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

Optimal. Leaf size=367 \[ \frac{4 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right ),\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{8 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 (e x)^{3/2} \left (a+b x^2\right )^{3/2} (a B+9 A b)}{9 a e^3}+\frac{4 (e x)^{3/2} \sqrt{a+b x^2} (a B+9 A b)}{15 e^3}+\frac{8 a \sqrt{e x} \sqrt{a+b x^2} (a B+9 A b)}{15 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}} \]

[Out]

(4*(9*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*e^3) + (8*a*(9*A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*Sqr
t[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) + (2*(9*A*b + a*B)*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*a*e^3) - (2*A*(a + b*x^2)
^(5/2))/(a*e*Sqrt[e*x]) - (8*a^(5/4)*(9*A*b + 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])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (
4*a^(5/4)*(9*A*b + 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])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.294509, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {453, 279, 329, 305, 220, 1196} \[ \frac{4 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}-\frac{8 a^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} (a B+9 A b) E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt [4]{a} \sqrt{e}}\right )|\frac{1}{2}\right )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{2 (e x)^{3/2} \left (a+b x^2\right )^{3/2} (a B+9 A b)}{9 a e^3}+\frac{4 (e x)^{3/2} \sqrt{a+b x^2} (a B+9 A b)}{15 e^3}+\frac{8 a \sqrt{e x} \sqrt{a+b x^2} (a B+9 A b)}{15 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(9*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(15*e^3) + (8*a*(9*A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*Sqr
t[b]*e^2*(Sqrt[a] + Sqrt[b]*x)) + (2*(9*A*b + a*B)*(e*x)^(3/2)*(a + b*x^2)^(3/2))/(9*a*e^3) - (2*A*(a + b*x^2)
^(5/2))/(a*e*Sqrt[e*x]) - (8*a^(5/4)*(9*A*b + 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])/(15*b^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (
4*a^(5/4)*(9*A*b + 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])/(15*b^(3/4)*e^(3/2)*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 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{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{(e x)^{3/2}} \, dx &=-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}}+\frac{(9 A b+a B) \int \sqrt{e x} \left (a+b x^2\right )^{3/2} \, dx}{a e^2}\\ &=\frac{2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}}+\frac{(2 (9 A b+a B)) \int \sqrt{e x} \sqrt{a+b x^2} \, dx}{3 e^2}\\ &=\frac{4 (9 A b+a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 e^3}+\frac{2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}}+\frac{(4 a (9 A b+a B)) \int \frac{\sqrt{e x}}{\sqrt{a+b x^2}} \, dx}{15 e^2}\\ &=\frac{4 (9 A b+a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 e^3}+\frac{2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}}+\frac{(8 a (9 A b+a B)) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 e^3}\\ &=\frac{4 (9 A b+a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 e^3}+\frac{2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}}+\frac{\left (8 a^{3/2} (9 A b+a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{15 \sqrt{b} e^2}-\frac{\left (8 a^{3/2} (9 A b+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 )}{15 \sqrt{b} e^2}\\ &=\frac{4 (9 A b+a B) (e x)^{3/2} \sqrt{a+b x^2}}{15 e^3}+\frac{8 a (9 A b+a B) \sqrt{e x} \sqrt{a+b x^2}}{15 \sqrt{b} e^2 \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{2 (9 A b+a B) (e x)^{3/2} \left (a+b x^2\right )^{3/2}}{9 a e^3}-\frac{2 A \left (a+b x^2\right )^{5/2}}{a e \sqrt{e x}}-\frac{8 a^{5/4} (9 A b+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 )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}+\frac{4 a^{5/4} (9 A b+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 )}{15 b^{3/4} e^{3/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0720828, size = 84, normalized size = 0.23 \[ \frac{2 x \sqrt{a+b x^2} \left (\frac{x^2 (a B+9 A b) \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )}{\sqrt{\frac{b x^2}{a}+1}}-\frac{3 A \left (a+b x^2\right )^2}{a}\right )}{3 (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[a + b*x^2]*((-3*A*(a + b*x^2)^2)/a + ((9*A*b + a*B)*x^2*Hypergeometric2F1[-3/2, 3/4, 7/4, -((b*x^2)/
a)])/Sqrt[1 + (b*x^2)/a]))/(3*(e*x)^(3/2))

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Maple [A]  time = 0.017, size = 421, normalized size = 1.2 \begin{align*}{\frac{2}{45\,be} \left ( 5\,B{x}^{6}{b}^{3}+108\,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 ){a}^{2}b-54\,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 ){a}^{2}b+12\,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 ){a}^{3}-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 ){a}^{3}+9\,A{x}^{4}{b}^{3}+16\,B{x}^{4}a{b}^{2}-36\,A{x}^{2}a{b}^{2}+11\,B{x}^{2}{a}^{2}b-45\,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)^(3/2),x)

[Out]

2/45*(5*B*x^6*b^3+108*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))*a^2*b-54*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)*El
lipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^2*b+12*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))*a^3-6*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^3+
9*A*x^4*b^3+16*B*x^4*a*b^2-36*A*x^2*a*b^2+11*B*x^2*a^2*b-45*A*a^2*b)/(b*x^2+a)^(1/2)/b/e/(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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Sympy [C]  time = 14.8636, size = 202, normalized size = 0.55 \begin{align*} \frac{A a^{\frac{3}{2}} \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{3}{2}} \sqrt{x} \Gamma \left (\frac{3}{4}\right )} + \frac{A \sqrt{a} b x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B a^{\frac{3}{2}} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{7}{4}\right )} + \frac{B \sqrt{a} b x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac{3}{2}} \Gamma \left (\frac{11}{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)**(3/2),x)

[Out]

A*a**(3/2)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*sqrt(x)*gamma(3/4)) +
A*sqrt(a)*b*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(7/4))
+ B*a**(3/2)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(7/4))
+ B*sqrt(a)*b*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(11/4
))

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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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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