3.379 \(\int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx\)

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

[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 = 0.30, 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 = {2024, 2032, 329, 305, 220, 1196} \[ \frac {14 b^2 x^{3/2} \left (b+c x^2\right )}{15 c^{5/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}+\frac {7 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}}-\frac {14 b^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{11/4} \sqrt {b x^2+c x^4}}-\frac {14 b \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^2}+\frac {2 x^{5/2} \sqrt {b x^2+c x^4}}{9 c} \]

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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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 329

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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 2024

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 0.04, size = 86, normalized size = 0.29 \[ \frac {2 x^{5/2} \left (7 b^2 \sqrt {\frac {c x^2}{b}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^2}{b}\right )-7 b^2-2 b c x^2+5 c^2 x^4\right )}{45 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \]

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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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2}} x^{\frac {7}{2}}}{c x^{2} + b}, x\right ) \]

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]

integral(sqrt(c*x^4 + b*x^2)*x^(7/2)/(c*x^2 + b), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {11}{2}}}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \]

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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maple [A]  time = 0.03, size = 217, normalized size = 0.73 \[ \frac {\left (10 c^{3} x^{6}-4 b \,c^{2} x^{4}-14 b^{2} c \,x^{2}+42 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, b^{3} \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-21 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, b^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {x}}{45 \sqrt {c \,x^{4}+b \,x^{2}}\, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)*(-1/(-b*c)^(1/2)*c*x)^(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
)*(-1/(-b*c)^(1/2)*c*x)^(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 \[ \int \frac {x^{\frac {11}{2}}}{\sqrt {c x^{4} + b x^{2}}}\,{d x} \]

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

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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{\frac {11}{2}}}{\sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \]

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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