3.1.89 \(\int x^7 (A+B x^2) \sqrt {b x^2+c x^4} \, dx\) [89]

Optimal. Leaf size=218 \[ \frac {7 b^3 (3 b B-4 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{384 c^4}+\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}-\frac {7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{11/2}} \]

[Out]

-7/384*b^2*(-4*A*c+3*B*b)*(c*x^4+b*x^2)^(3/2)/c^4+7/320*b*(-4*A*c+3*B*b)*x^2*(c*x^4+b*x^2)^(3/2)/c^3-1/40*(-4*
A*c+3*B*b)*x^4*(c*x^4+b*x^2)^(3/2)/c^2+1/12*B*x^6*(c*x^4+b*x^2)^(3/2)/c-7/1024*b^5*(-4*A*c+3*B*b)*arctanh(x^2*
c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(11/2)+7/1024*b^3*(-4*A*c+3*B*b)*(2*c*x^2+b)*(c*x^4+b*x^2)^(1/2)/c^5

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Rubi [A]
time = 0.25, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{11/2}}+\frac {7 b^3 \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4} (3 b B-4 A c)}{1024 c^5}-\frac {7 b^2 \left (b x^2+c x^4\right )^{3/2} (3 b B-4 A c)}{384 c^4}+\frac {7 b x^2 \left (b x^2+c x^4\right )^{3/2} (3 b B-4 A c)}{320 c^3}-\frac {x^4 \left (b x^2+c x^4\right )^{3/2} (3 b B-4 A c)}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

(7*b^3*(3*b*B - 4*A*c)*(b + 2*c*x^2)*Sqrt[b*x^2 + c*x^4])/(1024*c^5) - (7*b^2*(3*b*B - 4*A*c)*(b*x^2 + c*x^4)^
(3/2))/(384*c^4) + (7*b*(3*b*B - 4*A*c)*x^2*(b*x^2 + c*x^4)^(3/2))/(320*c^3) - ((3*b*B - 4*A*c)*x^4*(b*x^2 + c
*x^4)^(3/2))/(40*c^2) + (B*x^6*(b*x^2 + c*x^4)^(3/2))/(12*c) - (7*b^5*(3*b*B - 4*A*c)*ArcTanh[(Sqrt[c]*x^2)/Sq
rt[b*x^2 + c*x^4]])/(1024*c^(11/2))

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^7 \left (A+B x^2\right ) \sqrt {b x^2+c x^4} \, dx &=\frac {1}{2} \text {Subst}\left (\int x^3 (A+B x) \sqrt {b x+c x^2} \, dx,x,x^2\right )\\ &=\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}+\frac {\left (3 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right ) \text {Subst}\left (\int x^3 \sqrt {b x+c x^2} \, dx,x,x^2\right )}{12 c}\\ &=-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}+\frac {(7 b (3 b B-4 A c)) \text {Subst}\left (\int x^2 \sqrt {b x+c x^2} \, dx,x,x^2\right )}{80 c^2}\\ &=\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}-\frac {\left (7 b^2 (3 b B-4 A c)\right ) \text {Subst}\left (\int x \sqrt {b x+c x^2} \, dx,x,x^2\right )}{128 c^3}\\ &=-\frac {7 b^2 (3 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{384 c^4}+\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}+\frac {\left (7 b^3 (3 b B-4 A c)\right ) \text {Subst}\left (\int \sqrt {b x+c x^2} \, dx,x,x^2\right )}{256 c^4}\\ &=\frac {7 b^3 (3 b B-4 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{384 c^4}+\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}-\frac {\left (7 b^5 (3 b B-4 A c)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+c x^2}} \, dx,x,x^2\right )}{2048 c^5}\\ &=\frac {7 b^3 (3 b B-4 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{384 c^4}+\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}-\frac {\left (7 b^5 (3 b B-4 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 )}{1024 c^5}\\ &=\frac {7 b^3 (3 b B-4 A c) \left (b+2 c x^2\right ) \sqrt {b x^2+c x^4}}{1024 c^5}-\frac {7 b^2 (3 b B-4 A c) \left (b x^2+c x^4\right )^{3/2}}{384 c^4}+\frac {7 b (3 b B-4 A c) x^2 \left (b x^2+c x^4\right )^{3/2}}{320 c^3}-\frac {(3 b B-4 A c) x^4 \left (b x^2+c x^4\right )^{3/2}}{40 c^2}+\frac {B x^6 \left (b x^2+c x^4\right )^{3/2}}{12 c}-\frac {7 b^5 (3 b B-4 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{1024 c^{11/2}}\\ \end {align*}

