3.437 \(\int (a+b \log (c (d (e+f x)^p)^q))^3 \, dx\)

Optimal. Leaf size=121 \[ 6 a b^2 p^2 q^2 x-\frac{3 b p q (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{f}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{f}+\frac{6 b^3 p^2 q^2 (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f}-6 b^3 p^3 q^3 x \]

[Out]

6*a*b^2*p^2*q^2*x - 6*b^3*p^3*q^3*x + (6*b^3*p^2*q^2*(e + f*x)*Log[c*(d*(e + f*x)^p)^q])/f - (3*b*p*q*(e + f*x
)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2)/f + ((e + f*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3)/f

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Rubi [A]  time = 0.141472, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2389, 2296, 2295, 2445} \[ 6 a b^2 p^2 q^2 x-\frac{3 b p q (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{f}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{f}+\frac{6 b^3 p^2 q^2 (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f}-6 b^3 p^3 q^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d*(e + f*x)^p)^q])^3,x]

[Out]

6*a*b^2*p^2*q^2*x - 6*b^3*p^3*q^3*x + (6*b^3*p^2*q^2*(e + f*x)*Log[c*(d*(e + f*x)^p)^q])/f - (3*b*p*q*(e + f*x
)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2)/f + ((e + f*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3)/f

Rule 2389

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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

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

Rule 2445

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 )^3 \, dx &=\operatorname{Subst}\left (\int \left (a+b \log \left (c d^q (e+f x)^{p q}\right )\right )^3 \, dx,c d^q (e+f x)^{p q},c \left (d (e+f x)^p\right )^q\right )\\ &=\operatorname{Subst}\left (\frac{\operatorname{Subst}\left (\int \left (a+b \log \left (c d^q x^{p q}\right )\right )^3 \, 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 )\\ &=\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{f}-\operatorname{Subst}\left (\frac{(3 b p q) \operatorname{Subst}\left (\int \left (a+b \log \left (c d^q x^{p q}\right )\right )^2 \, 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 )\\ &=-\frac{3 b p q (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{f}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{f}+\operatorname{Subst}\left (\frac{\left (6 b^2 p^2 q^2\right ) \operatorname{Subst}\left (\int \left (a+b \log \left (c d^q x^{p q}\right )\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 )\\ &=6 a b^2 p^2 q^2 x-\frac{3 b p q (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{f}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{f}+\operatorname{Subst}\left (\frac{\left (6 b^3 p^2 q^2\right ) \operatorname{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 )\\ &=6 a b^2 p^2 q^2 x-6 b^3 p^3 q^3 x+\frac{6 b^3 p^2 q^2 (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )}{f}-\frac{3 b p q (e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2}{f}+\frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3}{f}\\ \end{align*}

Mathematica [A]  time = 0.0248021, size = 100, normalized size = 0.83 \[ \frac{(e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^3-3 b p q \left ((e+f x) \left (a+b \log \left (c \left (d (e+f x)^p\right )^q\right )\right )^2-2 b p q \left (f x (a-b p q)+b (e+f x) \log \left (c \left (d (e+f x)^p\right )^q\right )\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d*(e + f*x)^p)^q])^3,x]

[Out]

((e + f*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^3 - 3*b*p*q*((e + f*x)*(a + b*Log[c*(d*(e + f*x)^p)^q])^2 - 2*b*p*
q*(f*(a - b*p*q)*x + b*(e + f*x)*Log[c*(d*(e + f*x)^p)^q])))/f

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Maple [F]  time = 0.274, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\ln \left ( c \left ( d \left ( fx+e \right ) ^{p} \right ) ^{q} \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d*(f*x+e)^p)^q))^3,x)

[Out]

int((a+b*ln(c*(d*(f*x+e)^p)^q))^3,x)

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Maxima [B]  time = 1.12344, size = 428, normalized size = 3.54 \begin{align*} -3 \, a^{2} b f p q{\left (\frac{x}{f} - \frac{e \log \left (f x + e\right )}{f^{2}}\right )} + b^{3} x \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{3} + 3 \, a b^{2} x \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{2} + 3 \, a^{2} b x \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) - 3 \,{\left (2 \, f p q{\left (\frac{x}{f} - \frac{e \log \left (f x + e\right )}{f^{2}}\right )} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right ) + \frac{{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} p^{2} q^{2}}{f}\right )} a b^{2} -{\left (3 \, f p q{\left (\frac{x}{f} - \frac{e \log \left (f x + e\right )}{f^{2}}\right )} \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )^{2} -{\left (\frac{{\left (e \log \left (f x + e\right )^{3} + 3 \, e \log \left (f x + e\right )^{2} - 6 \, f x + 6 \, e \log \left (f x + e\right )\right )} p^{2} q^{2}}{f^{2}} - \frac{3 \,{\left (e \log \left (f x + e\right )^{2} - 2 \, f x + 2 \, e \log \left (f x + e\right )\right )} p q \log \left (\left ({\left (f x + e\right )}^{p} d\right )^{q} c\right )}{f^{2}}\right )} f p q\right )} b^{3} + a^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="maxima")

