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

Optimal. Leaf size=204 \[ \frac{5 c^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (11 b B-9 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{231 b^{13/4} \sqrt{b x^2+c x^4}}+\frac{10 c \sqrt{b x^2+c x^4} (11 b B-9 A c)}{231 b^3 x^{5/2}}-\frac{2 \sqrt{b x^2+c x^4} (11 b B-9 A c)}{77 b^2 x^{9/2}}-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}} \]

[Out]

(-2*A*Sqrt[b*x^2 + c*x^4])/(11*b*x^(13/2)) - (2*(11*b*B - 9*A*c)*Sqrt[b*x^2 + c*x^4])/(77*b^2*x^(9/2)) + (10*c
*(11*b*B - 9*A*c)*Sqrt[b*x^2 + c*x^4])/(231*b^3*x^(5/2)) + (5*c^(7/4)*(11*b*B - 9*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])/(231*b^(13/4)*
Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.311279, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {2038, 2025, 2032, 329, 220} \[ \frac{5 c^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (11 b B-9 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{231 b^{13/4} \sqrt{b x^2+c x^4}}+\frac{10 c \sqrt{b x^2+c x^4} (11 b B-9 A c)}{231 b^3 x^{5/2}}-\frac{2 \sqrt{b x^2+c x^4} (11 b B-9 A c)}{77 b^2 x^{9/2}}-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*Sqrt[b*x^2 + c*x^4])/(11*b*x^(13/2)) - (2*(11*b*B - 9*A*c)*Sqrt[b*x^2 + c*x^4])/(77*b^2*x^(9/2)) + (10*c
*(11*b*B - 9*A*c)*Sqrt[b*x^2 + c*x^4])/(231*b^3*x^(5/2)) + (5*c^(7/4)*(11*b*B - 9*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])/(231*b^(13/4)*
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 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{A+B x^2}{x^{11/2} \sqrt{b x^2+c x^4}} \, dx &=-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}}-\frac{\left (2 \left (-\frac{11 b B}{2}+\frac{9 A c}{2}\right )\right ) \int \frac{1}{x^{7/2} \sqrt{b x^2+c x^4}} \, dx}{11 b}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}}-\frac{2 (11 b B-9 A c) \sqrt{b x^2+c x^4}}{77 b^2 x^{9/2}}-\frac{(5 c (11 b B-9 A c)) \int \frac{1}{x^{3/2} \sqrt{b x^2+c x^4}} \, dx}{77 b^2}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}}-\frac{2 (11 b B-9 A c) \sqrt{b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac{10 c (11 b B-9 A c) \sqrt{b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac{\left (5 c^2 (11 b B-9 A c)\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{231 b^3}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}}-\frac{2 (11 b B-9 A c) \sqrt{b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac{10 c (11 b B-9 A c) \sqrt{b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac{\left (5 c^2 (11 b B-9 A c) x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{231 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}}-\frac{2 (11 b B-9 A c) \sqrt{b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac{10 c (11 b B-9 A c) \sqrt{b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac{\left (10 c^2 (11 b B-9 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 )}{231 b^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 A \sqrt{b x^2+c x^4}}{11 b x^{13/2}}-\frac{2 (11 b B-9 A c) \sqrt{b x^2+c x^4}}{77 b^2 x^{9/2}}+\frac{10 c (11 b B-9 A c) \sqrt{b x^2+c x^4}}{231 b^3 x^{5/2}}+\frac{5 c^{7/4} (11 b B-9 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 )}{231 b^{13/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0435078, size = 84, normalized size = 0.41 \[ -\frac{2 \left (x^2 \sqrt{\frac{c x^2}{b}+1} (11 b B-9 A c) \, _2F_1\left (-\frac{7}{4},\frac{1}{2};-\frac{3}{4};-\frac{c x^2}{b}\right )+7 A \left (b+c x^2\right )\right )}{77 b x^{9/2} \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(7*A*(b + c*x^2) + (11*b*B - 9*A*c)*x^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-7/4, 1/2, -3/4, -((c*x^2)/b
)]))/(77*b*x^(9/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.019, size = 274, normalized size = 1.3 \begin{align*} -{\frac{1}{231\,{b}^{3}} \left ( 45\,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}{x}^{5}{c}^{2}-55\,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}{x}^{5}bc+90\,A{c}^{3}{x}^{6}-110\,B{x}^{6}b{c}^{2}+36\,Ab{c}^{2}{x}^{4}-44\,B{x}^{4}{b}^{2}c-12\,A{b}^{2}c{x}^{2}+66\,B{x}^{2}{b}^{3}+42\,A{b}^{3} \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}}}}{x}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/231/(c*x^4+b*x^2)^(1/2)/x^(9/2)*(45*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)*x^5*c^2-55*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)*x^5*
b*c+90*A*c^3*x^6-110*B*x^6*b*c^2+36*A*b*c^2*x^4-44*B*x^4*b^2*c-12*A*b^2*c*x^2+66*B*x^2*b^3+42*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{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(11/2)/(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^(11/2)), 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^{10} + b x^{8}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(11/2)/(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^10 + b*x^8), 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**(11/2)/(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{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^(11/2)/(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^(11/2)), x)