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Mathematica [A]
time = 0.29, size = 187, normalized size = 0.86 \begin {gather*} \frac {x \left (\sqrt {c} x \left (b+c x^2\right ) \left (315 b^5 B-210 b^4 c \left (2 A+B x^2\right )+64 b c^4 x^6 \left (3 A+2 B x^2\right )+56 b^3 c^2 x^2 \left (5 A+3 B x^2\right )+256 c^5 x^8 \left (6 A+5 B x^2\right )-16 b^2 c^3 x^4 \left (14 A+9 B x^2\right )\right )+105 b^5 (3 b B-4 A c) \sqrt {b+c x^2} \log \left (-\sqrt {c} x+\sqrt {b+c x^2}\right )\right )}{15360 c^{11/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7*(A + B*x^2)*Sqrt[b*x^2 + c*x^4],x]

[Out]

(x*(Sqrt[c]*x*(b + c*x^2)*(315*b^5*B - 210*b^4*c*(2*A + B*x^2) + 64*b*c^4*x^6*(3*A + 2*B*x^2) + 56*b^3*c^2*x^2
*(5*A + 3*B*x^2) + 256*c^5*x^8*(6*A + 5*B*x^2) - 16*b^2*c^3*x^4*(14*A + 9*B*x^2)) + 105*b^5*(3*b*B - 4*A*c)*Sq
rt[b + c*x^2]*Log[-(Sqrt[c]*x) + Sqrt[b + c*x^2]]))/(15360*c^(11/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.41, size = 290, normalized size = 1.33

method	result	size
risch	\(-\frac {\left (-1280 B \,c^{5} x^{10}-1536 A \,c^{5} x^{8}-128 B b \,c^{4} x^{8}-192 A b \,c^{4} x^{6}+144 B \,b^{2} c^{3} x^{6}+224 A \,b^{2} c^{3} x^{4}-168 B \,b^{3} c^{2} x^{4}-280 A \,b^{3} c^{2} x^{2}+210 B \,b^{4} c \,x^{2}+420 A \,b^{4} c -315 b^{5} B \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{15360 c^{5}}+\frac {\left (\frac {7 b^{5} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) A}{256 c^{\frac {9}{2}}}-\frac {21 b^{6} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) B}{1024 c^{\frac {11}{2}}}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{x \sqrt {c \,x^{2}+b}}\)	\(207\)
default	\(\frac {\sqrt {x^{4} c +b \,x^{2}}\, \left (1280 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {9}{2}} x^{9}+1536 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {9}{2}} x^{7}-1152 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}} b \,x^{7}-1344 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {7}{2}} b \,x^{5}+1008 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} b^{2} x^{5}+1120 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {5}{2}} b^{2} x^{3}-840 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b^{3} x^{3}-840 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{\frac {3}{2}} b^{3} x +420 A \sqrt {c \,x^{2}+b}\, c^{\frac {3}{2}} b^{4} x +630 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {c}\, b^{4} x -315 B \sqrt {c \,x^{2}+b}\, \sqrt {c}\, b^{5} x +420 A \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{5} c -315 B \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+b}\right ) b^{6}\right )}{15360 x \sqrt {c \,x^{2}+b}\, c^{\frac {11}{2}}}\)	\(290\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15360*(c*x^4+b*x^2)^(1/2)*(1280*B*(c*x^2+b)^(3/2)*c^(9/2)*x^9+1536*A*(c*x^2+b)^(3/2)*c^(9/2)*x^7-1152*B*(c*x
^2+b)^(3/2)*c^(7/2)*b*x^7-1344*A*(c*x^2+b)^(3/2)*c^(7/2)*b*x^5+1008*B*(c*x^2+b)^(3/2)*c^(5/2)*b^2*x^5+1120*A*(
c*x^2+b)^(3/2)*c^(5/2)*b^2*x^3-840*B*(c*x^2+b)^(3/2)*c^(3/2)*b^3*x^3-840*A*(c*x^2+b)^(3/2)*c^(3/2)*b^3*x+420*A
*(c*x^2+b)^(1/2)*c^(3/2)*b^4*x+630*B*(c*x^2+b)^(3/2)*c^(1/2)*b^4*x-315*B*(c*x^2+b)^(1/2)*c^(1/2)*b^5*x+420*A*l
n(c^(1/2)*x+(c*x^2+b)^(1/2))*b^5*c-315*B*ln(c^(1/2)*x+(c*x^2+b)^(1/2))*b^6)/x/(c*x^2+b)^(1/2)/c^(11/2)