[Out]

-3*a^2*b*f*p*q*(x/f - e*log(f*x + e)/f^2) + b^3*x*log(((f*x + e)^p*d)^q*c)^3 + 3*a*b^2*x*log(((f*x + e)^p*d)^q
*c)^2 + 3*a^2*b*x*log(((f*x + e)^p*d)^q*c) - 3*(2*f*p*q*(x/f - e*log(f*x + e)/f^2)*log(((f*x + e)^p*d)^q*c) +
(e*log(f*x + e)^2 - 2*f*x + 2*e*log(f*x + e))*p^2*q^2/f)*a*b^2 - (3*f*p*q*(x/f - e*log(f*x + e)/f^2)*log(((f*x
+ e)^p*d)^q*c)^2 - ((e*log(f*x + e)^3 + 3*e*log(f*x + e)^2 - 6*f*x + 6*e*log(f*x + e))*p^2*q^2/f^2 - 3*(e*log
(f*x + e)^2 - 2*f*x + 2*e*log(f*x + e))*p*q*log(((f*x + e)^p*d)^q*c)/f^2)*f*p*q)*b^3 + a^3*x

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Fricas [B]  time = 2.07231, size = 1374, normalized size = 11.36 \begin{align*} \frac{b^{3} f q^{3} x \log \left (d\right )^{3} + b^{3} f x \log \left (c\right )^{3} +{\left (b^{3} f p^{3} q^{3} x + b^{3} e p^{3} q^{3}\right )} \log \left (f x + e\right )^{3} - 3 \,{\left (b^{3} f p q - a b^{2} f\right )} x \log \left (c\right )^{2} - 3 \,{\left (b^{3} e p^{3} q^{3} - a b^{2} e p^{2} q^{2} +{\left (b^{3} f p^{3} q^{3} - a b^{2} f p^{2} q^{2}\right )} x -{\left (b^{3} f p^{2} q^{2} x + b^{3} e p^{2} q^{2}\right )} \log \left (c\right ) -{\left (b^{3} f p^{2} q^{3} x + b^{3} e p^{2} q^{3}\right )} \log \left (d\right )\right )} \log \left (f x + e\right )^{2} + 3 \,{\left (2 \, b^{3} f p^{2} q^{2} - 2 \, a b^{2} f p q + a^{2} b f\right )} x \log \left (c\right ) + 3 \,{\left (b^{3} f q^{2} x \log \left (c\right ) -{\left (b^{3} f p q^{3} - a b^{2} f q^{2}\right )} x\right )} \log \left (d\right )^{2} -{\left (6 \, b^{3} f p^{3} q^{3} - 6 \, a b^{2} f p^{2} q^{2} + 3 \, a^{2} b f p q - a^{3} f\right )} x + 3 \,{\left (2 \, b^{3} e p^{3} q^{3} - 2 \, a b^{2} e p^{2} q^{2} + a^{2} b e p q +{\left (b^{3} f p q x + b^{3} e p q\right )} \log \left (c\right )^{2} +{\left (b^{3} f p q^{3} x + b^{3} e p q^{3}\right )} \log \left (d\right )^{2} +{\left (2 \, b^{3} f p^{3} q^{3} - 2 \, a b^{2} f p^{2} q^{2} + a^{2} b f p q\right )} x - 2 \,{\left (b^{3} e p^{2} q^{2} - a b^{2} e p q +{\left (b^{3} f p^{2} q^{2} - a b^{2} f p q\right )} x\right )} \log \left (c\right ) - 2 \,{\left (b^{3} e p^{2} q^{3} - a b^{2} e p q^{2} +{\left (b^{3} f p^{2} q^{3} - a b^{2} f p q^{2}\right )} x -{\left (b^{3} f p q^{2} x + b^{3} e p q^{2}\right )} \log \left (c\right )\right )} \log \left (d\right )\right )} \log \left (f x + e\right ) + 3 \,{\left (b^{3} f q x \log \left (c\right )^{2} - 2 \,{\left (b^{3} f p q^{2} - a b^{2} f q\right )} x \log \left (c\right ) +{\left (2 \, b^{3} f p^{2} q^{3} - 2 \, a b^{2} f p q^{2} + a^{2} b f q\right )} x\right )} \log \left (d\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="fricas")

