Integrand size = 15, antiderivative size = 71 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=-\frac {\sqrt {a+b x^3}}{6 x^6}-\frac {b \sqrt {a+b x^3}}{12 a x^3}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}} \] Output:
-1/6*(b*x^3+a)^(1/2)/x^6-1/12*b*(b*x^3+a)^(1/2)/a/x^3+1/12*b^2*arctanh((b* x^3+a)^(1/2)/a^(1/2))/a^(3/2)
Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=\frac {\left (-2 a-b x^3\right ) \sqrt {a+b x^3}}{12 a x^6}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}} \] Input:
Integrate[Sqrt[a + b*x^3]/x^7,x]
Output:
((-2*a - b*x^3)*Sqrt[a + b*x^3])/(12*a*x^6) + (b^2*ArcTanh[Sqrt[a + b*x^3] /Sqrt[a]])/(12*a^(3/2))
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {798, 51, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^3}}{x^7} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {\sqrt {b x^3+a}}{x^9}dx^3\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} b \int \frac {1}{x^6 \sqrt {b x^3+a}}dx^3-\frac {\sqrt {a+b x^3}}{2 x^6}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} b \left (-\frac {b \int \frac {1}{x^3 \sqrt {b x^3+a}}dx^3}{2 a}-\frac {\sqrt {a+b x^3}}{a x^3}\right )-\frac {\sqrt {a+b x^3}}{2 x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {x^6}{b}-\frac {a}{b}}d\sqrt {b x^3+a}}{a}-\frac {\sqrt {a+b x^3}}{a x^3}\right )-\frac {\sqrt {a+b x^3}}{2 x^6}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^3}}{a x^3}\right )-\frac {\sqrt {a+b x^3}}{2 x^6}\right )\) |
Input:
Int[Sqrt[a + b*x^3]/x^7,x]
Output:
(-1/2*Sqrt[a + b*x^3]/x^6 + (b*(-(Sqrt[a + b*x^3]/(a*x^3)) + (b*ArcTanh[Sq rt[a + b*x^3]/Sqrt[a]])/a^(3/2)))/4)/3
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]
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 = 0.50 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.70
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{3}+a}\, \left (b \,x^{3}+2 a \right )}{12 x^{6} a}+\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}\) | \(50\) |
default | \(-\frac {\sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {b \sqrt {b \,x^{3}+a}}{12 a \,x^{3}}+\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}\) | \(56\) |
elliptic | \(-\frac {\sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {b \sqrt {b \,x^{3}+a}}{12 a \,x^{3}}+\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}\) | \(56\) |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) b^{2} x^{6}-\sqrt {b \,x^{3}+a}\, \left (b \,x^{3} \sqrt {a}+2 a^{\frac {3}{2}}\right )}{12 a^{\frac {3}{2}} x^{6}}\) | \(56\) |
Input:
int((b*x^3+a)^(1/2)/x^7,x,method=_RETURNVERBOSE)
Output:
-1/12*(b*x^3+a)^(1/2)*(b*x^3+2*a)/x^6/a+1/12*b^2*arctanh((b*x^3+a)^(1/2)/a ^(1/2))/a^(3/2)
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=\left [\frac {\sqrt {a} b^{2} x^{6} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a^{2} x^{6}}, -\frac {\sqrt {-a} b^{2} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{3} + a}}\right ) + {\left (a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a^{2} x^{6}}\right ] \] Input:
integrate((b*x^3+a)^(1/2)/x^7,x, algorithm="fricas")
Output:
[1/24*(sqrt(a)*b^2*x^6*log((b*x^3 + 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) - 2*(a*b*x^3 + 2*a^2)*sqrt(b*x^3 + a))/(a^2*x^6), -1/12*(sqrt(-a)*b^2*x^6* arctan(sqrt(-a)/sqrt(b*x^3 + a)) + (a*b*x^3 + 2*a^2)*sqrt(b*x^3 + a))/(a^2 *x^6)]
Time = 2.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=- \frac {a}{6 \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {\sqrt {b}}{4 x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {b^{\frac {3}{2}}}{12 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{12 a^{\frac {3}{2}}} \] Input:
integrate((b*x**3+a)**(1/2)/x**7,x)
Output:
-a/(6*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) - sqrt(b)/(4*x**(9/2)*sqrt(a /(b*x**3) + 1)) - b**(3/2)/(12*a*x**(3/2)*sqrt(a/(b*x**3) + 1)) + b**2*asi nh(sqrt(a)/(sqrt(b)*x**(3/2)))/(12*a**(3/2))
Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=-\frac {b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{24 \, a^{\frac {3}{2}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b x^{3} + a} a b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{2} a - 2 \, {\left (b x^{3} + a\right )} a^{2} + a^{3}\right )}} \] Input:
integrate((b*x^3+a)^(1/2)/x^7,x, algorithm="maxima")
Output:
-1/24*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^( 3/2) - 1/12*((b*x^3 + a)^(3/2)*b^2 + sqrt(b*x^3 + a)*a*b^2)/((b*x^3 + a)^2 *a - 2*(b*x^3 + a)*a^2 + a^3)
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=-\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b x^{3} + a} a b^{3}}{a b^{2} x^{6}}}{12 \, b} \] Input:
integrate((b*x^3+a)^(1/2)/x^7,x, algorithm="giac")
Output:
-1/12*(b^3*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a) + ((b*x^3 + a)^(3 /2)*b^3 + sqrt(b*x^3 + a)*a*b^3)/(a*b^2*x^6))/b
Time = 0.55 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=\frac {b^2\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{24\,a^{3/2}}-\frac {\sqrt {b\,x^3+a}}{6\,x^6}-\frac {b\,\sqrt {b\,x^3+a}}{12\,a\,x^3} \] Input:
int((a + b*x^3)^(1/2)/x^7,x)
Output:
(b^2*log((((a + b*x^3)^(1/2) - a^(1/2))*((a + b*x^3)^(1/2) + a^(1/2))^3)/x ^6))/(24*a^(3/2)) - (a + b*x^3)^(1/2)/(6*x^6) - (b*(a + b*x^3)^(1/2))/(12* a*x^3)
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {a+b x^3}}{x^7} \, dx=\frac {-4 \sqrt {b \,x^{3}+a}\, a^{2}-2 \sqrt {b \,x^{3}+a}\, a b \,x^{3}-\sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {a}\right ) b^{2} x^{6}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {a}\right ) b^{2} x^{6}}{24 a^{2} x^{6}} \] Input:
int((b*x^3+a)^(1/2)/x^7,x)
Output:
( - 4*sqrt(a + b*x**3)*a**2 - 2*sqrt(a + b*x**3)*a*b*x**3 - sqrt(a)*log(sq rt(a + b*x**3) - sqrt(a))*b**2*x**6 + sqrt(a)*log(sqrt(a + b*x**3) + sqrt( a))*b**2*x**6)/(24*a**2*x**6)