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Maxima [A]
time = 0.30, size = 321, normalized size = 1.47 \begin {gather*} \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 )} A + \frac {1}{30720} \, {\left (\frac {2560 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{6}}{c} - \frac {2304 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b x^{4}}{c^{2}} + \frac {1260 \, \sqrt {c x^{4} + b x^{2}} b^{4} x^{2}}{c^{4}} + \frac {2016 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{2} x^{2}}{c^{3}} - \frac {315 \, b^{6} \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{c^{\frac {11}{2}}} + \frac {630 \, \sqrt {c x^{4} + b x^{2}} b^{5}}{c^{5}} - \frac {1680 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")

[Out]

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 + 5
60*(c*x^4 + b*x^2)^(3/2)*b^2/c^3)*A + 1/30720*(2560*(c*x^4 + b*x^2)^(3/2)*x^6/c - 2304*(c*x^4 + b*x^2)^(3/2)*b
*x^4/c^2 + 1260*sqrt(c*x^4 + b*x^2)*b^4*x^2/c^4 + 2016*(c*x^4 + b*x^2)^(3/2)*b^2*x^2/c^3 - 315*b^6*log(2*c*x^2
+ b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(11/2) + 630*sqrt(c*x^4 + b*x^2)*b^5/c^5 - 1680*(c*x^4 + b*x^2)^(3/2)*
b^3/c^4)*B

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Fricas [A]
time = 1.92, size = 368, normalized size = 1.69 \begin {gather*} \left [-\frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) - 2 \, {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{8} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{6} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30720 \, c^{6}}, \frac {105 \, {\left (3 \, B b^{6} - 4 \, A b^{5} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + {\left (1280 \, B c^{6} x^{10} + 128 \, {\left (B b c^{5} + 12 \, A c^{6}\right )} x^{8} + 315 \, B b^{5} c - 420 \, A b^{4} c^{2} - 48 \, {\left (3 \, B b^{2} c^{4} - 4 \, A b c^{5}\right )} x^{6} + 56 \, {\left (3 \, B b^{3} c^{3} - 4 \, A b^{2} c^{4}\right )} x^{4} - 70 \, {\left (3 \, B b^{4} c^{2} - 4 \, A b^{3} c^{3}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15360 \, c^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/30720*(105*(3*B*b^6 - 4*A*b^5*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(1280*B*c^6
*x^10 + 128*(B*b*c^5 + 12*A*c^6)*x^8 + 315*B*b^5*c - 420*A*b^4*c^2 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^6 + 56*(3*
B*b^3*c^3 - 4*A*b^2*c^4)*x^4 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^6, 1/15360*(105*(3*B
*b^6 - 4*A*b^5*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b)) + (1280*B*c^6*x^10 + 128*(B*b*c^5
+ 12*A*c^6)*x^8 + 315*B*b^5*c - 420*A*b^4*c^2 - 48*(3*B*b^2*c^4 - 4*A*b*c^5)*x^6 + 56*(3*B*b^3*c^3 - 4*A*b^2*c
^4)*x^4 - 70*(3*B*b^4*c^2 - 4*A*b^3*c^3)*x^2)*sqrt(c*x^4 + b*x^2))/c^6]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{7} \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**7*(B*x**2+A)*(c*x**4+b*x**2)**(1/2),x)

