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

Optimal. Leaf size=100 \[ \frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}+\frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c} \]

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Rubi [A]  time = 0.12, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1114, 701, 634, 618, 206, 628} \begin {gather*} \frac {\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^3}+\frac {b \left (b^2-3 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \sqrt {b^2-4 a c}}-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^2 + c*x^4),x]

[Out]

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

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 618

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, 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] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 93, normalized size = 0.93 \begin {gather*} \frac {-\frac {2 b \left (b^2-3 a c\right ) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\left (b^2-a c\right ) \log \left (a+b x^2+c x^4\right )+c x^2 \left (c x^2-2 b\right )}{4 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a + b*x^2 + c*x^4),x]

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 2.86, size = 313, normalized size = 3.13 \begin {gather*} \left [\frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} - 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x^{2} - {\left (b^{3} - 3 \, a b c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )}}, \frac {{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} x^{4} - 2 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} x^{2} + 2 \, {\left (b^{3} - 3 \, a b c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) + {\left (b^{4} - 5 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\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),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.56, size = 92, normalized size = 0.92 \begin {gather*} \frac {c x^{4} - 2 \, b x^{2}}{4 \, c^{2}} + \frac {{\left (b^{2} - a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} - \frac {{\left (b^{3} - 3 \, a b c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c*x^4 - 2*b*x^2)/c^2 + 1/4*(b^2 - a*c)*log(c*x^4 + b*x^2 + a)/c^3 - 1/2*(b^3 - 3*a*b*c)*arctan((2*c*x^2 +
b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c^3)

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maple [A]  time = 0.01, size = 142, normalized size = 1.42 \begin {gather*} \frac {x^{4}}{4 c}+\frac {3 a b \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{2}}-\frac {b^{3} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{3}}-\frac {b \,x^{2}}{2 c^{2}}-\frac {a \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{2}}+\frac {b^{2} \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 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),x)

[Out]

1/4/c*x^4-1/2*b/c^2*x^2-1/4/c^2*ln(c*x^4+b*x^2+a)*a+1/4/c^3*ln(c*x^4+b*x^2+a)*b^2+3/2/c^2/(4*a*c-b^2)^(1/2)*ar
ctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a*b-1/2/c^3/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^3

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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),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 positive or negative?

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mupad [B]  time = 4.40, size = 842, normalized size = 8.42 \begin {gather*} \frac {x^4}{4\,c}-\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {b\,x^2}{2\,c^2}+\frac {b\,\mathrm {atan}\left (\frac {2\,c^4\,\left (4\,a\,c-b^2\right )\,\left (\frac {\frac {b\,\left (3\,a\,c-b^2\right )\,\left (\frac {8\,a^2\,c^4-8\,a\,b^2\,c^3}{c^4}-\frac {8\,a\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )}{8\,c^3\,\sqrt {4\,a\,c-b^2}}-\frac {a\,b\,\left (3\,a\,c-b^2\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{c\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}}{a}-x^2\,\left (\frac {\frac {b\,\left (\frac {6\,b^3\,c^3-10\,a\,b\,c^4}{c^4}+\frac {4\,b\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (3\,a\,c-b^2\right )}{8\,c^3\,\sqrt {4\,a\,c-b^2}}+\frac {b^2\,\left (3\,a\,c-b^2\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,c\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}}{a}+\frac {b\,\left (\frac {2\,a^2\,b\,c^2-3\,a\,b^3\,c+b^5}{c^4}+\frac {\left (\frac {6\,b^3\,c^3-10\,a\,b\,c^4}{c^4}+\frac {4\,b\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {b^3\,{\left (3\,a\,c-b^2\right )}^2}{2\,c^4\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )+\frac {b\,\left (\frac {\left (\frac {8\,a^2\,c^4-8\,a\,b^2\,c^3}{c^4}-\frac {8\,a\,c^2\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{16\,a\,c^4-4\,b^2\,c^3}\right )\,\left (8\,a^2\,c^2-10\,a\,b^2\,c+2\,b^4\right )}{2\,\left (16\,a\,c^4-4\,b^2\,c^3\right )}-\frac {a^3\,c^2-2\,a^2\,b^2\,c+a\,b^4}{c^4}+\frac {a\,b^2\,{\left (3\,a\,c-b^2\right )}^2}{c^4\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )}{9\,a^2\,b^2\,c^2-6\,a\,b^4\,c+b^6}\right )\,\left (3\,a\,c-b^2\right )}{2\,c^3\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(a + b*x^2 + c*x^4),x)

