Integrand size = 15, antiderivative size = 95 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=\frac {5 b^2}{4 a^3 \sqrt {a+b x^3}}-\frac {1}{6 a x^6 \sqrt {a+b x^3}}+\frac {5 b}{12 a^2 x^3 \sqrt {a+b x^3}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
-5/4*b^2*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(7/2)+5/4*b^2/a^3/(b*x^3+a)^(1 /2)-1/6/a/x^6/(b*x^3+a)^(1/2)+5/12*b/a^2/x^3/(b*x^3+a)^(1/2)
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=\frac {-2 a^2+5 a b x^3+15 b^2 x^6}{12 a^3 x^6 \sqrt {a+b x^3}}-\frac {5 b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 a^{7/2}} \]
(-2*a^2 + 5*a*b*x^3 + 15*b^2*x^6)/(12*a^3*x^6*Sqrt[a + b*x^3]) - (5*b^2*Ar cTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(7/2))
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 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.
\(\displaystyle \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^9 \left (b x^3+a\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (-\frac {5 b \int \frac {1}{x^6 \left (b x^3+a\right )^{3/2}}dx^3}{4 a}-\frac {1}{2 a x^6 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x^3 \left (b x^3+a\right )^{3/2}}dx^3}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )}{4 a}-\frac {1}{2 a x^6 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{3} \left (-\frac {5 b \left (-\frac {3 b \left (\frac {\int \frac {1}{x^3 \sqrt {b x^3+a}}dx^3}{a}+\frac {2}{a \sqrt {a+b x^3}}\right )}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )}{4 a}-\frac {1}{2 a x^6 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (-\frac {5 b \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {x^6}{b}-\frac {a}{b}}d\sqrt {b x^3+a}}{a b}+\frac {2}{a \sqrt {a+b x^3}}\right )}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )}{4 a}-\frac {1}{2 a x^6 \sqrt {a+b x^3}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (-\frac {5 b \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+b x^3}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {1}{a x^3 \sqrt {a+b x^3}}\right )}{4 a}-\frac {1}{2 a x^6 \sqrt {a+b x^3}}\right )\) |
(-1/2*1/(a*x^6*Sqrt[a + b*x^3]) - (5*b*(-(1/(a*x^3*Sqrt[a + b*x^3])) - (3* b*(2/(a*Sqrt[a + b*x^3]) - (2*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/a^(3/2)))/ (2*a)))/(4*a))/3
3.5.27.3.1 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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 3.78 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(-\frac {5 \left (\sqrt {b \,x^{3}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) b^{2} x^{6}-b^{2} x^{6} \sqrt {a}-\frac {a^{\frac {3}{2}} b \,x^{3}}{3}+\frac {2 a^{\frac {5}{2}}}{15}\right )}{4 \sqrt {b \,x^{3}+a}\, a^{\frac {7}{2}} x^{6}}\) | \(74\) |
default | \(-\frac {\sqrt {b \,x^{3}+a}}{6 a^{2} x^{6}}+\frac {7 b \sqrt {b \,x^{3}+a}}{12 a^{3} x^{3}}+\frac {2 b^{2}}{3 a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {5 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\) | \(80\) |
elliptic | \(-\frac {\sqrt {b \,x^{3}+a}}{6 a^{2} x^{6}}+\frac {7 b \sqrt {b \,x^{3}+a}}{12 a^{3} x^{3}}+\frac {2 b^{2}}{3 a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {5 b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\) | \(80\) |
risch | \(-\frac {\sqrt {b \,x^{3}+a}\, \left (-7 b \,x^{3}+2 a \right )}{12 a^{3} x^{6}}+\frac {b^{2} \left (-\frac {14}{3 \sqrt {b \,x^{3}+a}}+15 a \left (\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\right )\right )}{8 a^{3}}\) | \(90\) |
-5/4*((b*x^3+a)^(1/2)*arctanh((b*x^3+a)^(1/2)/a^(1/2))*b^2*x^6-b^2*x^6*a^( 1/2)-1/3*a^(3/2)*b*x^3+2/15*a^(5/2))/(b*x^3+a)^(1/2)/a^(7/2)/x^6
Time = 0.27 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.14 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{9} + a b^{2} x^{6}\right )} \sqrt {a} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (15 \, a b^{2} x^{6} + 5 \, a^{2} b x^{3} - 2 \, a^{3}\right )} \sqrt {b x^{3} + a}}{24 \, {\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}, \frac {15 \, {\left (b^{3} x^{9} + a b^{2} x^{6}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{6} + 5 \, a^{2} b x^{3} - 2 \, a^{3}\right )} \sqrt {b x^{3} + a}}{12 \, {\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}\right ] \]
[1/24*(15*(b^3*x^9 + a*b^2*x^6)*sqrt(a)*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqr t(a) + 2*a)/x^3) + 2*(15*a*b^2*x^6 + 5*a^2*b*x^3 - 2*a^3)*sqrt(b*x^3 + a)) /(a^4*b*x^9 + a^5*x^6), 1/12*(15*(b^3*x^9 + a*b^2*x^6)*sqrt(-a)*arctan(sqr t(b*x^3 + a)*sqrt(-a)/a) + (15*a*b^2*x^6 + 5*a^2*b*x^3 - 2*a^3)*sqrt(b*x^3 + a))/(a^4*b*x^9 + a^5*x^6)]
Time = 3.77 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=- \frac {1}{6 a \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {5 \sqrt {b}}{12 a^{2} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {5 b^{\frac {3}{2}}}{4 a^{3} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {5 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 a^{\frac {7}{2}}} \]
-1/(6*a*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) + 5*sqrt(b)/(12*a**2*x**(9 /2)*sqrt(a/(b*x**3) + 1)) + 5*b**(3/2)/(4*a**3*x**(3/2)*sqrt(a/(b*x**3) + 1)) - 5*b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(4*a**(7/2))
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=\frac {15 \, {\left (b x^{3} + a\right )}^{2} b^{2} - 25 \, {\left (b x^{3} + a\right )} a b^{2} + 8 \, a^{2} b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x^{3} + a} a^{5}\right )}} + \frac {5 \, b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]
1/12*(15*(b*x^3 + a)^2*b^2 - 25*(b*x^3 + a)*a*b^2 + 8*a^2*b^2)/((b*x^3 + a )^(5/2)*a^3 - 2*(b*x^3 + a)^(3/2)*a^4 + sqrt(b*x^3 + a)*a^5) + 5/8*b^2*log ((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(7/2)
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=\frac {5 \, b^{2} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{3 \, \sqrt {b x^{3} + a} a^{3}} + \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x^{3} + a} a b^{2}}{12 \, a^{3} b^{2} x^{6}} \]
5/4*b^2*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2/3*b^2/(sqrt(b* x^3 + a)*a^3) + 1/12*(7*(b*x^3 + a)^(3/2)*b^2 - 9*sqrt(b*x^3 + a)*a*b^2)/( a^3*b^2*x^6)
Time = 6.11 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx=\frac {2\,b^2}{3\,a^3\,\sqrt {b\,x^3+a}}-\frac {\sqrt {b\,x^3+a}}{6\,a^2\,x^6}+\frac {5\,b^2\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{8\,a^{7/2}}+\frac {7\,b\,\sqrt {b\,x^3+a}}{12\,a^3\,x^3} \]