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

Optimal. Leaf size=121 \[ \frac{\left (-16 a c+15 b^2-10 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{48 c^3}-\frac{b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{7/2}}+\frac{x^4 \sqrt{a+b x^2+c x^4}}{6 c} \]

[Out]

(x^4*Sqrt[a + b*x^2 + c*x^4])/(6*c) + ((15*b^2 - 16*a*c - 10*b*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(48*c^3) - (b*(
5*b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(7/2))

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Rubi [A]  time = 0.114397, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1114, 742, 779, 621, 206} \[ \frac{\left (-16 a c+15 b^2-10 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{48 c^3}-\frac{b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{7/2}}+\frac{x^4 \sqrt{a+b x^2+c x^4}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^7/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

(x^4*Sqrt[a + b*x^2 + c*x^4])/(6*c) + ((15*b^2 - 16*a*c - 10*b*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(48*c^3) - (b*(
5*b^2 - 12*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(32*c^(7/2))

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 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 \frac{x^7}{\sqrt{a+b x^2+c x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )\\ &=\frac{x^4 \sqrt{a+b x^2+c x^4}}{6 c}+\frac{\operatorname{Subst}\left (\int \frac{x \left (-2 a-\frac{5 b x}{2}\right )}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{6 c}\\ &=\frac{x^4 \sqrt{a+b x^2+c x^4}}{6 c}+\frac{\left (15 b^2-16 a c-10 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{48 c^3}-\frac{\left (b \left (5 b^2-12 a c\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x+c x^2}} \, dx,x,x^2\right )}{32 c^3}\\ &=\frac{x^4 \sqrt{a+b x^2+c x^4}}{6 c}+\frac{\left (15 b^2-16 a c-10 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{48 c^3}-\frac{\left (b \left (5 b^2-12 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 )}{16 c^3}\\ &=\frac{x^4 \sqrt{a+b x^2+c x^4}}{6 c}+\frac{\left (15 b^2-16 a c-10 b c x^2\right ) \sqrt{a+b x^2+c x^4}}{48 c^3}-\frac{b \left (5 b^2-12 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{32 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0605068, size = 104, normalized size = 0.86 \[ \frac{2 \sqrt{c} \sqrt{a+b x^2+c x^4} \left (8 c \left (c x^4-2 a\right )+15 b^2-10 b c x^2\right )+\left (36 a b c-15 b^3\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{2 \sqrt{c} \sqrt{a+b x^2+c x^4}}\right )}{96 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4]*(15*b^2 - 10*b*c*x^2 + 8*c*(-2*a + c*x^4)) + (-15*b^3 + 36*a*b*c)*ArcTanh[(
b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(96*c^(7/2))

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Maple [A]  time = 0.179, size = 162, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}}{6\,c}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{5\,b{x}^{2}}{24\,{c}^{2}}\sqrt{c{x}^{4}+b{x}^{2}+a}}+{\frac{5\,{b}^{2}}{16\,{c}^{3}}\sqrt{c{x}^{4}+b{x}^{2}+a}}-{\frac{5\,{b}^{3}}{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{7}{2}}}}+{\frac{3\,ab}{8}\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}{3\,{c}^{2}}\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)^(1/2),x)

[Out]

1/6*x^4*(c*x^4+b*x^2+a)^(1/2)/c-5/24*b/c^2*x^2*(c*x^4+b*x^2+a)^(1/2)+5/16*b^2/c^3*(c*x^4+b*x^2+a)^(1/2)-5/32*b
^3/c^(7/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+3/8*b/c^(5/2)*a*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x
^2+a)^(1/2))-1/3*a/c^2*(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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68011, size = 567, normalized size = 4.69 \begin{align*} \left [-\frac{3 \,{\left (5 \, b^{3} - 12 \, a b c\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 (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c - 16 \, a c^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{192 \, c^{4}}, \frac{3 \,{\left (5 \, b^{3} - 12 \, a b c\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 (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c - 16 \, a c^{2}\right )} \sqrt{c x^{4} + b x^{2} + a}}{96 \, c^{4}}\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]

[-1/192*(3*(5*b^3 - 12*a*b*c)*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*(8*c^3*x^4 - 10*b*c^2*x^2 + 15*b^2*c - 16*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^4, 1/96*(3
*(5*b^3 - 12*a*b*c)*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*(8*c^3*x^4 - 10*b*c^2*x^2 + 15*b^2*c - 16*a*c^2)*sqrt(c*x^4 + b*x^2 + a))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{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)

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Giac [A]  time = 1.24593, size = 147, normalized size = 1.21 \begin{align*} \frac{1}{48} \, \sqrt{c x^{4} + b x^{2} + a}{\left (2 \, x^{2}{\left (\frac{4 \, x^{2}}{c} - \frac{5 \, b}{c^{2}}\right )} + \frac{15 \, b^{2} c - 16 \, a c^{2}}{c^{4}}\right )} + \frac{{\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x^{2} - \sqrt{c x^{4} + b x^{2} + a}\right )} \sqrt{c} - b \right |}\right )}{32 \, c^{\frac{9}{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]

1/48*sqrt(c*x^4 + b*x^2 + a)*(2*x^2*(4*x^2/c - 5*b/c^2) + (15*b^2*c - 16*a*c^2)/c^4) + 1/32*(5*b^3*c - 12*a*b*
c^2)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(9/2)