3.790 \(\int \frac{\sqrt{a+b x^2} (A+B x^2)}{(e x)^{7/2}} \, dx\)

Optimal. Leaf size=338 \[ \frac{2 \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt{a+b x^2}}-\frac{4 \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt{a+b x^2}}-\frac{2 \sqrt{a+b x^2} (5 a B+A b)}{5 a e^3 \sqrt{e x}}+\frac{4 \sqrt{b} \sqrt{e x} \sqrt{a+b x^2} (5 a B+A b)}{5 a e^4 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{3/2}}{5 a e (e x)^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.26434, antiderivative size = 338, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {453, 277, 329, 305, 220, 1196} \[ \frac{2 \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt{a+b x^2}}-\frac{4 \sqrt [4]{b} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (5 a B+A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{5 a^{3/4} e^{7/2} \sqrt{a+b x^2}}-\frac{2 \sqrt{a+b x^2} (5 a B+A b)}{5 a e^3 \sqrt{e x}}+\frac{4 \sqrt{b} \sqrt{e x} \sqrt{a+b x^2} (5 a B+A b)}{5 a e^4 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{3/2}}{5 a e (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x^2]*(A + B*x^2))/(e*x)^(7/2),x]

[Out]

(-2*(A*b + 5*a*B)*Sqrt[a + b*x^2])/(5*a*e^3*Sqrt[e*x]) + (4*Sqrt[b]*(A*b + 5*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(
5*a*e^4*(Sqrt[a] + Sqrt[b]*x)) - (2*A*(a + b*x^2)^(3/2))/(5*a*e*(e*x)^(5/2)) - (4*b^(1/4)*(A*b + 5*a*B)*(Sqrt[
a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[e*x])/(a^(1/4)*Sqrt
[e])], 1/2])/(5*a^(3/4)*e^(7/2)*Sqrt[a + b*x^2]) + (2*b^(1/4)*(A*b + 5*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[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(5*a^(3/4)*e^(7
/2)*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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 0.0593729, size = 95, normalized size = 0.28 \[ -\frac{2 x \sqrt{a+b x^2} \left (x^2 (5 a B+A b) \, _2F_1\left (-\frac{1}{2},-\frac{1}{4};\frac{3}{4};-\frac{b x^2}{a}\right )+A \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1}\right )}{5 a (e x)^{7/2} \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x^2]*(A + B*x^2))/(e*x)^(7/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.032, size = 417, normalized size = 1.2 \begin{align*}{\frac{2}{5\,{x}^{2}{e}^{3}a} \left ( 2\,A\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){x}^{2}ab-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 ){x}^{2}ab+10\,B\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){x}^{2}{a}^{2}-5\,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 ){x}^{2}{a}^{2}-2\,A{b}^{2}{x}^{4}-5\,B{x}^{4}ab-3\,aAb{x}^{2}-5\,B{x}^{2}{a}^{2}-A{a}^{2} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}{\frac{1}{\sqrt{ex}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(7/2),x)

[Out]

2/5/x^2*(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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-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))*x^2*a*b+10*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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))
^(1/2),1/2*2^(1/2))*x^2*a^2-5*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))*x^2*a^2-2*
A*b^2*x^4-5*B*x^4*a*b-3*a*A*b*x^2-5*B*x^2*a^2-A*a^2)/(b*x^2+a)^(1/2)/e^3/(e*x)^(1/2)/a

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{\left (e x\right )^{\frac{7}{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)/(e*x)^(7/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(7/2), x)

________________________________________________________________________________________

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} \sqrt{e x}}{e^{4} x^{4}}, 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)/(e*x)^(7/2),x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*sqrt(b*x^2 + a)*sqrt(e*x)/(e^4*x^4), x)

________________________________________________________________________________________

Sympy [C]  time = 109.453, size = 107, normalized size = 0.32 \begin{align*} \frac{A \sqrt{a} \Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, - \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{7}{2}} x^{\frac{5}{2}} \Gamma \left (- \frac{1}{4}\right )} + \frac{B \sqrt{a} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{7}{2}} \sqrt{x} \Gamma \left (\frac{3}{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)/(e*x)**(7/2),x)

[Out]

A*sqrt(a)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*x**(5/2)*gamma(-1/4))
+ B*sqrt(a)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(7/2)*sqrt(x)*gamma(3/4))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )} \sqrt{b x^{2} + a}}{\left (e x\right )^{\frac{7}{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)/(e*x)^(7/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/(e*x)^(7/2), x)