3.6.50 \(\int \frac {x^{-1+3 n}}{a+b x^n+c x^{2 n}} \, dx\) [550]

Optimal. Leaf size=87 \[ \frac {x^n}{c n}-\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c} n}-\frac {b \log \left (a+b x^n+c x^{2 n}\right )}{2 c^2 n} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1371, 717, 648, 632, 212, 642} \begin {gather*} -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^n}{\sqrt {b^2-4 a c}}\right )}{c^2 n \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^n+c x^{2 n}\right )}{2 c^2 n}+\frac {x^n}{c n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 3*n)/(a + b*x^n + c*x^(2*n)),x]

[Out]

x^n/(c*n) - ((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^n)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]*n) - (b*Log[a + b*
x^n + c*x^(2*n)])/(2*c^2*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 717

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

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

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Mathematica [A]
time = 0.15, size = 82, normalized size = 0.94 \begin {gather*} \frac {2 c x^n+\frac {2 \left (b^2-2 a c\right ) \tan ^{-1}\left (\frac {b+2 c x^n}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-b \log \left (a+x^n \left (b+c x^n\right )\right )}{2 c^2 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 3*n)/(a + b*x^n + c*x^(2*n)),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(663\) vs. \(2(81)=162\).
time = 0.09, size = 664, normalized size = 7.63

method	result	size
risch	\(-\frac {b \ln \left (x \right )}{c^{2}}+\frac {x^{n}}{c n}+\frac {4 n^{2} \ln \left (x \right ) a b c}{4 a \,c^{3} n^{2}-b^{2} c^{2} n^{2}}-\frac {n^{2} \ln \left (x \right ) b^{3}}{4 a \,c^{3} n^{2}-b^{2} c^{2} n^{2}}-\frac {2 \ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) a b}{\left (4 a c -b^{2}\right ) c n}+\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 \left (4 a c -b^{2}\right ) c^{2} n}+\frac {\ln \left (x^{n}-\frac {-2 a b c +b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 \left (4 a c -b^{2}\right ) c^{2} n}-\frac {2 \ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) a b}{\left (4 a c -b^{2}\right ) c n}+\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) b^{3}}{2 \left (4 a c -b^{2}\right ) c^{2} n}-\frac {\ln \left (x^{n}+\frac {2 a b c -b^{3}+\sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 c \left (2 a c -b^{2}\right )}\right ) \sqrt {-16 a^{3} c^{3}+20 a^{2} b^{2} c^{2}-8 a \,b^{4} c +b^{6}}}{2 \left (4 a c -b^{2}\right ) c^{2} n}\)	\(664\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+3*n)/(a+b*x^n+c*x^(2*n)),x,method=_RETURNVERBOSE)

[Out]

-b/c^2*ln(x)+x^n/c/n+4/(4*a*c^3*n^2-b^2*c^2*n^2)*n^2*ln(x)*a*b*c-1/(4*a*c^3*n^2-b^2*c^2*n^2)*n^2*ln(x)*b^3-2/(
4*a*c-b^2)/c/n*ln(x^n-1/2*(-2*a*b*c+b^3+(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*a*b+1
/2/(4*a*c-b^2)/c^2/n*ln(x^n-1/2*(-2*a*b*c+b^3+(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))
*b^3+1/2/(4*a*c-b^2)/c^2/n*ln(x^n-1/2*(-2*a*b*c+b^3+(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c
-b^2))*(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2)-2/(4*a*c-b^2)/c/n*ln(x^n+1/2*(2*a*b*c-b^3+(-16*a^3*c^3
+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*a*b+1/2/(4*a*c-b^2)/c^2/n*ln(x^n+1/2*(2*a*b*c-b^3+(-16*a^
3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*b^3-1/2/(4*a*c-b^2)/c^2/n*ln(x^n+1/2*(2*a*b*c-b^3+(-
16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(1/2))/c/(2*a*c-b^2))*(-16*a^3*c^3+20*a^2*b^2*c^2-8*a*b^4*c+b^6)^(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+3*n)/(a+b*x^n+c*x^(2*n)),x, algorithm="maxima")

[Out]

-b*log(x)/c^2 + x^n/(c*n) - integrate(-(a*b + (b^2 - a*c)*x^n)/(c^3*x*x^(2*n) + b*c^2*x*x^n + a*c^2*x), x)

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Fricas [A]
time = 0.39, size = 285, normalized size = 3.28 \begin {gather*} \left [-\frac {{\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} \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 ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{n} + {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} n}, -\frac {2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} \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 ) - 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{n} + {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{2 \, n} + b x^{n} + a\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n+c*x^(2*n)),x, algorithm="fricas")

[Out]

[-1/2*((b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*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)) - 2*(b^2*c - 4*a*c^2)*x^n + (b^3 - 4*a*b*c)*log(c*x^(2*n) + b*x^
n + a))/((b^2*c^2 - 4*a*c^3)*n), -1/2*(2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*sqrt(-b^2 + 4*a*c)*c*x^n
+ sqrt(-b^2 + 4*a*c)*b)/(b^2 - 4*a*c)) - 2*(b^2*c - 4*a*c^2)*x^n + (b^3 - 4*a*b*c)*log(c*x^(2*n) + b*x^n + a))
/((b^2*c^2 - 4*a*c^3)*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+3*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+3*n)/(a+b*x^n+c*x^(2*n)),x, algorithm="giac")

[Out]

integrate(x^(3*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^{3\,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^(3*n - 1)/(a + b*x^n + c*x^(2*n)),x)

[Out]

int(x^(3*n - 1)/(a + b*x^n + c*x^(2*n)), x)

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