3.3.29 \(\int \frac {(A+B x^2) \sqrt {b x^2+c x^4}}{x^{13/2}} \, dx\) [229]

Optimal. Leaf size=369 \[ \frac {4 c^{3/2} (3 b B-A c) x^{3/2} \left (b+c x^2\right )}{15 b^2 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {4 c (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b^2 x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}-\frac {4 c^{5/4} (3 b B-A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {b x^2+c x^4}}+\frac {2 c^{5/4} (3 b B-A c) x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {b x^2+c x^4}} \]

[Out]

-2/9*A*(c*x^4+b*x^2)^(3/2)/b/x^(15/2)+4/15*c^(3/2)*(-A*c+3*B*b)*x^(3/2)*(c*x^2+b)/b^2/(b^(1/2)+x*c^(1/2))/(c*x
^4+b*x^2)^(1/2)-2/15*(-A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^(7/2)-4/15*c*(-A*c+3*B*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^
(3/2)-4/15*c^(5/4)*(-A*c+3*B*b)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(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)/b^(7/4)/(c*x^4+b*x^2)^(1/2)+2/15*c^(5/4)*(-A*c+3*B*b)*x*(cos(2*arctan(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)/b^(7/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 369, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2063, 2045, 2050, 2057, 335, 311, 226, 1210} \begin {gather*} \frac {2 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-A c) F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {b x^2+c x^4}}-\frac {4 c^{5/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} (3 b B-A c) E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{7/4} \sqrt {b x^2+c x^4}}+\frac {4 c^{3/2} x^{3/2} \left (b+c x^2\right ) (3 b B-A c)}{15 b^2 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {4 c \sqrt {b x^2+c x^4} (3 b B-A c)}{15 b^2 x^{3/2}}-\frac {2 \sqrt {b x^2+c x^4} (3 b B-A c)}{15 b x^{7/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(13/2),x]

[Out]

(4*c^(3/2)*(3*b*B - A*c)*x^(3/2)*(b + c*x^2))/(15*b^2*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) - (2*(3*b*B -
A*c)*Sqrt[b*x^2 + c*x^4])/(15*b*x^(7/2)) - (4*c*(3*b*B - A*c)*Sqrt[b*x^2 + c*x^4])/(15*b^2*x^(3/2)) - (2*A*(b
*x^2 + c*x^4)^(3/2))/(9*b*x^(15/2)) - (4*c^(5/4)*(3*b*B - 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*b^(7/4)*Sqrt[b*x^2 + c*x^4]) + (2*
c^(5/4)*(3*b*B - 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])/(15*b^(7/4)*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 2045

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

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*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]) && LtQ[m + j*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 2063

