3.791 \(\int \frac{\sqrt{a+b x^2} (A+B x^2)}{x^{9/2}} \, dx\)

Optimal. Leaf size=152 \[ -\frac{2 b^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (A b-7 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{21 a^{5/4} \sqrt{a+b x^2}}+\frac{2 \sqrt{a+b x^2} (A b-7 a B)}{21 a x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{7 a x^{7/2}} \]

[Out]

(2*(A*b - 7*a*B)*Sqrt[a + b*x^2])/(21*a*x^(3/2)) - (2*A*(a + b*x^2)^(3/2))/(7*a*x^(7/2)) - (2*b^(3/4)*(A*b - 7
*a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(
1/4)], 1/2])/(21*a^(5/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.0930551, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {453, 277, 329, 220} \[ -\frac{2 b^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (A b-7 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{21 a^{5/4} \sqrt{a+b x^2}}+\frac{2 \sqrt{a+b x^2} (A b-7 a B)}{21 a x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{7 a x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - 7*a*B)*Sqrt[a + b*x^2])/(21*a*x^(3/2)) - (2*A*(a + b*x^2)^(3/2))/(7*a*x^(7/2)) - (2*b^(3/4)*(A*b - 7
*a*B)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(
1/4)], 1/2])/(21*a^(5/4)*Sqrt[a + b*x^2])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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{\sqrt{a+b x^2} \left (A+B x^2\right )}{x^{9/2}} \, dx &=-\frac{2 A \left (a+b x^2\right )^{3/2}}{7 a x^{7/2}}-\frac{\left (2 \left (\frac{A b}{2}-\frac{7 a B}{2}\right )\right ) \int \frac{\sqrt{a+b x^2}}{x^{5/2}} \, dx}{7 a}\\ &=\frac{2 (A b-7 a B) \sqrt{a+b x^2}}{21 a x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{7 a x^{7/2}}-\frac{(2 b (A b-7 a B)) \int \frac{1}{\sqrt{x} \sqrt{a+b x^2}} \, dx}{21 a}\\ &=\frac{2 (A b-7 a B) \sqrt{a+b x^2}}{21 a x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{7 a x^{7/2}}-\frac{(4 b (A b-7 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{21 a}\\ &=\frac{2 (A b-7 a B) \sqrt{a+b x^2}}{21 a x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{7 a x^{7/2}}-\frac{2 b^{3/4} (A b-7 a B) \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{21 a^{5/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0942077, size = 79, normalized size = 0.52 \[ \frac{2 \sqrt{a+b x^2} \left (\frac{x^2 (A b-7 a B) \, _2F_1\left (-\frac{3}{4},-\frac{1}{2};\frac{1}{4};-\frac{b x^2}{a}\right )}{\sqrt{\frac{b x^2}{a}+1}}-3 A \left (a+b x^2\right )\right )}{21 a x^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x^2]*(-3*A*(a + b*x^2) + ((A*b - 7*a*B)*x^2*Hypergeometric2F1[-3/4, -1/2, 1/4, -((b*x^2)/a)])/Sq
rt[1 + (b*x^2)/a]))/(21*a*x^(7/2))

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Maple [A]  time = 0.038, size = 242, normalized size = 1.6 \begin{align*} -{\frac{2}{21\,a} \left ( A\sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{2}\sqrt{{ \left ( -bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{ \left ( bx+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{-ab}{x}^{3}b-7\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ) \sqrt{-ab}{x}^{3}a+2\,A{b}^{2}{x}^{4}+7\,B{x}^{4}ab+5\,aAb{x}^{2}+7\,B{x}^{2}{a}^{2}+3\,A{a}^{2} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{x^{\frac{9}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^(9/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)/x^(9/2), x)

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Sympy [C]  time = 139.503, size = 97, normalized size = 0.64 \begin{align*} \frac{A \sqrt{a} \Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, - \frac{1}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 x^{\frac{7}{2}} \Gamma \left (- \frac{3}{4}\right )} + \frac{B \sqrt{a} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/x**(9/2),x)

[Out]

A*sqrt(a)*gamma(-7/4)*hyper((-7/4, -1/2), (-3/4,), b*x**2*exp_polar(I*pi)/a)/(2*x**(7/2)*gamma(-3/4)) + B*sqrt
(a)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*x**(3/2)*gamma(1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{x^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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