[Out]

(b^3*f*q^3*x*log(d)^3 + b^3*f*x*log(c)^3 + (b^3*f*p^3*q^3*x + b^3*e*p^3*q^3)*log(f*x + e)^3 - 3*(b^3*f*p*q - a
*b^2*f)*x*log(c)^2 - 3*(b^3*e*p^3*q^3 - a*b^2*e*p^2*q^2 + (b^3*f*p^3*q^3 - a*b^2*f*p^2*q^2)*x - (b^3*f*p^2*q^2
*x + b^3*e*p^2*q^2)*log(c) - (b^3*f*p^2*q^3*x + b^3*e*p^2*q^3)*log(d))*log(f*x + e)^2 + 3*(2*b^3*f*p^2*q^2 - 2
*a*b^2*f*p*q + a^2*b*f)*x*log(c) + 3*(b^3*f*q^2*x*log(c) - (b^3*f*p*q^3 - a*b^2*f*q^2)*x)*log(d)^2 - (6*b^3*f*
p^3*q^3 - 6*a*b^2*f*p^2*q^2 + 3*a^2*b*f*p*q - a^3*f)*x + 3*(2*b^3*e*p^3*q^3 - 2*a*b^2*e*p^2*q^2 + a^2*b*e*p*q
+ (b^3*f*p*q*x + b^3*e*p*q)*log(c)^2 + (b^3*f*p*q^3*x + b^3*e*p*q^3)*log(d)^2 + (2*b^3*f*p^3*q^3 - 2*a*b^2*f*p
^2*q^2 + a^2*b*f*p*q)*x - 2*(b^3*e*p^2*q^2 - a*b^2*e*p*q + (b^3*f*p^2*q^2 - a*b^2*f*p*q)*x)*log(c) - 2*(b^3*e*
p^2*q^3 - a*b^2*e*p*q^2 + (b^3*f*p^2*q^3 - a*b^2*f*p*q^2)*x - (b^3*f*p*q^2*x + b^3*e*p*q^2)*log(c))*log(d))*lo
g(f*x + e) + 3*(b^3*f*q*x*log(c)^2 - 2*(b^3*f*p*q^2 - a*b^2*f*q)*x*log(c) + (2*b^3*f*p^2*q^3 - 2*a*b^2*f*p*q^2
+ a^2*b*f*q)*x)*log(d))/f

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Sympy [A]  time = 9.30581, size = 1023, normalized size = 8.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d*(f*x+e)**p)**q))**3,x)

[Out]

Piecewise((a**3*x + 3*a**2*b*e*p*q*log(e + f*x)/f + 3*a**2*b*p*q*x*log(e + f*x) - 3*a**2*b*p*q*x + 3*a**2*b*q*
x*log(d) + 3*a**2*b*x*log(c) + 3*a*b**2*e*p**2*q**2*log(e + f*x)**2/f - 6*a*b**2*e*p**2*q**2*log(e + f*x)/f +
6*a*b**2*e*p*q**2*log(d)*log(e + f*x)/f + 6*a*b**2*e*p*q*log(c)*log(e + f*x)/f + 3*a*b**2*p**2*q**2*x*log(e +
f*x)**2 - 6*a*b**2*p**2*q**2*x*log(e + f*x) + 6*a*b**2*p**2*q**2*x + 6*a*b**2*p*q**2*x*log(d)*log(e + f*x) - 6
*a*b**2*p*q**2*x*log(d) + 6*a*b**2*p*q*x*log(c)*log(e + f*x) - 6*a*b**2*p*q*x*log(c) + 3*a*b**2*q**2*x*log(d)*
*2 + 6*a*b**2*q*x*log(c)*log(d) + 3*a*b**2*x*log(c)**2 + b**3*e*p**3*q**3*log(e + f*x)**3/f - 3*b**3*e*p**3*q*
*3*log(e + f*x)**2/f + 6*b**3*e*p**3*q**3*log(e + f*x)/f + 3*b**3*e*p**2*q**3*log(d)*log(e + f*x)**2/f - 6*b**
3*e*p**2*q**3*log(d)*log(e + f*x)/f + 3*b**3*e*p**2*q**2*log(c)*log(e + f*x)**2/f - 6*b**3*e*p**2*q**2*log(c)*
log(e + f*x)/f + 3*b**3*e*p*q**3*log(d)**2*log(e + f*x)/f + 6*b**3*e*p*q**2*log(c)*log(d)*log(e + f*x)/f + 3*b
**3*e*p*q*log(c)**2*log(e + f*x)/f + b**3*p**3*q**3*x*log(e + f*x)**3 - 3*b**3*p**3*q**3*x*log(e + f*x)**2 + 6
*b**3*p**3*q**3*x*log(e + f*x) - 6*b**3*p**3*q**3*x + 3*b**3*p**2*q**3*x*log(d)*log(e + f*x)**2 - 6*b**3*p**2*
q**3*x*log(d)*log(e + f*x) + 6*b**3*p**2*q**3*x*log(d) + 3*b**3*p**2*q**2*x*log(c)*log(e + f*x)**2 - 6*b**3*p*
*2*q**2*x*log(c)*log(e + f*x) + 6*b**3*p**2*q**2*x*log(c) + 3*b**3*p*q**3*x*log(d)**2*log(e + f*x) - 3*b**3*p*
q**3*x*log(d)**2 + 6*b**3*p*q**2*x*log(c)*log(d)*log(e + f*x) - 6*b**3*p*q**2*x*log(c)*log(d) + 3*b**3*p*q*x*l
og(c)**2*log(e + f*x) - 3*b**3*p*q*x*log(c)**2 + b**3*q**3*x*log(d)**3 + 3*b**3*q**2*x*log(c)*log(d)**2 + 3*b*
*3*q*x*log(c)**2*log(d) + b**3*x*log(c)**3, Ne(f, 0)), (x*(a + b*log(c*(d*e**p)**q))**3, True))

