Integrand size = 22, antiderivative size = 65 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {b n}{2 x}-\frac {b d n}{e \sqrt {x}}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \] Output:
1/2*b*n/x-b*d*n/e/x^(1/2)+b*d^2*n*ln(d+e/x^(1/2))/e^2-(a+b*ln(c*(d+e/x^(1/ 2))^n))/x
Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=-\frac {a}{x}+\frac {b n}{2 x}-\frac {b d n}{e \sqrt {x}}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \] Input:
Integrate[(a + b*Log[c*(d + e/Sqrt[x])^n])/x^2,x]
Output:
-(a/x) + (b*n)/(2*x) - (b*d*n)/(e*Sqrt[x]) + (b*d^2*n*Log[d + e/Sqrt[x]])/ e^2 - (b*Log[c*(d + e/Sqrt[x])^n])/x
Time = 0.43 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2904, 2842, 49, 2009}
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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle -2 \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{\sqrt {x}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle -2 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{2 x}-\frac {1}{2} b e n \int \frac {1}{\left (d+\frac {e}{\sqrt {x}}\right ) x}d\frac {1}{\sqrt {x}}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -2 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{2 x}-\frac {1}{2} b e n \int \left (\frac {d^2}{e^2 \left (d+\frac {e}{\sqrt {x}}\right )}-\frac {d}{e^2}+\frac {1}{e \sqrt {x}}\right )d\frac {1}{\sqrt {x}}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{2 x}-\frac {1}{2} b e n \left (\frac {d^2 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^3}-\frac {d}{e^2 \sqrt {x}}+\frac {1}{2 e x}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e/Sqrt[x])^n])/x^2,x]
Output:
-2*(-1/2*(b*e*n*(1/(2*e*x) - d/(e^2*Sqrt[x]) + (d^2*Log[d + e/Sqrt[x]])/e^ 3)) + (a + b*Log[c*(d + e/Sqrt[x])^n])/(2*x))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ ))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1))) Int[(f + g*x)^(q + 1)/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
Time = 1.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(-\frac {a}{x}-\frac {b \ln \left (c \,{\mathrm e}^{n \ln \left (d +\frac {e}{\sqrt {x}}\right )}\right )}{x}+\frac {b n}{2 x}+\frac {b \,d^{2} n \ln \left (d +\frac {e}{\sqrt {x}}\right )}{e^{2}}-\frac {b d n}{e \sqrt {x}}\) | \(63\) |
default | \(-\frac {a}{x}-\frac {b \ln \left (c \,{\mathrm e}^{n \ln \left (d +\frac {e}{\sqrt {x}}\right )}\right )}{x}+\frac {b n}{2 x}+\frac {b \,d^{2} n \ln \left (d +\frac {e}{\sqrt {x}}\right )}{e^{2}}-\frac {b d n}{e \sqrt {x}}\) | \(63\) |
Input:
int((a+b*ln(c*(d+e/x^(1/2))^n))/x^2,x,method=_RETURNVERBOSE)
Output:
-a/x-b/x*ln(c*exp(n*ln(d+e/x^(1/2))))+1/2*b*n/x+b*d^2*n*ln(d+e/x^(1/2))/e^ 2-b*d*n/e/x^(1/2)
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=-\frac {2 \, b d e n \sqrt {x} - b e^{2} n + 2 \, b e^{2} \log \left (c\right ) + 2 \, a e^{2} - 2 \, {\left (b d^{2} n x - b e^{2} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right )}{2 \, e^{2} x} \] Input:
integrate((a+b*log(c*(d+e/x^(1/2))^n))/x^2,x, algorithm="fricas")
Output:
-1/2*(2*b*d*e*n*sqrt(x) - b*e^2*n + 2*b*e^2*log(c) + 2*a*e^2 - 2*(b*d^2*n* x - b*e^2*n)*log((d*x + e*sqrt(x))/x))/(e^2*x)
Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (56) = 112\).
