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

Optimal. Leaf size=134 \[ \frac {\left (-8 a c+3 b^2-2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{2 c^2 \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {3 b \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{5/2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1114, 738, 779, 621, 206} \begin {gather*} \frac {\left (-8 a c+3 b^2-2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{2 c^2 \left (b^2-4 a c\right )}+\frac {x^4 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {3 b \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^4*(2*a + b*x^2))/((b^2 - 4*a*c)*Sqrt[a + b*x^2 + c*x^4]) + ((3*b^2 - 8*a*c - 2*b*c*x^2)*Sqrt[a + b*x^2 + c*
x^4])/(2*c^2*(b^2 - 4*a*c)) - (3*b*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*c^(5/2))

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 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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, 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] && LtQ[p, -1] && GtQ[m, 1] && 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 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]

Rubi steps

\begin {align*} \int \frac {x^7}{\left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {x^4 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}-\frac {\operatorname {Subst}\left (\int \frac {x (4 a+2 b x)}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{b^2-4 a c}\\ &=\frac {x^4 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (3 b^2-8 a c-2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{2 c^2 \left (b^2-4 a c\right )}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 c^2}\\ &=\frac {x^4 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (3 b^2-8 a c-2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{2 c^2 \left (b^2-4 a c\right )}-\frac {(3 b) \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 )}{2 c^2}\\ &=\frac {x^4 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}}+\frac {\left (3 b^2-8 a c-2 b c x^2\right ) \sqrt {a+b x^2+c x^4}}{2 c^2 \left (b^2-4 a c\right )}-\frac {3 b \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )}{4 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 137, normalized size = 1.02 \begin {gather*} \frac {\frac {2 \sqrt {c} \left (8 a^2 c+a \left (-3 b^2+10 b c x^2+4 c^2 x^4\right )-b^2 x^2 \left (3 b+c x^2\right )\right )}{\sqrt {a+b x^2+c x^4}}+3 b \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 )}{4 c^{5/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[c]*(8*a^2*c - b^2*x^2*(3*b + c*x^2) + a*(-3*b^2 + 10*b*c*x^2 + 4*c^2*x^4)))/Sqrt[a + b*x^2 + c*x^4] +
3*b*(b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^2)/(2*Sqrt[c]*Sqrt[a + b*x^2 + c*x^4])])/(4*c^(5/2)*(-b^2 + 4*a*c))

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IntegrateAlgebraic [A]  time = 0.58, size = 131, normalized size = 0.98 \begin {gather*} \frac {8 a^2 c-3 a b^2+10 a b c x^2+4 a c^2 x^4-3 b^3 x^2-b^2 c x^4}{2 c^2 \left (4 a c-b^2\right ) \sqrt {a+b x^2+c x^4}}+\frac {3 b \log \left (-2 c^{5/2} \sqrt {a+b x^2+c x^4}+b c^2+2 c^3 x^2\right )}{4 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b^2 + 8*a^2*c - 3*b^3*x^2 + 10*a*b*c*x^2 - b^2*c*x^4 + 4*a*c^2*x^4)/(2*c^2*(-b^2 + 4*a*c)*Sqrt[a + b*x^2
+ c*x^4]) + (3*b*Log[b*c^2 + 2*c^3*x^2 - 2*c^(5/2)*Sqrt[a + b*x^2 + c*x^4]])/(4*c^(5/2))

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fricas [A]  time = 1.58, size = 459, normalized size = 3.43 \begin {gather*} \left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{4} + a b^{3} - 4 \, a^{2} b c + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{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 ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} + 3 \, a b^{2} c - 8 \, a^{2} c^{2} + {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{8 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{2}\right )}}, \frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{4} + a b^{3} - 4 \, a^{2} b c + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{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 ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} + 3 \, a b^{2} c - 8 \, a^{2} c^{2} + {\left (3 \, b^{3} c - 10 \, a b c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{4 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{4} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{2}\right )}}\right ] \end {gather*}

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]

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

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giac [A]  time = 0.27, size = 154, normalized size = 1.15 \begin {gather*} \frac {{\left (\frac {{\left (b^{2} c - 4 \, a c^{2}\right )} x^{2}}{b^{2} c^{2} - 4 \, a c^{3}} + \frac {3 \, b^{3} - 10 \, a b c}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x^{2} + \frac {3 \, a b^{2} - 8 \, a^{2} c}{b^{2} c^{2} - 4 \, a c^{3}}}{2 \, \sqrt {c x^{4} + b x^{2} + a}} + \frac {3 \, b \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{4 \, c^{\frac {5}{2}}} \end {gather*}

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]

1/2*(((b^2*c - 4*a*c^2)*x^2/(b^2*c^2 - 4*a*c^3) + (3*b^3 - 10*a*b*c)/(b^2*c^2 - 4*a*c^3))*x^2 + (3*a*b^2 - 8*a
^2*c)/(b^2*c^2 - 4*a*c^3))/sqrt(c*x^4 + b*x^2 + a) + 3/4*b*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*
sqrt(c) - b))/c^(5/2)

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maple [B]  time = 0.02, size = 264, normalized size = 1.97 \begin {gather*} \frac {2 a b \,x^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}-\frac {3 b^{3} x^{2}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}+\frac {x^{4}}{2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}+\frac {a \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}-\frac {3 b^{4}}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{3}}+\frac {3 b \,x^{2}}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}-\frac {3 b \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{4 c^{\frac {5}{2}}}+\frac {a}{\sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2}}-\frac {3 b^{2}}{8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{3}} \end {gather*}

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]

1/2*x^4/c/(c*x^4+b*x^2+a)^(1/2)+3/4*b/c^2*x^2/(c*x^4+b*x^2+a)^(1/2)-3/8*b^2/c^3/(c*x^4+b*x^2+a)^(1/2)-3/4*b^3/
c^2/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*x^2-3/8*b^4/c^3/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)-3/4*b/c^(5/2)*ln((c*x^
2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+1/c^2*a/(c*x^4+b*x^2+a)^(1/2)+2/c*a*b/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2
)*x^2+1/c^2*a*b^2/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^7}{{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(a + b*x^2 + c*x^4)^(3/2),x)

[Out]

int(x^7/(a + b*x^2 + c*x^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7}}{\left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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