3.9.9 \(\int \frac {(e x)^{3/2} (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [809]

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

[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 = 0.08, 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 = {470, 294, 335, 226} \begin {gather*} \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 \text {ArcTan}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 294

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 335

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 470

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) \text {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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.10, size = 85, normalized size = 0.49 \begin {gather*} \frac {e \sqrt {e x} \left (-3 A b+5 a B+2 b B x^2+(3 A b-5 a B) \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )\right )}{3 b^2 \sqrt {a+b x^2}} \end {gather*}

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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Maple [A]
time = 0.12, size = 225, normalized size = 1.29

method	result	size
elliptic	\(\frac {\sqrt {e x}\, \sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {e^{2} x \left (A b -B a \right )}{b^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {2 B e \sqrt {b e \,x^{3}+a e x}}{3 b^{2}}+\frac {\left (\frac {\left (A b -B a \right ) e^{2}}{2 b^{2}}-\frac {B \,e^{2} a}{3 b^{2}}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{e x \sqrt {b \,x^{2}+a}}\)	\(223\)
default	\(\frac {e \sqrt {e x}\, \left (3 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, b -5 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a +4 b^{2} B \,x^{3}-6 A \,b^{2} x +10 B a b x \right )}{6 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(225\)
risch	\(\frac {2 B x \sqrt {b \,x^{2}+a}\, e^{2}}{3 b^{2} \sqrt {e x}}+\frac {\left (\frac {3 A \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{\sqrt {b e \,x^{3}+a e x}}-\frac {4 B a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}-3 a \left (A b -B a \right ) \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b e \,x^{3}+a e x}}\right )\right ) e^{2} \sqrt {\left (b \,x^{2}+a \right ) e x}}{3 b^{2} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(426\)

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,method=_RETURNVERBOSE)

[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)*(-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)*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)*(-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)*a+4*b^2*B*x^3-6*A*b^2*x+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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.30, size = 98, normalized size = 0.56 \begin {gather*} -\frac {{\left (5 \, B a^{2} - 3 \, A a b + {\left (5 \, B a b - 3 \, A b^{2}\right )} x^{2}\right )} \sqrt {b} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (2 \, B b^{2} x^{2} + 5 \, B a b - 3 \, A b^{2}\right )} \sqrt {b x^{2} + a} \sqrt {x} e^{\frac {3}{2}}}{3 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \end {gather*}

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]

-1/3*((5*B*a^2 - 3*A*a*b + (5*B*a*b - 3*A*b^2)*x^2)*sqrt(b)*e^(3/2)*weierstrassPInverse(-4*a/b, 0, x) - (2*B*b
^2*x^2 + 5*B*a*b - 3*A*b^2)*sqrt(b*x^2 + a)*sqrt(x)*e^(3/2))/(b^4*x^2 + a*b^3)

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Sympy [C] Result contains complex when optimal does not.
time = 12.58, size = 94, normalized size = 0.54 \begin {gather*} \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 )} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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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