3.936 \(\int x^7 (a+b x^2+c x^4)^{3/2} \, dx\)

Optimal. Leaf size=223 \[ \frac{\left (-16 a c+21 b^2-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac{b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac{3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{2048 c^5}-\frac{3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4096 c^{11/2}}+\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c} \]

[Out]

(3*b*(b^2 - 4*a*c)*(3*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(2048*c^5) - (b*(3*b^2 - 4*a*c)*(b +
2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(256*c^4) + (x^4*(a + b*x^2 + c*x^4)^(5/2))/(14*c) + ((21*b^2 - 16*a*c -
30*b*c*x^2)*(a + b*x^2 + c*x^4)^(5/2))/(560*c^3) - (3*b*(b^2 - 4*a*c)^2*(3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/
(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4096*c^(11/2))

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Rubi [A]  time = 0.211753, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {1114, 742, 779, 612, 621, 206} \[ \frac{\left (-16 a c+21 b^2-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac{b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac{3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{2048 c^5}-\frac{3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4096 c^{11/2}}+\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c} \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(3*b*(b^2 - 4*a*c)*(3*b^2 - 4*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(2048*c^5) - (b*(3*b^2 - 4*a*c)*(b +
2*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(256*c^4) + (x^4*(a + b*x^2 + c*x^4)^(5/2))/(14*c) + ((21*b^2 - 16*a*c -
30*b*c*x^2)*(a + b*x^2 + c*x^4)^(5/2))/(560*c^3) - (3*b*(b^2 - 4*a*c)^2*(3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/
(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4096*c^(11/2))

Rule 1114

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

Rule 742

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[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

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 621

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

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])

Rubi steps

\begin{align*} \int x^7 \left (a+b x^2+c x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^3 \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\operatorname{Subst}\left (\int x \left (-2 a-\frac{9 b x}{2}\right ) \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{14 c}\\ &=\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac{\left (b \left (3 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \left (a+b x+c x^2\right )^{3/2} \, dx,x,x^2\right )}{32 c^3}\\ &=-\frac{b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}+\frac{\left (3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{512 c^4}\\ &=\frac{3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{2048 c^5}-\frac{b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac{\left (3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{4096 c^5}\\ &=\frac{3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{2048 c^5}-\frac{b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac{\left (3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 c-x^2} \, dx,x,\frac{b+2 c x^2}{\sqrt{a+b x^2+c x^4}}\right )}{2048 c^5}\\ &=\frac{3 b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{2048 c^5}-\frac{b \left (3 b^2-4 a c\right ) \left (b+2 c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c}+\frac{\left (21 b^2-16 a c-30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{560 c^3}-\frac{3 b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{4096 c^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.251913, size = 192, normalized size = 0.86 \[ \frac{-\frac{\left (16 a c-21 b^2+30 b c x^2\right ) \left (a+b x^2+c x^4\right )^{5/2}}{40 c^2}+\frac{7 \left (4 a b c-3 b^3\right ) \left (2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4} \left (4 c \left (5 a+2 c x^4\right )-3 b^2+8 b c x^2\right )+3 \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )\right )}{2048 c^{9/2}}+x^4 \left (a+b x^2+c x^4\right )^{5/2}}{14 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^4*(a + b*x^2 + c*x^4)^(5/2) - ((-21*b^2 + 16*a*c + 30*b*c*x^2)*(a + b*x^2 + c*x^4)^(5/2))/(40*c^2) + (7*(-3
*b^3 + 4*a*b*c)*(2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4]*(-3*b^2 + 8*b*c*x^2 + 4*c*(5*a + 2*c*x^4)) +
3*(b^2 - 4*a*c)^2*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])]))/(2048*c^(9/2)))/(14*c)

