Integrand size = 22, antiderivative size = 109 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=-\frac {b e n}{6 d x^{3/2}}+\frac {b e^2 n}{4 d^2 x}-\frac {b e^3 n}{2 d^3 \sqrt {x}}+\frac {b e^4 n \log \left (d+e \sqrt {x}\right )}{2 d^4}-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{2 x^2}-\frac {b e^4 n \log (x)}{4 d^4} \] Output:
-1/6*b*e*n/d/x^(3/2)+1/4*b*e^2*n/d^2/x-1/2*b*e^3*n/d^3/x^(1/2)+1/2*b*e^4*n *ln(d+e*x^(1/2))/d^4-1/2*(a+b*ln(c*(d+e*x^(1/2))^n))/x^2-1/4*b*e^4*n*ln(x) /d^4
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{2 x^2}+\frac {1}{4} b e n \left (-\frac {2}{3 d x^{3/2}}+\frac {e}{d^2 x}-\frac {2 e^2}{d^3 \sqrt {x}}+\frac {2 e^3 \log \left (d+e \sqrt {x}\right )}{d^4}-\frac {e^3 \log (x)}{d^4}\right ) \] Input:
Integrate[(a + b*Log[c*(d + e*Sqrt[x])^n])/x^3,x]
Output:
-1/2*a/x^2 - (b*Log[c*(d + e*Sqrt[x])^n])/(2*x^2) + (b*e*n*(-2/(3*d*x^(3/2 )) + e/(d^2*x) - (2*e^2)/(d^3*Sqrt[x]) + (2*e^3*Log[d + e*Sqrt[x]])/d^4 - (e^3*Log[x])/d^4))/4
Time = 0.46 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2904, 2842, 54, 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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle 2 \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^{5/2}}d\sqrt {x}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle 2 \left (\frac {1}{4} b e n \int \frac {1}{\left (d+e \sqrt {x}\right ) x^2}d\sqrt {x}-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{4 x^2}\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle 2 \left (\frac {1}{4} b e n \int \left (\frac {e^4}{d^4 \left (d+e \sqrt {x}\right )}-\frac {e^3}{d^4 \sqrt {x}}+\frac {e^2}{d^3 x}-\frac {e}{d^2 x^{3/2}}+\frac {1}{d x^2}\right )d\sqrt {x}-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{4 x^2}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{4} b e n \left (\frac {e^3 \log \left (d+e \sqrt {x}\right )}{d^4}-\frac {e^3 \log \left (\sqrt {x}\right )}{d^4}-\frac {e^2}{d^3 \sqrt {x}}+\frac {e}{2 d^2 x}-\frac {1}{3 d x^{3/2}}\right )-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{4 x^2}\right )\) |
Input:
Int[(a + b*Log[c*(d + e*Sqrt[x])^n])/x^3,x]
Output:
2*(-1/4*(a + b*Log[c*(d + e*Sqrt[x])^n])/x^2 + (b*e*n*(-1/3*1/(d*x^(3/2)) + e/(2*d^2*x) - e^2/(d^3*Sqrt[x]) + (e^3*Log[d + e*Sqrt[x]])/d^4 - (e^3*Lo g[Sqrt[x]])/d^4))/4)
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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])
\[\int \frac {a +b \ln \left (c \left (d +e \sqrt {x}\right )^{n}\right )}{x^{3}}d x\]
Input:
int((a+b*ln(c*(d+e*x^(1/2))^n))/x^3,x)
Output:
int((a+b*ln(c*(d+e*x^(1/2))^n))/x^3,x)
Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=-\frac {6 \, b e^{4} n x^{2} \log \left (\sqrt {x}\right ) - 3 \, b d^{2} e^{2} n x + 6 \, b d^{4} \log \left (c\right ) + 6 \, a d^{4} - 6 \, {\left (b e^{4} n x^{2} - b d^{4} n\right )} \log \left (e \sqrt {x} + d\right ) + 2 \, {\left (3 \, b d e^{3} n x + b d^{3} e n\right )} \sqrt {x}}{12 \, d^{4} x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(1/2))^n))/x^3,x, algorithm="fricas")
Output:
-1/12*(6*b*e^4*n*x^2*log(sqrt(x)) - 3*b*d^2*e^2*n*x + 6*b*d^4*log(c) + 6*a *d^4 - 6*(b*e^4*n*x^2 - b*d^4*n)*log(e*sqrt(x) + d) + 2*(3*b*d*e^3*n*x + b *d^3*e*n)*sqrt(x))/(d^4*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (104) = 208\).
