3.3.58 \(\int \frac {x^{15/2} (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\) [258]

Optimal. Leaf size=377 \[ -\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}+\frac {7 b (11 b B-9 A c) x^{3/2} \left (b+c x^2\right )}{15 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}-\frac {7 b^{5/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{15/4} \sqrt {b x^2+c x^4}}+\frac {7 b^{5/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{30 c^{15/4} \sqrt {b x^2+c x^4}} \]

[Out]

-(-A*c+B*b)*x^(13/2)/b/c/(c*x^4+b*x^2)^(1/2)+7/15*b*(-9*A*c+11*B*b)*x^(3/2)*(c*x^2+b)/c^(7/2)/(b^(1/2)+x*c^(1/
2))/(c*x^4+b*x^2)^(1/2)+1/9*(-9*A*c+11*B*b)*x^(5/2)*(c*x^4+b*x^2)^(1/2)/b/c^2-7/45*(-9*A*c+11*B*b)*x^(1/2)*(c*
x^4+b*x^2)^(1/2)/c^3-7/15*b^(5/4)*(-9*A*c+11*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arc
tan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2)
)*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(15/4)/(c*x^4+b*x^2)^(1/2)+7/30*b^(5/4)*(-9*A*c+11*B*b)*x*(cos(2*a
rctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)
*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/c^(15/4)/(c*x^4+b*
x^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.32, antiderivative size = 377, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2062, 2049, 2057, 335, 311, 226, 1210} \begin {gather*} \frac {7 b^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-9 A c) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{30 c^{15/4} \sqrt {b x^2+c x^4}}-\frac {7 b^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (11 b B-9 A c) E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{15/4} \sqrt {b x^2+c x^4}}+\frac {7 b x^{3/2} \left (b+c x^2\right ) (11 b B-9 A c)}{15 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {7 \sqrt {x} \sqrt {b x^2+c x^4} (11 b B-9 A c)}{45 c^3}+\frac {x^{5/2} \sqrt {b x^2+c x^4} (11 b B-9 A c)}{9 b c^2}-\frac {x^{13/2} (b B-A c)}{b c \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(((b*B - A*c)*x^(13/2))/(b*c*Sqrt[b*x^2 + c*x^4])) + (7*b*(11*b*B - 9*A*c)*x^(3/2)*(b + c*x^2))/(15*c^(7/2)*(
Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (7*(11*b*B - 9*A*c)*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(45*c^3) + ((11*b
*B - 9*A*c)*x^(5/2)*Sqrt[b*x^2 + c*x^4])/(9*b*c^2) - (7*b^(5/4)*(11*b*B - 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[
(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*c^(15/4)*Sqrt[b*
x^2 + c*x^4]) + (7*b^(5/4)*(11*b*B - 9*A*c)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*
EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(30*c^(15/4)*Sqrt[b*x^2 + c*x^4])

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 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 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]

