3.237 \(\int (e x)^q (a+b \log (c (d x^m)^n))^3 \, dx\)
Optimal. Leaf size=135 \[ \frac{6 b^2 m^2 n^2 (e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (q+1)^3}+\frac{(e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3}{e (q+1)}-\frac{3 b m n (e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (q+1)^2}-\frac{6 b^3 m^3 n^3 (e x)^{q+1}}{e (q+1)^4} \]
[Out]
(-6*b^3*m^3*n^3*(e*x)^(1 + q))/(e*(1 + q)^4) + (6*b^2*m^2*n^2*(e*x)^(1 + q)*(a + b*Log[c*(d*x^m)^n]))/(e*(1 +
q)^3) - (3*b*m*n*(e*x)^(1 + q)*(a + b*Log[c*(d*x^m)^n])^2)/(e*(1 + q)^2) + ((e*x)^(1 + q)*(a + b*Log[c*(d*x^m)
^n])^3)/(e*(1 + q))
________________________________________________________________________________________
Rubi [A] time = 0.220854, antiderivative size = 135, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used =
{2305, 2304, 2445} \[ \frac{6 b^2 m^2 n^2 (e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (q+1)^3}+\frac{(e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3}{e (q+1)}-\frac{3 b m n (e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (q+1)^2}-\frac{6 b^3 m^3 n^3 (e x)^{q+1}}{e (q+1)^4} \]
Antiderivative was successfully verified.
[In]
Int[(e*x)^q*(a + b*Log[c*(d*x^m)^n])^3,x]
[Out]
(-6*b^3*m^3*n^3*(e*x)^(1 + q))/(e*(1 + q)^4) + (6*b^2*m^2*n^2*(e*x)^(1 + q)*(a + b*Log[c*(d*x^m)^n]))/(e*(1 +
q)^3) - (3*b*m*n*(e*x)^(1 + q)*(a + b*Log[c*(d*x^m)^n])^2)/(e*(1 + q)^2) + ((e*x)^(1 + q)*(a + b*Log[c*(d*x^m)
^n])^3)/(e*(1 + q))
Rule 2305
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]
Rule 2304
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]
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 (e x)^q \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3 \, dx &=\operatorname{Subst}\left (\int (e x)^q \left (a+b \log \left (c d^n x^{m n}\right )\right )^3 \, dx,c d^n x^{m n},c \left (d x^m\right )^n\right )\\ &=\frac{(e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3}{e (1+q)}-\operatorname{Subst}\left (\frac{(3 b m n) \int (e x)^q \left (a+b \log \left (c d^n x^{m n}\right )\right )^2 \, dx}{1+q},c d^n x^{m n},c \left (d x^m\right )^n\right )\\ &=-\frac{3 b m n (e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (1+q)^2}+\frac{(e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3}{e (1+q)}+\operatorname{Subst}\left (\frac{\left (6 b^2 m^2 n^2\right ) \int (e x)^q \left (a+b \log \left (c d^n x^{m n}\right )\right ) \, dx}{(1+q)^2},c d^n x^{m n},c \left (d x^m\right )^n\right )\\ &=-\frac{6 b^3 m^3 n^3 (e x)^{1+q}}{e (1+q)^4}+\frac{6 b^2 m^2 n^2 (e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (1+q)^3}-\frac{3 b m n (e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (1+q)^2}+\frac{(e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3}{e (1+q)}\\ \end{align*}
