3.244 \(\int \frac{x^{13/2} (A+B x^2)}{\sqrt{b x^2+c x^4}} \, dx\)

Optimal. Leaf size=243 \[ \frac{b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (13 b B-15 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{77 c^{17/4} \sqrt{b x^2+c x^4}}-\frac{2 b^2 \sqrt{b x^2+c x^4} (13 b B-15 A c)}{77 c^4 \sqrt{x}}-\frac{2 x^{7/2} \sqrt{b x^2+c x^4} (13 b B-15 A c)}{165 c^2}+\frac{6 b x^{3/2} \sqrt{b x^2+c x^4} (13 b B-15 A c)}{385 c^3}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c} \]

[Out]

(-2*b^2*(13*b*B - 15*A*c)*Sqrt[b*x^2 + c*x^4])/(77*c^4*Sqrt[x]) + (6*b*(13*b*B - 15*A*c)*x^(3/2)*Sqrt[b*x^2 +
c*x^4])/(385*c^3) - (2*(13*b*B - 15*A*c)*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(165*c^2) + (2*B*x^(11/2)*Sqrt[b*x^2 + c
*x^4])/(15*c) + (b^(11/4)*(13*b*B - 15*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])/(77*c^(17/4)*Sqrt[b*x^2 + c*x^4])

________________________________________________________________________________________

Rubi [A]  time = 0.367642, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 b^2 \sqrt{b x^2+c x^4} (13 b B-15 A c)}{77 c^4 \sqrt{x}}+\frac{b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (13 b B-15 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 c^{17/4} \sqrt{b x^2+c x^4}}-\frac{2 x^{7/2} \sqrt{b x^2+c x^4} (13 b B-15 A c)}{165 c^2}+\frac{6 b x^{3/2} \sqrt{b x^2+c x^4} (13 b B-15 A c)}{385 c^3}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^(13/2)*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(-2*b^2*(13*b*B - 15*A*c)*Sqrt[b*x^2 + c*x^4])/(77*c^4*Sqrt[x]) + (6*b*(13*b*B - 15*A*c)*x^(3/2)*Sqrt[b*x^2 +
c*x^4])/(385*c^3) - (2*(13*b*B - 15*A*c)*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(165*c^2) + (2*B*x^(11/2)*Sqrt[b*x^2 + c
*x^4])/(15*c) + (b^(11/4)*(13*b*B - 15*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])/(77*c^(17/4)*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^{13/2} \left (A+B x^2\right )}{\sqrt{b x^2+c x^4}} \, dx &=\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}-\frac{\left (2 \left (\frac{13 b B}{2}-\frac{15 A c}{2}\right )\right ) \int \frac{x^{13/2}}{\sqrt{b x^2+c x^4}} \, dx}{15 c}\\ &=-\frac{2 (13 b B-15 A c) x^{7/2} \sqrt{b x^2+c x^4}}{165 c^2}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}+\frac{(3 b (13 b B-15 A c)) \int \frac{x^{9/2}}{\sqrt{b x^2+c x^4}} \, dx}{55 c^2}\\ &=\frac{6 b (13 b B-15 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^3}-\frac{2 (13 b B-15 A c) x^{7/2} \sqrt{b x^2+c x^4}}{165 c^2}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}-\frac{\left (3 b^2 (13 b B-15 A c)\right ) \int \frac{x^{5/2}}{\sqrt{b x^2+c x^4}} \, dx}{77 c^3}\\ &=-\frac{2 b^2 (13 b B-15 A c) \sqrt{b x^2+c x^4}}{77 c^4 \sqrt{x}}+\frac{6 b (13 b B-15 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^3}-\frac{2 (13 b B-15 A c) x^{7/2} \sqrt{b x^2+c x^4}}{165 c^2}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}+\frac{\left (b^3 (13 b B-15 A c)\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{77 c^4}\\ &=-\frac{2 b^2 (13 b B-15 A c) \sqrt{b x^2+c x^4}}{77 c^4 \sqrt{x}}+\frac{6 b (13 b B-15 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^3}-\frac{2 (13 b B-15 A c) x^{7/2} \sqrt{b x^2+c x^4}}{165 c^2}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}+\frac{\left (b^3 (13 b B-15 A c) x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{77 c^4 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 b^2 (13 b B-15 A c) \sqrt{b x^2+c x^4}}{77 c^4 \sqrt{x}}+\frac{6 b (13 b B-15 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^3}-\frac{2 (13 b B-15 A c) x^{7/2} \sqrt{b x^2+c x^4}}{165 c^2}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}+\frac{\left (2 b^3 (13 b B-15 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 )}{77 c^4 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 b^2 (13 b B-15 A c) \sqrt{b x^2+c x^4}}{77 c^4 \sqrt{x}}+\frac{6 b (13 b B-15 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^3}-\frac{2 (13 b B-15 A c) x^{7/2} \sqrt{b x^2+c x^4}}{165 c^2}+\frac{2 B x^{11/2} \sqrt{b x^2+c x^4}}{15 c}+\frac{b^{11/4} (13 b B-15 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 )}{77 c^{17/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.158961, size = 143, normalized size = 0.59 \[ \frac{2 x^{3/2} \left (15 b^3 \sqrt{\frac{c x^2}{b}+1} (13 b B-15 A c) \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{b}\right )-\left (b+c x^2\right ) \left (-9 b^2 c \left (25 A+13 B x^2\right )+b c^2 x^2 \left (135 A+91 B x^2\right )-7 c^3 x^4 \left (15 A+11 B x^2\right )+195 b^3 B\right )\right )}{1155 c^4 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(13/2)*(A + B*x^2))/Sqrt[b*x^2 + c*x^4],x]

