3.4.79 \(\int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx\) [379]

Optimal. Leaf size=296 \[ \frac {14 b^2 x^{3/2} \left (b+c x^2\right )}{15 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {14 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {7 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}} \]

[Out]

14/15*b^2*x^(3/2)*(c*x^2+b)/c^(5/2)/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)+2/9*x^(5/2)*(c*x^4+b*x^2)^(1/2)/c-
14/45*b*x^(1/2)*(c*x^4+b*x^2)^(1/2)/c^2-14/15*b^(9/4)*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)))*EllipticE(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^(11/4)/(c*x^4+b*x^2)^(1/2)+7/15*b^(9/4)*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^(11/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2049, 2057, 335, 311, 226, 1210} \begin {gather*} \frac {7 b^{9/4} 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 )}{15 c^{11/4} \sqrt {b x^2+c x^4}}-\frac {14 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {14 b^2 x^{3/2} \left (b+c x^2\right )}{15 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(11/2)/Sqrt[b*x^2 + c*x^4],x]

[Out]

(14*b^2*x^(3/2)*(b + c*x^2))/(15*c^(5/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (14*b*Sqrt[x]*Sqrt[b*x^2
+ c*x^4])/(45*c^2) + (2*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(9*c) - (14*b^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*
x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*c^(11/4)*Sqrt[b*x^2 + c
*x^4]) + (7*b^(9/4)*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])/(15*c^(11/4)*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 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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 1210

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a +
c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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^{11/2}}{\sqrt {b x^2+c x^4}} \, dx &=\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {(7 b) \int \frac {x^{7/2}}{\sqrt {b x^2+c x^4}} \, dx}{9 c}\\ &=-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}+\frac {\left (7 b^2\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{15 c^2}\\ &=-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}+\frac {\left (7 b^2 x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{15 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}+\frac {\left (14 b^2 x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}+\frac {\left (14 b^{5/2} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^{5/2} \sqrt {b x^2+c x^4}}-\frac {\left (14 b^{5/2} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^{5/2} \sqrt {b x^2+c x^4}}\\ &=\frac {14 b^2 x^{3/2} \left (b+c x^2\right )}{15 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c}-\frac {14 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}}+\frac {7 b^{9/4} 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 )}{15 c^{11/4} \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 = 86, normalized size = 0.29 \begin {gather*} \frac {2 x^{5/2} \left (-7 b^2-2 b c x^2+5 c^2 x^4+7 b^2 \sqrt {1+\frac {c x^2}{b}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{b}\right )\right )}{45 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(11/2)/Sqrt[b*x^2 + c*x^4],x]

[Out]

(2*x^(5/2)*(-7*b^2 - 2*b*c*x^2 + 5*c^2*x^4 + 7*b^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/2, 3/4, 7/4, -((c*x
^2)/b)]))/(45*c^2*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.11, size = 217, normalized size = 0.73

method	result	size
default	\(\frac {\sqrt {x}\, \left (10 c^{3} x^{6}+42 b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-21 b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-4 b \,c^{2} x^{4}-14 b^{2} c \,x^{2}\right )}{45 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{3}}\)	\(217\)
risch	\(-\frac {2 x^{\frac {5}{2}} \left (-5 c \,x^{2}+7 b \right ) \left (c \,x^{2}+b \right )}{45 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {7 b^{2} \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 c^{3} \sqrt {c \,x^{3}+b x}\, \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(228\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/45/(c*x^4+b*x^2)^(1/2)*x^(1/2)/c^3*(10*c^3*x^6+42*b^3*((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)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)
,1/2*2^(1/2))-21*b^3*((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))-4*b*c^2*x^4-14*b^2*c*
x^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^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^(11/2)/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 = 62, normalized size = 0.21 \begin {gather*} -\frac {2 \, {\left (21 \, b^{2} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - \sqrt {c x^{4} + b x^{2}} {\left (5 \, c^{2} x^{2} - 7 \, b c\right )} \sqrt {x}\right )}}{45 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

-2/45*(21*b^2*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) - sqrt(c*x^4 + b*x^2)*(5*c
^2*x^2 - 7*b*c)*sqrt(x))/c^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{\frac {11}{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**(11/2)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**(11/2)/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^(11/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate(x^(11/2)/sqrt(c*x^4 + b*x^2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{11/2}}{\sqrt {c\,x^4+b\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^(11/2)/(b*x^2 + c*x^4)^(1/2), x)

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