Time = 101.14 (sec) , antiderivative size = 532, normalized size of antiderivative = 4.88 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=\begin {cases} - \frac {a + b \log {\left (0^{n} c \right )}}{2 x^{2}} & \text {for}\: d = 0 \wedge e = 0 \\- \frac {a}{2 x^{2}} - \frac {b n}{8 x^{2}} - \frac {b \log {\left (c \left (e \sqrt {x}\right )^{n} \right )}}{2 x^{2}} & \text {for}\: d = 0 \\- \frac {a + b \log {\left (0^{n} c \right )}}{2 x^{2}} & \text {for}\: d = - e \sqrt {x} \\- \frac {6 a d^{5} \sqrt {x}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {6 a d^{4} e x}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {6 b d^{5} \sqrt {x} \log {\left (c \left (d + e \sqrt {x}\right )^{n} \right )}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {2 b d^{4} e n x}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {6 b d^{4} e x \log {\left (c \left (d + e \sqrt {x}\right )^{n} \right )}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} + \frac {b d^{3} e^{2} n x^{\frac {3}{2}}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {3 b d^{2} e^{3} n x^{2}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {3 b d e^{4} n x^{\frac {5}{2}} \log {\left (x \right )}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {6 b d e^{4} n x^{\frac {5}{2}}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} + \frac {6 b d e^{4} x^{\frac {5}{2}} \log {\left (c \left (d + e \sqrt {x}\right )^{n} \right )}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} - \frac {3 b e^{5} n x^{3} \log {\left (x \right )}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} + \frac {6 b e^{5} x^{3} \log {\left (c \left (d + e \sqrt {x}\right )^{n} \right )}}{12 d^{5} x^{\frac {5}{2}} + 12 d^{4} e x^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*ln(c*(d+e*x**(1/2))**n))/x**3,x)
Output:
Piecewise((-(a + b*log(0**n*c))/(2*x**2), Eq(d, 0) & Eq(e, 0)), (-a/(2*x** 2) - b*n/(8*x**2) - b*log(c*(e*sqrt(x))**n)/(2*x**2), Eq(d, 0)), (-(a + b* log(0**n*c))/(2*x**2), Eq(d, -e*sqrt(x))), (-6*a*d**5*sqrt(x)/(12*d**5*x** (5/2) + 12*d**4*e*x**3) - 6*a*d**4*e*x/(12*d**5*x**(5/2) + 12*d**4*e*x**3) - 6*b*d**5*sqrt(x)*log(c*(d + e*sqrt(x))**n)/(12*d**5*x**(5/2) + 12*d**4* e*x**3) - 2*b*d**4*e*n*x/(12*d**5*x**(5/2) + 12*d**4*e*x**3) - 6*b*d**4*e* x*log(c*(d + e*sqrt(x))**n)/(12*d**5*x**(5/2) + 12*d**4*e*x**3) + b*d**3*e **2*n*x**(3/2)/(12*d**5*x**(5/2) + 12*d**4*e*x**3) - 3*b*d**2*e**3*n*x**2/ (12*d**5*x**(5/2) + 12*d**4*e*x**3) - 3*b*d*e**4*n*x**(5/2)*log(x)/(12*d** 5*x**(5/2) + 12*d**4*e*x**3) - 6*b*d*e**4*n*x**(5/2)/(12*d**5*x**(5/2) + 1 2*d**4*e*x**3) + 6*b*d*e**4*x**(5/2)*log(c*(d + e*sqrt(x))**n)/(12*d**5*x* *(5/2) + 12*d**4*e*x**3) - 3*b*e**5*n*x**3*log(x)/(12*d**5*x**(5/2) + 12*d **4*e*x**3) + 6*b*e**5*x**3*log(c*(d + e*sqrt(x))**n)/(12*d**5*x**(5/2) + 12*d**4*e*x**3), True))
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.77 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=\frac {1}{12} \, b e n {\left (\frac {6 \, e^{3} \log \left (e \sqrt {x} + d\right )}{d^{4}} - \frac {3 \, e^{3} \log \left (x\right )}{d^{4}} - \frac {6 \, e^{2} x - 3 \, d e \sqrt {x} + 2 \, d^{2}}{d^{3} x^{\frac {3}{2}}}\right )} - \frac {b \log \left ({\left (e \sqrt {x} + d\right )}^{n} c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \] Input:
integrate((a+b*log(c*(d+e*x^(1/2))^n))/x^3,x, algorithm="maxima")
Output:
1/12*b*e*n*(6*e^3*log(e*sqrt(x) + d)/d^4 - 3*e^3*log(x)/d^4 - (6*e^2*x - 3 *d*e*sqrt(x) + 2*d^2)/(d^3*x^(3/2))) - 1/2*b*log((e*sqrt(x) + d)^n*c)/x^2 - 1/2*a/x^2
Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (89) = 178\).
