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

Optimal. Leaf size=51 \[ \frac{(e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (q+1)}-\frac{b m n (e x)^{q+1}}{e (q+1)^2} \]

[Out]

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

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Rubi [A]  time = 0.0457418, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2304, 2445} \[ \frac{(e x)^{q+1} \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{e (q+1)}-\frac{b m n (e x)^{q+1}}{e (q+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.012254, size = 37, normalized size = 0.73 \[ \frac{x (e x)^q \left (a q+a+b (q+1) \log \left (c \left (d x^m\right )^n\right )-b m n\right )}{(q+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.086, 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 ) \, 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)),x)

[Out]

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

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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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.928524, size = 186, normalized size = 3.65 \begin{align*} \frac{{\left ({\left (b q + b\right )} x \log \left (c\right ) +{\left (b n q + b n\right )} x \log \left (d\right ) +{\left (b m n q + b m n\right )} x \log \left (x\right ) -{\left (b m n - a q - a\right )} x\right )} e^{\left (q \log \left (e\right ) + q \log \left (x\right )\right )}}{q^{2} + 2 \, q + 1} \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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 9.99231, size = 112, normalized size = 2.2 \begin{align*} a \left (\begin{cases} 0^{q} x & \text{for}\: e = 0 \\\frac{\begin{cases} \frac{\left (e x\right )^{q + 1}}{q + 1} & \text{for}\: q \neq -1 \\\log{\left (e x \right )} & \text{otherwise} \end{cases}}{e} & \text{otherwise} \end{cases}\right ) - b m n \left (\begin{cases} 0^{q} x & \text{for}\: \left (e = 0 \wedge q \neq -1\right ) \vee e = 0 \\\frac{\begin{cases} \frac{e e^{q} x x^{q}}{q + 1} & \text{for}\: q \neq -1 \\\log{\left (x \right )} & \text{otherwise} \end{cases}}{e q + e} & \text{for}\: q > -\infty \wedge q < \infty \wedge q \neq -1 \\\frac{\log{\left (e x \right )}^{2}}{2 e} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} 0^{q} x & \text{for}\: e = 0 \\\frac{\begin{cases} \frac{\left (e x\right )^{q + 1}}{q + 1} & \text{for}\: q \neq -1 \\\log{\left (e x \right )} & \text{otherwise} \end{cases}}{e} & \text{otherwise} \end{cases}\right ) \log{\left (c \left (d x^{m}\right )^{n} \right )} \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)),x)

[Out]

a*Piecewise((0**q*x, Eq(e, 0)), (Piecewise(((e*x)**(q + 1)/(q + 1), Ne(q, -1)), (log(e*x), True))/e, True)) -
b*m*n*Piecewise((0**q*x, Eq(e, 0) | (Eq(e, 0) & Ne(q, -1))), (Piecewise((e*e**q*x*x**q/(q + 1), Ne(q, -1)), (l
og(x), True))/(e*q + e), (q > -oo) & (q < oo) & Ne(q, -1)), (log(e*x)**2/(2*e), True)) + b*Piecewise((0**q*x,
Eq(e, 0)), (Piecewise(((e*x)**(q + 1)/(q + 1), Ne(q, -1)), (log(e*x), True))/e, True))*log(c*(d*x**m)**n)

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Giac [B]  time = 1.3412, size = 150, normalized size = 2.94 \begin{align*} \frac{b m n q x x^{q} e^{q} \log \left (x\right )}{q^{2} + 2 \, q + 1} + \frac{b m n x x^{q} e^{q} \log \left (x\right )}{q^{2} + 2 \, q + 1} - \frac{b m n x x^{q} e^{q}}{q^{2} + 2 \, q + 1} + \frac{b n x x^{q} e^{q} \log \left (d\right )}{q + 1} + \frac{b x x^{q} e^{q} \log \left (c\right )}{q + 1} + \frac{a x x^{q} e^{q}}{q + 1} \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)),x, algorithm="giac")

[Out]

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