[Out]

(2*x^(3/2)*(-((b + c*x^2)*(195*b^3*B - 7*c^3*x^4*(15*A + 11*B*x^2) - 9*b^2*c*(25*A + 13*B*x^2) + b*c^2*x^2*(13
5*A + 91*B*x^2))) + 15*b^3*(13*b*B - 15*A*c)*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)
]))/(1155*c^4*Sqrt[x^2*(b + c*x^2)])

________________________________________________________________________________________

Maple [A]  time = 0.025, size = 298, normalized size = 1.2 \begin{align*} -{\frac{1}{1155\,{c}^{5}}\sqrt{x} \left ( -154\,B{x}^{9}{c}^{5}-210\,A{x}^{7}{c}^{5}+28\,B{x}^{7}b{c}^{4}+225\,A\sqrt{-bc}\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 ){b}^{3}c+60\,A{x}^{5}b{c}^{4}-195\,B\sqrt{-bc}\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 ){b}^{4}-52\,B{x}^{5}{b}^{2}{c}^{3}-180\,A{x}^{3}{b}^{2}{c}^{3}+156\,B{x}^{3}{b}^{3}{c}^{2}-450\,Ax{b}^{3}{c}^{2}+390\,Bx{b}^{4}c \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^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x)

[Out]

-1/1155/(c*x^4+b*x^2)^(1/2)*x^(1/2)*(-154*B*x^9*c^5-210*A*x^7*c^5+28*B*x^7*b*c^4+225*A*(-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)*(-x*c/(-b*c)^(1/2))^(1/2)*Ellip
ticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^3*c+60*A*x^5*b*c^4-195*B*(-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)*(-x*c/(-b*c)^(1/2))^(1/2)*Ellipti
cF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^4-52*B*x^5*b^2*c^3-180*A*x^3*b^2*c^3+156*B*x^3*b^3*c
^2-450*A*x*b^3*c^2+390*B*x*b^4*c)/c^5

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{\frac{13}{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^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^(13/2)/sqrt(c*x^4 + b*x^2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{6} + A x^{4}\right )} \sqrt{c x^{4} + b x^{2}} \sqrt{x}}{c x^{2} + b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

integral((B*x^6 + A*x^4)*sqrt(c*x^4 + b*x^2)*sqrt(x)/(c*x^2 + b), x)

________________________________________________________________________________________

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**(13/2)*(B*x**2+A)/(c*x**4+b*x**2)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} x^{\frac{13}{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^(13/2)*(B*x^2+A)/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^(13/2)/sqrt(c*x^4 + b*x^2), x)