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Maple [B]  time = 0.174, size = 534, normalized size = 2.4 \begin{align*}{\frac{11\,ab{x}^{6}}{1120\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{31\,a{x}^{4}{b}^{2}}{2240\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{13\,a{b}^{3}{x}^{2}}{640\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{73\,{a}^{2}b{x}^{2}}{2240\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{9\,a{b}^{4}}{256\,{c}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{{a}^{3}}{35\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{21\,a{b}^{5}}{1024}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{9}{2}}}}-{\frac{15\,{a}^{2}{b}^{3}}{256}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{49\,{b}^{2}{a}^{2}}{640\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{a}^{3}b}{64}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{4\,a{x}^{8}}{35}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{9\,{b}^{6}}{2048\,{c}^{5}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{9\,{b}^{7}}{4096}\ln \left ({ \left ({\frac{b}{2}}+c{x}^{2} \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{4}+b{x}^{2}+a} \right ){c}^{-{\frac{11}{2}}}}+{\frac{{a}^{2}{x}^{4}}{70\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{{b}^{2}{x}^{8}}{560\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{9\,{b}^{3}{x}^{6}}{4480\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{4}{x}^{4}}{1280\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{3\,{b}^{5}{x}^{2}}{1024\,{c}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{c{x}^{12}}{14}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,b{x}^{10}}{56}\sqrt{c{x}^{4}+b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(c*x^4+b*x^2+a)^(3/2),x)

[Out]

11/1120*a*b*x^6/c*(c*x^4+b*x^2+a)^(1/2)-31/2240*a*b^2/c^2*x^4*(c*x^4+b*x^2+a)^(1/2)+13/640*a*b^3/c^3*x^2*(c*x^
4+b*x^2+a)^(1/2)-73/2240*a^2*b/c^2*x^2*(c*x^4+b*x^2+a)^(1/2)-9/256*a*b^4/c^4*(c*x^4+b*x^2+a)^(1/2)-1/35*a^3/c^
2*(c*x^4+b*x^2+a)^(1/2)+21/1024*a*b^5/c^(9/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))-15/256*a^2*b^3/c
^(7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+49/640*a^2*b^2/c^3*(c*x^4+b*x^2+a)^(1/2)+3/64*a^3*b/c^(
5/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+4/35*a*x^8*(c*x^4+b*x^2+a)^(1/2)+9/2048*b^6/c^5*(c*x^4+b*
x^2+a)^(1/2)-9/4096*b^7/c^(11/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+1/70*a^2*x^4/c*(c*x^4+b*x^2+a
)^(1/2)+1/560*b^2*x^8/c*(c*x^4+b*x^2+a)^(1/2)-9/4480*b^3/c^2*x^6*(c*x^4+b*x^2+a)^(1/2)+3/1280*b^4/c^3*x^4*(c*x
^4+b*x^2+a)^(1/2)-3/1024*b^5/c^4*x^2*(c*x^4+b*x^2+a)^(1/2)+1/14*c*x^12*(c*x^4+b*x^2+a)^(1/2)+5/56*b*x^10*(c*x^
4+b*x^2+a)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.75869, size = 1265, normalized size = 5.67 \begin{align*} \left [-\frac{105 \,{\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (5120 \, c^{7} x^{12} + 6400 \, b c^{6} x^{10} + 128 \,{\left (b^{2} c^{5} + 64 \, a c^{6}\right )} x^{8} + 315 \, b^{6} c - 2520 \, a b^{4} c^{2} + 5488 \, a^{2} b^{2} c^{3} - 2048 \, a^{3} c^{4} - 16 \,{\left (9 \, b^{3} c^{4} - 44 \, a b c^{5}\right )} x^{6} + 8 \,{\left (21 \, b^{4} c^{3} - 124 \, a b^{2} c^{4} + 128 \, a^{2} c^{5}\right )} x^{4} - 2 \,{\left (105 \, b^{5} c^{2} - 728 \, a b^{3} c^{3} + 1168 \, a^{2} b c^{4}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{286720 \, c^{6}}, \frac{105 \,{\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (2 \, c x^{2} + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \,{\left (5120 \, c^{7} x^{12} + 6400 \, b c^{6} x^{10} + 128 \,{\left (b^{2} c^{5} + 64 \, a c^{6}\right )} x^{8} + 315 \, b^{6} c - 2520 \, a b^{4} c^{2} + 5488 \, a^{2} b^{2} c^{3} - 2048 \, a^{3} c^{4} - 16 \,{\left (9 \, b^{3} c^{4} - 44 \, a b c^{5}\right )} x^{6} + 8 \,{\left (21 \, b^{4} c^{3} - 124 \, a b^{2} c^{4} + 128 \, a^{2} c^{5}\right )} x^{4} - 2 \,{\left (105 \, b^{5} c^{2} - 728 \, a b^{3} c^{3} + 1168 \, a^{2} b c^{4}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{143360 \, c^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/286720*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*c^3)*sqrt(c)*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2
- 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) - 4*(5120*c^7*x^12 + 6400*b*c^6*x^10 + 128*(b^2*c^5
+ 64*a*c^6)*x^8 + 315*b^6*c - 2520*a*b^4*c^2 + 5488*a^2*b^2*c^3 - 2048*a^3*c^4 - 16*(9*b^3*c^4 - 44*a*b*c^5)*
x^6 + 8*(21*b^4*c^3 - 124*a*b^2*c^4 + 128*a^2*c^5)*x^4 - 2*(105*b^5*c^2 - 728*a*b^3*c^3 + 1168*a^2*b*c^4)*x^2)
*sqrt(c*x^4 + b*x^2 + a))/c^6, 1/143360*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*c^3)*sqrt(-c)*arc
tan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) + 2*(5120*c^7*x^12 + 6400*b*
c^6*x^10 + 128*(b^2*c^5 + 64*a*c^6)*x^8 + 315*b^6*c - 2520*a*b^4*c^2 + 5488*a^2*b^2*c^3 - 2048*a^3*c^4 - 16*(9
*b^3*c^4 - 44*a*b*c^5)*x^6 + 8*(21*b^4*c^3 - 124*a*b^2*c^4 + 128*a^2*c^5)*x^4 - 2*(105*b^5*c^2 - 728*a*b^3*c^3
+ 1168*a^2*b*c^4)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^6]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{7} \left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(x**7*(a + b*x**2 + c*x**4)**(3/2), x)

