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

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

[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)

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Rubi [A]  time = 0.13, 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 = {2305, 2304, 2445} \[ \frac {(e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{e (q+1)}-\frac {2 b m n (e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (q+1)^2}+\frac {2 b^2 m^2 n^2 (e x)^{q+1}}{e (q+1)^3} \]

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 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 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 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 )^2 \, dx &=\operatorname {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)}-\operatorname {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*}

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Mathematica [A]  time = 0.04, size = 90, normalized size = 0.97 \[ \frac {x (e x)^q \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )^2}{q+1}-\frac {2 b m n x^{-q} (e x)^q \left (\frac {x^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{q+1}-\frac {b m n x^{q+1}}{(q+1)^2}\right )}{q+1} \]

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)

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fricas [B]  time = 0.59, size = 391, normalized size = 4.20 \[ \frac {{\left ({\left (b^{2} q^{2} + 2 \, b^{2} q + b^{2}\right )} x \log \relax (c)^{2} + {\left (b^{2} n^{2} q^{2} + 2 \, b^{2} n^{2} q + b^{2} n^{2}\right )} x \log \relax (d)^{2} + {\left (b^{2} m^{2} n^{2} q^{2} + 2 \, b^{2} m^{2} n^{2} q + b^{2} m^{2} n^{2}\right )} x \log \relax (x)^{2} - 2 \, {\left (b^{2} m n - a b q^{2} - a b + {\left (b^{2} m n - 2 \, a b\right )} q\right )} x \log \relax (c) + {\left (2 \, b^{2} m^{2} n^{2} - 2 \, a b m n + a^{2} q^{2} + a^{2} - 2 \, {\left (a b m n - a^{2}\right )} q\right )} x + 2 \, {\left ({\left (b^{2} n q^{2} + 2 \, b^{2} n q + b^{2} n\right )} x \log \relax (c) - {\left (b^{2} m n^{2} - a b n q^{2} - a b n + {\left (b^{2} m n^{2} - 2 \, a b n\right )} q\right )} x\right )} \log \relax (d) + 2 \, {\left ({\left (b^{2} m n q^{2} + 2 \, b^{2} m n q + b^{2} m n\right )} x \log \relax (c) + {\left (b^{2} m n^{2} q^{2} + 2 \, b^{2} m n^{2} q + b^{2} m n^{2}\right )} x \log \relax (d) - {\left (b^{2} m^{2} n^{2} - a b m n q^{2} - a b m n + {\left (b^{2} m^{2} n^{2} - 2 \, a b m n\right )} q\right )} x\right )} \log \relax (x)\right )} e^{\left (q \log \relax (e) + q \log \relax (x)\right )}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} \]

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(e) + q*log(x))/(q^3 + 3*q^2 + 3*q + 1)

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giac [B]  time = 0.35, size = 561, normalized size = 6.03 \[ \frac {b^{2} m^{2} n^{2} q^{2} x x^{q} e^{q} \log \relax (x)^{2}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} + \frac {2 \, b^{2} m^{2} n^{2} q x x^{q} e^{q} \log \relax (x)^{2}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} - \frac {2 \, b^{2} m^{2} n^{2} q x x^{q} e^{q} \log \relax (x)}{q^{3} + 3 \, q^{2} + 3 \, q + 1} + \frac {2 \, b^{2} m n^{2} q x x^{q} e^{q} \log \relax (d) \log \relax (x)}{q^{2} + 2 \, q + 1} + \frac {b^{2} m^{2} n^{2} x x^{q} e^{q} \log \relax (x)^{2}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} - \frac {2 \, b^{2} m^{2} n^{2} x x^{q} e^{q} \log \relax (x)}{q^{3} + 3 \, q^{2} + 3 \, q + 1} + \frac {2 \, b^{2} m n q x x^{q} e^{q} \log \relax (c) \log \relax (x)}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} m n^{2} x x^{q} e^{q} \log \relax (d) \log \relax (x)}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} m^{2} n^{2} x x^{q} e^{q}}{q^{3} + 3 \, q^{2} + 3 \, q + 1} - \frac {2 \, b^{2} m n^{2} x x^{q} e^{q} \log \relax (d)}{q^{2} + 2 \, q + 1} + \frac {b^{2} n^{2} x x^{q} e^{q} \log \relax (d)^{2}}{q + 1} + \frac {2 \, a b m n q x x^{q} e^{q} \log \relax (x)}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} m n x x^{q} e^{q} \log \relax (c) \log \relax (x)}{q^{2} + 2 \, q + 1} - \frac {2 \, b^{2} m n x x^{q} e^{q} \log \relax (c)}{q^{2} + 2 \, q + 1} + \frac {2 \, b^{2} n x x^{q} e^{q} \log \relax (c) \log \relax (d)}{q + 1} + \frac {2 \, a b m n x x^{q} e^{q} \log \relax (x)}{q^{2} + 2 \, q + 1} - \frac {2 \, a b m n x x^{q} e^{q}}{q^{2} + 2 \, q + 1} + \frac {b^{2} x x^{q} e^{q} \log \relax (c)^{2}}{q + 1} + \frac {2 \, a b n x x^{q} e^{q} \log \relax (d)}{q + 1} + \frac {2 \, a b x x^{q} e^{q} \log \relax (c)}{q + 1} + \frac {a^{2} x x^{q} e^{q}}{q + 1} \]

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)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \left (d \,x^{m}\right )^{n}\right )+a \right )^{2} \left (e x \right )^{q}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 149, normalized size = 1.60 \[ -\frac {2 \, a b e^{q} m n x x^{q}}{{\left (q + 1\right )}^{2}} + 2 \, {\left (\frac {e^{q} m^{2} n^{2} x x^{q}}{{\left (q + 1\right )}^{3}} - \frac {e^{q} m n x x^{q} \log \left (\left (d x^{m}\right )^{n} c\right )}{{\left (q + 1\right )}^{2}}\right )} b^{2} + \frac {\left (e x\right )^{q + 1} b^{2} \log \left (\left (d x^{m}\right )^{n} c\right )^{2}}{e {\left (q + 1\right )}} + \frac {2 \, \left (e x\right )^{q + 1} a b \log \left (\left (d x^{m}\right )^{n} c\right )}{e {\left (q + 1\right )}} + \frac {\left (e x\right )^{q + 1} a^{2}}{e {\left (q + 1\right )}} \]

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]

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

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

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)

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

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)

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