3.798 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{(e x)^{5/2}} \, dx\)

Optimal. Leaf size=210 \[ \frac{4 a^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 a B+7 A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{21 \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (3 a B+7 A b)}{21 a e^3}+\frac{4 \sqrt{e x} \sqrt{a+b x^2} (3 a B+7 A b)}{21 e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.136615, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {453, 279, 329, 220} \[ \frac{4 a^{3/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (3 a B+7 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{21 \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}+\frac{2 \sqrt{e x} \left (a+b x^2\right )^{3/2} (3 a B+7 A b)}{21 a e^3}+\frac{4 \sqrt{e x} \sqrt{a+b x^2} (3 a B+7 A b)}{21 e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

(4*(7*A*b + 3*a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(21*e^3) + (2*(7*A*b + 3*a*B)*Sqrt[e*x]*(a + b*x^2)^(3/2))/(21*a
*e^3) - (2*A*(a + b*x^2)^(5/2))/(3*a*e*(e*x)^(3/2)) + (4*a^(3/4)*(7*A*b + 3*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])/(21*b^(1/4
)*e^(5/2)*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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{(e x)^{5/2}} \, dx &=-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac{(7 A b+3 a B) \int \frac{\left (a+b x^2\right )^{3/2}}{\sqrt{e x}} \, dx}{3 a e^2}\\ &=\frac{2 (7 A b+3 a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac{(2 (7 A b+3 a B)) \int \frac{\sqrt{a+b x^2}}{\sqrt{e x}} \, dx}{7 e^2}\\ &=\frac{4 (7 A b+3 a B) \sqrt{e x} \sqrt{a+b x^2}}{21 e^3}+\frac{2 (7 A b+3 a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac{(4 a (7 A b+3 a B)) \int \frac{1}{\sqrt{e x} \sqrt{a+b x^2}} \, dx}{21 e^2}\\ &=\frac{4 (7 A b+3 a B) \sqrt{e x} \sqrt{a+b x^2}}{21 e^3}+\frac{2 (7 A b+3 a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac{(8 a (7 A b+3 a B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{21 e^3}\\ &=\frac{4 (7 A b+3 a B) \sqrt{e x} \sqrt{a+b x^2}}{21 e^3}+\frac{2 (7 A b+3 a B) \sqrt{e x} \left (a+b x^2\right )^{3/2}}{21 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac{4 a^{3/4} (7 A b+3 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 )}{21 \sqrt [4]{b} e^{5/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0760289, size = 85, normalized size = 0.4 \[ \frac{2 x \sqrt{a+b x^2} \left (\frac{x^2 (3 a B+7 A b) \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{b x^2}{a}\right )}{\sqrt{\frac{b x^2}{a}+1}}-\frac{A \left (a+b x^2\right )^2}{a}\right )}{3 (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/(e*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.015, size = 255, normalized size = 1.2 \begin{align*}{\frac{2}{21\,bx{e}^{2}} \left ( 14\,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}xab+6\,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{a}^{2}+3\,B{x}^{6}{b}^{3}+7\,A{x}^{4}{b}^{3}+12\,B{x}^{4}a{b}^{2}+9\,B{x}^{2}{a}^{2}b-7\,A{a}^{2}b \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)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x)

[Out]

2/21/(b*x^2+a)^(1/2)/x*(14*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*a*b+6*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*a^2+3*B*x^6*b^3+7
*A*x^4*b^3+12*B*x^4*a*b^2+9*B*x^2*a^2*b-7*A*a^2*b)/b/e^2/(e*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b x^{4} +{\left (B a + A b\right )} x^{2} + A a\right )} \sqrt{b x^{2} + a} \sqrt{e x}}{e^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 25.1801, size = 202, normalized size = 0.96 \begin{align*} \frac{A a^{\frac{3}{2}} \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 e^{\frac{5}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{1}{4}\right )} + \frac{A \sqrt{a} b \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{B a^{\frac{3}{2}} \sqrt{x} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right )} + \frac{B \sqrt{a} b x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{5}{2}} \Gamma \left (\frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/(e*x)**(5/2),x)

[Out]

A*a**(3/2)*gamma(-3/4)*hyper((-3/4, -1/2), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*x**(3/2)*gamma(1/4))
+ A*sqrt(a)*b*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*gamma(5/4))
+ B*a**(3/2)*sqrt(x)*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*gamma(5/4)) +
B*sqrt(a)*b*x**(5/2)*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(5/2)*gamma(9/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (e x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/(e*x)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^(3/2)/(e*x)^(5/2), x)