3.353 \(\int x^{5/2} \sqrt {b x^2+c x^4} \, dx\)

Optimal. Leaf size=176 \[ \frac {10 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}}-\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c} \]

[Out]

4/77*b*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c+2/11*x^(7/2)*(c*x^4+b*x^2)^(1/2)-20/231*b^2*(c*x^4+b*x^2)^(1/2)/c^2/x^(1/
2)+10/231*b^(11/4)*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)))*E
llipticF(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^(9/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2021, 2024, 2032, 329, 220} \[ -\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {10 b^{11/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{231 c^{9/4} \sqrt {b x^2+c x^4}}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

(-20*b^2*Sqrt[b*x^2 + c*x^4])/(231*c^2*Sqrt[x]) + (4*b*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(77*c) + (2*x^(7/2)*Sqrt[b
*x^2 + c*x^4])/11 + (10*b^(11/4)*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])/(231*c^(9/4)*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 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 2021

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 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 x^{5/2} \sqrt {b x^2+c x^4} \, dx &=\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {1}{11} (2 b) \int \frac {x^{9/2}}{\sqrt {b x^2+c x^4}} \, dx\\ &=\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}-\frac {\left (10 b^2\right ) \int \frac {x^{5/2}}{\sqrt {b x^2+c x^4}} \, dx}{77 c}\\ &=-\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {\left (10 b^3\right ) \int \frac {\sqrt {x}}{\sqrt {b x^2+c x^4}} \, dx}{231 c^2}\\ &=-\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {\left (10 b^3 x \sqrt {b+c x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {b+c x^2}} \, dx}{231 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {\left (20 b^3 x \sqrt {b+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{231 c^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {20 b^2 \sqrt {b x^2+c x^4}}{231 c^2 \sqrt {x}}+\frac {4 b x^{3/2} \sqrt {b x^2+c x^4}}{77 c}+\frac {2}{11} x^{7/2} \sqrt {b x^2+c x^4}+\frac {10 b^{11/4} 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 )}{231 c^{9/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 102, normalized size = 0.58 \[ \frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {\frac {c x^2}{b}+1} \left (-5 b^2+2 b c x^2+7 c^2 x^4\right )+5 b^2 \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\frac {c x^2}{b}\right )\right )}{77 c^2 \sqrt {x} \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*(Sqrt[1 + (c*x^2)/b]*(-5*b^2 + 2*b*c*x^2 + 7*c^2*x^4) + 5*b^2*Hypergeometric2F1[-1/2,
1/4, 5/4, -((c*x^2)/b)]))/(77*c^2*Sqrt[x]*Sqrt[1 + (c*x^2)/b])

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {c x^{4} + b x^{2}} x^{\frac {5}{2}}, x\right ) \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{4} + b x^{2}} x^{\frac {5}{2}}\,{d x} \]

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(sqrt(c*x^4 + b*x^2)*x^(5/2), x)

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maple [A]  time = 0.03, size = 157, normalized size = 0.89 \[ \frac {2 \sqrt {c \,x^{4}+b \,x^{2}}\, \left (21 c^{4} x^{7}+27 b \,c^{3} x^{5}-4 b^{2} c^{2} x^{3}-10 b^{3} c x +5 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, b^{3} \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )\right )}{231 \left (c \,x^{2}+b \right ) c^{3} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)*(21*x^7*c^4+5*b^3*(-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)*(-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))+27*x^5*b*c^3-4*x^3*b^2*c^2-10*x*b^3*c)/c^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{4} + b x^{2}} x^{\frac {5}{2}}\,{d x} \]

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(sqrt(c*x^4 + b*x^2)*x^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{5/2}\,\sqrt {c\,x^4+b\,x^2} \,d x \]

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

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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