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

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

[Out]

-1/32*b*(-12*a*c+5*b^2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x^4+b*x^2+a)^(1/2))/c^(7/2)+1/6*x^4*(c*x^4+b*x^2+a)
^(1/2)/c+1/48*(-10*b*c*x^2-16*a*c+15*b^2)*(c*x^4+b*x^2+a)^(1/2)/c^3

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Rubi [A]  time = 0.11, antiderivative size = 121, 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, 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 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 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 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}{\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*}

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Mathematica [A]  time = 0.06, size = 104, normalized size = 0.86 \[ \frac {\left (36 a b c-15 b^3\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right )+2 \sqrt {c} \sqrt {a+b x^2+c x^4} \left (8 c \left (c x^4-2 a\right )+15 b^2-10 b c x^2\right )}{96 c^{7/2}} \]

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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fricas [A]  time = 0.79, size = 241, normalized size = 1.99 \[ \left [-\frac {3 \, {\left (5 \, b^{3} - 12 \, a b c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c - 16 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{192 \, c^{4}}, \frac {3 \, {\left (5 \, b^{3} - 12 \, a b c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{4} - 10 \, b c^{2} x^{2} + 15 \, b^{2} c - 16 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{96 \, c^{4}}\right ] \]

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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giac [A]  time = 0.21, size = 103, normalized size = 0.85 \[ \frac {1}{48} \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, x^{2} {\left (\frac {4 \, x^{2}}{c} - \frac {5 \, b}{c^{2}}\right )} + \frac {15 \, b^{2} - 16 \, a c}{c^{3}}\right )} + \frac {{\left (5 \, b^{3} - 12 \, a b c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right )}{32 \, c^{\frac {7}{2}}} \]

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 - 16*a*c)/c^3) + 1/32*(5*b^3 - 12*a*b*c)*log
(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - b))/c^(7/2)

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maple [A]  time = 0.02, size = 162, normalized size = 1.34 \[ \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{4}}{6 c}-\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b \,x^{2}}{24 c^{2}}+\frac {3 a b \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}-\frac {5 b^{3} \ln \left (\frac {c \,x^{2}+\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{32 c^{\frac {7}{2}}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, a}{3 c^{2}}+\frac {5 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{16 c^{3}} \]

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((c*x^2+1/2*b)/c^(1/2)+(c*x^4+b*x^2+a)^(1/2))+3/8*b/c^(5/2)*a*ln((c*x^2+1/2*b)/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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 \[ \int \frac {x^7}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

int(x^7/(a + b*x^2 + c*x^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{7}}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \]

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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