Time = 47.49 (sec) , antiderivative size = 391, normalized size of antiderivative = 6.02 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\begin {cases} - \frac {a + b \log {\left (0^{n} c \right )}}{x} & \text {for}\: \left (d = 0 \vee d = - \frac {e}{\sqrt {x}}\right ) \wedge \left (d = - \frac {e}{\sqrt {x}} \vee e = 0\right ) \\- \frac {a + b \log {\left (c d^{n} \right )}}{x} & \text {for}\: e = 0 \\- \frac {2 a d e^{2} x^{3}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 a e^{3} x^{\frac {5}{2}}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} + \frac {2 b d^{3} x^{4} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 b d^{2} e n x^{\frac {7}{2}}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} + \frac {2 b d^{2} e x^{\frac {7}{2}} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {b d e^{2} n x^{3}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 b d e^{2} x^{3} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} + \frac {b e^{3} n x^{\frac {5}{2}}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 b e^{3} x^{\frac {5}{2}} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*ln(c*(d+e/x**(1/2))**n))/x**2,x)
Output:
Piecewise((-(a + b*log(0**n*c))/x, (Eq(d, 0) | Eq(d, -e/sqrt(x))) & (Eq(e, 0) | Eq(d, -e/sqrt(x)))), (-(a + b*log(c*d**n))/x, Eq(e, 0)), (-2*a*d*e** 2*x**3/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*a*e**3*x**(5/2)/(2*d*e**2*x** 4 + 2*e**3*x**(7/2)) + 2*b*d**3*x**4*log(c*(d + e/sqrt(x))**n)/(2*d*e**2*x **4 + 2*e**3*x**(7/2)) - 2*b*d**2*e*n*x**(7/2)/(2*d*e**2*x**4 + 2*e**3*x** (7/2)) + 2*b*d**2*e*x**(7/2)*log(c*(d + e/sqrt(x))**n)/(2*d*e**2*x**4 + 2* e**3*x**(7/2)) - b*d*e**2*n*x**3/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*b*d *e**2*x**3*log(c*(d + e/sqrt(x))**n)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) + b *e**3*n*x**(5/2)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*b*e**3*x**(5/2)*log (c*(d + e/sqrt(x))**n)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)), True))
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {1}{2} \, b e n {\left (\frac {2 \, d^{2} \log \left (d \sqrt {x} + e\right )}{e^{3}} - \frac {d^{2} \log \left (x\right )}{e^{3}} - \frac {2 \, d \sqrt {x} - e}{e^{2} x}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )}{x} - \frac {a}{x} \] Input:
integrate((a+b*log(c*(d+e/x^(1/2))^n))/x^2,x, algorithm="maxima")
Output:
1/2*b*e*n*(2*d^2*log(d*sqrt(x) + e)/e^3 - d^2*log(x)/e^3 - (2*d*sqrt(x) - e)/(e^2*x)) - b*log(c*(d + e/sqrt(x))^n)/x - a/x
Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (57) = 114\).
Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.78 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (d \sqrt {x} + e\right )} b d n}{e \sqrt {x}} - \frac {{\left (d \sqrt {x} + e\right )}^{2} b n}{e x}\right )} \log \left (\frac {d \sqrt {x} + e}{\sqrt {x}}\right ) + \frac {{\left (b n - 2 \, b \log \left (c\right ) - 2 \, a\right )} {\left (d \sqrt {x} + e\right )}^{2}}{e x} - \frac {4 \, {\left (b d n - b d \log \left (c\right ) - a d\right )} {\left (d \sqrt {x} + e\right )}}{e \sqrt {x}}}{2 \, e} \] Input:
integrate((a+b*log(c*(d+e/x^(1/2))^n))/x^2,x, algorithm="giac")
Output:
1/2*(2*(2*(d*sqrt(x) + e)*b*d*n/(e*sqrt(x)) - (d*sqrt(x) + e)^2*b*n/(e*x)) *log((d*sqrt(x) + e)/sqrt(x)) + (b*n - 2*b*log(c) - 2*a)*(d*sqrt(x) + e)^2 /(e*x) - 4*(b*d*n - b*d*log(c) - a*d)*(d*sqrt(x) + e)/(e*sqrt(x)))/e
Time = 14.60 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {b\,n}{2\,x}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}{x}-\frac {b\,d\,n}{e\,\sqrt {x}}+\frac {b\,d^2\,n\,\ln \left (d+\frac {e}{\sqrt {x}}\right )}{e^2} \] Input:
int((a + b*log(c*(d + e/x^(1/2))^n))/x^2,x)
Output:
(b*n)/(2*x) - a/x - (b*log(c*(d + e/x^(1/2))^n))/x - (b*d*n)/(e*x^(1/2)) + (b*d^2*n*log(d + e/x^(1/2)))/e^2
Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {-2 \sqrt {x}\, b d e n +2 \,\mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{n} c}{x^{\frac {n}{2}}}\right ) b \,d^{2} x -2 \,\mathrm {log}\left (\frac {\left (\sqrt {x}\, d +e \right )^{n} c}{x^{\frac {n}{2}}}\right ) b \,e^{2}-2 a \,e^{2}+b \,e^{2} n}{2 e^{2} x} \] Input:
int((a+b*log(c*(d+e/x^(1/2))^n))/x^2,x)
Output:
( - 2*sqrt(x)*b*d*e*n + 2*log(((sqrt(x)*d + e)**n*c)/x**(n/2))*b*d**2*x - 2*log(((sqrt(x)*d + e)**n*c)/x**(n/2))*b*e**2 - 2*a*e**2 + b*e**2*n)/(2*e* *2*x)