3.4.99 \(\int \frac {\sqrt {x}}{(b x^2+c x^4)^{3/2}} \, dx\) [399]

Optimal. Leaf size=145 \[ \frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}-\frac {5 c^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{6 b^{9/4} \sqrt {b x^2+c x^4}} \]

[Out]

1/b/x^(1/2)/(c*x^4+b*x^2)^(1/2)-5/3*(c*x^4+b*x^2)^(1/2)/b^2/x^(5/2)-5/6*c^(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)/b^(9/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2048, 2050, 2057, 335, 226} \begin {gather*} -\frac {5 c^{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 )}{6 b^{9/4} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}+\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(b*x^2 + c*x^4)^(3/2),x]

[Out]

1/(b*Sqrt[x]*Sqrt[b*x^2 + c*x^4]) - (5*Sqrt[b*x^2 + c*x^4])/(3*b^2*x^(5/2)) - (5*c^(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])/(6*b^(9/4)*S
qrt[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 2048

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(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]) && LtQ[p, -1]

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*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]) && LtQ[m + j*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 {\sqrt {x}}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}+\frac {5 \int \frac {1}{x^{3/2} \sqrt {b x^2+c x^4}} \, dx}{2 b}\\ &=\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}-\frac {(5 c) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{6 b^2}\\ &=\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}-\frac {\left (5 c x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{6 b^2 \sqrt {b x^2+c x^4}}\\ &=\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}-\frac {\left (5 c x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{3 b^2 \sqrt {b x^2+c x^4}}\\ &=\frac {1}{b \sqrt {x} \sqrt {b x^2+c x^4}}-\frac {5 \sqrt {b x^2+c x^4}}{3 b^2 x^{5/2}}-\frac {5 c^{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 )}{6 b^{9/4} \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 = 10.02, size = 60, normalized size = 0.41 \begin {gather*} -\frac {2 \sqrt {1+\frac {c x^2}{b}} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {c x^2}{b}\right )}{3 b \sqrt {x} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-3/4, 3/2, 1/4, -((c*x^2)/b)])/(3*b*Sqrt[x]*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.12, size = 127, normalized size = 0.88

method	result	size
default	\(-\frac {x^{\frac {3}{2}} \left (c \,x^{2}+b \right ) \left (5 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-b c}\, x +10 c \,x^{2}+4 b \right )}{6 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} b^{2}}\)	\(127\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right )}{3 b^{2} \sqrt {x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {c \left (\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}+3 b \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{3 b^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/(c*x^4+b*x^2)^(3/2)*x^(3/2)*(c*x^2+b)*(5*((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))*(-b*c)^(1/2)*x+10*c*x^2+4*b)/b^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^(1/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 73, normalized size = 0.50 \begin {gather*} -\frac {5 \, {\left (c x^{5} + b x^{3}\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + \sqrt {c x^{4} + b x^{2}} {\left (5 \, c x^{2} + 2 \, b\right )} \sqrt {x}}{3 \, {\left (b^{2} c x^{5} + b^{3} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(5*(c*x^5 + b*x^3)*sqrt(c)*weierstrassPInverse(-4*b/c, 0, x) + sqrt(c*x^4 + b*x^2)*(5*c*x^2 + 2*b)*sqrt(x
))/(b^2*c*x^5 + b^3*x^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^(1/2)/(b*x^2 + c*x^4)^(3/2), x)

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