Integrand size = 21, antiderivative size = 146 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}-\frac {4 c \sqrt {b x^2+c x^4}}{21 b x^{5/2}}-\frac {2 c^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{21 b^{5/4} \sqrt {b x^2+c x^4}} \]
-2/7*(c*x^4+b*x^2)^(1/2)/x^(9/2)-4/21*c*(c*x^4+b*x^2)^(1/2)/b/x^(5/2)-2/21 *c^(7/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c ^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))), 1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/b ^(5/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.39 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},-\frac {1}{2},-\frac {3}{4},-\frac {c x^2}{b}\right )}{7 x^{9/2} \sqrt {1+\frac {c x^2}{b}}} \]
(-2*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-7/4, -1/2, -3/4, -((c*x^2)/b) ])/(7*x^(9/2)*Sqrt[1 + (c*x^2)/b])
Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1425, 1430, 1431, 266, 761}
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.
\(\displaystyle \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {2}{7} c \int \frac {1}{x^{3/2} \sqrt {c x^4+b x^2}}dx-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle \frac {2}{7} c \left (-\frac {c \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {2}{7} c \left (-\frac {c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2}{7} c \left (-\frac {2 c x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 b \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2}{7} c \left (-\frac {c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 b^{5/4} \sqrt {b x^2+c x^4}}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^{5/2}}\right )-\frac {2 \sqrt {b x^2+c x^4}}{7 x^{9/2}}\) |
(-2*Sqrt[b*x^2 + c*x^4])/(7*x^(9/2)) + (2*c*((-2*Sqrt[b*x^2 + c*x^4])/(3*b *x^(5/2)) - (c^(3/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + S qrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*b^(5/ 4)*Sqrt[b*x^2 + c*x^4])))/7
3.4.61.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[m + 2*p + 1, 0 ]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p) Int[(d*x)^(m + 2*p)*(b + c *x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p]
Time = 0.34 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97
method | result | size |
default | \(-\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (\sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}+2 c^{2} x^{4}+5 b c \,x^{2}+3 b^{2}\right )}{21 x^{\frac {9}{2}} \left (c \,x^{2}+b \right ) b}\) | \(142\) |
risch | \(-\frac {2 \left (2 c \,x^{2}+3 b \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{21 x^{\frac {9}{2}} b}-\frac {2 c \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{21 b \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(178\) |
-2/21*(c*x^4+b*x^2)^(1/2)/x^(9/2)/(c*x^2+b)*((-b*c)^(1/2)*((c*x+(-b*c)^(1/ 2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*( -x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2) ,1/2*2^(1/2))*c*x^3+2*c^2*x^4+5*b*c*x^2+3*b^2)/b
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=-\frac {2 \, {\left (2 \, c^{\frac {3}{2}} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} {\left (2 \, c x^{2} + 3 \, b\right )} \sqrt {x}\right )}}{21 \, b x^{5}} \]
-2/21*(2*c^(3/2)*x^5*weierstrassPInverse(-4*b/c, 0, x) + sqrt(c*x^4 + b*x^ 2)*(2*c*x^2 + 3*b)*sqrt(x))/(b*x^5)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{\frac {11}{2}}}\, dx \]
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {11}{2}}} \,d x } \]
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {11}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{11/2}} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^2}}{x^{11/2}} \,d x \]