3.6.100 \(\int \frac {(d x)^m}{(a+b x^n+c x^{2 n})^2} \, dx\) [600]

Optimal. Leaf size=328 \[ \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )}+\frac {c \left (\frac {4 a c (1+m-2 n)-b^2 (1+m-n)}{\sqrt {b^2-4 a c}}-b (1+m-n)\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) d (1+m) n}-\frac {c \left (4 a c (1+m-2 n)-b^2 (1+m-n)+b \sqrt {b^2-4 a c} (1+m-n)\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m) n} \]

[Out]

(d*x)^(1+m)*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/d/n/(a+b*x^n+c*x^(2*n))+c*(d*x)^(1+m)*hypergeom([1, (1+m)/n],[(
1+m+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(-b*(1+m-n)+(4*a*c*(1+m-2*n)-b^2*(1+m-n))/(-4*a*c+b^2)^(1/2))/a/(-4
*a*c+b^2)/d/(1+m)/n/(b-(-4*a*c+b^2)^(1/2))-c*(d*x)^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-2*c*x^n/(b+(-4*a*
c+b^2)^(1/2)))*(4*a*c*(1+m-2*n)-b^2*(1+m-n)+b*(1+m-n)*(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)^(3/2)/d/(1+m)/n/(b+(-
4*a*c+b^2)^(1/2))

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Rubi [A]
time = 0.68, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1398, 1574, 371} \begin {gather*} \frac {c (d x)^{m+1} \left (\frac {4 a c (m-2 n+1)-b^2 (m-n+1)}{\sqrt {b^2-4 a c}}-b (m-n+1)\right ) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a d (m+1) n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (d x)^{m+1} \left (b (m-n+1) \sqrt {b^2-4 a c}+4 a c (m-2 n+1)-\left (b^2 (m-n+1)\right )\right ) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a d (m+1) n \left (b^2-4 a c\right )^{3/2} \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(d x)^{m+1} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*x)^(1 + m)*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*d*n*(a + b*x^n + c*x^(2*n))) + (c*((4*a*c*(1 + m - 2*
n) - b^2*(1 + m - n))/Sqrt[b^2 - 4*a*c] - b*(1 + m - n))*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m
+ n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*d*(1 + m)*n) - (c*(4*a*c
*(1 + m - 2*n) - b^2*(1 + m - n) + b*Sqrt[b^2 - 4*a*c]*(1 + m - n))*(d*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)
/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4*a*c])*d*(1 +
m)*n)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1398

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(d*x)^(m + 1))*(
b^2 - 2*a*c + b*c*x^n)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*n*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(a*n*(p +
1)*(b^2 - 4*a*c)), Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(n*(p + 1) + m + 1) - 2*a*c*(m + 2*n*(
p + 1) + 1) + b*c*(2*n*p + 3*n + m + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ
[b^2 - 4*a*c, 0] && ILtQ[p + 1, 0]

Rule 1574

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0])

