Integrand size = 21, antiderivative size = 118 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}+\frac {2 b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {b x^2+c x^4}} \] Output:
2/3*(c*x^4+b*x^2)^(1/2)/x^(1/2)+2/3*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^(1/4)/(c*x^4+b*x^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.84 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )}{\sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \] Input:
Integrate[Sqrt[b*x^2 + c*x^4]/x^(3/2),x]
Output:
(2*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)])/ (Sqrt[x]*Sqrt[1 + (c*x^2)/b])
Time = 0.40 (sec) , antiderivative size = 118, 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 = {1426, 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.
\(\displaystyle \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx\) |
\(\Big \downarrow \) 1426 |
\(\displaystyle \frac {2}{3} b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\) |
\(\Big \downarrow \) 1431 |
\(\displaystyle \frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {4 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}\) |
Input:
Int[Sqrt[b*x^2 + c*x^4]/x^(3/2),x]
Output:
(2*Sqrt[b*x^2 + c*x^4])/(3*Sqrt[x]) + (2*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^(1/4)*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*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 4*p + 1))), x] + Simp[2*b*(p/(d^2 *(m + 4*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 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 = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10
method | result | size |
default | \(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (b \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+c^{2} x^{3}+b c x \right )}{3 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c}\) | \(130\) |
risch | \(\frac {2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3 \sqrt {x}}+\frac {2 b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\) | \(165\) |
Input:
int((c*x^4+b*x^2)^(1/2)/x^(3/2),x,method=_RETURNVERBOSE)
Output:
2/3*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)*(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))+c^2*x^3+b*c*x)/c
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \, {\left (2 \, b \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} c \sqrt {x}\right )}}{3 \, c x} \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="fricas")
Output:
2/3*(2*b*sqrt(c)*x*weierstrassPInverse(-4*b/c, 0, x) + sqrt(c*x^4 + b*x^2) *c*sqrt(x))/(c*x)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{\frac {3}{2}}}\, dx \] Input:
integrate((c*x**4+b*x**2)**(1/2)/x**(3/2),x)
Output:
Integral(sqrt(x**2*(b + c*x**2))/x**(3/2), x)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {3}{2}}} \,d x } \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="maxima")
Output:
integrate(sqrt(c*x^4 + b*x^2)/x^(3/2), x)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {3}{2}}} \,d x } \] Input:
integrate((c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="giac")
Output:
integrate(sqrt(c*x^4 + b*x^2)/x^(3/2), x)
Timed out. \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^2}}{x^{3/2}} \,d x \] Input:
int((b*x^2 + c*x^4)^(1/2)/x^(3/2),x)
Output:
int((b*x^2 + c*x^4)^(1/2)/x^(3/2), x)
\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \sqrt {x}\, \sqrt {c \,x^{2}+b}}{3}+\frac {2 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c \,x^{3}+b x}d x \right ) b}{3} \] Input:
int((c*x^4+b*x^2)^(1/2)/x^(3/2),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