3.8.92 \(\int \frac {\sqrt {a+b x^2} (A+B x^2)}{x^{11/2}} \, dx\) [792]

Optimal. Leaf size=331 \[ \frac {2 (A b-3 a B) \sqrt {a+b x^2}}{15 a x^{5/2}}+\frac {4 b (A b-3 a B) \sqrt {a+b x^2}}{15 a^2 \sqrt {x}}-\frac {4 b^{3/2} (A b-3 a B) \sqrt {x} \sqrt {a+b x^2}}{15 a^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}+\frac {4 b^{5/4} (A b-3 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{7/4} \sqrt {a+b x^2}}-\frac {2 b^{5/4} (A b-3 a B) \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{7/4} \sqrt {a+b x^2}} \]

[Out]

-2/9*A*(b*x^2+a)^(3/2)/a/x^(9/2)+2/15*(A*b-3*B*a)*(b*x^2+a)^(1/2)/a/x^(5/2)+4/15*b*(A*b-3*B*a)*(b*x^2+a)^(1/2)
/a^2/x^(1/2)-4/15*b^(3/2)*(A*b-3*B*a)*x^(1/2)*(b*x^2+a)^(1/2)/a^2/(a^(1/2)+x*b^(1/2))+4/15*b^(5/4)*(A*b-3*B*a)
*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticE(sin(2*arcta
n(b^(1/4)*x^(1/2)/a^(1/4))),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^(7/4)/(
b*x^2+a)^(1/2)-2/15*b^(5/4)*(A*b-3*B*a)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*
x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),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^(7/4)/(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {464, 283, 331, 335, 311, 226, 1210} \begin {gather*} -\frac {2 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (A b-3 a B) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{7/4} \sqrt {a+b x^2}}+\frac {4 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} (A b-3 a B) E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{7/4} \sqrt {a+b x^2}}-\frac {4 b^{3/2} \sqrt {x} \sqrt {a+b x^2} (A b-3 a B)}{15 a^2 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {4 b \sqrt {a+b x^2} (A b-3 a B)}{15 a^2 \sqrt {x}}+\frac {2 \sqrt {a+b x^2} (A b-3 a B)}{15 a x^{5/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x^2]*(A + B*x^2))/x^(11/2),x]

[Out]

(2*(A*b - 3*a*B)*Sqrt[a + b*x^2])/(15*a*x^(5/2)) + (4*b*(A*b - 3*a*B)*Sqrt[a + b*x^2])/(15*a^2*Sqrt[x]) - (4*b
^(3/2)*(A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x^2])/(15*a^2*(Sqrt[a] + Sqrt[b]*x)) - (2*A*(a + b*x^2)^(3/2))/(9*a*x^
(9/2)) + (4*b^(5/4)*(A*b - 3*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[x])/a^(1/4)], 1/2])/(15*a^(7/4)*Sqrt[a + b*x^2]) - (2*b^(5/4)*(A*b - 3*a*B)*(Sqrt[a] + Sq
rt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(15*a^
(7/4)*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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 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 )}{x^{11/2}} \, dx &=-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}-\frac {\left (2 \left (\frac {3 A b}{2}-\frac {9 a B}{2}\right )\right ) \int \frac {\sqrt {a+b x^2}}{x^{7/2}} \, dx}{9 a}\\ &=\frac {2 (A b-3 a B) \sqrt {a+b x^2}}{15 a x^{5/2}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}-\frac {(2 b (A b-3 a B)) \int \frac {1}{x^{3/2} \sqrt {a+b x^2}} \, dx}{15 a}\\ &=\frac {2 (A b-3 a B) \sqrt {a+b x^2}}{15 a x^{5/2}}+\frac {4 b (A b-3 a B) \sqrt {a+b x^2}}{15 a^2 \sqrt {x}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}-\frac {\left (2 b^2 (A b-3 a B)\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2}} \, dx}{15 a^2}\\ &=\frac {2 (A b-3 a B) \sqrt {a+b x^2}}{15 a x^{5/2}}+\frac {4 b (A b-3 a B) \sqrt {a+b x^2}}{15 a^2 \sqrt {x}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}-\frac {\left (4 b^2 (A b-3 a B)\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{15 a^2}\\ &=\frac {2 (A b-3 a B) \sqrt {a+b x^2}}{15 a x^{5/2}}+\frac {4 b (A b-3 a B) \sqrt {a+b x^2}}{15 a^2 \sqrt {x}}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}-\frac {\left (4 b^{3/2} (A b-3 a B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{15 a^{3/2}}+\frac {\left (4 b^{3/2} (A b-3 a B)\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{15 a^{3/2}}\\ &=\frac {2 (A b-3 a B) \sqrt {a+b x^2}}{15 a x^{5/2}}+\frac {4 b (A b-3 a B) \sqrt {a+b x^2}}{15 a^2 \sqrt {x}}-\frac {4 b^{3/2} (A b-3 a B) \sqrt {x} \sqrt {a+b x^2}}{15 a^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 A \left (a+b x^2\right )^{3/2}}{9 a x^{9/2}}+\frac {4 b^{5/4} (A b-3 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 {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{7/4} \sqrt {a+b x^2}}-\frac {2 b^{5/4} (A b-3 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 {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 a^{7/4} \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 = 10.08, size = 80, normalized size = 0.24 \begin {gather*} \frac {2 \sqrt {a+b x^2} \left (-5 A \left (a+b x^2\right )+\frac {3 (A b-3 a B) x^2 \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {b x^2}{a}\right )}{\sqrt {1+\frac {b x^2}{a}}}\right )}{45 a x^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x^2]*(A + B*x^2))/x^(11/2),x]

