Integrand size = 18, antiderivative size = 130 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=-\frac {b p}{15 a x^{5/2}}+\frac {b^2 p}{12 a^2 x^2}-\frac {b^3 p}{9 a^3 x^{3/2}}+\frac {b^4 p}{6 a^4 x}-\frac {b^5 p}{3 a^5 \sqrt {x}}+\frac {b^6 p \log \left (a+b \sqrt {x}\right )}{3 a^6}-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{3 x^3}-\frac {b^6 p \log (x)}{6 a^6} \]
-1/15*b*p/a/x^(5/2)+1/12*b^2*p/a^2/x^2-1/9*b^3*p/a^3/x^(3/2)+1/6*b^4*p/a^4 /x-1/6*b^6*p*ln(x)/a^6+1/3*b^6*p*ln(a+b*x^(1/2))/a^6-1/3*ln(c*(a+b*x^(1/2) )^p)/x^3-1/3*b^5*p/a^5/x^(1/2)
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {a b p \sqrt {x} \left (-12 a^4+15 a^3 b \sqrt {x}-20 a^2 b^2 x+30 a b^3 x^{3/2}-60 b^4 x^2\right )+60 b^6 p x^3 \log \left (a+b \sqrt {x}\right )-60 a^6 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-30 b^6 p x^3 \log (x)}{180 a^6 x^3} \]
(a*b*p*Sqrt[x]*(-12*a^4 + 15*a^3*b*Sqrt[x] - 20*a^2*b^2*x + 30*a*b^3*x^(3/ 2) - 60*b^4*x^2) + 60*b^6*p*x^3*Log[a + b*Sqrt[x]] - 60*a^6*Log[c*(a + b*S qrt[x])^p] - 30*b^6*p*x^3*Log[x])/(180*a^6*x^3)
Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.
\(\displaystyle \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle 2 \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^{7/2}}d\sqrt {x}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle 2 \left (\frac {1}{6} b p \int \frac {1}{\left (a+b \sqrt {x}\right ) x^3}d\sqrt {x}-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{6 x^3}\right )\) |
\(\Big \downarrow \) 54 |
\(\displaystyle 2 \left (\frac {1}{6} b p \int \left (\frac {b^6}{a^6 \left (a+b \sqrt {x}\right )}-\frac {b^5}{a^6 \sqrt {x}}+\frac {b^4}{a^5 x}-\frac {b^3}{a^4 x^{3/2}}+\frac {b^2}{a^3 x^2}-\frac {b}{a^2 x^{5/2}}+\frac {1}{a x^3}\right )d\sqrt {x}-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{6 x^3}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{6} b p \left (\frac {b^5 \log \left (a+b \sqrt {x}\right )}{a^6}-\frac {b^5 \log \left (\sqrt {x}\right )}{a^6}-\frac {b^4}{a^5 \sqrt {x}}+\frac {b^3}{2 a^4 x}-\frac {b^2}{3 a^3 x^{3/2}}+\frac {b}{4 a^2 x^2}-\frac {1}{5 a x^{5/2}}\right )-\frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{6 x^3}\right )\) |
2*(-1/6*Log[c*(a + b*Sqrt[x])^p]/x^3 + (b*p*(-1/5*1/(a*x^(5/2)) + b/(4*a^2 *x^2) - b^2/(3*a^3*x^(3/2)) + b^3/(2*a^4*x) - b^4/(a^5*Sqrt[x]) + (b^5*Log [a + b*Sqrt[x]])/a^6 - (b^5*Log[Sqrt[x]])/a^6))/6)
3.1.53.3.1 Defintions of rubi rules used
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])
Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.76
method | result | size |
parts | \(-\frac {\ln \left (c \left (a +b \sqrt {x}\right )^{p}\right )}{3 x^{3}}+\frac {p b \left (\frac {2 b^{5} \ln \left (a +b \sqrt {x}\right )}{a^{6}}-\frac {2}{5 a \,x^{\frac {5}{2}}}-\frac {2 b^{4}}{a^{5} \sqrt {x}}-\frac {2 b^{2}}{3 a^{3} x^{\frac {3}{2}}}-\frac {b^{5} \ln \left (x \right )}{a^{6}}+\frac {b^{3}}{a^{4} x}+\frac {b}{2 a^{2} x^{2}}\right )}{6}\) | \(99\) |
-1/3*ln(c*(a+b*x^(1/2))^p)/x^3+1/6*p*b*(2/a^6*b^5*ln(a+b*x^(1/2))-2/5/a/x^ (5/2)-2/a^5*b^4/x^(1/2)-2/3*b^2/a^3/x^(3/2)-1/a^6*b^5*ln(x)+b^3/a^4/x+1/2* b/a^2/x^2)
Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.84 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=-\frac {60 \, b^{6} p x^{3} \log \left (\sqrt {x}\right ) - 30 \, a^{2} b^{4} p x^{2} - 15 \, a^{4} b^{2} p x + 60 \, a^{6} \log \left (c\right ) - 60 \, {\left (b^{6} p x^{3} - a^{6} p\right )} \log \left (b \sqrt {x} + a\right ) + 4 \, {\left (15 \, a b^{5} p x^{2} + 5 \, a^{3} b^{3} p x + 3 \, a^{5} b p\right )} \sqrt {x}}{180 \, a^{6} x^{3}} \]
-1/180*(60*b^6*p*x^3*log(sqrt(x)) - 30*a^2*b^4*p*x^2 - 15*a^4*b^2*p*x + 60 *a^6*log(c) - 60*(b^6*p*x^3 - a^6*p)*log(b*sqrt(x) + a) + 4*(15*a*b^5*p*x^ 2 + 5*a^3*b^3*p*x + 3*a^5*b*p)*sqrt(x))/(a^6*x^3)
Timed out. \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {1}{180} \, b p {\left (\frac {60 \, b^{5} \log \left (b \sqrt {x} + a\right )}{a^{6}} - \frac {30 \, b^{5} \log \left (x\right )}{a^{6}} - \frac {60 \, b^{4} x^{2} - 30 \, a b^{3} x^{\frac {3}{2}} + 20 \, a^{2} b^{2} x - 15 \, a^{3} b \sqrt {x} + 12 \, a^{4}}{a^{5} x^{\frac {5}{2}}}\right )} - \frac {\log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right )}{3 \, x^{3}} \]
1/180*b*p*(60*b^5*log(b*sqrt(x) + a)/a^6 - 30*b^5*log(x)/a^6 - (60*b^4*x^2 - 30*a*b^3*x^(3/2) + 20*a^2*b^2*x - 15*a^3*b*sqrt(x) + 12*a^4)/(a^5*x^(5/ 2))) - 1/3*log((b*sqrt(x) + a)^p*c)/x^3
Leaf count of result is larger than twice the leaf count of optimal. 324 vs. \(2 (104) = 208\).
