Integrand size = 13, antiderivative size = 87 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=-\frac {\sqrt {a+b x}}{3 x^3}-\frac {b \sqrt {a+b x}}{12 a x^2}+\frac {b^2 \sqrt {a+b x}}{8 a^2 x}-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
-1/8*b^3*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(5/2)-1/3*(b*x+a)^(1/2)/x^3-1/12 *b*(b*x+a)^(1/2)/a/x^2+1/8*b^2*(b*x+a)^(1/2)/a^2/x
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=-\frac {\sqrt {a+b x} \left (8 a^2+2 a b x-3 b^2 x^2\right )}{24 a^2 x^3}-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{8 a^{5/2}} \]
-1/24*(Sqrt[a + b*x]*(8*a^2 + 2*a*b*x - 3*b^2*x^2))/(a^2*x^3) - (b^3*ArcTa nh[Sqrt[a + b*x]/Sqrt[a]])/(8*a^(5/2))
Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {51, 52, 52, 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.
\(\displaystyle \int \frac {\sqrt {a+b x}}{x^4} \, dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{6} b \int \frac {1}{x^3 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x}}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x^2 \sqrt {a+b x}}dx}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )}{4 a}-\frac {\sqrt {a+b x}}{2 a x^2}\right )-\frac {\sqrt {a+b x}}{3 x^3}\) |
-1/3*Sqrt[a + b*x]/x^3 + (b*(-1/2*Sqrt[a + b*x]/(a*x^2) - (3*b*(-(Sqrt[a + b*x]/(a*x)) + (b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)))/(4*a)))/6
3.3.91.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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]
Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (-3 b^{2} x^{2}+2 a b x +8 a^{2}\right )}{24 x^{3} a^{2}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {5}{2}}}\) | \(56\) |
pseudoelliptic | \(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} x^{3}-\sqrt {b x +a}\, \left (\sqrt {a}\, b^{2} x^{2}-\frac {2 a^{\frac {3}{2}} b x}{3}-\frac {8 a^{\frac {5}{2}}}{3}\right )}{8 a^{\frac {5}{2}} x^{3}}\) | \(61\) |
derivativedivides | \(2 b^{3} \left (-\frac {-\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6 a}+\frac {\sqrt {b x +a}}{16}}{b^{3} x^{3}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )\) | \(66\) |
default | \(2 b^{3} \left (-\frac {-\frac {\left (b x +a \right )^{\frac {5}{2}}}{16 a^{2}}+\frac {\left (b x +a \right )^{\frac {3}{2}}}{6 a}+\frac {\sqrt {b x +a}}{16}}{b^{3} x^{3}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{16 a^{\frac {5}{2}}}\right )\) | \(66\) |
-1/24*(b*x+a)^(1/2)*(-3*b^2*x^2+2*a*b*x+8*a^2)/x^3/a^2-1/8*b^3*arctanh((b* x+a)^(1/2)/a^(1/2))/a^(5/2)
Time = 0.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.67 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=\left [\frac {3 \, \sqrt {a} b^{3} x^{3} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{48 \, a^{3} x^{3}}, \frac {3 \, \sqrt {-a} b^{3} x^{3} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x + a}}{24 \, a^{3} x^{3}}\right ] \]
[1/48*(3*sqrt(a)*b^3*x^3*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2* (3*a*b^2*x^2 - 2*a^2*b*x - 8*a^3)*sqrt(b*x + a))/(a^3*x^3), 1/24*(3*sqrt(- a)*b^3*x^3*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (3*a*b^2*x^2 - 2*a^2*b*x - 8 *a^3)*sqrt(b*x + a))/(a^3*x^3)]
Time = 6.52 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.40 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=- \frac {a}{3 \sqrt {b} x^{\frac {7}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {5 \sqrt {b}}{12 x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {b^{\frac {3}{2}}}{24 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {b^{\frac {5}{2}}}{8 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{8 a^{\frac {5}{2}}} \]
-a/(3*sqrt(b)*x**(7/2)*sqrt(a/(b*x) + 1)) - 5*sqrt(b)/(12*x**(5/2)*sqrt(a/ (b*x) + 1)) + b**(3/2)/(24*a*x**(3/2)*sqrt(a/(b*x) + 1)) + b**(5/2)/(8*a** 2*sqrt(x)*sqrt(a/(b*x) + 1)) - b**3*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(8*a* *(5/2))
Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=\frac {b^{3} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{16 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{3} - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 3 \, \sqrt {b x + a} a^{2} b^{3}}{24 \, {\left ({\left (b x + a\right )}^{3} a^{2} - 3 \, {\left (b x + a\right )}^{2} a^{3} + 3 \, {\left (b x + a\right )} a^{4} - a^{5}\right )}} \]
1/16*b^3*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(5/2) + 1/24*(3*(b*x + a)^(5/2)*b^3 - 8*(b*x + a)^(3/2)*a*b^3 - 3*sqrt(b*x + a)* a^2*b^3)/((b*x + a)^3*a^2 - 3*(b*x + a)^2*a^3 + 3*(b*x + a)*a^4 - a^5)
Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.97 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x + a} a^{2} b^{4}}{a^{2} b^{3} x^{3}}}{24 \, b} \]
1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*(b*x + a)^( 5/2)*b^4 - 8*(b*x + a)^(3/2)*a*b^4 - 3*sqrt(b*x + a)*a^2*b^4)/(a^2*b^3*x^3 ))/b
Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {a+b x}}{x^4} \, dx=\frac {{\left (a+b\,x\right )}^{5/2}}{8\,a^2\,x^3}-\frac {{\left (a+b\,x\right )}^{3/2}}{3\,a\,x^3}-\frac {\sqrt {a+b\,x}}{8\,x^3}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{8\,a^{5/2}} \]