3.920 \(\int x^7 \sqrt{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=171 \[ \frac{\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}-\frac{b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^4}+\frac{b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{9/2}}+\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c} \]

[Out]

-(b*(7*b^2 - 12*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*c^4) + (x^4*(a + b*x^2 + c*x^4)^(3/2))/(10*c)
+ ((35*b^2 - 32*a*c - 42*b*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(480*c^3) + (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*Ar
cTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(512*c^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.155271, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{\left (-32 a c+35 b^2-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}-\frac{b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^4}+\frac{b \left (7 b^2-12 a c\right ) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{512 c^{9/2}}+\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

Int[x^7*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

-(b*(7*b^2 - 12*a*c)*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(256*c^4) + (x^4*(a + b*x^2 + c*x^4)^(3/2))/(10*c)
+ ((35*b^2 - 32*a*c - 42*b*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(480*c^3) + (b*(7*b^2 - 12*a*c)*(b^2 - 4*a*c)*Ar
cTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(512*c^(9/2))

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 \sqrt{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^3 \sqrt{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac{\operatorname{Subst}\left (\int x \left (-2 a-\frac{7 b x}{2}\right ) \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{10 c}\\ &=\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac{\left (35 b^2-32 a c-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}-\frac{\left (b \left (7 b^2-12 a c\right )\right ) \operatorname{Subst}\left (\int \sqrt{a+b x+c x^2} \, dx,x,x^2\right )}{64 c^3}\\ &=-\frac{b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac{\left (35 b^2-32 a c-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}+\frac{\left (b \left (7 b^2-12 a c\right ) \left (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 )}{512 c^4}\\ &=-\frac{b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac{\left (35 b^2-32 a c-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}+\frac{\left (b \left (7 b^2-12 a c\right ) \left (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 )}{256 c^4}\\ &=-\frac{b \left (7 b^2-12 a c\right ) \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}}{256 c^4}+\frac{x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c}+\frac{\left (35 b^2-32 a c-42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{480 c^3}+\frac{b \left (7 b^2-12 a c\right ) \left (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 )}{512 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.149221, size = 164, normalized size = 0.96 \[ \frac{-\frac{\left (32 a c-35 b^2+42 b c x^2\right ) \left (a+b x^2+c x^4\right )^{3/2}}{48 c^2}+\frac{5 \left (12 a b c-7 b^3\right ) \left (2 \sqrt{c} \left (b+2 c x^2\right ) \sqrt{a+b x^2+c x^4}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )\right )}{256 c^{7/2}}+x^4 \left (a+b x^2+c x^4\right )^{3/2}}{10 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7*Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(x^4*(a + b*x^2 + c*x^4)^(3/2) - ((-35*b^2 + 32*a*c + 42*b*c*x^2)*(a + b*x^2 + c*x^4)^(3/2))/(48*c^2) + (5*(-7
*b^3 + 12*a*b*c)*(2*Sqrt[c]*(b + 2*c*x^2)*Sqrt[a + b*x^2 + c*x^4] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqr
t[c]*Sqrt[a + b*x^2 + c*x^4])]))/(256*c^(7/2)))/(10*c)

________________________________________________________________________________________

Maple [A]  time = 0.175, size = 296, normalized size = 1.7 \begin{align*}{\frac{{x}^{4}}{10\,c} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{7\,b{x}^{2}}{80\,{c}^{2}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{b}^{2}}{96\,{c}^{3}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{3}{x}^{2}}{128\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{7\,{b}^{4}}{256\,{c}^{4}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{5\,a{b}^{3}}{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{7}{2}}}}+{\frac{7\,{b}^{5}}{512}\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{3\,ab{x}^{2}}{32\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,{b}^{2}a}{64\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{3\,b{a}^{2}}{32}\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{a}{15\,{c}^{2}} \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(c*x^4+b*x^2+a)^(1/2),x)

[Out]

1/10*x^4*(c*x^4+b*x^2+a)^(3/2)/c-7/80*b/c^2*x^2*(c*x^4+b*x^2+a)^(3/2)+7/96*b^2/c^3*(c*x^4+b*x^2+a)^(3/2)-7/128
*b^3/c^3*(c*x^4+b*x^2+a)^(1/2)*x^2-7/256*b^4/c^4*(c*x^4+b*x^2+a)^(1/2)-5/64*b^3/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/
2)+(c*x^4+b*x^2+a)^(1/2))*a+7/512*b^5/c^(9/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+3/32*b/c^2*a*(c*
x^4+b*x^2+a)^(1/2)*x^2+3/64*b^2/c^3*a*(c*x^4+b*x^2+a)^(1/2)+3/32*b/c^(5/2)*a^2*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4
+b*x^2+a)^(1/2))-1/15*a/c^2*(c*x^4+b*x^2+a)^(3/2)

________________________________________________________________________________________

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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.64125, size = 859, normalized size = 5.02 \begin{align*} \left [\frac{15 \,{\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\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 (384 \, c^{5} x^{8} + 48 \, b c^{4} x^{6} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} - 8 \,{\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{4} + 2 \,{\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{15360 \, c^{5}}, -\frac{15 \,{\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\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 (384 \, c^{5} x^{8} + 48 \, b c^{4} x^{6} - 105 \, b^{4} c + 460 \, a b^{2} c^{2} - 256 \, a^{2} c^{3} - 8 \,{\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} x^{4} + 2 \,{\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} x^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{7680 \, c^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15360*(15*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*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*(384*c^5*x^8 + 48*b*c^4*x^6 - 105*b^4*c + 460*a*b^2*c^2 - 256*a^2*c
^3 - 8*(7*b^2*c^3 - 16*a*c^4)*x^4 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^5, -1/7680*(1
5*(7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*sqrt(-c)*arctan(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*(384*c^5*x^8 + 48*b*c^4*x^6 - 105*b^4*c + 460*a*b^2*c^2 - 256*a^2*c^3 - 8*(7*b^2*c^3
- 16*a*c^4)*x^4 + 2*(35*b^3*c^2 - 116*a*b*c^3)*x^2)*sqrt(c*x^4 + b*x^2 + a))/c^5]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(c*x**4+b*x**2+a)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.38191, size = 248, normalized size = 1.45 \begin{align*} \frac{1}{3840} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, x^{2} + \frac{b}{c}\right )} x^{2} - \frac{7 \, b^{2} c^{5} - 16 \, a c^{6}}{c^{7}}\right )} x^{2} + \frac{35 \, b^{3} c^{4} - 116 \, a b c^{5}}{c^{7}}\right )} x^{2} - \frac{105 \, b^{4} c^{3} - 460 \, a b^{2} c^{4} + 256 \, a^{2} c^{5}}{c^{7}}\right )} - \frac{{\left (7 \, b^{5} c^{3} - 40 \, a b^{3} c^{4} + 48 \, a^{2} b c^{5}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{512 \, c^{\frac{15}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3840*sqrt(c*x^4 + b*x^2 + a)*(2*(4*(6*(8*x^2 + b/c)*x^2 - (7*b^2*c^5 - 16*a*c^6)/c^7)*x^2 + (35*b^3*c^4 - 11
6*a*b*c^5)/c^7)*x^2 - (105*b^4*c^3 - 460*a*b^2*c^4 + 256*a^2*c^5)/c^7) - 1/512*(7*b^5*c^3 - 40*a*b^3*c^4 + 48*
a^2*b*c^5)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(15/2)