Integrand size = 21, antiderivative size = 121 \[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}} \] Output:
2/3*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)-1/3*b^(3/4)*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^(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 = 10.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.58 \[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 x^{3/2} \left (b+c x^2-b \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{3 c \sqrt {x^2 \left (b+c x^2\right )}} \] Input:
Integrate[x^(5/2)/Sqrt[b*x^2 + c*x^4],x]
Output:
(2*x^(3/2)*(b + c*x^2 - b*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(3*c*Sqrt[x^2*(b + c*x^2)])
Time = 0.43 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1429, 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.
\(\displaystyle \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1429 |
\(\displaystyle \frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\) |
Input:
Int[x^(5/2)/Sqrt[b*x^2 + c*x^4],x]
Output:
(2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(3/4)*x*(Sqrt[b] + Sqrt[c]*x)*S qrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[ x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])
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^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]
Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {\sqrt {x}\, \left (b \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-2 c^{2} x^{3}-2 b c x \right )}{3 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{2}}\) | \(123\) |
risch | \(\frac {2 x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{3 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c^{2} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(166\) |
Input:
int(x^(5/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3/(c*x^4+b*x^2)^(1/2)*x^(1/2)*(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)*(-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))-2*c^2*x^3-2*b*c*x)/c^2
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.36 \[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=-\frac {2 \, {\left (b \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) - \sqrt {c x^{4} + b x^{2}} c \sqrt {x}\right )}}{3 \, c^{2} x} \] Input:
integrate(x^(5/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
Output:
-2/3*(b*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) - sqrt(c*x^4 + b*x^2)* c*sqrt(x))/(c^2*x)
\[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{\frac {5}{2}}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:
integrate(x**(5/2)/(c*x**4+b*x**2)**(1/2),x)
Output:
Integral(x**(5/2)/sqrt(x**2*(b + c*x**2)), x)
\[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {x^{\frac {5}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \] Input:
integrate(x^(5/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(x^(5/2)/sqrt(c*x^4 + b*x^2), x)
\[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\int { \frac {x^{\frac {5}{2}}}{\sqrt {c x^{4} + b x^{2}}} \,d x } \] Input:
integrate(x^(5/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
Output:
integrate(x^(5/2)/sqrt(c*x^4 + b*x^2), x)
Timed out. \[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\int \frac {x^{5/2}}{\sqrt {c\,x^4+b\,x^2}} \,d x \] Input:
int(x^(5/2)/(b*x^2 + c*x^4)^(1/2),x)
Output:
int(x^(5/2)/(b*x^2 + c*x^4)^(1/2), x)
\[ \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx=\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}-\left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b}{3 c} \] Input:
int(x^(5/2)/(c*x^4+b*x^2)^(1/2),x)
Output:
(2*sqrt(x)*sqrt(b + c*x**2) - int((sqrt(x)*sqrt(b + c*x**2))/(b*x + c*x**3 ),x)*b)/(3*c)