3.8.100 \(\int \frac {(e x)^{5/2} (A+B x^2)}{\sqrt {a+b x^2}} \, dx\) [800]

Optimal. Leaf size=338 \[ \frac {2 (9 A b-7 a B) e (e x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {2 a (9 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^2}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 a^{5/4} (9 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}}-\frac {a^{5/4} (9 A b-7 a B) e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}} \]

[Out]

2/45*(9*A*b-7*B*a)*e*(e*x)^(3/2)*(b*x^2+a)^(1/2)/b^2+2/9*B*(e*x)^(7/2)*(b*x^2+a)^(1/2)/b/e-2/15*a*(9*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))+2/15*a^(5/4)*(9*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(s
in(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)-1/15*a^(5/4)*(9*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]
time = 0.18, antiderivative size = 338, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {470, 327, 335, 311, 226, 1210} \begin {gather*} -\frac {a^{5/4} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (9 A b-7 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 )}{15 b^{11/4} \sqrt {a+b x^2}}+\frac {2 a^{5/4} e^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (9 A b-7 a B) E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^2}}-\frac {2 a e^2 \sqrt {e x} \sqrt {a+b x^2} (9 A b-7 a B)}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 e (e x)^{3/2} \sqrt {a+b x^2} (9 A b-7 a B)}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(9*A*b - 7*a*B)*e*(e*x)^(3/2)*Sqrt[a + b*x^2])/(45*b^2) + (2*B*(e*x)^(7/2)*Sqrt[a + b*x^2])/(9*b*e) - (2*a*
(9*A*b - 7*a*B)*e^2*Sqrt[e*x]*Sqrt[a + b*x^2])/(15*b^(5/2)*(Sqrt[a] + Sqrt[b]*x)) + (2*a^(5/4)*(9*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])/(15*b^(11/4)*Sqrt[a + b*x^2]) - (a^(5/4)*(9*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]
)/(15*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 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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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]

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*} \int \frac {(e x)^{5/2} \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx &=\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {\left (2 \left (-\frac {9 A b}{2}+\frac {7 a B}{2}\right )\right ) \int \frac {(e x)^{5/2}}{\sqrt {a+b x^2}} \, dx}{9 b}\\ &=\frac {2 (9 A b-7 a B) e (e x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {\left (a (9 A b-7 a B) e^2\right ) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{15 b^2}\\ &=\frac {2 (9 A b-7 a B) e (e x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {(2 a (9 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 )}{15 b^2}\\ &=\frac {2 (9 A b-7 a B) e (e x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {\left (2 a^{3/2} (9 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 )}{15 b^{5/2}}+\frac {\left (2 a^{3/2} (9 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 )}{15 b^{5/2}}\\ &=\frac {2 (9 A b-7 a B) e (e x)^{3/2} \sqrt {a+b x^2}}{45 b^2}+\frac {2 B (e x)^{7/2} \sqrt {a+b x^2}}{9 b e}-\frac {2 a (9 A b-7 a B) e^2 \sqrt {e x} \sqrt {a+b x^2}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {2 a^{5/4} (9 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 )}{15 b^{11/4} \sqrt {a+b x^2}}-\frac {a^{5/4} (9 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 )}{15 b^{11/4} \sqrt {a+b x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.10, size = 96, normalized size = 0.28 \begin {gather*} \frac {2 e (e x)^{3/2} \left (-\left (\left (a+b x^2\right ) \left (-9 A b+7 a B-5 b B x^2\right )\right )+a (-9 A b+7 a B) \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{45 b^2 \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(e*x)^(3/2)*(-((a + b*x^2)*(-9*A*b + 7*a*B - 5*b*B*x^2)) + a*(-9*A*b + 7*a*B)*Sqrt[1 + (b*x^2)/a]*Hyperge
ometric2F1[1/2, 3/4, 7/4, -((b*x^2)/a)]))/(45*b^2*Sqrt[a + b*x^2])

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 417, normalized size = 1.23

method	result	size
risch	\(\frac {2 x^{2} \left (5 b B \,x^{2}+9 A b -7 B a \right ) \sqrt {b \,x^{2}+a}\, e^{3}}{45 b^{2} \sqrt {e x}}-\frac {a \left (9 A b -7 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) e^{3} \sqrt {\left (b \,x^{2}+a \right ) e x}}{15 b^{3} \sqrt {b e \,x^{3}+a e x}\, \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(242\)
elliptic	\(\frac {\sqrt {e x}\, \sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (\frac {2 B \,e^{2} x^{3} \sqrt {b e \,x^{3}+a e x}}{9 b}+\frac {2 \left (A \,e^{3}-\frac {7 B \,e^{3} a}{9 b}\right ) x \sqrt {b e \,x^{3}+a e x}}{5 b e}-\frac {3 \left (A \,e^{3}-\frac {7 B \,e^{3} a}{9 b}\right ) a \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{5 b^{2} \sqrt {b e \,x^{3}+a e x}}\right )}{e x \sqrt {b \,x^{2}+a}}\)	\(275\)
default	\(-\frac {e^{2} \sqrt {e x}\, \left (-10 B \,b^{3} x^{6}+54 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} b -27 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} b -42 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{3}+21 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{3}-18 A \,b^{3} x^{4}+4 B a \,b^{2} x^{4}-18 A a \,b^{2} x^{2}+14 B \,a^{2} b \,x^{2}\right )}{45 x \sqrt {b \,x^{2}+a}\, b^{3}}\)	\(417\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(B*x^2+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/45*e^2/x*(e*x)^(1/2)/(b*x^2+a)^(1/2)/b^3*(-10*B*b^3*x^6+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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2
))^(1/2),1/2*2^(1/2))*a^2*b-27*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^2*b-42*
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+21*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^3-18*A*b^3*x^4+4*B*a*b^2*x^4-18*A*a*b^2*x^2+14*B*a^2*b*x^2)

________________________________________________________________________________________

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)^(5/2)*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.39, size = 83, normalized size = 0.25 \begin {gather*} -\frac {2 \, {\left (3 \, {\left (7 \, B a^{2} - 9 \, A a b\right )} \sqrt {b} e^{\frac {5}{2}} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) - {\left (5 \, B b^{2} x^{3} - {\left (7 \, B a b - 9 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a} \sqrt {x} e^{\frac {5}{2}}\right )}}{45 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

-2/45*(3*(7*B*a^2 - 9*A*a*b)*sqrt(b)*e^(5/2)*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)) - (
5*B*b^2*x^3 - (7*B*a*b - 9*A*b^2)*x)*sqrt(b*x^2 + a)*sqrt(x)*e^(5/2))/b^3

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 14.89, size = 94, normalized size = 0.28 \begin {gather*} \frac {A e^{\frac {5}{2}} 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 \sqrt {a} \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 {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(B*x**2+A)/(b*x**2+a)**(1/2),x)

[Out]

A*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(11/4)) +
B*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((1/2, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(15/4))

________________________________________________________________________________________

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)^(5/2)*(B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^{5/2}}{\sqrt {b\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(e*x)^(5/2))/(a + b*x^2)^(1/2),x)

[Out]

int(((A + B*x^2)*(e*x)^(5/2))/(a + b*x^2)^(1/2), x)

________________________________________________________________________________________