Rule 2049

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

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2062

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

Rubi steps

\begin {align*} \int \frac {x^{15/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}+\frac {\left (\frac {11 b B}{2}-\frac {9 A c}{2}\right ) \int \frac {x^{11/2}}{\sqrt {b x^2+c x^4}} \, dx}{b c}\\ &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}-\frac {(7 (11 b B-9 A c)) \int \frac {x^{7/2}}{\sqrt {b x^2+c x^4}} \, dx}{18 c^2}\\ &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}+\frac {(7 b (11 b B-9 A c)) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{30 c^3}\\ &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}+\frac {\left (7 b (11 b B-9 A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{30 c^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}+\frac {\left (7 b (11 b B-9 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^3 \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}+\frac {\left (7 b^{3/2} (11 b B-9 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^{7/2} \sqrt {b x^2+c x^4}}-\frac {\left (7 b^{3/2} (11 b B-9 A c) x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 c^{7/2} \sqrt {b x^2+c x^4}}\\ &=-\frac {(b B-A c) x^{13/2}}{b c \sqrt {b x^2+c x^4}}+\frac {7 b (11 b B-9 A c) x^{3/2} \left (b+c x^2\right )}{15 c^{7/2} \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {7 (11 b B-9 A c) \sqrt {x} \sqrt {b x^2+c x^4}}{45 c^3}+\frac {(11 b B-9 A c) x^{5/2} \sqrt {b x^2+c x^4}}{9 b c^2}-\frac {7 b^{5/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 c^{15/4} \sqrt {b x^2+c x^4}}+\frac {7 b^{5/4} (11 b B-9 A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{30 c^{15/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.11, size = 110, normalized size = 0.29 \begin {gather*} \frac {2 x^{5/2} \left (77 b^2 B+c^2 x^2 \left (9 A+5 B x^2\right )-b c \left (63 A+11 B x^2\right )+7 b (-11 b B+9 A c) \sqrt {1+\frac {c x^2}{b}} \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\frac {c x^2}{b}\right )\right )}{45 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(2*x^(5/2)*(77*b^2*B + c^2*x^2*(9*A + 5*B*x^2) - b*c*(63*A + 11*B*x^2) + 7*b*(-11*b*B + 9*A*c)*Sqrt[1 + (c*x^2
)/b]*Hypergeometric2F1[3/4, 3/2, 7/4, -((c*x^2)/b)]))/(45*c^3*Sqrt[x^2*(b + c*x^2)])

________________________________________________________________________________________

Maple [A]
time = 0.43, size = 420, normalized size = 1.11

method	result	size
default	\(-\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (-20 B \,c^{3} x^{6}+378 A \,b^{2} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-189 A \,b^{2} c \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-462 B \,b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+231 B \,b^{3} \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )-36 A \,c^{3} x^{4}+44 B b \,c^{2} x^{4}-126 A b \,c^{2} x^{2}+154 B \,b^{2} c \,x^{2}\right )}{90 \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} c^{4}}\)	\(420\)
risch	\(\frac {2 x^{\frac {5}{2}} \left (5 B c \,x^{2}+9 A c -16 B b \right ) \left (c \,x^{2}+b \right )}{45 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}-\frac {b \left (\frac {\left (24 A c -31 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{c \sqrt {c \,x^{3}+b x}}-15 b \left (A c -B b \right ) \left (\frac {x^{2}}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}-\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(440\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/90/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(-20*B*c^3*x^6+378*A*b^2*c*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)
*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*
c)^(1/2))^(1/2),1/2*2^(1/2))-189*A*b^2*c*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/
(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-4
62*B*b^3*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)
^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))+231*B*b^3*((c*x+(-b*c)^(1/2))/(-b
*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-
b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-36*A*c^3*x^4+44*B*b*c^2*x^4-126*A*b*c^2*x^2+154*B*b^2*c*x^2)/c^4

________________________________________________________________________________________

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(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*x^(15/2)/(c*x^4 + b*x^2)^(3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.62, size = 134, normalized size = 0.36 \begin {gather*} -\frac {21 \, {\left (11 \, B b^{3} - 9 \, A b^{2} c + {\left (11 \, B b^{2} c - 9 \, A b c^{2}\right )} x^{2}\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) - {\left (10 \, B c^{3} x^{4} - 77 \, B b^{2} c + 63 \, A b c^{2} - 2 \, {\left (11 \, B b c^{2} - 9 \, A c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{45 \, {\left (c^{5} x^{2} + b c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

-1/45*(21*(11*B*b^3 - 9*A*b^2*c + (11*B*b^2*c - 9*A*b*c^2)*x^2)*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrass
PInverse(-4*b/c, 0, x)) - (10*B*c^3*x^4 - 77*B*b^2*c + 63*A*b*c^2 - 2*(11*B*b*c^2 - 9*A*c^3)*x^2)*sqrt(c*x^4 +
b*x^2)*sqrt(x))/(c^5*x^2 + b*c^4)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(15/2)*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7771 deep

________________________________________________________________________________________

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(x^(15/2)*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*x^(15/2)/(c*x^4 + b*x^2)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^{15/2}\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x)

[Out]

int((x^(15/2)*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2), x)

________________________________________________________________________________________