3.9.17 \(\int \frac {(e x)^{3/2} (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\) [817]

Optimal. Leaf size=185 \[ \frac {(A b-a B) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/2}}-\frac {(A b+5 a B) e \sqrt {e x}}{6 a b^2 \sqrt {a+b x^2}}+\frac {(A b+5 a B) e^{3/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{12 a^{5/4} b^{9/4} \sqrt {a+b x^2}} \]

[Out]

1/3*(A*b-B*a)*(e*x)^(5/2)/a/b/e/(b*x^2+a)^(3/2)-1/6*(A*b+5*B*a)*e*(e*x)^(1/2)/a/b^2/(b*x^2+a)^(1/2)+1/12*(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^(5/4)/b^(9/4)/(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 185, 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 = {468, 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}} (5 a B+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 )}{12 a^{5/4} b^{9/4} \sqrt {a+b x^2}}-\frac {e \sqrt {e x} (5 a B+A b)}{6 a b^2 \sqrt {a+b x^2}}+\frac {(e x)^{5/2} (A b-a B)}{3 a b e \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*(e*x)^(5/2))/(3*a*b*e*(a + b*x^2)^(3/2)) - ((A*b + 5*a*B)*e*Sqrt[e*x])/(6*a*b^2*Sqrt[a + b*x^2])
+ ((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])/(12*a^(5/4)*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 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{5/2}} \, dx &=\frac {(A b-a B) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/2}}+\frac {\left (\frac {A b}{2}+\frac {5 a B}{2}\right ) \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{3 a b}\\ &=\frac {(A b-a B) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/2}}-\frac {(A b+5 a B) e \sqrt {e x}}{6 a b^2 \sqrt {a+b x^2}}+\frac {\left ((A b+5 a B) e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt {a+b x^2}} \, dx}{12 a b^2}\\ &=\frac {(A b-a B) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/2}}-\frac {(A b+5 a B) e \sqrt {e x}}{6 a b^2 \sqrt {a+b x^2}}+\frac {((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 )}{6 a b^2}\\ &=\frac {(A b-a B) (e x)^{5/2}}{3 a b e \left (a+b x^2\right )^{3/2}}-\frac {(A b+5 a B) e \sqrt {e x}}{6 a b^2 \sqrt {a+b x^2}}+\frac {(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 )}{12 a^{5/4} 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 = 10.14, size = 105, normalized size = 0.57 \begin {gather*} \frac {e \sqrt {e x} \left (-5 a^2 B+A b^2 x^2-a b \left (A+7 B x^2\right )+(A b+5 a B) \left (a+b x^2\right ) \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 )}{6 a b^2 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[e*x]*(-5*a^2*B + A*b^2*x^2 - a*b*(A + 7*B*x^2) + (A*b + 5*a*B)*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hyperge
ometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)]))/(6*a*b^2*(a + b*x^2)^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(428\) vs. \(2(192)=384\).
time = 0.12, size = 429, normalized size = 2.32

method	result	size
elliptic	\(\frac {\sqrt {e x}\, \sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {e \left (A b -B a \right ) \sqrt {b e \,x^{3}+a e x}}{3 b^{4} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {e^{2} x \left (A b -7 B a \right )}{6 b^{2} a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {\left (\frac {B \,e^{2}}{b^{2}}+\frac {e^{2} \left (A b -7 B a \right )}{12 b^{2} a}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{e x \sqrt {b \,x^{2}+a}}\)	\(245\)
default	\(\frac {\left (A \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, b^{2} x^{2}+5 B \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, a b \,x^{2}+A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a b +5 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, a^{2}+2 A \,b^{3} x^{3}-14 B a \,b^{2} x^{3}-2 A a \,b^{2} x -10 B \,a^{2} b x \right ) e \sqrt {e x}}{12 x a \,b^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\)	\(429\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(B*x^2+A)/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/12*(A*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*((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)*b^2*x^2+5*B*EllipticF
(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*((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)*a*b*x^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)*EllipticF(((b*x+(-a*b)^
(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*a*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^2+2*A*b^3*x^3-14*B*a*b^2*x^3-2*A*a*b^2*x-10*B*a^2*b*x)*e/x*(e*x)^(1/2)/a/b^3
/(b*x^2+a)^(3/2)

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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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.26, size = 144, normalized size = 0.78 \begin {gather*} \frac {{\left ({\left (5 \, B a b^{2} + A b^{3}\right )} x^{4} + 5 \, B a^{3} + A a^{2} b + 2 \, {\left (5 \, B a^{2} b + A a b^{2}\right )} x^{2}\right )} \sqrt {b} e^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) - {\left (5 \, B a^{2} b + A a b^{2} + {\left (7 \, B a b^{2} - A b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {x} e^{\frac {3}{2}}}{6 \, {\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

1/6*(((5*B*a*b^2 + A*b^3)*x^4 + 5*B*a^3 + A*a^2*b + 2*(5*B*a^2*b + A*a*b^2)*x^2)*sqrt(b)*e^(3/2)*weierstrassPI
nverse(-4*a/b, 0, x) - (5*B*a^2*b + A*a*b^2 + (7*B*a*b^2 - A*b^3)*x^2)*sqrt(b*x^2 + a)*sqrt(x)*e^(3/2))/(a*b^5
*x^4 + 2*a^2*b^4*x^2 + a^3*b^3)

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Sympy [C] Result contains complex when optimal does not.
time = 60.39, size = 94, normalized size = 0.51 \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 {5}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{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 {9}{4}, \frac {5}{2} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{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)**(5/2),x)

[Out]

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

[Out]

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

[Out]

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

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