Mathematica [A] time = 0.0550658, size = 91, normalized size = 0.67 \[ \frac{x (e x)^q \left (\left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^3-\frac{3 b m n \left ((q+1)^2 \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2+2 b m n \left (b m n-(q+1) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )\right )\right )}{(q+1)^3}\right )}{q+1} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*x)^q*(a + b*Log[c*(d*x^m)^n])^3,x]
[Out]
(x*(e*x)^q*((a + b*Log[c*(d*x^m)^n])^3 - (3*b*m*n*((1 + q)^2*(a + b*Log[c*(d*x^m)^n])^2 + 2*b*m*n*(b*m*n - (1
+ q)*(a + b*Log[c*(d*x^m)^n]))))/(1 + q)^3))/(1 + q)
________________________________________________________________________________________
Maple [F] time = 0.126, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{q} \left ( a+b\ln \left ( c \left ( d{x}^{m} \right ) ^{n} \right ) \right ) ^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x)^q*(a+b*ln(c*(d*x^m)^n))^3,x)
[Out]
int((e*x)^q*(a+b*ln(c*(d*x^m)^n))^3,x)
________________________________________________________________________________________
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((e*x)^q*(a+b*log(c*(d*x^m)^n))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 1.01005, size = 2808, normalized size = 20.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)^q*(a+b*log(c*(d*x^m)^n))^3,x, algorithm="fricas")
[Out]
((b^3*q^3 + 3*b^3*q^2 + 3*b^3*q + b^3)*x*log(c)^3 + (b^3*n^3*q^3 + 3*b^3*n^3*q^2 + 3*b^3*n^3*q + b^3*n^3)*x*lo
g(d)^3 + (b^3*m^3*n^3*q^3 + 3*b^3*m^3*n^3*q^2 + 3*b^3*m^3*n^3*q + b^3*m^3*n^3)*x*log(x)^3 + 3*(a*b^2*q^3 - b^3
*m*n + a*b^2 - (b^3*m*n - 3*a*b^2)*q^2 - (2*b^3*m*n - 3*a*b^2)*q)*x*log(c)^2 + 3*(2*b^3*m^2*n^2 + a^2*b*q^3 -
2*a*b^2*m*n + a^2*b - (2*a*b^2*m*n - 3*a^2*b)*q^2 + (2*b^3*m^2*n^2 - 4*a*b^2*m*n + 3*a^2*b)*q)*x*log(c) + 3*((
b^3*n^2*q^3 + 3*b^3*n^2*q^2 + 3*b^3*n^2*q + b^3*n^2)*x*log(c) + (a*b^2*n^2*q^3 - b^3*m*n^3 + a*b^2*n^2 - (b^3*
m*n^3 - 3*a*b^2*n^2)*q^2 - (2*b^3*m*n^3 - 3*a*b^2*n^2)*q)*x)*log(d)^2 + 3*((b^3*m^2*n^2*q^3 + 3*b^3*m^2*n^2*q^
2 + 3*b^3*m^2*n^2*q + b^3*m^2*n^2)*x*log(c) + (b^3*m^2*n^3*q^3 + 3*b^3*m^2*n^3*q^2 + 3*b^3*m^2*n^3*q + b^3*m^2
*n^3)*x*log(d) + (a*b^2*m^2*n^2*q^3 - b^3*m^3*n^3 + a*b^2*m^2*n^2 - (b^3*m^3*n^3 - 3*a*b^2*m^2*n^2)*q^2 - (2*b
^3*m^3*n^3 - 3*a*b^2*m^2*n^2)*q)*x)*log(x)^2 - (6*b^3*m^3*n^3 - 6*a*b^2*m^2*n^2 - a^3*q^3 + 3*a^2*b*m*n - a^3
