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

Optimal. Leaf size=174 \[ \frac {e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{6 \sqrt [4]{a} b^{9/4} \sqrt {a+b x^2}}-\frac {e \sqrt {e x} (3 A b-5 a B)}{3 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{5/2}}{3 b e \sqrt {a+b x^2}} \]

[Out]

2/3*B*(e*x)^(5/2)/b/e/(b*x^2+a)^(1/2)-1/3*(3*A*b-5*B*a)*e*(e*x)^(1/2)/b^2/(b*x^2+a)^(1/2)+1/6*(3*A*b-5*B*a)*e^
(3/2)*(cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^
(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(e*x)^(1/2)/a^(1/4)/e^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+
a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(1/4)/b^(9/4)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {459, 288, 329, 220} \[ \frac {e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (3 A b-5 a B) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{6 \sqrt [4]{a} b^{9/4} \sqrt {a+b x^2}}-\frac {e \sqrt {e x} (3 A b-5 a B)}{3 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{5/2}}{3 b e \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*A*b - 5*a*B)*e*Sqrt[e*x])/(3*b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(5/2))/(3*b*e*Sqrt[a + b*x^2]) + ((3*A*b -
5*a*B)*e^(3/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sq
rt[e*x])/(a^(1/4)*Sqrt[e])], 1/2])/(6*a^(1/4)*b^(9/4)*Sqrt[a + b*x^2])

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 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 459

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

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac {2 B (e x)^{5/2}}{3 b e \sqrt {a+b x^2}}-\frac {\left (2 \left (-\frac {3 A b}{2}+\frac {5 a B}{2}\right )\right ) \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac {(3 A b-5 a B) e \sqrt {e x}}{3 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{5/2}}{3 b e \sqrt {a+b x^2}}+\frac {\left ((3 A b-5 a B) e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{6 b^2}\\ &=-\frac {(3 A b-5 a B) e \sqrt {e x}}{3 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{5/2}}{3 b e \sqrt {a+b x^2}}+\frac {((3 A b-5 a B) e) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{3 b^2}\\ &=-\frac {(3 A b-5 a B) e \sqrt {e x}}{3 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{5/2}}{3 b e \sqrt {a+b x^2}}+\frac {(3 A b-5 a B) e^{3/2} \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 )}{6 \sqrt [4]{a} b^{9/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 85, normalized size = 0.49 \[ \frac {e \sqrt {e x} \left (\sqrt {\frac {b x^2}{a}+1} (3 A b-5 a B) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )+5 a B-3 A b+2 b B x^2\right )}{3 b^2 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B e x^{3} + A e x\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 225, normalized size = 1.29 \[ \frac {\sqrt {e x}\, \left (4 B \,b^{2} x^{3}-6 A \,b^{2} x +10 B a b x +3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, A b \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, B a \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) e}{6 \sqrt {b \,x^{2}+a}\, b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*e/x*(e*x)^(1/2)*(3*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)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*b-
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)*(-1/(-a*b)^(1/2)*
b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*a+4*B*b^2*x^3-6*A*x*b^2
+10*B*a*b*x)/(b*x^2+a)^(1/2)/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {3}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^{3/2}}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 38.48, size = 94, normalized size = 0.54 \[ \frac {A e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {B e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 3/2), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(9/4)) + B
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((3/2, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(13/4))

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