Time = 0.13 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.28 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=-\frac {\frac {6 \, b e^{5} n \log \left (e \sqrt {x} + d\right )}{{\left (e \sqrt {x} + d\right )}^{4} - 4 \, {\left (e \sqrt {x} + d\right )}^{3} d + 6 \, {\left (e \sqrt {x} + d\right )}^{2} d^{2} - 4 \, {\left (e \sqrt {x} + d\right )} d^{3} + d^{4}} - \frac {6 \, b e^{5} n \log \left (e \sqrt {x} + d\right )}{d^{4}} + \frac {6 \, b e^{5} n \log \left (e \sqrt {x}\right )}{d^{4}} + \frac {6 \, {\left (e \sqrt {x} + d\right )}^{3} b e^{5} n - 21 \, {\left (e \sqrt {x} + d\right )}^{2} b d e^{5} n + 26 \, {\left (e \sqrt {x} + d\right )} b d^{2} e^{5} n - 11 \, b d^{3} e^{5} n + 6 \, b d^{3} e^{5} \log \left (c\right ) + 6 \, a d^{3} e^{5}}{{\left (e \sqrt {x} + d\right )}^{4} d^{3} - 4 \, {\left (e \sqrt {x} + d\right )}^{3} d^{4} + 6 \, {\left (e \sqrt {x} + d\right )}^{2} d^{5} - 4 \, {\left (e \sqrt {x} + d\right )} d^{6} + d^{7}}}{12 \, e} \] Input:
integrate((a+b*log(c*(d+e*x^(1/2))^n))/x^3,x, algorithm="giac")
Output:
-1/12*(6*b*e^5*n*log(e*sqrt(x) + d)/((e*sqrt(x) + d)^4 - 4*(e*sqrt(x) + d) ^3*d + 6*(e*sqrt(x) + d)^2*d^2 - 4*(e*sqrt(x) + d)*d^3 + d^4) - 6*b*e^5*n* log(e*sqrt(x) + d)/d^4 + 6*b*e^5*n*log(e*sqrt(x))/d^4 + (6*(e*sqrt(x) + d) ^3*b*e^5*n - 21*(e*sqrt(x) + d)^2*b*d*e^5*n + 26*(e*sqrt(x) + d)*b*d^2*e^5 *n - 11*b*d^3*e^5*n + 6*b*d^3*e^5*log(c) + 6*a*d^3*e^5)/((e*sqrt(x) + d)^4 *d^3 - 4*(e*sqrt(x) + d)^3*d^4 + 6*(e*sqrt(x) + d)^2*d^5 - 4*(e*sqrt(x) + d)*d^6 + d^7))/e
Time = 15.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.76 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=\frac {b\,e^4\,n\,\mathrm {atanh}\left (\frac {2\,e\,\sqrt {x}}{d}+1\right )}{d^4}-\frac {\frac {b\,e\,n}{3\,d}+\frac {b\,e^3\,n\,x}{d^3}-\frac {b\,e^2\,n\,\sqrt {x}}{2\,d^2}}{2\,x^{3/2}}-\frac {b\,\ln \left (c\,{\left (d+e\,\sqrt {x}\right )}^n\right )}{2\,x^2}-\frac {a}{2\,x^2} \] Input:
int((a + b*log(c*(d + e*x^(1/2))^n))/x^3,x)
Output:
(b*e^4*n*atanh((2*e*x^(1/2))/d + 1))/d^4 - ((b*e*n)/(3*d) + (b*e^3*n*x)/d^ 3 - (b*e^2*n*x^(1/2))/(2*d^2))/(2*x^(3/2)) - (b*log(c*(d + e*x^(1/2))^n))/ (2*x^2) - a/(2*x^2)
Time = 0.21 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^3} \, dx=\frac {-2 \sqrt {x}\, b \,d^{3} e n -6 \sqrt {x}\, b d \,e^{3} n x -6 \,\mathrm {log}\left (\sqrt {x}\right ) b \,e^{4} n \,x^{2}-6 \,\mathrm {log}\left (\left (\sqrt {x}\, e +d \right )^{n} c \right ) b \,d^{4}+6 \,\mathrm {log}\left (\left (\sqrt {x}\, e +d \right )^{n} c \right ) b \,e^{4} x^{2}-6 a \,d^{4}+3 b \,d^{2} e^{2} n x}{12 d^{4} x^{2}} \] Input:
int((a+b*log(c*(d+e*x^(1/2))^n))/x^3,x)
Output:
( - 2*sqrt(x)*b*d**3*e*n - 6*sqrt(x)*b*d*e**3*n*x - 6*log(sqrt(x))*b*e**4* n*x**2 - 6*log((sqrt(x)*e + d)**n*c)*b*d**4 + 6*log((sqrt(x)*e + d)**n*c)* b*e**4*x**2 - 6*a*d**4 + 3*b*d**2*e**2*n*x)/(12*d**4*x**2)