Time = 0.31 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.49 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=-\frac {\frac {60 \, b^{7} p \log \left (b \sqrt {x} + a\right )}{{\left (b \sqrt {x} + a\right )}^{6} - 6 \, {\left (b \sqrt {x} + a\right )}^{5} a + 15 \, {\left (b \sqrt {x} + a\right )}^{4} a^{2} - 20 \, {\left (b \sqrt {x} + a\right )}^{3} a^{3} + 15 \, {\left (b \sqrt {x} + a\right )}^{2} a^{4} - 6 \, {\left (b \sqrt {x} + a\right )} a^{5} + a^{6}} - \frac {60 \, b^{7} p \log \left (b \sqrt {x} + a\right )}{a^{6}} + \frac {60 \, b^{7} p \log \left (b \sqrt {x}\right )}{a^{6}} + \frac {60 \, {\left (b \sqrt {x} + a\right )}^{5} b^{7} p - 330 \, {\left (b \sqrt {x} + a\right )}^{4} a b^{7} p + 740 \, {\left (b \sqrt {x} + a\right )}^{3} a^{2} b^{7} p - 855 \, {\left (b \sqrt {x} + a\right )}^{2} a^{3} b^{7} p + 522 \, {\left (b \sqrt {x} + a\right )} a^{4} b^{7} p - 137 \, a^{5} b^{7} p + 60 \, a^{5} b^{7} \log \left (c\right )}{{\left (b \sqrt {x} + a\right )}^{6} a^{5} - 6 \, {\left (b \sqrt {x} + a\right )}^{5} a^{6} + 15 \, {\left (b \sqrt {x} + a\right )}^{4} a^{7} - 20 \, {\left (b \sqrt {x} + a\right )}^{3} a^{8} + 15 \, {\left (b \sqrt {x} + a\right )}^{2} a^{9} - 6 \, {\left (b \sqrt {x} + a\right )} a^{10} + a^{11}}}{180 \, b} \]
-1/180*(60*b^7*p*log(b*sqrt(x) + a)/((b*sqrt(x) + a)^6 - 6*(b*sqrt(x) + a) ^5*a + 15*(b*sqrt(x) + a)^4*a^2 - 20*(b*sqrt(x) + a)^3*a^3 + 15*(b*sqrt(x) + a)^2*a^4 - 6*(b*sqrt(x) + a)*a^5 + a^6) - 60*b^7*p*log(b*sqrt(x) + a)/a ^6 + 60*b^7*p*log(b*sqrt(x))/a^6 + (60*(b*sqrt(x) + a)^5*b^7*p - 330*(b*sq rt(x) + a)^4*a*b^7*p + 740*(b*sqrt(x) + a)^3*a^2*b^7*p - 855*(b*sqrt(x) + a)^2*a^3*b^7*p + 522*(b*sqrt(x) + a)*a^4*b^7*p - 137*a^5*b^7*p + 60*a^5*b^ 7*log(c))/((b*sqrt(x) + a)^6*a^5 - 6*(b*sqrt(x) + a)^5*a^6 + 15*(b*sqrt(x) + a)^4*a^7 - 20*(b*sqrt(x) + a)^3*a^8 + 15*(b*sqrt(x) + a)^2*a^9 - 6*(b*s qrt(x) + a)*a^10 + a^11))/b
Time = 1.75 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.75 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x^4} \, dx=\frac {2\,b^6\,p\,\mathrm {atanh}\left (\frac {2\,b\,\sqrt {x}}{a}+1\right )}{3\,a^6}-\frac {\frac {b\,p}{5\,a}-\frac {b^2\,p\,\sqrt {x}}{4\,a^2}+\frac {b^5\,p\,x^2}{a^5}-\frac {b^4\,p\,x^{3/2}}{2\,a^4}+\frac {b^3\,p\,x}{3\,a^3}}{3\,x^{5/2}}-\frac {\ln \left (c\,{\left (a+b\,\sqrt {x}\right )}^p\right )}{3\,x^3} \]