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Dist[(a*d*(m + j*p + 1
) - b*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)), Int[(e*x)^(m + n)*(a*x^j + b*x^(j + n))^p, x], x] /; F
reeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m
+ j*p, -1] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && (GtQ[e, 0] || IntegersQ[j,
n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x^2\right ) \sqrt {b x^2+c x^4}}{x^{13/2}} \, dx &=-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}-\frac {\left (2 \left (-\frac {9 b B}{2}+\frac {3 A c}{2}\right )\right ) \int \frac {\sqrt {b x^2+c x^4}}{x^{9/2}} \, dx}{9 b}\\ &=-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}+\frac {(2 c (3 b B-A c)) \int \frac {1}{\sqrt {x} \sqrt {b x^2+c x^4}} \, dx}{15 b}\\ &=-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {4 c (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b^2 x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}+\frac {\left (2 c^2 (3 b B-A c)\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{15 b^2}\\ &=-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {4 c (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b^2 x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}+\frac {\left (2 c^2 (3 b B-A c) x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{15 b^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {4 c (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b^2 x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}+\frac {\left (4 c^2 (3 b B-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 b^2 \sqrt {b x^2+c x^4}}\\ &=-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {4 c (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b^2 x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}+\frac {\left (4 c^{3/2} (3 b B-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 b^{3/2} \sqrt {b x^2+c x^4}}-\frac {\left (4 c^{3/2} (3 b B-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 b^{3/2} \sqrt {b x^2+c x^4}}\\ &=\frac {4 c^{3/2} (3 b B-A c) x^{3/2} \left (b+c x^2\right )}{15 b^2 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {2 (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b x^{7/2}}-\frac {4 c (3 b B-A c) \sqrt {b x^2+c x^4}}{15 b^2 x^{3/2}}-\frac {2 A \left (b x^2+c x^4\right )^{3/2}}{9 b x^{15/2}}-\frac {4 c^{5/4} (3 b B-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 b^{7/4} \sqrt {b x^2+c x^4}}+\frac {2 c^{5/4} (3 b B-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 )}{15 b^{7/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 99, normalized size = 0.27 \begin {gather*} -\frac {2 \sqrt {x^2 \left (b+c x^2\right )} \left (5 A \left (b+c x^2\right ) \sqrt {1+\frac {c x^2}{b}}+3 (3 b B-A c) x^2 \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {c x^2}{b}\right )\right )}{45 b x^{11/2} \sqrt {1+\frac {c x^2}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*Sqrt[b*x^2 + c*x^4])/x^(13/2),x]

[Out]

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

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Maple [A]
time = 0.39, size = 452, normalized size = 1.22

method	result	size
risch	\(-\frac {2 \left (-6 A \,c^{2} x^{4}+18 x^{4} b B c +2 A b c \,x^{2}+9 b^{2} B \,x^{2}+5 b^{2} A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{45 x^{\frac {11}{2}} b^{2}}-\frac {2 c \left (A c -3 B b \right ) \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 b^{2} \sqrt {c \,x^{3}+b x}\, x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}\)	\(267\)
default	\(-\frac {2 \sqrt {x^{4} c +b \,x^{2}}\, \left (6 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-3 A \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-18 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c \,x^{4}+9 B \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b^{2} c \,x^{4}-6 A \,c^{3} x^{6}+18 x^{6} B b \,c^{2}-4 A b \,c^{2} x^{4}+27 x^{4} B \,b^{2} c +7 A \,b^{2} c \,x^{2}+9 x^{2} B \,b^{3}+5 A \,b^{3}\right )}{45 x^{\frac {11}{2}} \left (c \,x^{2}+b \right ) b^{2}}\)	\(452\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(13/2),x,method=_RETURNVERBOSE)

[Out]

-2/45*(c*x^4+b*x^2)^(1/2)/x^(11/2)/(c*x^2+b)*(6*A*((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))*b*c^2*x^4-3*A*((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))*b*c^2*x^4-18*B*((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))*b^2*c*x^4+9*B*((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))*b^2*c*x^4-6*A*c^3*x^6+18*x^6*B*b*c^2-4*A*b*c^2*x^4+27*x^4*B*b^2*c+7*A*b^2*c
*x^2+9*x^2*B*b^3+5*A*b^3)/b^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)*(c*x^4+b*x^2)^(1/2)/x^(13/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(13/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.55, size = 103, normalized size = 0.28 \begin {gather*} -\frac {2 \, {\left (6 \, {\left (3 \, B b c - A c^{2}\right )} \sqrt {c} x^{6} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (6 \, {\left (3 \, B b c - A c^{2}\right )} x^{4} + 5 \, A b^{2} + {\left (9 \, B b^{2} + 2 \, A b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}\right )}}{45 \, b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x^(13/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )}{x^{\frac {13}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x**(13/2),x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x**(13/2), x)

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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)*(c*x^4+b*x^2)^(1/2)/x^(13/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2)*(B*x^2 + A)/x^(13/2), 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 {c\,x^4+b\,x^2}}{x^{13/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(13/2),x)

[Out]

int(((A + B*x^2)*(b*x^2 + c*x^4)^(1/2))/x^(13/2), x)

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