3.255 \(\int \frac{A+B x^2}{x^{9/2} \sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=369 \[ -\frac{c^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 b B-7 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{2 c^{3/2} x^{3/2} \left (b+c x^2\right ) (9 b B-7 A c)}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{2 c^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 b B-7 A c) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{2 c \sqrt{b x^2+c x^4} (9 b B-7 A c)}{15 b^3 x^{3/2}}-\frac{2 \sqrt{b x^2+c x^4} (9 b B-7 A c)}{45 b^2 x^{7/2}}-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}} \]

[Out]

(-2*c^(3/2)*(9*b*B - 7*A*c)*x^(3/2)*(b + c*x^2))/(15*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (2*A*Sqr
t[b*x^2 + c*x^4])/(9*b*x^(11/2)) - (2*(9*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(45*b^2*x^(7/2)) + (2*c*(9*b*B - 7*
A*c)*Sqrt[b*x^2 + c*x^4])/(15*b^3*x^(3/2)) + (2*c^(5/4)*(9*b*B - 7*A*c)*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*b^(11/4)*Sqrt[b*x^2 + c*x
^4]) - (c^(5/4)*(9*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])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.440989, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2038, 2025, 2032, 329, 305, 220, 1196} \[ -\frac{2 c^{3/2} x^{3/2} \left (b+c x^2\right ) (9 b B-7 A c)}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{c^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 b B-7 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{2 c^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (9 b B-7 A c) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}+\frac{2 c \sqrt{b x^2+c x^4} (9 b B-7 A c)}{15 b^3 x^{3/2}}-\frac{2 \sqrt{b x^2+c x^4} (9 b B-7 A c)}{45 b^2 x^{7/2}}-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(-2*c^(3/2)*(9*b*B - 7*A*c)*x^(3/2)*(b + c*x^2))/(15*b^3*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (2*A*Sqr
t[b*x^2 + c*x^4])/(9*b*x^(11/2)) - (2*(9*b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(45*b^2*x^(7/2)) + (2*c*(9*b*B - 7*
A*c)*Sqrt[b*x^2 + c*x^4])/(15*b^3*x^(3/2)) + (2*c^(5/4)*(9*b*B - 7*A*c)*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*b^(11/4)*Sqrt[b*x^2 + c*x
^4]) - (c^(5/4)*(9*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])/(15*b^(11/4)*Sqrt[b*x^2 + c*x^4])

Rule 2038

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[(c*e^(j - 1)*(e*x)^(m - j + 1)*(a*x^j + b*x^(j + n))^(p + 1))/(a*(m + j*p + 1)), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, -(n*p) - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rule 2025

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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{A+B x^2}{x^{9/2} \sqrt{b x^2+c x^4}} \, dx &=-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{\left (2 \left (-\frac{9 b B}{2}+\frac{7 A c}{2}\right )\right ) \int \frac{1}{x^{5/2} \sqrt{b x^2+c x^4}} \, dx}{9 b}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{2 (9 b B-7 A c) \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}-\frac{(c (9 b B-7 A c)) \int \frac{1}{\sqrt{x} \sqrt{b x^2+c x^4}} \, dx}{15 b^2}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{2 (9 b B-7 A c) \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac{2 c (9 b B-7 A c) \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}-\frac{\left (c^2 (9 b B-7 A c)\right ) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx}{15 b^3}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{2 (9 b B-7 A c) \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac{2 c (9 b B-7 A c) \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}-\frac{\left (c^2 (9 b B-7 A c) x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{15 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{2 (9 b B-7 A c) \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac{2 c (9 b B-7 A c) \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}-\frac{\left (2 c^2 (9 b B-7 A c) 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 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{2 (9 b B-7 A c) \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac{2 c (9 b B-7 A c) \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}-\frac{\left (2 c^{3/2} (9 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 )}{15 b^{5/2} \sqrt{b x^2+c x^4}}+\frac{\left (2 c^{3/2} (9 b B-7 A c) 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 b^{5/2} \sqrt{b x^2+c x^4}}\\ &=-\frac{2 c^{3/2} (9 b B-7 A c) x^{3/2} \left (b+c x^2\right )}{15 b^3 \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}-\frac{2 A \sqrt{b x^2+c x^4}}{9 b x^{11/2}}-\frac{2 (9 b B-7 A c) \sqrt{b x^2+c x^4}}{45 b^2 x^{7/2}}+\frac{2 c (9 b B-7 A c) \sqrt{b x^2+c x^4}}{15 b^3 x^{3/2}}+\frac{2 c^{5/4} (9 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}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{11/4} \sqrt{b x^2+c x^4}}-\frac{c^{5/4} (9 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 )}{15 b^{11/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.044351, size = 84, normalized size = 0.23 \[ -\frac{2 \left (x^2 \sqrt{\frac{c x^2}{b}+1} (9 b B-7 A c) \, _2F_1\left (-\frac{5}{4},\frac{1}{2};-\frac{1}{4};-\frac{c x^2}{b}\right )+5 A \left (b+c x^2\right )\right )}{45 b x^{7/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^(9/2)*Sqrt[b*x^2 + c*x^4]),x]

[Out]

(-2*(5*A*(b + c*x^2) + (9*b*B - 7*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-5/4, 1/2, -1/4, -((c*x^2)/b)
]))/(45*b*x^(7/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.02, size = 443, normalized size = 1.2 \begin{align*}{\frac{1}{45\,{b}^{3}} \left ( 42\,A\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{4}b{c}^{2}-21\,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 ){x}^{4}b{c}^{2}-54\,B\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ){x}^{4}{b}^{2}c+27\,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 ){x}^{4}{b}^{2}c-42\,A{c}^{3}{x}^{6}+54\,B{x}^{6}b{c}^{2}-28\,Ab{c}^{2}{x}^{4}+36\,B{x}^{4}{b}^{2}c+4\,A{b}^{2}c{x}^{2}-18\,B{x}^{2}{b}^{3}-10\,A{b}^{3} \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x)

[Out]

1/45/(c*x^4+b*x^2)^(1/2)/x^(7/2)*(42*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)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4
*b*c^2-21*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))*x^4*b*c^2-54*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)*EllipticE(
((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*x^4*b^2*c+27*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))*x^4*b^2*c-42*A*c^3*x^6+54*B*x^6*b*c^2-28*A*b*c^2*x^4+36*B*x^4*b^2*c+4*A*b^2*c*x^2-18
*B*x^2*b^3-10*A*b^3)/b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2)*x^(9/2)), 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^{9} + b x^{7}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(9/2)/(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^9 + b*x^7), 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((B*x**2+A)/x**(9/2)/(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{B x^{2} + A}{\sqrt{c x^{4} + b x^{2}} x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)/(sqrt(c*x^4 + b*x^2)*x^(9/2)), x)