Integrand size = 21, antiderivative size = 352 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{5 x^5}-\frac {3 b \left (c x^2\right )^{5/2} \sqrt {a+b \left (c x^2\right )^{3/2}}}{20 a c x^7}-\frac {3^{3/4} \sqrt {2+\sqrt {3}} b^{5/3} \left (c x^2\right )^{5/2} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}+b^{2/3} c x^2-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}}\right ),-7-4 \sqrt {3}\right )}{20 a x^5 \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}} \]
-1/5*(a+b*(c*x^2)^(3/2))^(1/2)/x^5-3/20*b*(c*x^2)^(5/2)*(a+b*(c*x^2)^(3/2) )^(1/2)/a/c/x^7-1/20*3^(3/4)*b^(5/3)*(c*x^2)^(5/2)*EllipticF((a^(1/3)*(1-3 ^(1/2))+b^(1/3)*(c*x^2)^(1/2))/(a^(1/3)*(1+3^(1/2))+b^(1/3)*(c*x^2)^(1/2)) ,I*3^(1/2)+2*I)*(a^(1/3)+b^(1/3)*(c*x^2)^(1/2))*(1/2*6^(1/2)+1/2*2^(1/2))* ((a^(2/3)+b^(2/3)*c*x^2-a^(1/3)*b^(1/3)*(c*x^2)^(1/2))/(a^(1/3)*(1+3^(1/2) )+b^(1/3)*(c*x^2)^(1/2))^2)^(1/2)/a/x^5/(a+b*(c*x^2)^(3/2))^(1/2)/(a^(1/3) *(a^(1/3)+b^(1/3)*(c*x^2)^(1/2))/(a^(1/3)*(1+3^(1/2))+b^(1/3)*(c*x^2)^(1/2 ))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.20 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},-\frac {1}{2},-\frac {2}{3},-\frac {b \left (c x^2\right )^{3/2}}{a}\right )}{5 x^5 \sqrt {1+\frac {b \left (c x^2\right )^{3/2}}{a}}} \]
-1/5*(Sqrt[a + b*(c*x^2)^(3/2)]*Hypergeometric2F1[-5/3, -1/2, -2/3, -((b*( c*x^2)^(3/2))/a)])/(x^5*Sqrt[1 + (b*(c*x^2)^(3/2))/a])
Time = 0.31 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {892, 809, 847, 759}
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.
\(\displaystyle \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx\) |
\(\Big \downarrow \) 892 |
\(\displaystyle \frac {\left (c x^2\right )^{5/2} \int \frac {\sqrt {b \left (c x^2\right )^{3/2}+a}}{c^3 x^6}d\sqrt {c x^2}}{x^5}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle \frac {\left (c x^2\right )^{5/2} \left (\frac {3}{10} b \int \frac {1}{\left (c x^2\right )^{3/2} \sqrt {b \left (c x^2\right )^{3/2}+a}}d\sqrt {c x^2}-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{5 \left (c x^2\right )^{5/2}}\right )}{x^5}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {\left (c x^2\right )^{5/2} \left (\frac {3}{10} b \left (-\frac {b \int \frac {1}{\sqrt {b \left (c x^2\right )^{3/2}+a}}d\sqrt {c x^2}}{4 a}-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{2 a c x^2}\right )-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{5 \left (c x^2\right )^{5/2}}\right )}{x^5}\) |
\(\Big \downarrow \) 759 |
\(\displaystyle \frac {\left (c x^2\right )^{5/2} \left (\frac {3}{10} b \left (-\frac {\sqrt {2+\sqrt {3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sqrt {c x^2}+b^{2/3} c x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{b} \sqrt {c x^2}+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} \sqrt {c x^2}+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} a \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} \sqrt {c x^2}\right )^2}} \sqrt {a+b \left (c x^2\right )^{3/2}}}-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{2 a c x^2}\right )-\frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{5 \left (c x^2\right )^{5/2}}\right )}{x^5}\) |
((c*x^2)^(5/2)*(-1/5*Sqrt[a + b*(c*x^2)^(3/2)]/(c*x^2)^(5/2) + (3*b*(-1/2* Sqrt[a + b*(c*x^2)^(3/2)]/(a*c*x^2) - (Sqrt[2 + Sqrt[3]]*b^(2/3)*(a^(1/3) + b^(1/3)*Sqrt[c*x^2])*Sqrt[(a^(2/3) + b^(2/3)*c*x^2 - a^(1/3)*b^(1/3)*Sqr t[c*x^2])/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])^2]*EllipticF[ArcSi n[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])/((1 + Sqrt[3])*a^(1/3) + b ^(1/3)*Sqrt[c*x^2])], -7 - 4*Sqrt[3]])/(2*3^(1/4)*a*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*Sqrt[c*x^2]))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*Sqrt[c*x^2])^2]* Sqrt[a + b*(c*x^2)^(3/2)])))/10))/x^5
3.30.50.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[a, b, c, n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1)) Subst[Int[x^m*(a + b *x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x ] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
\[\int \frac {\sqrt {a +b \left (c \,x^{2}\right )^{\frac {3}{2}}}}{x^{6}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.27 \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=-\frac {3 \, \sqrt {c x^{2}} \sqrt {\frac {\sqrt {c x^{2}} b c}{x}} b c x^{4} {\rm weierstrassPInverse}\left (0, -\frac {4 \, \sqrt {c x^{2}} a}{b c^{2} x}, x\right ) + {\left (3 \, \sqrt {c x^{2}} b c x^{2} + 4 \, a\right )} \sqrt {\sqrt {c x^{2}} b c x^{2} + a}}{20 \, a x^{5}} \]
-1/20*(3*sqrt(c*x^2)*sqrt(sqrt(c*x^2)*b*c/x)*b*c*x^4*weierstrassPInverse(0 , -4*sqrt(c*x^2)*a/(b*c^2*x), x) + (3*sqrt(c*x^2)*b*c*x^2 + 4*a)*sqrt(sqrt (c*x^2)*b*c*x^2 + a))/(a*x^5)
\[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=\int \frac {\sqrt {a + b \left (c x^{2}\right )^{\frac {3}{2}}}}{x^{6}}\, dx \]
\[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=\int { \frac {\sqrt {\left (c x^{2}\right )^{\frac {3}{2}} b + a}}{x^{6}} \,d x } \]
\[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=\int { \frac {\sqrt {\left (c x^{2}\right )^{\frac {3}{2}} b + a}}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b \left (c x^2\right )^{3/2}}}{x^6} \, dx=\int \frac {\sqrt {a+b\,{\left (c\,x^2\right )}^{3/2}}}{x^6} \,d x \]