Integrand size = 13, antiderivative size = 87 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 b^2}{4 a^3 \sqrt {a+b x}}-\frac {1}{2 a x^2 \sqrt {a+b x}}+\frac {5 b}{4 a^2 x \sqrt {a+b x}}-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
-15/4*b^2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(7/2)+15/4*b^2/a^3/(b*x+a)^(1/2 )-1/2/a/x^2/(b*x+a)^(1/2)+5/4*b/a^2/x/(b*x+a)^(1/2)
Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {-2 a^2+5 a b x+15 b^2 x^2}{4 a^3 x^2 \sqrt {a+b x}}-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
(-2*a^2 + 5*a*b*x + 15*b^2*x^2)/(4*a^3*x^2*Sqrt[a + b*x]) - (15*b^2*ArcTan h[Sqrt[a + b*x]/Sqrt[a]])/(4*a^(7/2))
Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {52, 52, 61, 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.
\(\displaystyle \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5 b \int \frac {1}{x^2 (a+b x)^{3/2}}dx}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {5 b \left (-\frac {3 b \int \frac {1}{x (a+b x)^{3/2}}dx}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {5 b \left (-\frac {3 b \left (\frac {\int \frac {1}{x \sqrt {a+b x}}dx}{a}+\frac {2}{a \sqrt {a+b x}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {5 b \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a b}+\frac {2}{a \sqrt {a+b x}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {5 b \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+b x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x \sqrt {a+b x}}\right )}{4 a}-\frac {1}{2 a x^2 \sqrt {a+b x}}\) |
-1/2*1/(a*x^2*Sqrt[a + b*x]) - (5*b*(-(1/(a*x*Sqrt[a + b*x])) - (3*b*(2/(a *Sqrt[a + b*x]) - (2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)))/(2*a)))/(4* a)
3.4.50.3.1 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 + 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] :> 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] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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]
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.69
method | result | size |
risch | \(-\frac {\sqrt {b x +a}\, \left (-7 b x +2 a \right )}{4 a^{3} x^{2}}+\frac {b^{2} \left (-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {16}{\sqrt {b x +a}}\right )}{8 a^{3}}\) | \(60\) |
pseudoelliptic | \(-\frac {15 \left (\sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-\sqrt {a}\, b^{2} x^{2}-\frac {a^{\frac {3}{2}} b x}{3}+\frac {2 a^{\frac {5}{2}}}{15}\right )}{4 \sqrt {b x +a}\, a^{\frac {7}{2}} x^{2}}\) | \(66\) |
derivativedivides | \(2 b^{2} \left (\frac {1}{a^{3} \sqrt {b x +a}}-\frac {\frac {-\frac {7 \left (b x +a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\) | \(68\) |
default | \(2 b^{2} \left (\frac {1}{a^{3} \sqrt {b x +a}}-\frac {\frac {-\frac {7 \left (b x +a \right )^{\frac {3}{2}}}{8}+\frac {9 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{3}}\right )\) | \(68\) |
-1/4*(b*x+a)^(1/2)*(-7*b*x+2*a)/a^3/x^2+1/8*b^2/a^3*(-30*arctanh((b*x+a)^( 1/2)/a^(1/2))/a^(1/2)+16/(b*x+a)^(1/2))
Time = 0.24 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.17 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {a} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{8 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}, \frac {15 \, {\left (b^{3} x^{3} + a b^{2} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{4 \, {\left (a^{4} b x^{3} + a^{5} x^{2}\right )}}\right ] \]
[1/8*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(a)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2), 1/4*(15*(b^3*x^3 + a*b^2*x^2)*sqrt(-a)*arctan(sqrt(b*x + a)*s qrt(-a)/a) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x + a))/(a^4*b*x^3 + a^5*x^2)]
Time = 5.55 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=- \frac {1}{2 a \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {5 \sqrt {b}}{4 a^{2} x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} + \frac {15 b^{\frac {3}{2}}}{4 a^{3} \sqrt {x} \sqrt {\frac {a}{b x} + 1}} - \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {7}{2}}} \]
-1/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2)*s qrt(a/(b*x) + 1)) + 15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x) + 1)) - 15*b* *2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2))
Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 \, {\left (b x + a\right )}^{2} b^{2} - 25 \, {\left (b x + a\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (b x + a\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x + a} a^{5}\right )}} + \frac {15 \, b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]
1/4*(15*(b*x + a)^2*b^2 - 25*(b*x + a)*a*b^2 + 8*a^2*b^2)/((b*x + a)^(5/2) *a^3 - 2*(b*x + a)^(3/2)*a^4 + sqrt(b*x + a)*a^5) + 15/8*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(7/2)
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3}} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \]
15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2*b^2/(sqrt(b*x + a)*a^3) + 1/4*(7*(b*x + a)^(3/2)*b^2 - 9*sqrt(b*x + a)*a*b^2)/(a^3*b^2*x^ 2)
Time = 0.07 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 (a+b x)^{3/2}} \, dx=\frac {\frac {2\,b^2}{a}+\frac {15\,b^2\,{\left (a+b\,x\right )}^2}{4\,a^3}-\frac {25\,b^2\,\left (a+b\,x\right )}{4\,a^2}}{{\left (a+b\,x\right )}^{5/2}-2\,a\,{\left (a+b\,x\right )}^{3/2}+a^2\,\sqrt {a+b\,x}}-\frac {15\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{4\,a^{7/2}} \]