Integrand size = 21, antiderivative size = 296 \[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {14 b^2 x^{3/2} \left (b+c x^2\right )}{15 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {14 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {7 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}} \] Output:
14/15*b^2*x^(3/2)*(c*x^2+b)/c^(5/2)/(b^(1/2)+c^(1/2)*x)/(c*x^4+b*x^2)^(1/2 )-14/45*b*x^(1/2)*(c*x^4+b*x^2)^(1/2)/c^2+2/9*x^(5/2)*(c*x^4+b*x^2)^(1/2)/ c-14/15*b^(9/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1 /2)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))/c^(11/4) /(c*x^4+b*x^2)^(1/2)+7/15*b^(9/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x^2+b)/(b^(1/2 )+c^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)),1/ 2*2^(1/2))/c^(11/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.29 \[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{5/2} \left (-7 b^2-2 b c x^2+5 c^2 x^4+7 b^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {c x^2}{b}\right )\right )}{45 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x^(11/2)/Sqrt[b*x^2 + c*x^4],x]
Output:
(2*x^(5/2)*(-7*b^2 - 2*b*c*x^2 + 5*c^2*x^4 + 7*b^2*Sqrt[1 + (c*x^2)/b]*Hyp ergeometric2F1[1/2, 3/4, 7/4, -((c*x^2)/b)]))/(45*c^2*Sqrt[x^2*(b + c*x^2) ])
Time = 0.71 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1429, 1429, 1431, 266, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \int \frac {x^{7/2}}{\sqrt {c x^4+b x^2}}dx}{9 c}\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b \int \frac {x^{3/2}}{\sqrt {c x^4+b x^2}}dx}{5 c}\right )}{9 c}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {3 b x \sqrt {b+c x^2} \int \frac {\sqrt {x}}{\sqrt {c x^2+b}}dx}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \int \frac {x}{\sqrt {c x^2+b}}d\sqrt {x}}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\sqrt {b} \int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {b} \sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt {b} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{\sqrt {c x^2+b}}d\sqrt {x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {7 b \left (\frac {2 \sqrt {x} \sqrt {b x^2+c x^4}}{5 c}-\frac {6 b x \sqrt {b+c x^2} \left (\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{2 c^{3/4} \sqrt {b+c x^2}}-\frac {\frac {\sqrt [4]{b} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {b+c x^2}}-\frac {\sqrt {x} \sqrt {b+c x^2}}{\sqrt {b}+\sqrt {c} x}}{\sqrt {c}}\right )}{5 c \sqrt {b x^2+c x^4}}\right )}{9 c}\) |
Input:
Int[x^(11/2)/Sqrt[b*x^2 + c*x^4],x]
Output:
(2*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(9*c) - (7*b*((2*Sqrt[x]*Sqrt[b*x^2 + c*x^ 4])/(5*c) - (6*b*x*Sqrt[b + c*x^2]*(-((-((Sqrt[x]*Sqrt[b + c*x^2])/(Sqrt[b ] + Sqrt[c]*x)) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(c^( 1/4)*Sqrt[b + c*x^2]))/Sqrt[c]) + (b^(1/4)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1 /4)], 1/2])/(2*c^(3/4)*Sqrt[b + c*x^2])))/(5*c*Sqrt[b*x^2 + c*x^4])))/(9*c )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*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] && GtQ[m + 2*p - 1, 0] && NeQ[m + 4*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]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 0.95 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.73
method | result | size |
default | \(\frac {\sqrt {x}\, \left (10 c^{3} x^{6}+42 b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-21 b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-4 b \,c^{2} x^{4}-14 b^{2} c \,x^{2}\right )}{45 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{3}}\) | \(217\) |
risch | \(-\frac {2 x^{\frac {5}{2}} \left (-5 c \,x^{2}+7 b \right ) \left (c \,x^{2}+b \right )}{45 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {7 b^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 c^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(228\) |
Input:
int(x^(11/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/45/(c*x^4+b*x^2)^(1/2)*x^(1/2)/c^3*(10*c^3*x^6+42*b^3*((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)*(-c /(-b*c)^(1/2)*x)^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1 /2*2^(1/2))-21*b^3*((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)*(-c/(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c* x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-4*b*c^2*x^4-14*b^2*c*x^2)
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.21 \[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left (21 \, b^{2} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - \sqrt {c x^{4} + b x^{2}} {\left (5 \, c^{2} x^{2} - 7 \, b c\right )} \sqrt {x}\right )}}{45 \, c^{3}} \] Input:
integrate(x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
-2/45*(21*b^2*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/ c, 0, x)) - sqrt(c*x^4 + b*x^2)*(5*c^2*x^2 - 7*b*c)*sqrt(x))/c^3
\[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{\frac {11}{2}}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(x**(11/2)/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**(11/2)/sqrt(x**2*(b + c*x**2)), x)
\[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {x^{\frac {11}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \] Input:
integrate(x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(x^(11/2)/sqrt(c*x^4 + b*x^2), x)
\[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {x^{\frac {11}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \] Input:
integrate(x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
integrate(x^(11/2)/sqrt(c*x^4 + b*x^2), x)
Timed out. \[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{11/2}}{\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int(x^(11/2)/(b*x^2 + c*x^4)^(1/2),x)
Output:
int(x^(11/2)/(b*x^2 + c*x^4)^(1/2), x)
\[ \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {-14 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b x +10 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, c \,x^{3}+21 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{2}+b}d x \right ) b^{2}}{45 c^{2}} \] Input:
int(x^(11/2)/(c*x^4+b*x^2)^(1/2),x)
Output:
( - 14*sqrt(x)*sqrt(b + c*x**2)*b*x + 10*sqrt(x)*sqrt(b + c*x**2)*c*x**3 + 21*int((sqrt(x)*sqrt(b + c*x**2))/(b + c*x**2),x)*b**2)/(45*c**2)