3.8.88 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{(e x)^{3/2}} \, dx\) [788]

Optimal. Leaf size=333 \[ \frac {2 (5 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{5 a e^3}+\frac {4 (5 A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{5 \sqrt {b} e^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}}-\frac {4 \sqrt [4]{a} (5 A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{3/4} e^{3/2} \sqrt {a+b x^2}}+\frac {2 \sqrt [4]{a} (5 A b+a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt [4]{a} \sqrt {e}}\right )|\frac {1}{2}\right )}{5 b^{3/4} e^{3/2} \sqrt {a+b x^2}} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 333, 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 = {464, 285, 335, 311, 226, 1210} \begin {gather*} \frac {2 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (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 )}{5 b^{3/4} e^{3/2} \sqrt {a+b x^2}}-\frac {4 \sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (a B+5 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 )}{5 b^{3/4} e^{3/2} \sqrt {a+b x^2}}+\frac {2 (e x)^{3/2} \sqrt {a+b x^2} (a B+5 A b)}{5 a e^3}+\frac {4 \sqrt {e x} \sqrt {a+b x^2} (a B+5 A b)}{5 \sqrt {b} e^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*A*b + a*B)*(e*x)^(3/2)*Sqrt[a + b*x^2])/(5*a*e^3) + (4*(5*A*b + a*B)*Sqrt[e*x]*Sqrt[a + b*x^2])/(5*Sqrt[
b]*e^2*(Sqrt[a] + Sqrt[b]*x)) - (2*A*(a + b*x^2)^(3/2))/(a*e*Sqrt[e*x]) - (4*a^(1/4)*(5*A*b + a*B)*(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^(3/4)*e^(3/2)*Sqrt[a + b*x^2]) + (2*a^(1/4)*(5*A*b + a*B)*(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])/(5*b^(3/4)*e^(3/2)*S
qrt[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 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 464

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

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 {\sqrt {a+b x^2} \left (A+B x^2\right )}{(e x)^{3/2}} \, dx &=-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}}+\frac {(5 A b+a B) \int \sqrt {e x} \sqrt {a+b x^2} \, dx}{a e^2}\\ &=\frac {2 (5 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{5 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}}+\frac {(2 (5 A b+a B)) \int \frac {\sqrt {e x}}{\sqrt {a+b x^2}} \, dx}{5 e^2}\\ &=\frac {2 (5 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{5 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}}+\frac {(4 (5 A b+a B)) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 e^3}\\ &=\frac {2 (5 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{5 a e^3}-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}}+\frac {\left (4 \sqrt {a} (5 A b+a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{5 \sqrt {b} e^2}-\frac {\left (4 \sqrt {a} (5 A b+a B)\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 \sqrt {b} e^2}\\ &=\frac {2 (5 A b+a B) (e x)^{3/2} \sqrt {a+b x^2}}{5 a e^3}+\frac {4 (5 A b+a B) \sqrt {e x} \sqrt {a+b x^2}}{5 \sqrt {b} e^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{3/2}}{a e \sqrt {e x}}-\frac {4 \sqrt [4]{a} (5 A b+a B) \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^{3/4} e^{3/2} \sqrt {a+b x^2}}+\frac {2 \sqrt [4]{a} (5 A b+a B) \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 )}{5 b^{3/4} e^{3/2} \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 = 9.54, size = 96, normalized size = 0.29 \begin {gather*} \frac {2 x \sqrt {a+b x^2} \left (-3 A \left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}}+(5 A b+a B) x^2 \, _2F_1\left (-\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^2}{a}\right )\right )}{3 a (e x)^{3/2} \sqrt {1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 391, normalized size = 1.17

method	result	size
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-B \,x^{2}+5 A \right )}{5 e \sqrt {e x}}+\frac {\left (2 A b +\frac {2 B a}{5}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) \sqrt {\left (b \,x^{2}+a \right ) e x}}{b \sqrt {b e \,x^{3}+a e x}\, e \sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(228\)
elliptic	\(\frac {\sqrt {\left (b \,x^{2}+a \right ) e x}\, \left (-\frac {2 \left (b e \,x^{2}+a e \right ) A}{e^{2} \sqrt {x \left (b e \,x^{2}+a e \right )}}+\frac {2 B x \sqrt {b e \,x^{3}+a e x}}{5 e^{2}}+\frac {\left (\frac {A b +B a}{e}+\frac {b A}{e}-\frac {3 B a}{5 e}\right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{b \sqrt {b e \,x^{3}+a e x}}\right )}{\sqrt {e x}\, \sqrt {b \,x^{2}+a}}\)	\(263\)
default	\(\frac {4 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b -2 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b +\frac {4 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}}{5}-\frac {2 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}}{5}+\frac {2 b^{2} B \,x^{4}}{5}-2 A \,b^{2} x^{2}+\frac {2 B a b \,x^{2}}{5}-2 a b A}{\sqrt {b \,x^{2}+a}\, b e \sqrt {e x}}\)	\(391\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(3/2),x, algorithm="maxima")

[Out]

e^(-3/2)*integrate((B*x^2 + A)*sqrt(b*x^2 + a)/x^(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 = 67, normalized size = 0.20 \begin {gather*} -\frac {2 \, {\left (2 \, {\left (B a + 5 \, A b\right )} \sqrt {b} x {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) - {\left (B b x^{2} - 5 \, A b\right )} \sqrt {b x^{2} + a} \sqrt {x}\right )} e^{\left (-\frac {3}{2}\right )}}{5 \, b x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.88, size = 100, normalized size = 0.30 \begin {gather*} \frac {A \sqrt {a} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} + \frac {B \sqrt {a} x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(b*x**2+a)**(1/2)/(e*x)**(3/2),x)

[Out]

A*sqrt(a)*gamma(-1/4)*hyper((-1/2, -1/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*sqrt(x)*gamma(3/4)) +
B*sqrt(a)*x**(3/2)*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b*x**2*exp_polar(I*pi)/a)/(2*e**(3/2)*gamma(7/4))

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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((B*x^2+A)*(b*x^2+a)^(1/2)/(e*x)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/(e*x)^(3/2),x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/(e*x)^(3/2), x)

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