\(\int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx\) [357]

Optimal result

Integrand size = 21, antiderivative size = 118 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}+\frac {2 b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {b x^2+c x^4}} \]

[Out]

2/3*(c*x^4+b*x^2)^(1/2)/x^(1/2)+2/3*b^(3/4)*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)/c^(1/4)/(c*x^4+b*x^2)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2046, 2057, 335, 226} \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 \sqrt [4]{c} \sqrt {b x^2+c x^4}}+\frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}} \]

[In]

Int[Sqrt[b*x^2 + c*x^4]/x^(3/2),x]

[Out]

(2*Sqrt[b*x^2 + c*x^4])/(3*Sqrt[x]) + (2*b^(3/4)*x*(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])/(3*c^(1/4)*Sqrt[b*x^2 + c*x^4])

Rule 226

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]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2046

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a*x^j + b
*x^n)^p/(c*(m + n*p + 1))), x] + Dist[a*(n - j)*(p/(c^j*(m + n*p + 1))), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}+\frac {1}{3} (2 b) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx \\ & = \frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}+\frac {\left (2 b x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{3 \sqrt {b x^2+c x^4}} \\ & = \frac {2 \sqrt {b x^2+c x^4}}{3 \sqrt {x}}+\frac {\left (4 b x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{3 \sqrt {b x^2+c x^4}} \\ & = \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}} F\left (2 \tan ^{-1}\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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.64 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-\frac {c x^2}{b}\right )}{\sqrt {x} \sqrt {1+\frac {c x^2}{b}}} \]

[In]

Integrate[Sqrt[b*x^2 + c*x^4]/x^(3/2),x]

[Out]

(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])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.10

method	result	size
default	\(\frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (b \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+c^{2} x^{3}+b c x \right )}{3 x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) c}\)	\(130\)
risch	\(\frac {2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3 \sqrt {x}}+\frac {2 b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(165\)

[In]

int((c*x^4+b*x^2)^(1/2)/x^(3/2),x,method=_RETURNVERBOSE)

[Out]

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)*(-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^2*x^3+b*c*x)/c

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.36 \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\frac {2 \, {\left (2 \, b \sqrt {c} x {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} c \sqrt {x}\right )}}{3 \, c x} \]

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )}}{x^{\frac {3}{2}}}\, dx \]

[In]

integrate((c*x**4+b*x**2)**(1/2)/x**(3/2),x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))/x**(3/2), x)

Maxima [F]

\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^(3/2), x)

Giac [F]

\[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{2}}}{x^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)/x^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b x^2+c x^4}}{x^{3/2}} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^2}}{x^{3/2}} \,d x \]

[In]

int((b*x^2 + c*x^4)^(1/2)/x^(3/2),x)

[Out]

int((b*x^2 + c*x^4)^(1/2)/x^(3/2), x)