = 1.56 \begin {gather*} a x + b \left (\begin {cases} \frac {e \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )}}{f} - p q x + x \log {\left (c \left (d \left (e + f x\right )^{p}\right )^{q} \right )} & \text {for}\: f \neq 0 \\x \log {\left (c \left (d e^{p}\right )^{q} \right )} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x + b*Piecewise((e*log(c*(d*(e + f*x)**p)**q)/f - p*q*x + x*log(c*(d*(e + f*x)**p)**q), Ne(f, 0)), (x*log(c*
(d*e**p)**q), True))

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Giac [A]
time = 4.99, size = 64, normalized size = 1.88 \begin {gather*} {\left (\frac {{\left (f x + e\right )} p q \log \left (f x + e\right )}{f} - \frac {{\left (f x + e\right )} p q}{f} + \frac {{\left (f x + e\right )} q \log \left (d\right )}{f} + \frac {{\left (f x + e\right )} \log \left (c\right )}{f}\right )} b + a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.22, size = 41, normalized size = 1.21 \begin {gather*} x\,\left (a-b\,p\,q\right )+b\,x\,\ln \left (c\,{\left (d\,{\left (e+f\,x\right )}^p\right )}^q\right )+\frac {b\,e\,p\,q\,\ln \left (e+f\,x\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*log(c*(d*(e + f*x)^p)^q),x)

[Out]

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

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