[Out]

(2*Sqrt[a + b*x^2]*(-5*A*(a + b*x^2) + (3*(A*b - 3*a*B)*x^2*Hypergeometric2F1[-5/4, -1/2, -1/4, -((b*x^2)/a)])
/Sqrt[1 + (b*x^2)/a]))/(45*a*x^(9/2))

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Maple [A]
time = 0.12, size = 439, normalized size = 1.33

method	result	size
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}\, \left (-6 A \,b^{2} x^{4}+18 B a b \,x^{4}+2 a A b \,x^{2}+9 B \,a^{2} x^{2}+5 a^{2} A \right )}{45 x^{\frac {9}{2}} a^{2}}-\frac {2 b \left (A b -3 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) \sqrt {x \left (b \,x^{2}+a \right )}}{15 a^{2} \sqrt {b \,x^{3}+a x}\, \sqrt {x}\, \sqrt {b \,x^{2}+a}}\)	\(250\)
elliptic	\(\frac {\sqrt {x \left (b \,x^{2}+a \right )}\, \left (-\frac {2 A \sqrt {b \,x^{3}+a x}}{9 x^{5}}-\frac {2 \left (2 A b +9 B a \right ) \sqrt {b \,x^{3}+a x}}{45 a \,x^{3}}+\frac {4 \left (b \,x^{2}+a \right ) b \left (A b -3 B a \right )}{15 a^{2} \sqrt {x \left (b \,x^{2}+a \right )}}-\frac {2 b \left (A b -3 B a \right ) \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{15 a^{2} \sqrt {b \,x^{3}+a x}}\right )}{\sqrt {x}\, \sqrt {b \,x^{2}+a}}\)	\(270\)
default	\(-\frac {2 \left (6 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a \,b^{2} x^{4}-3 A \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a \,b^{2} x^{4}-18 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} b \,x^{4}+9 B \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2} b \,x^{4}-6 A \,b^{3} x^{6}+18 B a \,b^{2} x^{6}-4 A a \,b^{2} x^{4}+27 B \,a^{2} b \,x^{4}+7 A \,a^{2} b \,x^{2}+9 B \,a^{3} x^{2}+5 A \,a^{3}\right )}{45 \sqrt {b \,x^{2}+a}\, x^{\frac {9}{2}} a^{2}}\)	\(439\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(b*x^2+a)^(1/2)/x^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/45*(6*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^2*x^4-3*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^2*x^4-18*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*x^4+9*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*x^4-6*A*b^3*x^6+18*B*a*b^2*x^6-4*A*a*b^2*x^4+27*B*a^2*b*x^4+7*A*a^2*b*x^2+9*B*a^3*x^2+5*A*a^3)/(b*x^2+a)^(1/
2)/x^(9/2)/a^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)/x^(11/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/x^(11/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.24, size = 99, normalized size = 0.30 \begin {gather*} -\frac {2 \, {\left (6 \, {\left (3 \, B a b - A b^{2}\right )} \sqrt {b} x^{5} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left (6 \, {\left (3 \, B a b - A b^{2}\right )} x^{4} + 5 \, A a^{2} + {\left (9 \, B a^{2} + 2 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {x}\right )}}{45 \, a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(b*x^2+a)^(1/2)/x^(11/2),x, algorithm="fricas")

[Out]

-2/45*(6*(3*B*a*b - A*b^2)*sqrt(b)*x^5*weierstrassZeta(-4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)) + (6*(3*B
*a*b - A*b^2)*x^4 + 5*A*a^2 + (9*B*a^2 + 2*A*a*b)*x^2)*sqrt(b*x^2 + a)*sqrt(x))/(a^2*x^5)

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Sympy [C] Result contains complex when optimal does not.
time = 32.69, size = 100, normalized size = 0.30 \begin {gather*} \frac {A \sqrt {a} \Gamma \left (- \frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {9}{4}, - \frac {1}{2} \\ - \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 x^{\frac {9}{2}} \Gamma \left (- \frac {5}{4}\right )} + \frac {B \sqrt {a} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 x^{\frac {5}{2}} \Gamma \left (- \frac {1}{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)/x**(11/2),x)

[Out]

A*sqrt(a)*gamma(-9/4)*hyper((-9/4, -1/2), (-5/4,), b*x**2*exp_polar(I*pi)/a)/(2*x**(9/2)*gamma(-5/4)) + B*sqrt
(a)*gamma(-5/4)*hyper((-5/4, -1/2), (-1/4,), b*x**2*exp_polar(I*pi)/a)/(2*x**(5/2)*gamma(-1/4))

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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)/x^(11/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*sqrt(b*x^2 + a)/x^(11/2), 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}}{x^{11/2}} \,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))/x^(11/2),x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(1/2))/x^(11/2), x)

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