3.3.38 \(\int (e x)^q (a+b \log (c (d x^m)^n))^2 \, dx\) [238]

Optimal. Leaf size=93 \[ \frac {2 b^2 m^2 n^2 (e x)^{1+q}}{e (1+q)^3}-\frac {2 b m n (e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (1+q)^2}+\frac {(e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (1+q)} \]

[Out]

2*b^2*m^2*n^2*(e*x)^(1+q)/e/(1+q)^3-2*b*m*n*(e*x)^(1+q)*(a+b*ln(c*(d*x^m)^n))/e/(1+q)^2+(e*x)^(1+q)*(a+b*ln(c*
(d*x^m)^n))^2/e/(1+q)

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2342, 2341, 2495} \begin {gather*} \frac {(e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (q+1)}-\frac {2 b m n (e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (q+1)^2}+\frac {2 b^2 m^2 n^2 (e x)^{q+1}}{e (q+1)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*x)^q*(a + b*Log[c*(d*x^m)^n])^2,x]

[Out]

(2*b^2*m^2*n^2*(e*x)^(1 + q))/(e*(1 + q)^3) - (2*b*m*n*(e*x)^(1 + q)*(a + b*Log[c*(d*x^m)^n]))/(e*(1 + q)^2) +
 ((e*x)^(1 + q)*(a + b*Log[c*(d*x^m)^n])^2)/(e*(1 + q))

Rule 2341

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 2342

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 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 (e x)^q \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2 \, dx &=\text {Subst}\left (\int (e x)^q \left (a+b \log \left (c d^n x^{m n}\right )\right )^2 \, 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 )^2}{e (1+q)}-\text {Subst}\left (\frac {(2 b m n) \int (e x)^q \left (a+b \log \left (c d^n x^{m n}\right )\right ) \, dx}{1+q},c d^n x^{m n},c \left (d x^m\right )^n\right )\\ &=\frac {2 b^2 m^2 n^2 (e x)^{1+q}}{e (1+q)^3}-\frac {2 b m n (e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (1+q)^2}+\frac {(e x)^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (1+q)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 90, normalized size = 0.97 \begin {gather*} \frac {x (e x)^q \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{1+q}-\frac {2 b m n x^{-q} (e x)^q \left (-\frac {b m n x^{1+q}}{(1+q)^2}+\frac {x^{1+q} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{1+q}\right )}{1+q} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^q*(a + b*Log[c*(d*x^m)^n])^2,x]

[Out]

(x*(e*x)^q*(a + b*Log[c*(d*x^m)^n])^2)/(1 + q) - (2*b*m*n*(e*x)^q*(-((b*m*n*x^(1 + q))/(1 + q)^2) + (x^(1 + q)
*(a + b*Log[c*(d*x^m)^n]))/(1 + q)))/((1 + q)*x^q)

________________________________________________________________________________________

Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (e x \right )^{q} \left (a +b \ln \left (c \left (d \,x^{m}\right )^{n}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^q*(a+b*ln(c*(d*x^m)^n))^2,x)

[Out]

int((e*x)^q*(a+b*ln(c*(d*x^m)^n))^2,x)

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 152, normalized size = 1.63 \begin {gather*} \frac {\left (x e\right )^{q + 1} b^{2} e^{\left (-1\right )} \log \left (\left (d x^{m}\right )^{n} c\right )^{2}}{q + 1} - \frac {2 \, a b m n x e^{\left (q \log \left (x\right ) + q\right )}}{{\left (q + 1\right )}^{2}} + \frac {2 \, \left (x e\right )^{q + 1} a b e^{\left (-1\right )} \log \left (\left (d x^{m}\right )^{n} c\right )}{q + 1} + 2 \, {\left (\frac {m^{2} n^{2} x e^{\left (q \log \left (x\right ) + q\right )}}{{\left (q + 1\right )}^{3}} - \frac {m n x e^{\left (q \log \left (x\right ) + q\right )} \log \left (\left (d x^{m}\right )^{n} c\right )}{{\left (q + 1\right )}^{2}}\right )} b^{2} + \frac {\left (x e\right )^{q + 1} a^{2} e^{\left (-1\right )}}{q + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^q*(a+b*log(c*(d*x^m)^n))^2,x, algorithm="maxima")

[Out]