+ 3*(a^2*b*m*n - a^3)*q^2 - 3*(2*a*b^2*m^2*n^2 - 2*a^2*b*m*n + a^3)*q)*x + 3*((b^3*n*q^3 + 3*b^3*n*q^2 + 3*b^3
*n*q + b^3*n)*x*log(c)^2 + 2*(a*b^2*n*q^3 - b^3*m*n^2 + a*b^2*n - (b^3*m*n^2 - 3*a*b^2*n)*q^2 - (2*b^3*m*n^2 -
3*a*b^2*n)*q)*x*log(c) + (2*b^3*m^2*n^3 + a^2*b*n*q^3 - 2*a*b^2*m*n^2 + a^2*b*n - (2*a*b^2*m*n^2 - 3*a^2*b*n)
*q^2 + (2*b^3*m^2*n^3 - 4*a*b^2*m*n^2 + 3*a^2*b*n)*q)*x)*log(d) + 3*((b^3*m*n*q^3 + 3*b^3*m*n*q^2 + 3*b^3*m*n*
q + b^3*m*n)*x*log(c)^2 + (b^3*m*n^3*q^3 + 3*b^3*m*n^3*q^2 + 3*b^3*m*n^3*q + b^3*m*n^3)*x*log(d)^2 + 2*(a*b^2*
m*n*q^3 - b^3*m^2*n^2 + a*b^2*m*n - (b^3*m^2*n^2 - 3*a*b^2*m*n)*q^2 - (2*b^3*m^2*n^2 - 3*a*b^2*m*n)*q)*x*log(c
) + (2*b^3*m^3*n^3 + a^2*b*m*n*q^3 - 2*a*b^2*m^2*n^2 + a^2*b*m*n - (2*a*b^2*m^2*n^2 - 3*a^2*b*m*n)*q^2 + (2*b^
3*m^3*n^3 - 4*a*b^2*m^2*n^2 + 3*a^2*b*m*n)*q)*x + 2*((b^3*m*n^2*q^3 + 3*b^3*m*n^2*q^2 + 3*b^3*m*n^2*q + b^3*m*
n^2)*x*log(c) + (a*b^2*m*n^2*q^3 - b^3*m^2*n^3 + a*b^2*m*n^2 - (b^3*m^2*n^3 - 3*a*b^2*m*n^2)*q^2 - (2*b^3*m^2*
n^3 - 3*a*b^2*m*n^2)*q)*x)*log(d))*log(x))*e^(q*log(e) + q*log(x))/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e x\right )^{q} \left (a + b \log{\left (c \left (d x^{m}\right )^{n} \right )}\right )^{3}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)**q*(a+b*ln(c*(d*x**m)**n))**3,x)
[Out]
Integral((e*x)**q*(a + b*log(c*(d*x**m)**n))**3, x)
________________________________________________________________________________________
Giac [B] time = 1.39869, size = 2445, normalized size = 18.11 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)^q*(a+b*log(c*(d*x^m)^n))^3,x, algorithm="giac")
[Out]
b^3*m^3*n^3*q^3*x*x^q*e^q*log(x)^3/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) + 3*b^3*m^3*n^3*q^2*x*x^q*e^q*log(x)^3/(q^4
+ 4*q^3 + 6*q^2 + 4*q + 1) - 3*b^3*m^3*n^3*q^2*x*x^q*e^q*log(x)^2/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) + 3*b^3*m^2
*n^3*q^2*x*x^q*e^q*log(d)*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) + 3*b^3*m^3*n^3*q*x*x^q*e^q*log(x)^3/(q^4 + 4*q^3 +
6*q^2 + 4*q + 1) - 6*b^3*m^3*n^3*q*x*x^q*e^q*log(x)^2/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) + 3*b^3*m^2*n^2*q^2*x*x
^q*e^q*log(c)*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) + 6*b^3*m^2*n^3*q*x*x^q*e^q*log(d)*log(x)^2/(q^3 + 3*q^2 + 3*q
+ 1) + b^3*m^3*n^3*x*x^q*e^q*log(x)^3/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) + 6*b^3*m^3*n^3*q*x*x^q*e^q*log(x)/(q^4
+ 4*q^3 + 6*q^2 + 4*q + 1) - 6*b^3*m^2*n^3*q*x*x^q*e^q*log(d)*log(x)/(q^3 + 3*q^2 + 3*q + 1) + 3*b^3*m*n^3*q*x