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Giac [A]  time = 1.36957, size = 374, normalized size = 1.68 \begin{align*} \frac{1}{71680} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (10 \,{\left (4 \, c x^{2} + 5 \, b\right )} x^{2} + \frac{b^{2} c^{10} + 64 \, a c^{11}}{c^{11}}\right )} x^{2} - \frac{9 \, b^{3} c^{9} - 44 \, a b c^{10}}{c^{11}}\right )} x^{2} + \frac{21 \, b^{4} c^{8} - 124 \, a b^{2} c^{9} + 128 \, a^{2} c^{10}}{c^{11}}\right )} x^{2} - \frac{105 \, b^{5} c^{7} - 728 \, a b^{3} c^{8} + 1168 \, a^{2} b c^{9}}{c^{11}}\right )} x^{2} + \frac{315 \, b^{6} c^{6} - 2520 \, a b^{4} c^{7} + 5488 \, a^{2} b^{2} c^{8} - 2048 \, a^{3} c^{9}}{c^{11}}\right )} + \frac{3 \,{\left (3 \, b^{7} c^{6} - 28 \, a b^{5} c^{7} + 80 \, a^{2} b^{3} c^{8} - 64 \, a^{3} b c^{9}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{4096 \, c^{\frac{23}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/71680*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(2*(8*(10*(4*c*x^2 + 5*b)*x^2 + (b^2*c^10 + 64*a*c^11)/c^11)*x^2 - (9*b^
3*c^9 - 44*a*b*c^10)/c^11)*x^2 + (21*b^4*c^8 - 124*a*b^2*c^9 + 128*a^2*c^10)/c^11)*x^2 - (105*b^5*c^7 - 728*a*
b^3*c^8 + 1168*a^2*b*c^9)/c^11)*x^2 + (315*b^6*c^6 - 2520*a*b^4*c^7 + 5488*a^2*b^2*c^8 - 2048*a^3*c^9)/c^11) +
3/4096*(3*b^7*c^6 - 28*a*b^5*c^7 + 80*a^2*b^3*c^8 - 64*a^3*b*c^9)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^
2 + a))*sqrt(c) - b))/c^(23/2)