(x*e)^(q + 1)*b^2*e^(-1)*log((d*x^m)^n*c)^2/(q + 1) - 2*a*b*m*n*x*e^(q*log(x) + q)/(q + 1)^2 + 2*(x*e)^(q + 1)
*a*b*e^(-1)*log((d*x^m)^n*c)/(q + 1) + 2*(m^2*n^2*x*e^(q*log(x) + q)/(q + 1)^3 - m*n*x*e^(q*log(x) + q)*log((d
*x^m)^n*c)/(q + 1)^2)*b^2 + (x*e)^(q + 1)*a^2*e^(-1)/(q + 1)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (93) = 186\).
time = 0.39, size = 388, normalized size = 4.17 \begin {gather*} \frac {{\left ({\left (b^{2} q^{2} + 2 \, b^{2} q + b^{2}\right )} x \log \left (c\right )^{2} + {\left (b^{2} n^{2} q^{2} + 2 \, b^{2} n^{2} q + b^{2} n^{2}\right )} x \log \left (d\right )^{2} + {\left (b^{2} m^{2} n^{2} q^{2} + 2 \, b^{2} m^{2} n^{2} q + b^{2} m^{2} n^{2}\right )} x \log \left (x\right )^{2} - 2 \, {\left (b^{2} m n - a b q^{2} - a b + {\left (b^{2} m n - 2 \, a b\right )} q\right )} x \log \left (c\right ) + {\left (2 \, b^{2} m^{2} n^{2} - 2 \, a b m n + a^{2} q^{2} + a^{2} - 2 \, {\left (a b m n - a^{2}\right )} q\right )} x + 2 \, {\left ({\left (b^{2} n q^{2} + 2 \, b^{2} n q + b^{2} n\right )} x \log \left (c\right ) - {\left (b^{2} m n^{2} - a b n q^{2} - a b n + {\left (b^{2} m n^{2} - 2 \, a b n\right )} q\right )} x\right )} \log \left (d\right ) + 2 \, {\left ({\left (b^{2} m n q^{2} + 2 \, b^{2} m n q + b^{2} m n\right )} x \log \left (c\right ) + {\left (b^{2} m n^{2} q^{2} + 2 \, b^{2} m n^{2} q + b^{2} m n^{2}\right )} x \log \left (d\right ) - {\left (b^{2} m^{2} n^{2} - a b m n q^{2} - a b m n + {\left (b^{2} m^{2} n^{2} - 2 \, a b m n\right )} q\right )} x\right )} \log \left (x\right )\right )} e^{\left (q \log \left (x\right ) + q\right )}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^q*(a+b*log(c*(d*x^m)^n))^2,x, algorithm="fricas")

[Out]

((b^2*q^2 + 2*b^2*q + b^2)*x*log(c)^2 + (b^2*n^2*q^2 + 2*b^2*n^2*q + b^2*n^2)*x*log(d)^2 + (b^2*m^2*n^2*q^2 +
2*b^2*m^2*n^2*q + b^2*m^2*n^2)*x*log(x)^2 - 2*(b^2*m*n - a*b*q^2 - a*b + (b^2*m*n - 2*a*b)*q)*x*log(c) + (2*b^
2*m^2*n^2 - 2*a*b*m*n + a^2*q^2 + a^2 - 2*(a*b*m*n - a^2)*q)*x + 2*((b^2*n*q^2 + 2*b^2*n*q + b^2*n)*x*log(c) -
 (b^2*m*n^2 - a*b*n*q^2 - a*b*n + (b^2*m*n^2 - 2*a*b*n)*q)*x)*log(d) + 2*((b^2*m*n*q^2 + 2*b^2*m*n*q + b^2*m*n
)*x*log(c) + (b^2*m*n^2*q^2 + 2*b^2*m*n^2*q + b^2*m*n^2)*x*log(d) - (b^2*m^2*n^2 - a*b*m*n*q^2 - a*b*m*n + (b^
2*m^2*n^2 - 2*a*b*m*n)*q)*x)*log(x))*e^(q*log(x) + q)/(q^3 + 3*q^2 + 3*q + 1)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e x\right )^{q} \left (a + b \log {\left (c \left (d x^{m}\right )^{n} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**q*(a+b*ln(c*(d*x**m)**n))**2,x)

[Out]

