3.4.82 \(\int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx\) [382]

Optimal. Leaf size=121 \[ \frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}} \]

[Out]

2/3*(c*x^4+b*x^2)^(1/2)/c/x^(1/2)-1/3*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^(5/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2049, 2057, 335, 226} \begin {gather*} \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}} F\left (2 \text {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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/Sqrt[b*x^2 + c*x^4],x]

[Out]

(2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (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^(5/4)*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 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 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*} \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 \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{3 c}\\ &=\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {\left (b x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{3 c \sqrt {b x^2+c x^4}}\\ &=\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {\left (2 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 c \sqrt {b x^2+c x^4}}\\ &=\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}} F\left (2 \tan ^{-1}\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}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 70, normalized size = 0.58 \begin {gather*} \frac {2 x^{3/2} \left (b+c x^2-b \sqrt {1+\frac {c x^2}{b}} \, _2F_1\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 )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/Sqrt[b*x^2 + c*x^4],x]

[Out]

(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)])

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Maple [A]
time = 0.12, size = 123, normalized size = 1.02

method	result	size
default	\(-\frac {\sqrt {x}\, \left (b \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-2 c^{2} x^{3}-2 b c x \right )}{3 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{2}}\)	\(123\)
risch	\(\frac {2 x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{3 c \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 c^{2} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(5/2)/sqrt(c*x^4 + b*x^2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.08, size = 43, normalized size = 0.36 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

-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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {5}{2}}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**(5/2)/sqrt(x**2*(b + c*x**2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^(5/2)/sqrt(c*x^4 + b*x^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}}{\sqrt {c\,x^4+b\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^(5/2)/(b*x^2 + c*x^4)^(1/2), x)

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