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Giac [B]  time = 1.31845, size = 1110, normalized size = 9.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d*(f*x+e)^p)^q))^3,x, algorithm="giac")

[Out]

(f*x + e)*b^3*p^3*q^3*log(f*x + e)^3/f - 3*(f*x + e)*b^3*p^3*q^3*log(f*x + e)^2/f + 3*(f*x + e)*b^3*p^2*q^3*lo
g(f*x + e)^2*log(d)/f + 6*(f*x + e)*b^3*p^3*q^3*log(f*x + e)/f + 3*(f*x + e)*b^3*p^2*q^2*log(f*x + e)^2*log(c)
/f - 6*(f*x + e)*b^3*p^2*q^3*log(f*x + e)*log(d)/f + 3*(f*x + e)*b^3*p*q^3*log(f*x + e)*log(d)^2/f - 6*(f*x +
e)*b^3*p^3*q^3/f + 3*(f*x + e)*a*b^2*p^2*q^2*log(f*x + e)^2/f - 6*(f*x + e)*b^3*p^2*q^2*log(f*x + e)*log(c)/f
+ 6*(f*x + e)*b^3*p^2*q^3*log(d)/f + 6*(f*x + e)*b^3*p*q^2*log(f*x + e)*log(c)*log(d)/f - 3*(f*x + e)*b^3*p*q^
3*log(d)^2/f + (f*x + e)*b^3*q^3*log(d)^3/f - 6*(f*x + e)*a*b^2*p^2*q^2*log(f*x + e)/f + 6*(f*x + e)*b^3*p^2*q
^2*log(c)/f + 3*(f*x + e)*b^3*p*q*log(f*x + e)*log(c)^2/f + 6*(f*x + e)*a*b^2*p*q^2*log(f*x + e)*log(d)/f - 6*
(f*x + e)*b^3*p*q^2*log(c)*log(d)/f + 3*(f*x + e)*b^3*q^2*log(c)*log(d)^2/f + 6*(f*x + e)*a*b^2*p^2*q^2/f + 6*
(f*x + e)*a*b^2*p*q*log(f*x + e)*log(c)/f - 3*(f*x + e)*b^3*p*q*log(c)^2/f - 6*(f*x + e)*a*b^2*p*q^2*log(d)/f
+ 3*(f*x + e)*b^3*q*log(c)^2*log(d)/f + 3*(f*x + e)*a*b^2*q^2*log(d)^2/f + 3*(f*x + e)*a^2*b*p*q*log(f*x + e)/
f - 6*(f*x + e)*a*b^2*p*q*log(c)/f + (f*x + e)*b^3*log(c)^3/f + 6*(f*x + e)*a*b^2*q*log(c)*log(d)/f - 3*(f*x +
e)*a^2*b*p*q/f + 3*(f*x + e)*a*b^2*log(c)^2/f + 3*(f*x + e)*a^2*b*q*log(d)/f + 3*(f*x + e)*a^2*b*log(c)/f + (
f*x + e)*a^3/f