[Out]

Integral(x**7*sqrt(x**2*(b + c*x**2))*(A + B*x**2), x)

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Giac [A]
time = 1.04, size = 245, normalized size = 1.12 \begin {gather*} \frac {1}{15360} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (10 \, B x^{2} \mathrm {sgn}\left (x\right ) + \frac {B b c^{9} \mathrm {sgn}\left (x\right ) + 12 \, A c^{10} \mathrm {sgn}\left (x\right )}{c^{10}}\right )} x^{2} - \frac {3 \, {\left (3 \, B b^{2} c^{8} \mathrm {sgn}\left (x\right ) - 4 \, A b c^{9} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} + \frac {7 \, {\left (3 \, B b^{3} c^{7} \mathrm {sgn}\left (x\right ) - 4 \, A b^{2} c^{8} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} - \frac {35 \, {\left (3 \, B b^{4} c^{6} \mathrm {sgn}\left (x\right ) - 4 \, A b^{3} c^{7} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} x^{2} + \frac {105 \, {\left (3 \, B b^{5} c^{5} \mathrm {sgn}\left (x\right ) - 4 \, A b^{4} c^{6} \mathrm {sgn}\left (x\right )\right )}}{c^{10}}\right )} \sqrt {c x^{2} + b} x + \frac {7 \, {\left (3 \, B b^{6} \mathrm {sgn}\left (x\right ) - 4 \, A b^{5} c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{1024 \, c^{\frac {11}{2}}} - \frac {7 \, {\left (3 \, B b^{6} \log \left ({\left | b \right |}\right ) - 4 \, A b^{5} c \log \left ({\left | b \right |}\right )\right )} \mathrm {sgn}\left (x\right )}{2048 \, c^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^2+A)*(c*x^4+b*x^2)^(1/2),x, algorithm="giac")

[Out]

1/15360*(2*(4*(2*(8*(10*B*x^2*sgn(x) + (B*b*c^9*sgn(x) + 12*A*c^10*sgn(x))/c^10)*x^2 - 3*(3*B*b^2*c^8*sgn(x) -
4*A*b*c^9*sgn(x))/c^10)*x^2 + 7*(3*B*b^3*c^7*sgn(x) - 4*A*b^2*c^8*sgn(x))/c^10)*x^2 - 35*(3*B*b^4*c^6*sgn(x)
- 4*A*b^3*c^7*sgn(x))/c^10)*x^2 + 105*(3*B*b^5*c^5*sgn(x) - 4*A*b^4*c^6*sgn(x))/c^10)*sqrt(c*x^2 + b)*x + 7/10
24*(3*B*b^6*sgn(x) - 4*A*b^5*c*sgn(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(11/2) - 7/2048*(3*B*b^6*log(a
bs(b)) - 4*A*b^5*c*log(abs(b)))*sgn(x)/c^(11/2)

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Mupad [B]
time = 1.47, size = 289, normalized size = 1.33 \begin {gather*} \frac {A\,x^4\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{10\,c}+\frac {B\,x^6\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{12\,c}-\frac {3\,B\,b\,\left (\frac {7\,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 )}{10\,c}+\frac {x^4\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{5\,c}\right )}{8\,c}+\frac {7\,A\,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^7*(A + B*x^2)*(b*x^2 + c*x^4)^(1/2),x)

[Out]

(A*x^4*(b*x^2 + c*x^4)^(3/2))/(10*c) + (B*x^6*(b*x^2 + c*x^4)^(3/2))/(12*c) - (3*B*b*((7*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)))/(10*c) + (x^4*(b*x^2 + c*x^4)^(3/2))/(5*c)))/(
8*c) + (7*A*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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