3.248 \(\int \frac{x^{5/2} (A+B x^2)}{\sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=167 \[ \frac{b^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (5 b B-7 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{21 c^{9/4} \sqrt{b x^2+c x^4}}-\frac{2 \sqrt{b x^2+c x^4} (5 b B-7 A c)}{21 c^2 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c} \]

[Out]

(-2*(5*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(21*c^2*Sqrt[x]) + (2*B*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(7*c) + (b^(3/4)
*(5*b*B - 7*A*c)*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])/(21*c^(9/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.255178, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2039, 2024, 2032, 329, 220} \[ \frac{b^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (5 b B-7 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{21 c^{9/4} \sqrt{b x^2+c x^4}}-\frac{2 \sqrt{b x^2+c x^4} (5 b B-7 A c)}{21 c^2 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(5*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(21*c^2*Sqrt[x]) + (2*B*x^(3/2)*Sqrt[b*x^2 + c*x^4])/(7*c) + (b^(3/4)
*(5*b*B - 7*A*c)*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])/(21*c^(9/4)*Sqrt[b*x^2 + c*x^4])

Rule 2039

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(d*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(b*(m + n + p*(j + n) + 1)), x] - Dist[(a*d*(m
+ j*p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)), Int[(e*x)^m*(a*x^j + b*x^(j + n))^p, x
], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[
m + n + p*(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])

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]

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 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]

Rubi steps

\begin{align*} \int \frac{x^{5/2} \left (A+B x^2\right )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c}-\frac{\left (2 \left (\frac{5 b B}{2}-\frac{7 A c}{2}\right )\right ) \int \frac{x^{5/2}}{\sqrt{b x^2+c x^4}} \, dx}{7 c}\\ &=-\frac{2 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{21 c^2 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c}+\frac{(b (5 b B-7 A c)) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{21 c^2}\\ &=-\frac{2 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{21 c^2 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c}+\frac{\left (b (5 b B-7 A c) x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{21 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{21 c^2 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c}+\frac{\left (2 b (5 b B-7 A c) x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{21 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 (5 b B-7 A c) \sqrt{b x^2+c x^4}}{21 c^2 \sqrt{x}}+\frac{2 B x^{3/2} \sqrt{b x^2+c x^4}}{7 c}+\frac{b^{3/4} (5 b B-7 A c) 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 )}{21 c^{9/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.111283, size = 97, normalized size = 0.58 \[ \frac{2 x^{3/2} \left (b \sqrt{\frac{c x^2}{b}+1} (5 b B-7 A c) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{b}\right )-\left (b+c x^2\right ) \left (-7 A c+5 b B-3 B c x^2\right )\right )}{21 c^2 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(-((b + c*x^2)*(5*b*B - 7*A*c - 3*B*c*x^2)) + b*(5*b*B - 7*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2
F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(21*c^2*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.015, size = 248, normalized size = 1.5 \begin{align*} -{\frac{1}{21\,{c}^{3}}\sqrt{x} \left ( 7\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{-bc}bc-5\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) \sqrt{-bc}{b}^{2}-6\,B{c}^{3}{x}^{5}-14\,A{x}^{3}{c}^{3}+4\,B{x}^{3}b{c}^{2}-14\,Ab{c}^{2}x+10\,B{b}^{2}cx \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21/(c*x^4+b*x^2)^(1/2)*x^(1/2)*(7*A*((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)*b*c-5*B*((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)*b^2-6*B*c^3
*x^5-14*A*x^3*c^3+4*B*x^3*b*c^2-14*A*b*c^2*x+10*B*b^2*c*x)/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{\frac{5}{2}}}{\sqrt{c x^{4} + b x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2}}{\left (B x^{2} + A\right )} \sqrt{x}}{c x^{2} + b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{\frac{5}{2}}}{\sqrt{c x^{4} + b x^{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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