3.793 \(\int \frac{\sqrt{a+b x^2} (A+B x^2)}{x^{13/2}} \, dx\)

Optimal. Leaf size=187 \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 A b-11 a B) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{231 a^{9/4} \sqrt{a+b x^2}}+\frac{4 b \sqrt{a+b x^2} (5 A b-11 a B)}{231 a^2 x^{3/2}}+\frac{2 \sqrt{a+b x^2} (5 A b-11 a B)}{77 a x^{7/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}} \]

[Out]

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

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Rubi [A]  time = 0.112352, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {453, 277, 325, 329, 220} \[ \frac{2 b^{7/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 A b-11 a B) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{231 a^{9/4} \sqrt{a+b x^2}}+\frac{4 b \sqrt{a+b x^2} (5 A b-11 a B)}{231 a^2 x^{3/2}}+\frac{2 \sqrt{a+b x^2} (5 A b-11 a B)}{77 a x^{7/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*A*b - 11*a*B)*Sqrt[a + b*x^2])/(77*a*x^(7/2)) + (4*b*(5*A*b - 11*a*B)*Sqrt[a + b*x^2])/(231*a^2*x^(3/2))
- (2*A*(a + b*x^2)^(3/2))/(11*a*x^(11/2)) + (2*b^(7/4)*(5*A*b - 11*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])/(231*a^(9/4)*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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^{13/2}} \, dx &=-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}-\frac{\left (2 \left (\frac{5 A b}{2}-\frac{11 a B}{2}\right )\right ) \int \frac{\sqrt{a+b x^2}}{x^{9/2}} \, dx}{11 a}\\ &=\frac{2 (5 A b-11 a B) \sqrt{a+b x^2}}{77 a x^{7/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}-\frac{(2 b (5 A b-11 a B)) \int \frac{1}{x^{5/2} \sqrt{a+b x^2}} \, dx}{77 a}\\ &=\frac{2 (5 A b-11 a B) \sqrt{a+b x^2}}{77 a x^{7/2}}+\frac{4 b (5 A b-11 a B) \sqrt{a+b x^2}}{231 a^2 x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}+\frac{\left (2 b^2 (5 A b-11 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x^2}} \, dx}{231 a^2}\\ &=\frac{2 (5 A b-11 a B) \sqrt{a+b x^2}}{77 a x^{7/2}}+\frac{4 b (5 A b-11 a B) \sqrt{a+b x^2}}{231 a^2 x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}+\frac{\left (4 b^2 (5 A b-11 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\sqrt{x}\right )}{231 a^2}\\ &=\frac{2 (5 A b-11 a B) \sqrt{a+b x^2}}{77 a x^{7/2}}+\frac{4 b (5 A b-11 a B) \sqrt{a+b x^2}}{231 a^2 x^{3/2}}-\frac{2 A \left (a+b x^2\right )^{3/2}}{11 a x^{11/2}}+\frac{2 b^{7/4} (5 A b-11 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 )}{231 a^{9/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.100848, size = 80, normalized size = 0.43 \[ \frac{2 \sqrt{a+b x^2} \left (\frac{x^2 (5 A b-11 a B) \, _2F_1\left (-\frac{7}{4},-\frac{1}{2};-\frac{3}{4};-\frac{b x^2}{a}\right )}{\sqrt{\frac{b x^2}{a}+1}}-7 A \left (a+b x^2\right )\right )}{77 a x^{11/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x^2]*(-7*A*(a + b*x^2) + ((5*A*b - 11*a*B)*x^2*Hypergeometric2F1[-7/4, -1/2, -3/4, -((b*x^2)/a)]
)/Sqrt[1 + (b*x^2)/a]))/(77*a*x^(11/2))

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Maple [A]  time = 0.028, size = 270, normalized size = 1.4 \begin{align*}{\frac{2}{231\,{a}^{2}} \left ( 5\,A\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}^{5}{b}^{2}-11\,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}^{5}ab+10\,A{x}^{6}{b}^{3}-22\,B{x}^{6}a{b}^{2}+4\,A{x}^{4}a{b}^{2}-55\,B{x}^{4}{a}^{2}b-27\,A{x}^{2}{a}^{2}b-33\,B{x}^{2}{a}^{3}-21\,A{a}^{3} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{x}^{-{\frac{11}{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^(13/2),x)

[Out]

2/231/(b*x^2+a)^(1/2)*(5*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^
5*b^2-11*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^5*a*b+10*A*x^6*b
^3-22*B*x^6*a*b^2+4*A*x^4*a*b^2-55*B*x^4*a^2*b-27*A*x^2*a^2*b-33*B*x^2*a^3-21*A*a^3)/x^(11/2)/a^2

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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{13}{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^(13/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/x^(13/2), 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{13}{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^(13/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)/x^(13/2), 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)*(b*x**2+a)**(1/2)/x**(13/2),x)

[Out]

Timed out

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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{13}{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^(13/2),x, algorithm="giac")

[Out]

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