*x^q*e^q*log(d)^2*log(x)/(q^2 + 2*q + 1) - 3*b^3*m^3*n^3*x*x^q*e^q*log(x)^2/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) +
3*a*b^2*m^2*n^2*q^2*x*x^q*e^q*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) + 6*b^3*m^2*n^2*q*x*x^q*e^q*log(c)*log(x)^2/(q^
3 + 3*q^2 + 3*q + 1) + 3*b^3*m^2*n^3*x*x^q*e^q*log(d)*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) + 6*b^3*m^3*n^3*x*x^q*e
^q*log(x)/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) - 6*b^3*m^2*n^2*q*x*x^q*e^q*log(c)*log(x)/(q^3 + 3*q^2 + 3*q + 1) -
6*b^3*m^2*n^3*x*x^q*e^q*log(d)*log(x)/(q^3 + 3*q^2 + 3*q + 1) + 6*b^3*m*n^2*q*x*x^q*e^q*log(c)*log(d)*log(x)/(
q^2 + 2*q + 1) + 3*b^3*m*n^3*x*x^q*e^q*log(d)^2*log(x)/(q^2 + 2*q + 1) + 6*a*b^2*m^2*n^2*q*x*x^q*e^q*log(x)^2/
(q^3 + 3*q^2 + 3*q + 1) + 3*b^3*m^2*n^2*x*x^q*e^q*log(c)*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) - 6*b^3*m^3*n^3*x*x^
q*e^q/(q^4 + 4*q^3 + 6*q^2 + 4*q + 1) + 6*b^3*m^2*n^3*x*x^q*e^q*log(d)/(q^3 + 3*q^2 + 3*q + 1) - 3*b^3*m*n^3*x
*x^q*e^q*log(d)^2/(q^2 + 2*q + 1) + b^3*n^3*x*x^q*e^q*log(d)^3/(q + 1) - 6*a*b^2*m^2*n^2*q*x*x^q*e^q*log(x)/(q
^3 + 3*q^2 + 3*q + 1) - 6*b^3*m^2*n^2*x*x^q*e^q*log(c)*log(x)/(q^3 + 3*q^2 + 3*q + 1) + 3*b^3*m*n*q*x*x^q*e^q*
log(c)^2*log(x)/(q^2 + 2*q + 1) + 6*a*b^2*m*n^2*q*x*x^q*e^q*log(d)*log(x)/(q^2 + 2*q + 1) + 6*b^3*m*n^2*x*x^q*
e^q*log(c)*log(d)*log(x)/(q^2 + 2*q + 1) + 3*a*b^2*m^2*n^2*x*x^q*e^q*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) + 6*b^3*
m^2*n^2*x*x^q*e^q*log(c)/(q^3 + 3*q^2 + 3*q + 1) - 6*b^3*m*n^2*x*x^q*e^q*log(c)*log(d)/(q^2 + 2*q + 1) + 3*b^3
*n^2*x*x^q*e^q*log(c)*log(d)^2/(q + 1) - 6*a*b^2*m^2*n^2*x*x^q*e^q*log(x)/(q^3 + 3*q^2 + 3*q + 1) + 6*a*b^2*m*
n*q*x*x^q*e^q*log(c)*log(x)/(q^2 + 2*q + 1) + 3*b^3*m*n*x*x^q*e^q*log(c)^2*log(x)/(q^2 + 2*q + 1) + 6*a*b^2*m*
n^2*x*x^q*e^q*log(d)*log(x)/(q^2 + 2*q + 1) + 6*a*b^2*m^2*n^2*x*x^q*e^q/(q^3 + 3*q^2 + 3*q + 1) - 3*b^3*m*n*x*
x^q*e^q*log(c)^2/(q^2 + 2*q + 1) - 6*a*b^2*m*n^2*x*x^q*e^q*log(d)/(q^2 + 2*q + 1) + 3*b^3*n*x*x^q*e^q*log(c)^2
*log(d)/(q + 1) + 3*a*b^2*n^2*x*x^q*e^q*log(d)^2/(q + 1) + 3*a^2*b*m*n*q*x*x^q*e^q*log(x)/(q^2 + 2*q + 1) + 6*
a*b^2*m*n*x*x^q*e^q*log(c)*log(x)/(q^2 + 2*q + 1) - 6*a*b^2*m*n*x*x^q*e^q*log(c)/(q^2 + 2*q + 1) + b^3*x*x^q*e
^q*log(c)^3/(q + 1) + 6*a*b^2*n*x*x^q*e^q*log(c)*log(d)/(q + 1) + 3*a^2*b*m*n*x*x^q*e^q*log(x)/(q^2 + 2*q + 1)
- 3*a^2*b*m*n*x*x^q*e^q/(q^2 + 2*q + 1) + 3*a*b^2*x*x^q*e^q*log(c)^2/(q + 1) + 3*a^2*b*n*x*x^q*e^q*log(d)/(q
+ 1) + 3*a^2*b*x*x^q*e^q*log(c)/(q + 1) + a^3*x*x^q*e^q/(q + 1)