\(\int (a+b \log (c \sqrt {d+e x}))^p \, dx\) [32]

Optimal result

Integrand size = 18, antiderivative size = 88 \[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\frac {2^{-p} e^{-\frac {2 a}{b}} \Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \sqrt {d+e x}\right )\right )}{b}\right ) \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \left (-\frac {a+b \log \left (c \sqrt {d+e x}\right )}{b}\right )^{-p}}{c^2 e} \]

[Out]

GAMMA(p+1,-2*(a+b*ln(c*(e*x+d)^(1/2)))/b)*(a+b*ln(c*(e*x+d)^(1/2)))^p/(2^p)/c^2/e/exp(2*a/b)/(((-a-b*ln(c*(e*x
+d)^(1/2)))/b)^p)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2436, 2336, 2212} \[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\frac {2^{-p} e^{-\frac {2 a}{b}} \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \left (-\frac {a+b \log \left (c \sqrt {d+e x}\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {2 \left (a+b \log \left (c \sqrt {d+e x}\right )\right )}{b}\right )}{c^2 e} \]

[In]

Int[(a + b*Log[c*Sqrt[d + e*x]])^p,x]

[Out]

(Gamma[1 + p, (-2*(a + b*Log[c*Sqrt[d + e*x]]))/b]*(a + b*Log[c*Sqrt[d + e*x]])^p)/(2^p*c^2*e*E^((2*a)/b)*(-((
a + b*Log[c*Sqrt[d + e*x]])/b))^p)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2336

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[1/(n*c^(1/n)), Subst[Int[E^(x/n)*(a + b*x)^p
, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Rule 2436

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (a+b \log \left (c \sqrt {x}\right )\right )^p \, dx,x,d+e x\right )}{e} \\ & = \frac {2 \text {Subst}\left (\int e^{2 x} (a+b x)^p \, dx,x,\log \left (c \sqrt {d+e x}\right )\right )}{c^2 e} \\ & = \frac {2^{-p} e^{-\frac {2 a}{b}} \Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \sqrt {d+e x}\right )\right )}{b}\right ) \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \left (-\frac {a+b \log \left (c \sqrt {d+e x}\right )}{b}\right )^{-p}}{c^2 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\frac {2^{-p} e^{-\frac {2 a}{b}} \Gamma \left (1+p,-\frac {2 \left (a+b \log \left (c \sqrt {d+e x}\right )\right )}{b}\right ) \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \left (-\frac {a+b \log \left (c \sqrt {d+e x}\right )}{b}\right )^{-p}}{c^2 e} \]

[In]

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

[Out]

(Gamma[1 + p, (-2*(a + b*Log[c*Sqrt[d + e*x]]))/b]*(a + b*Log[c*Sqrt[d + e*x]])^p)/(2^p*c^2*e*E^((2*a)/b)*(-((
a + b*Log[c*Sqrt[d + e*x]])/b))^p)

Maple [F]

\[\int \left (a +b \ln \left (c \sqrt {e x +d}\right )\right )^{p}d x\]

[In]

int((a+b*ln(c*(e*x+d)^(1/2)))^p,x)

[Out]

int((a+b*ln(c*(e*x+d)^(1/2)))^p,x)

Fricas [F]

\[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\int { {\left (b \log \left (\sqrt {e x + d} c\right ) + a\right )}^{p} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^(1/2)))^p,x, algorithm="fricas")

[Out]

integral((b*log(sqrt(e*x + d)*c) + a)^p, x)

Sympy [F]

\[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\int \left (a + b \log {\left (c \sqrt {d + e x} \right )}\right )^{p}\, dx \]

[In]

integrate((a+b*ln(c*(e*x+d)**(1/2)))**p,x)

[Out]

Integral((a + b*log(c*sqrt(d + e*x)))**p, x)

Maxima [A] (verification not implemented)

none

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.67 \[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=-\frac {2 \, {\left (b \log \left (\sqrt {e x + d} c\right ) + a\right )}^{p + 1} e^{\left (-\frac {2 \, a}{b}\right )} E_{-p}\left (-\frac {2 \, {\left (b \log \left (\sqrt {e x + d} c\right ) + a\right )}}{b}\right )}{b c^{2} e} \]

[In]

integrate((a+b*log(c*(e*x+d)^(1/2)))^p,x, algorithm="maxima")

[Out]

-2*(b*log(sqrt(e*x + d)*c) + a)^(p + 1)*e^(-2*a/b)*exp_integral_e(-p, -2*(b*log(sqrt(e*x + d)*c) + a)/b)/(b*c^
2*e)

Giac [F]

\[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\int { {\left (b \log \left (\sqrt {e x + d} c\right ) + a\right )}^{p} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^(1/2)))^p,x, algorithm="giac")

[Out]

integrate((b*log(sqrt(e*x + d)*c) + a)^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \log \left (c \sqrt {d+e x}\right )\right )^p \, dx=\int {\left (a+b\,\ln \left (c\,\sqrt {d+e\,x}\right )\right )}^p \,d x \]

[In]

int((a + b*log(c*(d + e*x)^(1/2)))^p,x)

[Out]

int((a + b*log(c*(d + e*x)^(1/2)))^p, x)