Integral((e*x)**q*(a + b*log(c*(d*x**m)**n))**2, x)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 561 vs. \(2 (93) = 186\).
time = 3.12, size = 561, normalized size = 6.03 \begin {gather*} \frac {b^{2} m^{2} n^{2} q^{2} x x^{q} e^{q} \log \left (x\right )^{2}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} + \frac {2 \, b^{2} m^{2} n^{2} q x x^{q} e^{q} \log \left (x\right )^{2}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} - \frac {2 \, b^{2} m^{2} n^{2} q x x^{q} e^{q} \log \left (x\right )}{q^{3} + 3 \, q^{2} + 3 \, q + 1} + \frac {2 \, b^{2} m n^{2} q x x^{q} e^{q} \log \left (d\right ) \log \left (x\right )}{q^{2} + 2 \, q + 1} + \frac {b^{2} m^{2} n^{2} x x^{q} e^{q} \log \left (x\right )^{2}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} - \frac {2 \, b^{2} m^{2} n^{2} x x^{q} e^{q} \log \left (x\right )}{q^{3} + 3 \, q^{2} + 3 \, q + 1} + \frac {2 \, b^{2} m n q x x^{q} e^{q} \log \left (c\right ) \log \left (x\right )}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} m n^{2} x x^{q} e^{q} \log \left (d\right ) \log \left (x\right )}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} m^{2} n^{2} x x^{q} e^{q}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} - \frac {2 \, b^{2} m n^{2} x x^{q} e^{q} \log \left (d\right )}{q^{2} + 2 \, q + 1} + \frac {b^{2} n^{2} x x^{q} e^{q} \log \left (d\right )^{2}}{q + 1} + \frac {2 \, a b m n q x x^{q} e^{q} \log \left (x\right )}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} m n x x^{q} e^{q} \log \left (c\right ) \log \left (x\right )}{q^{2} + 2 \, q + 1} - \frac {2 \, b^{2} m n x x^{q} e^{q} \log \left (c\right )}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} n x x^{q} e^{q} \log \left (c\right ) \log \left (d\right )}{q + 1} + \frac {2 \, a b m n x x^{q} e^{q} \log \left (x\right )}{q^{2} + 2 \, q + 1} - \frac {2 \, a b m n x x^{q} e^{q}}{q^{2} + 2 \, q + 1} + \frac {b^{2} x x^{q} e^{q} \log \left (c\right )^{2}}{q + 1} + \frac {2 \, a b n x x^{q} e^{q} \log \left (d\right )}{q + 1} + \frac {2 \, a b x x^{q} e^{q} \log \left (c\right )}{q + 1} + \frac {a^{2} x x^{q} e^{q}}{q + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^q*(a+b*log(c*(d*x^m)^n))^2,x, algorithm="giac")

[Out]

b^2*m^2*n^2*q^2*x*x^q*e^q*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) + 2*b^2*m^2*n^2*q*x*x^q*e^q*log(x)^2/(q^3 + 3*q^2 +
 3*q + 1) - 2*b^2*m^2*n^2*q*x*x^q*e^q*log(x)/(q^3 + 3*q^2 + 3*q + 1) + 2*b^2*m*n^2*q*x*x^q*e^q*log(d)*log(x)/(
q^2 + 2*q + 1) + b^2*m^2*n^2*x*x^q*e^q*log(x)^2/(q^3 + 3*q^2 + 3*q + 1) - 2*b^2*m^2*n^2*x*x^q*e^q*log(x)/(q^3
+ 3*q^2 + 3*q + 1) + 2*b^2*m*n*q*x*x^q*e^q*log(c)*log(x)/(q^2 + 2*q + 1) + 2*b^2*m*n^2*x*x^q*e^q*log(d)*log(x)
/(q^2 + 2*q + 1) + 2*b^2*m^2*n^2*x*x^q*e^q/(q^3 + 3*q^2 + 3*q + 1) - 2*b^2*m*n^2*x*x^q*e^q*log(d)/(q^2 + 2*q +
 1) + b^2*n^2*x*x^q*e^q*log(d)^2/(q + 1) + 2*a*b*m*n*q*x*x^q*e^q*log(x)/(q^2 + 2*q + 1) + 2*b^2*m*n*x*x^q*e^q*
log(c)*log(x)/(q^2 + 2*q + 1) - 2*b^2*m*n*x*x^q*e^q*log(c)/(q^2 + 2*q + 1) + 2*b^2*n*x*x^q*e^q*log(c)*log(d)/(
q + 1) + 2*a*b*m*n*x*x^q*e^q*log(x)/(q^2 + 2*q + 1) - 2*a*b*m*n*x*x^q*e^q/(q^2 + 2*q + 1) + b^2*x*x^q*e^q*log(
c)^2/(q + 1) + 2*a*b*n*x*x^q*e^q*log(d)/(q + 1) + 2*a*b*x*x^q*e^q*log(c)/(q + 1) + a^2*x*x^q*e^q/(q + 1)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e\,x\right )}^q\,{\left (a+b\,\ln \left (c\,{\left (d\,x^m\right )}^n\right )\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^q*(a + b*log(c*(d*x^m)^n))^2,x)

[Out]

int((e*x)^q*(a + b*log(c*(d*x^m)^n))^2, x)

________________________________________________________________________________________