\(\int \frac {(e x)^{5/2} (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [808]

Optimal result

Integrand size = 26, antiderivative size = 337 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {3 (5 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^2}}{5 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{11/4} \sqrt {a+b x^2}}+\frac {3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{10 b^{11/4} \sqrt {a+b x^2}} \]

[Out]

-1/5*(5*A*b-7*B*a)*e*(e*x)^(3/2)/b^2/(b*x^2+a)^(1/2)+2/5*B*(e*x)^(7/2)/b/e/(b*x^2+a)^(1/2)+3/5*(5*A*b-7*B*a)*e
^2*(e*x)^(1/2)*(b*x^2+a)^(1/2)/b^(5/2)/(a^(1/2)+x*b^(1/2))-3/5*a^(1/4)*(5*A*b-7*B*a)*e^(5/2)*(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)))*EllipticE(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)/b^(11/4)/(b*x^2+a)^(1/2)+3/10*a^(1/4)*(5*A*b-7*B*a)*e^(5/2)*(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)/b^(11/4)/(b*x^2
+a)^(1/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {470, 294, 335, 311, 226, 1210} \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {3 \sqrt [4]{a} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-7 a B) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right ),\frac {1}{2}\right )}{10 b^{11/4} \sqrt {a+b x^2}}-\frac {3 \sqrt [4]{a} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (5 A b-7 a B) E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{11/4} \sqrt {a+b x^2}}+\frac {3 e^2 \sqrt {e x} \sqrt {a+b x^2} (5 A b-7 a B)}{5 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {e (e x)^{3/2} (5 A b-7 a B)}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}} \]

[In]

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

[Out]

-1/5*((5*A*b - 7*a*B)*e*(e*x)^(3/2))/(b^2*Sqrt[a + b*x^2]) + (2*B*(e*x)^(7/2))/(5*b*e*Sqrt[a + b*x^2]) + (3*(5
*A*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) - (3*a^(1/4)*(5*A*b - 7*a*B)*e^
(5/2)*(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])/(5*b^(11/4)*Sqrt[a + b*x^2]) + (3*a^(1/4)*(5*A*b - 7*a*B)*e^(5/2)*(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])/
(10*b^(11/4)*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 311

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 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]

Rule 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = \frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}-\frac {\left (2 \left (-\frac {5 A b}{2}+\frac {7 a B}{2}\right )\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{5 b} \\ & = -\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {\left (3 (5 A b-7 a B) e^2\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{10 b^2} \\ & = -\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {(3 (5 A b-7 a B) e) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 b^2} \\ & = -\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {\left (3 \sqrt {a} (5 A b-7 a B) e^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 b^{5/2}}-\frac {\left (3 \sqrt {a} (5 A b-7 a B) e^2\right ) \text {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 )}{5 b^{5/2}} \\ & = -\frac {(5 A b-7 a B) e (e x)^{3/2}}{5 b^2 \sqrt {a+b x^2}}+\frac {2 B (e x)^{7/2}}{5 b e \sqrt {a+b x^2}}+\frac {3 (5 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^2}}{5 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \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 )}{5 b^{11/4} \sqrt {a+b x^2}}+\frac {3 \sqrt [4]{a} (5 A b-7 a B) e^{5/2} \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 )}{10 b^{11/4} \sqrt {a+b x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.25 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {2 e (e x)^{3/2} \left (5 A b-7 a B+b B x^2+(-5 A b+7 a B) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{5 b^2 \sqrt {a+b x^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.73 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.82

method	result	size
elliptic	\(\frac {\sqrt {e x}\, \sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {e^{3} x^{2} \left (A b -B a \right )}{b^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}+\frac {2 B \,e^{2} x \sqrt {b e \,x^{3}+a e x}}{5 b^{2}}+\frac {\left (\frac {3 \left (A b -B a \right ) e^{3}}{2 b^{2}}-\frac {3 B \,e^{3} a}{5 b^{2}}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{e x \sqrt {b \,x^{2}+a}}\)	\(278\)
default	\(\frac {e^{2} \sqrt {e x}\, \left (30 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, a b -15 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, a b -42 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, E\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, a^{2}+21 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, F\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {2}\, a^{2}+4 b^{2} B \,x^{4}-10 A \,b^{2} x^{2}+14 B a b \,x^{2}\right )}{10 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(391\)
risch	\(\frac {2 B \,x^{2} \sqrt {b \,x^{2}+a}\, e^{3}}{5 b^{2} \sqrt {e x}}+\frac {\left (\frac {\left (5 A b -8 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{b \sqrt {b e \,x^{3}+a e x}}-5 a \left (A b -B a \right ) \left (\frac {x^{2}}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b e x}}-\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, E\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, F\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{2 a b \sqrt {b e \,x^{3}+a e x}}\right )\right ) e^{3} \sqrt {\left (b \,x^{2}+a \right ) e x}}{5 b^{2} \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(428\)

[In]

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

[Out]

1/e/x*(e*x)^(1/2)/(b*x^2+a)^(1/2)*((b*x^2+a)*e*x)^(1/2)*(-1/b^2*e^3*x^2*(A*b-B*a)/((x^2+a/b)*b*e*x)^(1/2)+2/5*
B/b^2*e^2*x*(b*e*x^3+a*e*x)^(1/2)+(3/2*(A*b-B*a)*e^3/b^2-3/5*B/b^2*e^3*a)*(-a*b)^(1/2)/b*((x+(-a*b)^(1/2)/b)/(
-a*b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-x/(-a*b)^(1/2)*b)^(1/2)/(b*e*x^3+a*e*x)^(1
/2)*(-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*Ellipti
cF(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.36 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {3 \, {\left ({\left (7 \, B a b - 5 \, A b^{2}\right )} e^{2} x^{2} + {\left (7 \, B a^{2} - 5 \, A a b\right )} e^{2}\right )} \sqrt {b e} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left (2 \, B b^{2} e^{2} x^{3} + {\left (7 \, B a b - 5 \, A b^{2}\right )} e^{2} x\right )} \sqrt {b x^{2} + a} \sqrt {e x}}{5 \, {\left (b^{4} x^{2} + a b^{3}\right )}} \]

[In]

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

[Out]

1/5*(3*((7*B*a*b - 5*A*b^2)*e^2*x^2 + (7*B*a^2 - 5*A*a*b)*e^2)*sqrt(b*e)*weierstrassZeta(-4*a/b, 0, weierstras
sPInverse(-4*a/b, 0, x)) + (2*B*b^2*e^2*x^3 + (7*B*a*b - 5*A*b^2)*e^2*x)*sqrt(b*x^2 + a)*sqrt(e*x))/(b^4*x^2 +
a*b^3)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 51.92 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.28 \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} + \frac {B e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {15}{4}\right )} \]

[In]

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

[Out]

A*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((3/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(11/4)) +
B*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((3/2, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*gamma(15/4
))

Maxima [F]

\[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{2} + A\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^{5/2}}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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