Integrand size = 20, antiderivative size = 69 \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=-\frac {4 b n \sqrt {d+e x}}{e}+\frac {4 b \sqrt {d} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{e}+\frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e} \]
4*b*n*arctanh((e*x+d)^(1/2)/d^(1/2))*d^(1/2)/e-4*b*n*(e*x+d)^(1/2)/e+2*(a+ b*ln(c*x^n))*(e*x+d)^(1/2)/e
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=\frac {4 b \sqrt {d} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (a-2 b n+b \log \left (c x^n\right )\right )}{e} \]
(4*b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(a - 2*b*n + b*Log[c*x^n]))/e
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2756, 60, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 2756 |
\(\displaystyle \frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {2 b n \int \frac {\sqrt {d+e x}}{x}dx}{e}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {2 b n \left (d \int \frac {1}{x \sqrt {d+e x}}dx+2 \sqrt {d+e x}\right )}{e}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {2 b n \left (\frac {2 d \int \frac {1}{\frac {d+e x}{e}-\frac {d}{e}}d\sqrt {d+e x}}{e}+2 \sqrt {d+e x}\right )}{e}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {2 b n \left (2 \sqrt {d+e x}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{e}\) |
(-2*b*n*(2*Sqrt[d + e*x] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))/e + (2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e
3.2.47.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Simp[b*n*(p/(e*(q + 1))) Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (IntegersQ[2*p, 2*q] && !IGtQ[q, 0]) || (EqQ[p, 2] & & NeQ[q, 1]))
Time = 0.40 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {2 \sqrt {e x +d}\, a +2 b \left (\ln \left (c \,x^{n}\right ) \sqrt {e x +d}+2 n \left (-\sqrt {e x +d}+\sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )\right )\right )}{e}\) | \(62\) |
default | \(\frac {2 \sqrt {e x +d}\, a +2 b \left (\ln \left (c \,x^{n}\right ) \sqrt {e x +d}+2 n \left (-\sqrt {e x +d}+\sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )\right )\right )}{e}\) | \(62\) |
parts | \(\frac {2 a \sqrt {e x +d}}{e}+\frac {2 b \left (\ln \left (c \,x^{n}\right ) \sqrt {e x +d}-2 n \left (\sqrt {e x +d}-\sqrt {d}\, \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )\right )\right )}{e}\) | \(64\) |
2/e*((e*x+d)^(1/2)*a+b*(ln(c*x^n)*(e*x+d)^(1/2)+2*n*(-(e*x+d)^(1/2)+d^(1/2 )*arctanh((e*x+d)^(1/2)/d^(1/2)))))
Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.68 \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=\left [\frac {2 \, {\left (b \sqrt {d} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (b n \log \left (x\right ) - 2 \, b n + b \log \left (c\right ) + a\right )} \sqrt {e x + d}\right )}}{e}, -\frac {2 \, {\left (2 \, b \sqrt {-d} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) - {\left (b n \log \left (x\right ) - 2 \, b n + b \log \left (c\right ) + a\right )} \sqrt {e x + d}\right )}}{e}\right ] \]
[2*(b*sqrt(d)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (b*n*log(x) - 2*b*n + b*log(c) + a)*sqrt(e*x + d))/e, -2*(2*b*sqrt(-d)*n*arctan(sqrt( e*x + d)*sqrt(-d)/d) - (b*n*log(x) - 2*b*n + b*log(c) + a)*sqrt(e*x + d))/ e]
Time = 2.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.90 \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=a \left (\begin {cases} \frac {2 \sqrt {d + e x}}{e} & \text {for}\: e \neq 0 \\\frac {x}{\sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {4 \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e} + \frac {4 d}{e^{\frac {3}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} + \frac {4 \sqrt {x}}{\sqrt {e} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x}{\sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 \sqrt {d + e x}}{e} & \text {for}\: e \neq 0 \\\frac {x}{\sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((2*sqrt(d + e*x)/e, Ne(e, 0)), (x/sqrt(d), True)) - b*n*Piecew ise((-4*sqrt(d)*asinh(sqrt(d)/(sqrt(e)*sqrt(x)))/e + 4*d/(e**(3/2)*sqrt(x) *sqrt(d/(e*x) + 1)) + 4*sqrt(x)/(sqrt(e)*sqrt(d/(e*x) + 1)), (e > -oo) & ( e < oo) & Ne(e, 0)), (x/sqrt(d), True)) + b*Piecewise((2*sqrt(d + e*x)/e, Ne(e, 0)), (x/sqrt(d), True))*log(c*x**n)
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.19 \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=-\frac {2 \, {\left (\sqrt {d} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right ) + 2 \, \sqrt {e x + d}\right )} b n}{e} + \frac {2 \, \sqrt {e x + d} b \log \left (c x^{n}\right )}{e} + \frac {2 \, \sqrt {e x + d} a}{e} \]
-2*(sqrt(d)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d))) + 2*s qrt(e*x + d))*b*n/e + 2*sqrt(e*x + d)*b*log(c*x^n)/e + 2*sqrt(e*x + d)*a/e
Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=-\frac {2 \, {\left ({\left (\frac {2 \, d \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \sqrt {e x + d} \log \left (x\right ) + 2 \, \sqrt {e x + d}\right )} b n - \sqrt {e x + d} b \log \left (c\right ) - \sqrt {e x + d} a\right )}}{e} \]
-2*((2*d*arctan(sqrt(e*x + d)/sqrt(-d))/sqrt(-d) - sqrt(e*x + d)*log(x) + 2*sqrt(e*x + d))*b*n - sqrt(e*x + d)*b*log(c) - sqrt(e*x + d)*a)/e
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{\sqrt {d+e x}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{\sqrt {d+e\,x}} \,d x \]