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

Optimal. Leaf size=243 \[ -\frac{2 b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (3 b B-5 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{231 c^{13/4} \sqrt{b x^2+c x^4}}+\frac{4 b^2 \sqrt{b x^2+c x^4} (3 b B-5 A c)}{231 c^3 \sqrt{x}}-\frac{4 b x^{3/2} \sqrt{b x^2+c x^4} (3 b B-5 A c)}{385 c^2}-\frac{2 x^{7/2} \sqrt{b x^2+c x^4} (3 b B-5 A c)}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c} \]

[Out]

(4*b^2*(3*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^3*Sqrt[x]) - (4*b*(3*b*B - 5*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^
4])/(385*c^2) - (2*(3*b*B - 5*A*c)*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(55*c) + (2*B*x^(3/2)*(b*x^2 + c*x^4)^(3/2))/(
15*c) - (2*b^(11/4)*(3*b*B - 5*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*Elliptic
F[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(13/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.390354, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2039, 2021, 2024, 2032, 329, 220} \[ \frac{4 b^2 \sqrt{b x^2+c x^4} (3 b B-5 A c)}{231 c^3 \sqrt{x}}-\frac{2 b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (3 b B-5 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{231 c^{13/4} \sqrt{b x^2+c x^4}}-\frac{4 b x^{3/2} \sqrt{b x^2+c x^4} (3 b B-5 A c)}{385 c^2}-\frac{2 x^{7/2} \sqrt{b x^2+c x^4} (3 b B-5 A c)}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*(3*b*B - 5*A*c)*Sqrt[b*x^2 + c*x^4])/(231*c^3*Sqrt[x]) - (4*b*(3*b*B - 5*A*c)*x^(3/2)*Sqrt[b*x^2 + c*x^
4])/(385*c^2) - (2*(3*b*B - 5*A*c)*x^(7/2)*Sqrt[b*x^2 + c*x^4])/(55*c) + (2*B*x^(3/2)*(b*x^2 + c*x^4)^(3/2))/(
15*c) - (2*b^(11/4)*(3*b*B - 5*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*Elliptic
F[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(231*c^(13/4)*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 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]

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 x^{5/2} \left (A+B x^2\right ) \sqrt{b x^2+c x^4} \, dx &=\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac{\left (2 \left (\frac{9 b B}{2}-\frac{15 A c}{2}\right )\right ) \int x^{5/2} \sqrt{b x^2+c x^4} \, dx}{15 c}\\ &=-\frac{2 (3 b B-5 A c) x^{7/2} \sqrt{b x^2+c x^4}}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac{(2 b (3 b B-5 A c)) \int \frac{x^{9/2}}{\sqrt{b x^2+c x^4}} \, dx}{55 c}\\ &=-\frac{4 b (3 b B-5 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^2}-\frac{2 (3 b B-5 A c) x^{7/2} \sqrt{b x^2+c x^4}}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}+\frac{\left (2 b^2 (3 b B-5 A c)\right ) \int \frac{x^{5/2}}{\sqrt{b x^2+c x^4}} \, dx}{77 c^2}\\ &=\frac{4 b^2 (3 b B-5 A c) \sqrt{b x^2+c x^4}}{231 c^3 \sqrt{x}}-\frac{4 b (3 b B-5 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^2}-\frac{2 (3 b B-5 A c) x^{7/2} \sqrt{b x^2+c x^4}}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac{\left (2 b^3 (3 b B-5 A c)\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{231 c^3}\\ &=\frac{4 b^2 (3 b B-5 A c) \sqrt{b x^2+c x^4}}{231 c^3 \sqrt{x}}-\frac{4 b (3 b B-5 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^2}-\frac{2 (3 b B-5 A c) x^{7/2} \sqrt{b x^2+c x^4}}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac{\left (2 b^3 (3 b B-5 A c) x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{231 c^3 \sqrt{b x^2+c x^4}}\\ &=\frac{4 b^2 (3 b B-5 A c) \sqrt{b x^2+c x^4}}{231 c^3 \sqrt{x}}-\frac{4 b (3 b B-5 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^2}-\frac{2 (3 b B-5 A c) x^{7/2} \sqrt{b x^2+c x^4}}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac{\left (4 b^3 (3 b B-5 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 c^3 \sqrt{b x^2+c x^4}}\\ &=\frac{4 b^2 (3 b B-5 A c) \sqrt{b x^2+c x^4}}{231 c^3 \sqrt{x}}-\frac{4 b (3 b B-5 A c) x^{3/2} \sqrt{b x^2+c x^4}}{385 c^2}-\frac{2 (3 b B-5 A c) x^{7/2} \sqrt{b x^2+c x^4}}{55 c}+\frac{2 B x^{3/2} \left (b x^2+c x^4\right )^{3/2}}{15 c}-\frac{2 b^{11/4} (3 b B-5 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 c^{13/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.161978, size = 136, normalized size = 0.56 \[ \frac{2 \sqrt{x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right ) \sqrt{\frac{c x^2}{b}+1} \left (-3 b c \left (25 A+21 B x^2\right )+7 c^2 x^2 \left (15 A+11 B x^2\right )+45 b^2 B\right )+15 b^2 (5 A c-3 b B) \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{b}\right )\right )}{1155 c^3 \sqrt{x} \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)*Sqrt[1 + (c*x^2)/b]*(45*b^2*B + 7*c^2*x^2*(15*A + 11*B*x^2) - 3*b*c*(25*
A + 21*B*x^2)) + 15*b^2*(-3*b*B + 5*A*c)*Hypergeometric2F1[-1/2, 1/4, 5/4, -((c*x^2)/b)]))/(1155*c^3*Sqrt[x]*S
qrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.053, size = 307, normalized size = 1.3 \begin{align*}{\frac{2}{ \left ( 1155\,c{x}^{2}+1155\,b \right ){c}^{4}}\sqrt{c{x}^{4}+b{x}^{2}} \left ( 77\,B{x}^{9}{c}^{5}+105\,A{x}^{7}{c}^{5}+91\,B{x}^{7}b{c}^{4}+25\,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+135\,A{x}^{5}b{c}^{4}-15\,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}-4\,B{x}^{5}{b}^{2}{c}^{3}-20\,A{x}^{3}{b}^{2}{c}^{3}+12\,B{x}^{3}{b}^{3}{c}^{2}-50\,Ax{b}^{3}{c}^{2}+30\,Bx{b}^{4}c \right ){x}^{-{\frac{3}{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]

2/1155*(c*x^4+b*x^2)^(1/2)/x^(3/2)/(c*x^2+b)*(77*B*x^9*c^5+105*A*x^7*c^5+91*B*x^7*b*c^4+25*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)
*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^3*c+135*A*x^5*b*c^4-15*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)*E
llipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b^4-4*B*x^5*b^2*c^3-20*A*x^3*b^2*c^3+12*B*x^3*b^
3*c^2-50*A*x*b^3*c^2+30*B*x*b^4*c)/c^4

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

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