Rubi steps

\begin {align*} \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx &=\frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )}-\frac {\int \frac {(d x)^m \left (-2 a c (1+m-2 n)+b^2 (1+m-n)+b c (1+m-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )}-\frac {\int \left (\frac {\left (b c (1+m-n)+\frac {c \left (b^2-4 a c+b^2 m-4 a c m-b^2 n+8 a c n\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n}+\frac {\left (b c (1+m-n)-\frac {c \left (b^2-4 a c+b^2 m-4 a c m-b^2 n+8 a c n\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n}\right ) \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )}+\frac {\left (c \left (4 a c (1+m-2 n)-b^2 (1+m-n)-b \sqrt {b^2-4 a c} (1+m-n)\right )\right ) \int \frac {(d x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (c \left (4 a c (1+m-2 n)-b^2 (1+m-n)+b \sqrt {b^2-4 a c} (1+m-n)\right )\right ) \int \frac {(d x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}\\ &=\frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )}+\frac {c \left (4 a c (1+m-2 n)-b^2 (1+m-n)-b \sqrt {b^2-4 a c} (1+m-n)\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) d (1+m) n}-\frac {c \left (4 a c (1+m-2 n)-b^2 (1+m-n)+b \sqrt {b^2-4 a c} (1+m-n)\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m) n}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1890\) vs. \(2(328)=656\).
time = 4.36, size = 1890, normalized size = 5.76 \begin {gather*} \frac {x (d x)^m \left (\frac {2 b^2 c}{a \sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m)}-\frac {8 c^2}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m)}+\frac {2 b^2 c}{a \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) (1+m)}+\frac {8 c^2}{\left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) (1+m)}-\frac {2 b^2 c}{a \sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {4 c^2}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {4 c^2}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {2 b^2 c}{a \left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) (1+m) n}-\frac {2 b^2 c m}{a \sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {4 c^2 m}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {4 c^2 m}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {2 b^2 c m}{a \left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) (1+m) n}-\frac {b^2}{a n \left (a+x^n \left (b+c x^n\right )\right )}+\frac {2 c}{n \left (a+x^n \left (b+c x^n\right )\right )}-\frac {b c x^n}{a n \left (a+x^n \left (b+c x^n\right )\right )}+\frac {2^{-\frac {1+m}{n}} c \left (4 a c \sqrt {b^2-4 a c} (1+m-2 n)+4 a b c (1+m-n)-b^2 \sqrt {b^2-4 a c} (1+m-n)+b^3 (-1-m+n)\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};1-\frac {1+m}{n};\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{a \sqrt {b^2-4 a c} \left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {2^{-\frac {1+m}{n}} b c (-1-m+n) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};1-\frac {1+m}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{a \sqrt {b^2-4 a c} (1+m) n}-\frac {2^{\frac {-1-m+n}{n}} b^2 c \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};\frac {-1-m+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{a \sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m)}+\frac {2^{-\frac {1+m-3 n}{n}} c^2 \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};\frac {-1-m+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m)}+\frac {2^{\frac {-1-m+n}{n}} b^2 c \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};\frac {-1-m+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{a \sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}-\frac {2^{-\frac {1+m-2 n}{n}} c^2 \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};\frac {-1-m+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}+\frac {2^{\frac {-1-m+n}{n}} b^2 c m \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};\frac {-1-m+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{a \sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}-\frac {2^{-\frac {1+m-2 n}{n}} c^2 m \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};\frac {-1-m+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) (1+m) n}\right )}{-b^2+4 a c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^m*((2*b^2*c)/(a*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)) - (8*c^2)/(Sqrt[b^2 - 4*a*c]*(b +
Sqrt[b^2 - 4*a*c])*(1 + m)) + (2*b^2*c)/(a*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 + m)) + (8*c^2)/((-b^2 + 4*a
*c + b*Sqrt[b^2 - 4*a*c])*(1 + m)) - (2*b^2*c)/(a*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n) + (4*c^
2)/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n) + (4*c^2)/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 + m
)*n) + (2*b^2*c)/(a*(-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*(1 + m)*n) - (2*b^2*c*m)/(a*Sqrt[b^2 - 4*a*c]*(b + Sq
rt[b^2 - 4*a*c])*(1 + m)*n) + (4*c^2*m)/(Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n) + (4*c^2*m)/((b^
2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 + m)*n) + (2*b^2*c*m)/(a*(-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*(1 + m)*n) -
b^2/(a*n*(a + x^n*(b + c*x^n))) + (2*c)/(n*(a + x^n*(b + c*x^n))) - (b*c*x^n)/(a*n*(a + x^n*(b + c*x^n))) + (
c*(4*a*c*Sqrt[b^2 - 4*a*c]*(1 + m - 2*n) + 4*a*b*c*(1 + m - n) - b^2*Sqrt[b^2 - 4*a*c]*(1 + m - n) + b^3*(-1 -
m + n))*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 -
4*a*c] + 2*c*x^n)])/(2^((1 + m)/n)*a*Sqrt[b^2 - 4*a*c]*(-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*(1 + m)*n*((c*x^n)
/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n)) + (b*c*(-1 - m + n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)
/n), 1 - (1 + m)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(2^((1 + m)/n)*a*Sqrt[b^2 - 4*
a*c]*(1 + m)*n*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n)) - (2^((-1 - m + n)/n)*b^2*c*Hypergeome
tric2F1[-((1 + m)/n), -((1 + m)/n), (-1 - m + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]
)/(a*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n)
) + (c^2*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), (-1 - m + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 -
4*a*c] + 2*c*x^n)])/(2^((1 + m - 3*n)/n)*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*((c*x^n)/(b + Sqrt
[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n)) + (2^((-1 - m + n)/n)*b^2*c*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n
), (-1 - m + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(a*Sqrt[b^2 - 4*a*c]*(b + Sqrt[
b^2 - 4*a*c])*(1 + m)*n*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n)) - (c^2*Hypergeometric2F1[-((1
+ m)/n), -((1 + m)/n), (-1 - m + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(2^((1 + m
- 2*n)/n)*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1
+ m)/n)) + (2^((-1 - m + n)/n)*b^2*c*m*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), (-1 - m + n)/n, (b + Sqr
t[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(a*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n*((c
*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n)) - (c^2*m*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), (
-1 - m + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)])/(2^((1 + m - 2*n)/n)*Sqrt[b^2 - 4*a
*c]*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^((1 + m)/n))))/(-b^2 + 4*a*c
)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (d x \right )^{m}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((d*x)^m/(a+b*x^n+c*x^(2*n))^2,x)

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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((d*x)^m/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")

[Out]

(b*c*d^m*x*e^(m*log(x) + n*log(x)) + (b^2*d^m - 2*a*c*d^m)*x*x^m)/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*
c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n) + integrate(-(b*c*d^m*(m - n + 1)*e^(m*log(x) + n*log(x)) + (b^2
*d^m*(m - n + 1) - 2*a*c*d^m*(m - 2*n + 1))*x^m)/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) +
(a*b^3*n - 4*a^2*b*c*n)*x^n), x)

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Fricas [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((d*x)^m/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")

[Out]

integral((d*x)^m/(c^2*x^(4*n) + b^2*x^(2*n) + 2*a*b*x^n + a^2 + 2*(b*c*x^n + a*c)*x^(2*n)), x)

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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((d*x)**m/(a+b*x**n+c*x**(2*n))**2,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((d*x)^m/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")

[Out]

integrate((d*x)^m/(c*x^(2*n) + b*x^n + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(a + b*x^n + c*x^(2*n))^2,x)

[Out]

int((d*x)^m/(a + b*x^n + c*x^(2*n))^2, x)

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