3.6.51 \(\int \frac {x^{-1+2 n}}{a+b x^n+c x^{2 n}} \, dx\) [551]

Optimal. Leaf size=68 \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} n}+\frac {\log \left (a+b x^n+c x^{2 n}\right )}{2 c n} \]

[Out]

1/2*ln(a+b*x^n+c*x^(2*n))/c/n+b*arctanh((b+2*c*x^n)/(-4*a*c+b^2)^(1/2))/c/n/(-4*a*c+b^2)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1371, 648, 632, 212, 642} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c n \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x^n+c x^{2 n}\right )}{2 c n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)/(a + b*x^n + c*x^(2*n)),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 648

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 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 66, normalized size = 0.97 \begin {gather*} \frac {-\frac {2 b \tan ^{-1}\left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}+\log \left (a+x^n \left (b+c x^n\right )\right )}{2 c n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)/(a + b*x^n + c*x^(2*n)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(401\) vs. \(2(62)=124\).
time = 0.07, size = 402, normalized size = 5.91

method	result	size
risch	\(\frac {\ln \left (x \right )}{c}-\frac {4 n^{2} \ln \left (x \right ) a c}{4 a \,c^{2} n^{2}-b^{2} c \,n^{2}}+\frac {n^{2} \ln \left (x \right ) b^{2}}{4 a \,c^{2} n^{2}-b^{2} c \,n^{2}}+\frac {2 \ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) a}{\left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) b^{2}}{2 c \left (4 a c -b^{2}\right ) n}+\frac {\ln \left (x^{n}-\frac {-b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 c \left (4 a c -b^{2}\right ) n}+\frac {2 \ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) a}{\left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) b^{2}}{2 c \left (4 a c -b^{2}\right ) n}-\frac {\ln \left (x^{n}+\frac {b^{2}+\sqrt {-4 a \,b^{2} c +b^{4}}}{2 b c}\right ) \sqrt {-4 a \,b^{2} c +b^{4}}}{2 c \left (4 a c -b^{2}\right ) n}\)	\(402\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+2*n)/(a+b*x^n+c*x^(2*n)),x,method=_RETURNVERBOSE)

[Out]

1/c*ln(x)-4/(4*a*c^2*n^2-b^2*c*n^2)*n^2*ln(x)*a*c+1/(4*a*c^2*n^2-b^2*c*n^2)*n^2*ln(x)*b^2+2/(4*a*c-b^2)/n*ln(x
^n-1/2*(-b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*a-1/2/c/(4*a*c-b^2)/n*ln(x^n-1/2*(-b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*
b^2+1/2/c/(4*a*c-b^2)/n*ln(x^n-1/2*(-b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*(-4*a*b^2*c+b^4)^(1/2)+2/(4*a*c-b^2)/n*l
n(x^n+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*a-1/2/c/(4*a*c-b^2)/n*ln(x^n+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)
*b^2-1/2/c/(4*a*c-b^2)/n*ln(x^n+1/2*(b^2+(-4*a*b^2*c+b^4)^(1/2))/b/c)*(-4*a*b^2*c+b^4)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

log(x)/c - integrate((b*x^n + a)/(c^2*x*x^(2*n) + b*c*x*x^n + a*c*x), x)

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Fricas [A]
time = 0.36, size = 231, normalized size = 3.40 \begin {gather*} \left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{2 \, n} + b^{2} - 2 \, a c + 2 \, {\left (b c + \sqrt {b^{2} - 4 \, a c} c\right )} x^{n} + \sqrt {b^{2} - 4 \, a c} b}{c x^{2 \, n} + b x^{n} + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} n}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {2 \, \sqrt {-b^{2} + 4 \, a c} c x^{n} + \sqrt {-b^{2} + 4 \, a c} b}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(b^2 - 4*a*c)*b*log((2*c^2*x^(2*n) + b^2 - 2*a*c + 2*(b*c + sqrt(b^2 - 4*a*c)*c)*x^n + sqrt(b^2 - 4*
a*c)*b)/(c*x^(2*n) + b*x^n + a)) + (b^2 - 4*a*c)*log(c*x^(2*n) + b*x^n + a))/((b^2*c - 4*a*c^2)*n), 1/2*(2*sqr
t(-b^2 + 4*a*c)*b*arctan(-(2*sqrt(-b^2 + 4*a*c)*c*x^n + sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) + (b^2 - 4*a*c)*l
og(c*x^(2*n) + b*x^n + a))/((b^2*c - 4*a*c^2)*n)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)/(a+b*x**n+c*x**(2*n)),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(2*n - 1)/(c*x^(2*n) + b*x^n + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{2\,n-1}}{a+b\,x^n+c\,x^{2\,n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n - 1)/(a + b*x^n + c*x^(2*n)),x)

[Out]

int(x^(2*n - 1)/(a + b*x^n + c*x^(2*n)), x)

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