3.90 \(\int x^5 (A+B x^2) \sqrt {b x^2+c x^4} \, dx\)

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

[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 = 0.33, 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 = {2034, 794, 670, 640, 612, 620, 206} \[ -\frac {b^2 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (7 b B-10 A c)}{256 c^4}+\frac {b^4 (7 b B-10 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{256 c^{9/2}}+\frac {b \left (b x^2+c x^4\right )^{3/2} (7 b B-10 A c)}{96 c^3}-\frac {x^2 \left (b x^2+c x^4\right )^{3/2} (7 b B-10 A c)}{80 c^2}+\frac {B x^4 \left (b x^2+c x^4\right )^{3/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(7*b*B - 10*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(256*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 206

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

Rule 612

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 620

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 640

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 670

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 794

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 2034

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} \operatorname {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 ) \operatorname {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)) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.30, size = 173, normalized size = 0.96 \[ \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (15 b^{7/2} (7 b B-10 A c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )+\sqrt {c} x \sqrt {\frac {c x^2}{b}+1} \left (10 b^3 c \left (15 A+7 B x^2\right )-4 b^2 c^2 x^2 \left (25 A+14 B x^2\right )+16 b c^3 x^4 \left (5 A+3 B x^2\right )+96 c^4 x^6 \left (5 A+4 B x^2\right )-105 b^4 B\right )\right )}{3840 c^{9/2} x \sqrt {\frac {c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/b]*(-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^(7/2)*(7*b*B - 10*A*c)*A
rcSinh[(Sqrt[c]*x)/Sqrt[b]]))/(3840*c^(9/2)*x*Sqrt[1 + (c*x^2)/b])

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fricas [A]  time = 0.90, size = 321, normalized size = 1.77 \[ \left [-\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (384 \, B c^{5} x^{8} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{6} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{7680 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 10 \, A b^{4} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) - {\left (384 \, B c^{5} x^{8} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{6} - 105 \, B b^{4} c + 150 \, A b^{3} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 10 \, A b c^{4}\right )} x^{4} + 10 \, {\left (7 \, B b^{3} c^{2} - 10 \, A b^{2} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{3840 \, c^{5}}\right ] \]

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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giac [A]  time = 0.20, size = 211, normalized size = 1.17 \[ \frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x^{2} \mathrm {sgn}\relax (x) + \frac {B b c^{7} \mathrm {sgn}\relax (x) + 10 \, A c^{8} \mathrm {sgn}\relax (x)}{c^{8}}\right )} x^{2} - \frac {7 \, B b^{2} c^{6} \mathrm {sgn}\relax (x) - 10 \, A b c^{7} \mathrm {sgn}\relax (x)}{c^{8}}\right )} x^{2} + \frac {5 \, {\left (7 \, B b^{3} c^{5} \mathrm {sgn}\relax (x) - 10 \, A b^{2} c^{6} \mathrm {sgn}\relax (x)\right )}}{c^{8}}\right )} x^{2} - \frac {15 \, {\left (7 \, B b^{4} c^{4} \mathrm {sgn}\relax (x) - 10 \, A b^{3} c^{5} \mathrm {sgn}\relax (x)\right )}}{c^{8}}\right )} \sqrt {c x^{2} + b} x - \frac {{\left (7 \, B b^{5} \mathrm {sgn}\relax (x) - 10 \, A b^{4} c \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{256 \, c^{\frac {9}{2}}} + \frac {{\left (7 \, B b^{5} \log \left ({\left | b \right |}\right ) - 10 \, A b^{4} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\relax (x)}{512 \, c^{\frac {9}{2}}} \]

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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maple [A]  time = 0.06, size = 248, normalized size = 1.37 \[ \frac {\sqrt {c \,x^{4}+b \,x^{2}}\, \left (384 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,c^{\frac {7}{2}} x^{7}+480 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,c^{\frac {7}{2}} x^{5}-336 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B b \,c^{\frac {5}{2}} x^{5}-400 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A b \,c^{\frac {5}{2}} x^{3}+280 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,b^{2} c^{\frac {3}{2}} x^{3}-150 A \,b^{4} c \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )+105 B \,b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right )-150 \sqrt {c \,x^{2}+b}\, A \,b^{3} c^{\frac {3}{2}} x +105 \sqrt {c \,x^{2}+b}\, B \,b^{4} \sqrt {c}\, x +300 \left (c \,x^{2}+b \right )^{\frac {3}{2}} A \,b^{2} c^{\frac {3}{2}} x -210 \left (c \,x^{2}+b \right )^{\frac {3}{2}} B \,b^{3} \sqrt {c}\, x \right )}{3840 \sqrt {c \,x^{2}+b}\, c^{\frac {9}{2}} x} \]

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)

[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)*x^5*b-400*A*(c*x^2+b)^(3/2)*c^(5/2)*x^3*b+280*B*(c*x^2+b)^(3/2)*c^(3/2)*x^3*b^2+300*A*(c*x^2+b
)^(3/2)*c^(3/2)*x*b^2-210*B*(c*x^2+b)^(3/2)*c^(1/2)*x*b^3-150*A*(c*x^2+b)^(1/2)*c^(3/2)*x*b^3+105*B*(c*x^2+b)^
(1/2)*c^(1/2)*x*b^4-150*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 = 1.51, size = 273, normalized size = 1.51 \[ \frac {1}{768} \, {\left (\frac {60 \, \sqrt {c x^{4} + b x^{2}} b^{2} x^{2}}{c^{2}} + \frac {96 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{2}}{c} - \frac {15 \, b^{4} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {7}{2}}} + \frac {30 \, \sqrt {c x^{4} + b x^{2}} b^{3}}{c^{3}} - \frac {80 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b}{c^{2}}\right )} A + \frac {1}{7680} \, {\left (\frac {768 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{4}}{c} - \frac {420 \, \sqrt {c x^{4} + b x^{2}} b^{3} x^{2}}{c^{3}} - \frac {672 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{2}}{c^{2}} + \frac {105 \, b^{5} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {9}{2}}} - \frac {210 \, \sqrt {c x^{4} + b x^{2}} b^{4}}{c^{4}} + \frac {560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2}}{c^{3}}\right )} B \]

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

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

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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