3.1.90 \(\int x^5 (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [90]

Optimal. Leaf size=181 \[ -\frac {b^2 (7 b B-10 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^4}+\frac {b (7 b B-10 A c) \left (b x^2+c x^4\right )^{3/2}}{96 c^3}-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {b^4 (7 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{9/2}} \]

[Out]

1/96*b*(-10*A*c+7*B*b)*(c*x^4+b*x^2)^(3/2)/c^3-1/80*(-10*A*c+7*B*b)*x^2*(c*x^4+b*x^2)^(3/2)/c^2+1/10*B*x^4*(c*
x^4+b*x^2)^(3/2)/c+1/256*b^4*(-10*A*c+7*B*b)*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(9/2)-1/256*b^2*(-10*A
*c+7*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^4

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Rubi [A]
time = 0.22, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {2059, 808, 684, 654, 626, 634, 212} \begin {gather*} \frac {b^4 (7 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{9/2}}-\frac {b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (7 b B-10 A c)}{256 c^4}+\frac {b \left (b x^2+c x^4\right )^{3/2} (7 b B-10 A c)}{96 c^3}-\frac {x^2 \left (b x^2+c x^4\right )^{3/2} (7 b B-10 A c)}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

-1/256*(b^2*(7*b*B - 10*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/c^4 + (b*(7*b*B - 10*A*c)*(b*x^2 + c*x^4)^(3/2
))/(96*c^3) - ((7*b*B - 10*A*c)*x^2*(b*x^2 + c*x^4)^(3/2))/(80*c^2) + (B*x^4*(b*x^2 + c*x^4)^(3/2))/(10*c) + (
b^4*(7*b*B - 10*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(256*c^(9/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 808

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 2059

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int x^5 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \text {Subst}\left (\int x^2 (A+B x) \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {\left (2 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \text {Subst}\left (\int x^2 \sqrt {b x+c x^2} \, dx,x,x^2\right )}{10 c}\\ &=-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {(b (7 b B-10 A c)) \text {Subst}\left (\int x \sqrt {b x+c x^2} \, dx,x,x^2\right )}{32 c^2}\\ &=\frac {b (7 b B-10 A c) \left (b x^2+c x^4\right )^{3/2}}{96 c^3}-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}-\frac {\left (b^2 (7 b B-10 A c)\right ) \text {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{64 c^3}\\ &=-\frac {b^2 (7 b B-10 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^4}+\frac {b (7 b B-10 A c) \left (b x^2+c x^4\right )^{3/2}}{96 c^3}-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {\left (b^4 (7 b B-10 A c)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{512 c^4}\\ &=-\frac {b^2 (7 b B-10 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^4}+\frac {b (7 b B-10 A c) \left (b x^2+c x^4\right )^{3/2}}{96 c^3}-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {\left (b^4 (7 b B-10 A c)\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^4}\\ &=-\frac {b^2 (7 b B-10 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{256 c^4}+\frac {b (7 b B-10 A c) \left (b x^2+c x^4\right )^{3/2}}{96 c^3}-\frac {(7 b B-10 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c}+\frac {b^4 (7 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 168, normalized size = 0.93 \begin {gather*} \frac {x \left (-\sqrt {c} x \left (b+c x^2\right ) \left (105 b^4 B-16 b c^3 x^4 \left (5 A+3 B x^2\right )-96 c^4 x^6 \left (5 A+4 B x^2\right )-10 b^3 c \left (15 A+7 B x^2\right )+4 b^2 c^2 x^2 \left (25 A+14 B x^2\right )\right )-15 b^4 (7 b B-10 A c) \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{3840 c^{9/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*(-(Sqrt[c]*x*(b + c*x^2)*(105*b^4*B - 16*b*c^3*x^4*(5*A + 3*B*x^2) - 96*c^4*x^6*(5*A + 4*B*x^2) - 10*b^3*c*
(15*A + 7*B*x^2) + 4*b^2*c^2*x^2*(25*A + 14*B*x^2))) - 15*b^4*(7*b*B - 10*A*c)*Sqrt[b + c*x^2]*Log[-(Sqrt[c]*x
) + Sqrt[b + c*x^2]]))/(3840*c^(9/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.38, size = 248, normalized size = 1.37

