Integrand size = 21, antiderivative size = 97 \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=-\frac {\sqrt {a+b \sqrt {c x^3}}}{3 x^3}-\frac {b c \sqrt {a+b \sqrt {c x^3}}}{6 a \sqrt {c x^3}}+\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^3}}}{\sqrt {a}}\right )}{6 a^{3/2}} \]
1/6*b^2*c*arctanh((a+b*(c*x^3)^(1/2))^(1/2)/a^(1/2))/a^(3/2)-1/3*(a+b*(c*x ^3)^(1/2))^(1/2)/x^3-1/6*b*c*(a+b*(c*x^3)^(1/2))^(1/2)/a/(c*x^3)^(1/2)
Time = 1.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=-\frac {\sqrt {a+b \sqrt {c x^3}} \left (2 a+b \sqrt {c x^3}\right )}{6 a x^3}+\frac {b^2 c \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^3}}}{\sqrt {a}}\right )}{6 a^{3/2}} \]
-1/6*(Sqrt[a + b*Sqrt[c*x^3]]*(2*a + b*Sqrt[c*x^3]))/(a*x^3) + (b^2*c*ArcT anh[Sqrt[a + b*Sqrt[c*x^3]]/Sqrt[a]])/(6*a^(3/2))
Time = 0.23 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {893, 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.
\(\displaystyle \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx\) |
\(\Big \downarrow \) 893 |
\(\displaystyle \int \frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{x^4}dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {2}{3} \int \frac {\sqrt {b \sqrt {c} x^{3/2}+a}}{x^{9/2}}dx^{3/2}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {2}{3} \left (\frac {1}{4} b \sqrt {c} \int \frac {1}{x^3 \sqrt {b \sqrt {c} x^{3/2}+a}}dx^{3/2}-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{2 x^3}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {2}{3} \left (\frac {1}{4} b \sqrt {c} \left (-\frac {b \sqrt {c} \int \frac {1}{x^{3/2} \sqrt {b \sqrt {c} x^{3/2}+a}}dx^{3/2}}{2 a}-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{a x^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{2 x^3}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2}{3} \left (\frac {1}{4} b \sqrt {c} \left (-\frac {\int \frac {1}{\frac {x^3}{b \sqrt {c}}-\frac {a}{b \sqrt {c}}}d\sqrt {b \sqrt {c} x^{3/2}+a}}{a}-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{a x^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{2 x^3}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2}{3} \left (\frac {1}{4} b \sqrt {c} \left (\frac {b \sqrt {c} \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{a x^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c} x^{3/2}}}{2 x^3}\right )\) |
(2*(-1/2*Sqrt[a + b*Sqrt[c]*x^(3/2)]/x^3 + (b*Sqrt[c]*(-(Sqrt[a + b*Sqrt[c ]*x^(3/2)]/(a*x^(3/2))) + (b*Sqrt[c]*ArcTanh[Sqrt[a + b*Sqrt[c]*x^(3/2)]/S qrt[a]])/a^(3/2)))/4))/3
3.30.63.3.1 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]
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]]
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo l] :> With[{k = Denominator[n]}, Subst[Int[(d*x)^m*(a + b*c^n*x^(n*q))^p, x ], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b, c, d, m, p, q}, x] && FractionQ[n]
Time = 4.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84
method | result | size |
default | \(-\frac {-b^{2} \operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {c \,x^{3}}}}{\sqrt {a}}\right ) c \,x^{3} a +\sqrt {c \,x^{3}}\, b \sqrt {a +b \sqrt {c \,x^{3}}}\, a^{\frac {3}{2}}+2 \sqrt {a +b \sqrt {c \,x^{3}}}\, a^{\frac {5}{2}}}{6 x^{3} a^{\frac {5}{2}}}\) | \(81\) |
-1/6*(-b^2*arctanh((a+b*(c*x^3)^(1/2))^(1/2)/a^(1/2))*c*x^3*a+(c*x^3)^(1/2 )*b*(a+b*(c*x^3)^(1/2))^(1/2)*a^(3/2)+2*(a+b*(c*x^3)^(1/2))^(1/2)*a^(5/2)) /x^3/a^(5/2)
Timed out. \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=\int \frac {\sqrt {a + b \sqrt {c x^{3}}}}{x^{4}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=-\frac {1}{12} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {\sqrt {c x^{3}} b + a} - \sqrt {a}}{\sqrt {\sqrt {c x^{3}} b + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (\sqrt {c x^{3}} b + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {\sqrt {c x^{3}} b + a} a b^{2}\right )}}{{\left (\sqrt {c x^{3}} b + a\right )}^{2} a - 2 \, {\left (\sqrt {c x^{3}} b + a\right )} a^{2} + a^{3}}\right )} c \]
-1/12*(b^2*log((sqrt(sqrt(c*x^3)*b + a) - sqrt(a))/(sqrt(sqrt(c*x^3)*b + a ) + sqrt(a)))/a^(3/2) + 2*((sqrt(c*x^3)*b + a)^(3/2)*b^2 + sqrt(sqrt(c*x^3 )*b + a)*a*b^2)/((sqrt(c*x^3)*b + a)^2*a - 2*(sqrt(c*x^3)*b + a)*a^2 + a^3 ))*c
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=-\frac {{\left (\frac {b^{3} c^{3} \arctan \left (\frac {\sqrt {\sqrt {c x} b c^{2} x + a c^{2}}}{\sqrt {-a} c}\right )}{\sqrt {-a} a} + \frac {\sqrt {\sqrt {c x} b c^{2} x + a c^{2}} a b^{3} c^{6} + {\left (\sqrt {c x} b c^{2} x + a c^{2}\right )}^{\frac {3}{2}} b^{3} c^{4}}{a b^{2} c^{5} x^{3}}\right )} {\left | c \right |}}{6 \, b c^{3}} \]
-1/6*(b^3*c^3*arctan(sqrt(sqrt(c*x)*b*c^2*x + a*c^2)/(sqrt(-a)*c))/(sqrt(- a)*a) + (sqrt(sqrt(c*x)*b*c^2*x + a*c^2)*a*b^3*c^6 + (sqrt(c*x)*b*c^2*x + a*c^2)^(3/2)*b^3*c^4)/(a*b^2*c^5*x^3))*abs(c)/(b*c^3)
Timed out. \[ \int \frac {\sqrt {a+b \sqrt {c x^3}}}{x^4} \, dx=\int \frac {\sqrt {a+b\,\sqrt {c\,x^3}}}{x^4} \,d x \]