3.224 \(\int \frac{(A+B x^2) \sqrt{b x^2+c x^4}}{x^{3/2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{2 b^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (b B-7 A c) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{21 c^{5/4} \sqrt{b x^2+c x^4}}-\frac{2 \sqrt{b x^2+c x^4} (b B-7 A c)}{21 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.254143, antiderivative size = 165, 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, 2021, 2032, 329, 220} \[ -\frac{2 b^{3/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} (b B-7 A c) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{21 c^{5/4} \sqrt{b x^2+c x^4}}-\frac{2 \sqrt{b x^2+c x^4} (b B-7 A c)}{21 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(3/2),x]

[Out]

(-2*(b*B - 7*A*c)*Sqrt[b*x^2 + c*x^4])/(21*c*Sqrt[x]) + (2*B*(b*x^2 + c*x^4)^(3/2))/(7*c*x^(5/2)) - (2*b^(3/4)
*(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)*S
qrt[x])/b^(1/4)], 1/2])/(21*c^(5/4)*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 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{\left (A+B x^2\right ) \sqrt{b x^2+c x^4}}{x^{3/2}} \, dx &=\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac{\left (2 \left (\frac{b B}{2}-\frac{7 A c}{2}\right )\right ) \int \frac{\sqrt{b x^2+c x^4}}{x^{3/2}} \, dx}{7 c}\\ &=-\frac{2 (b B-7 A c) \sqrt{b x^2+c x^4}}{21 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac{(2 b (b B-7 A c)) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{21 c}\\ &=-\frac{2 (b B-7 A c) \sqrt{b x^2+c x^4}}{21 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac{\left (2 b (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 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 (b B-7 A c) \sqrt{b x^2+c x^4}}{21 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac{\left (4 b (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 \sqrt{b x^2+c x^4}}\\ &=-\frac{2 (b B-7 A c) \sqrt{b x^2+c x^4}}{21 c \sqrt{x}}+\frac{2 B \left (b x^2+c x^4\right )^{3/2}}{7 c x^{5/2}}-\frac{2 b^{3/4} (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^{5/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0503193, size = 94, normalized size = 0.57 \[ \frac{2 \sqrt{x^2 \left (b+c x^2\right )} \left ((7 A c-b B) \, _2F_1\left (-\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{b}\right )+B \sqrt{\frac{c x^2}{b}+1} \left (b+c x^2\right )\right )}{7 c \sqrt{x} \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(3/2),x]

[Out]

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

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Maple [A]  time = 0.02, size = 257, normalized size = 1.6 \begin{align*}{\frac{2}{ \left ( 21\,c{x}^{2}+21\,b \right ){c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}} \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-B\sqrt{{ \left ( cx+\sqrt{-bc} \right ){\frac{1}{\sqrt{-bc}}}}}\sqrt{2}\sqrt{{ \left ( -cx+\sqrt{-bc} \right ){\frac{1}{\sqrt{-bc}}}}}\sqrt{-{cx{\frac{1}{\sqrt{-bc}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( cx+\sqrt{-bc} \right ){\frac{1}{\sqrt{-bc}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-bc}{b}^{2}+3\,B{c}^{3}{x}^{5}+7\,A{x}^{3}{c}^{3}+5\,B{x}^{3}b{c}^{2}+7\,Ab{c}^{2}x+2\,B{b}^{2}cx \right ){x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{2}}{\left (B x^{2} + A\right )}}{x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(3/2), 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 )}}{x^{\frac{3}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(3/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**(3/2),x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{2}}{\left (B x^{2} + A\right )}}{x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(3/2), x)