method	result	size
risch	\(\frac {\left (384 B \,c^{4} x^{8}+480 A \,c^{4} x^{6}+48 B b \,c^{3} x^{6}+80 A b \,c^{3} x^{4}-56 B \,b^{2} c^{2} x^{4}-100 A \,b^{2} c^{2} x^{2}+70 B \,b^{3} c \,x^{2}+150 A \,b^{3} c -105 B \,b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{3840 c^{4}}+\frac {\left (-\frac {5 b^{4} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) A}{128 c^{\frac {7}{2}}}+\frac {7 b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) B}{256 c^{\frac {9}{2}}}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(183\)
default	\(\frac {\sqrt {x^{4} c +b \,x^{2}}\, \left (384 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}} x^{7}+480 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}} x^{5}-336 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} b \,x^{5}-400 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} b \,x^{3}+280 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b^{2} x^{3}+300 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b^{2} x -210 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b^{3} x -150 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{3} x +105 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{4} x -150 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{4} c +105 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5}\right )}{3840 x \sqrt {c \,x^{2}+b}\, c^{\frac {9}{2}}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3840*(c*x^4+b*x^2)^(1/2)*(384*B*(c*x^2+b)^(3/2)*c^(7/2)*x^7+480*A*(c*x^2+b)^(3/2)*c^(7/2)*x^5-336*B*(c*x^2+b
)^(3/2)*c^(5/2)*b*x^5-400*A*(c*x^2+b)^(3/2)*c^(5/2)*b*x^3+280*B*(c*x^2+b)^(3/2)*c^(3/2)*b^2*x^3+300*A*(c*x^2+b
)^(3/2)*c^(3/2)*b^2*x-210*B*(c*x^2+b)^(3/2)*c^(1/2)*b^3*x-150*A*(c*x^2+b)^(1/2)*c^(3/2)*b^3*x+105*B*(c*x^2+b)^
(1/2)*c^(1/2)*b^4*x-150*A*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^4*c+105*B*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^5)/x/(c*x^
2+b)^(1/2)/c^(9/2)

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Maxima [A]
time = 0.28, size = 273, normalized size = 1.51 \begin {gather*} \frac {1}{768} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c^{2}} + \frac {96 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{c} - \frac {15 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c^{2}}\right )} A + \frac {1}{7680} \, {\left (\frac {768 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{4}}{c} - \frac {420 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{3}} - \frac {672 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c^{2}} + \frac {105 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} - \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{4}} + \frac {560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/768*(60*sqrt(c*x^4 + b*x^2)*b^2*x^2/c^2 + 96*(c*x^4 + b*x^2)^(3/2)*x^2/c - 15*b^4*log(2*c*x^2 + b + 2*sqrt(c
*x^4 + b*x^2)*sqrt(c))/c^(7/2) + 30*sqrt(c*x^4 + b*x^2)*b^3/c^3 - 80*(c*x^4 + b*x^2)^(3/2)*b/c^2)*A + 1/7680*(
768*(c*x^4 + b*x^2)^(3/2)*x^4/c - 420*sqrt(c*x^4 + b*x^2)*b^3*x^2/c^3 - 672*(c*x^4 + b*x^2)^(3/2)*b*x^2/c^2 +
105*b^5*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(9/2) - 210*sqrt(c*x^4 + b*x^2)*b^4/c^4 + 560*(c*x^
4 + b*x^2)^(3/2)*b^2/c^3)*B