[Out]

x^4/(4*c) - (log(a + b*x^2 + c*x^4)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) - (b*x^2)/(2*
c^2) + (b*atan((2*c^4*(4*a*c - b^2)*(((b*(3*a*c - b^2)*((8*a^2*c^4 - 8*a*b^2*c^3)/c^4 - (8*a*c^2*(2*b^4 + 8*a^
2*c^2 - 10*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3)))/(8*c^3*(4*a*c - b^2)^(1/2)) - (a*b*(3*a*c - b^2)*(2*b^4 + 8*a^2*
c^2 - 10*a*b^2*c))/(c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a - x^2*(((b*((6*b^3*c^3 - 10*a*b*c^4)/c^4
+ (4*b*c^2*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(3*a*c - b^2))/(8*c^3*(4*a*c - b^2)^(1/2)
) + (b^2*(3*a*c - b^2)*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*c*(4*a*c - b^2)^(1/2)*(16*a*c^4 - 4*b^2*c^3)))/a +
(b*((b^5 + 2*a^2*b*c^2 - 3*a*b^3*c)/c^4 + (((6*b^3*c^3 - 10*a*b*c^4)/c^4 + (4*b*c^2*(2*b^4 + 8*a^2*c^2 - 10*a
*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a*c^4 - 4*b^2*c^3)) - (b^3*(3*a*c -
b^2)^2)/(2*c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))) + (b*((((8*a^2*c^4 - 8*a*b^2*c^3)/c^4 - (8*a*c^2*(2
*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(16*a*c^4 - 4*b^2*c^3))*(2*b^4 + 8*a^2*c^2 - 10*a*b^2*c))/(2*(16*a*c^4 - 4*b^2
*c^3)) - (a*b^4 + a^3*c^2 - 2*a^2*b^2*c)/c^4 + (a*b^2*(3*a*c - b^2)^2)/(c^4*(4*a*c - b^2))))/(2*a*(4*a*c - b^2
)^(1/2))))/(b^6 + 9*a^2*b^2*c^2 - 6*a*b^4*c))*(3*a*c - b^2))/(2*c^3*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 2.91, size = 391, normalized size = 3.91 \begin {gather*} - \frac {b x^{2}}{2 c^{2}} + \left (- \frac {b \sqrt {- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) \log {\left (x^{2} + \frac {2 a^{2} c - a b^{2} + 8 a c^{3} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) \log {\left (x^{2} + \frac {2 a^{2} c - a b^{2} + 8 a c^{3} \left (\frac {b \sqrt {- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right ) - 2 b^{2} c^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}} \left (3 a c - b^{2}\right )}{4 c^{3} \left (4 a c - b^{2}\right )} - \frac {a c - b^{2}}{4 c^{3}}\right )}{3 a b c - b^{3}} \right )} + \frac {x^{4}}{4 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*x**2/(2*c**2) + (-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4*c**3))*log
(x**2 + (2*a**2*c - a*b**2 + 8*a*c**3*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c -
b**2)/(4*c**3)) - 2*b**2*c**2*(-b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4
*c**3)))/(3*a*b*c - b**3)) + (b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b**2)/(4*c
**3))*log(x**2 + (2*a**2*c - a*b**2 + 8*a*c**3*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) -
(a*c - b**2)/(4*c**3)) - 2*b**2*c**2*(b*sqrt(-4*a*c + b**2)*(3*a*c - b**2)/(4*c**3*(4*a*c - b**2)) - (a*c - b
**2)/(4*c**3)))/(3*a*b*c - b**3)) + x**4/(4*c)

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