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Fricas [A]
time = 2.72, size = 321, normalized size = 1.77 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{8} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{6} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{7680 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (384 \, B c^{5} x^{8} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{6} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3840 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/7680*(15*(7*B*b^5 - 10*A*b^4*c)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(384*B*c^5*x
^8 + 48*(B*b*c^4 + 10*A*c^5)*x^6 - 105*B*b^4*c + 150*A*b^3*c^2 - 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 10*(7*B*b^
3*c^2 - 10*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^5, -1/3840*(15*(7*B*b^5 - 10*A*b^4*c)*sqrt(-c)*arctan(sqrt(c
*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) - (384*B*c^5*x^8 + 48*(B*b*c^4 + 10*A*c^5)*x^6 - 105*B*b^4*c + 150*A*b^3*c
^2 - 8*(7*B*b^2*c^3 - 10*A*b*c^4)*x^4 + 10*(7*B*b^3*c^2 - 10*A*b^2*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^5]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{5} \sqrt {x^{2} \left (b + c x^{2}\right )} \left (A + B x^{2}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**5*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)

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Giac [A]
time = 0.77, size = 211, normalized size = 1.17 \begin {gather*} \frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c^{7} \mathrm {sgn}\left (x\right ) + 10 \, A c^{8} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} - \frac {7 \, B b^{2} c^{6} \mathrm {sgn}\left (x\right ) - 10 \, A b c^{7} \mathrm {sgn}\left (x\right )}{c^{8}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{3} c^{5} \mathrm {sgn}\left (x\right ) - 10 \, A b^{2} c^{6} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{4} c^{4} \mathrm {sgn}\left (x\right ) - 10 \, A b^{3} c^{5} \mathrm {sgn}\left (x\right )\right )}}{c^{8}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (7 \, B b^{5} \mathrm {sgn}\left (x\right ) - 10 \, A b^{4} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{256 \, c^{\frac {9}{2}}} + \frac {{\left (7 \, B b^{5} \log \left ({\left | b \right |}\right ) - 10 \, A b^{4} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{512 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/3840*(2*(4*(6*(8*B*x^2*sgn(x) + (B*b*c^7*sgn(x) + 10*A*c^8*sgn(x))/c^8)*x^2 - (7*B*b^2*c^6*sgn(x) - 10*A*b*c
^7*sgn(x))/c^8)*x^2 + 5*(7*B*b^3*c^5*sgn(x) - 10*A*b^2*c^6*sgn(x))/c^8)*x^2 - 15*(7*B*b^4*c^4*sgn(x) - 10*A*b^
3*c^5*sgn(x))/c^8)*sqrt(c*x^2 + b)*x - 1/256*(7*B*b^5*sgn(x) - 10*A*b^4*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*
x^2 + b)))/c^(9/2) + 1/512*(7*B*b^5*log(abs(b)) - 10*A*b^4*c*log(abs(b)))*sgn(x)/c^(9/2)

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Mupad [B]
time = 0.89, size = 233, normalized size = 1.29 \begin {gather*} \frac {A\,x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{8\,c}-\frac {5\,A\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{16\,c}+\frac {B\,x^4\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{10\,c}+\frac {7\,B\,b\,\left (\frac {5\,b\,\left (\frac {b^3\,\ln \left (b+2\,c\,x^2+2\,\sqrt {c}\,\left |x\right |\,\sqrt {c\,x^2+b}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^4+b\,x^2}\,\left (-3\,b^2+2\,b\,c\,x^2+8\,c^2\,x^4\right )}{24\,c^2}\right )}{8\,c}-\frac {x^2\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{4\,c}\right )}{20\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)

[Out]

(A*x^2*(b*x^2 + c*x^4)^(3/2))/(8*c) - (5*A*b*((b^3*log(b + 2*c*x^2 + 2*c^(1/2)*abs(x)*(b + c*x^2)^(1/2)))/(16*
c^(5/2)) + ((b*x^2 + c*x^4)^(1/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(24*c^2)))/(16*c) + (B*x^4*(b*x^2 + c*x^4)^
(3/2))/(10*c) + (7*B*b*((5*b*((b^3*log(b + 2*c*x^2 + 2*c^(1/2)*abs(x)*(b + c*x^2)^(1/2)))/(16*c^(5/2)) + ((b*x
^2 + c*x^4)^(1/2)*(8*c^2*x^4 - 3*b^2 + 2*b*c*x^2))/(24*c^2)))/(8*c) - (x^2*(b*x^2 + c*x^4)^(3/2))/(4*c)))/(20*
c)

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