3.600 \(\int \frac {(d x)^m}{(a+b x^n+c x^{2 n})^2} \, dx\)

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.96, 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 = {1384, 1560, 364} \[ \frac {c (d x)^{m+1} \left (\frac {4 a c (m-2 n+1)-b^2 (m-n+1)}{\sqrt {b^2-4 a c}}-b (m-n+1)\right ) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a d (m+1) n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (d x)^{m+1} \left (b (m-n+1) \sqrt {b^2-4 a c}+4 a c (m-2 n+1)+b^2 (-(m-n+1))\right ) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a d (m+1) n \left (b^2-4 a c\right )^{3/2} \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(d x)^{m+1} \left (-2 a c+b^2+b c x^n\right )}{a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \]

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1384

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 1560

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*}

________________________________________________________________________________________

Mathematica [B]  time = 6.36, size = 3515, normalized size = 10.72 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(d*x)^m*(-b^2 + 2*a*c - b*c*x^n))/(a*(-b^2 + 4*a*c)*n*(a + b*x^n + c*x^(2*n))) - (b*c*x^(1 + n)*(d*x)^m*(x^
n)^((1 + m)/n - (1 + m + n)/n)*(-(((x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometri
c2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c]
)/c + x^n))])/Sqrt[b^2 - 4*a*c]) + ((x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometr
ic2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c
])/c + x^n))])/Sqrt[b^2 - 4*a*c]))/(a*(-b^2 + 4*a*c)*(1 + m)) + (b*c*x^(1 + n)*(d*x)^m*(x^n)^((1 + m)/n - (1 +
m + n)/n)*(-(((x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -
((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))])/Sqrt[b
^2 - 4*a*c]) + ((x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n),
-((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))])/Sqrt[
b^2 - 4*a*c]))/(a*(-b^2 + 4*a*c)*(1 + m)*n) + (b*c*m*x^(1 + n)*(d*x)^m*(x^n)^((1 + m)/n - (1 + m + n)/n)*(-(((
x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 -
(1 + m)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))])/Sqrt[b^2 - 4*a*c]) + (
(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1
- (1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))])/Sqrt[b^2 - 4*a*c]))/(
a*(-b^2 + 4*a*c)*(1 + m)*n) + (b^2*x*(d*x)^m*((1 - (x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/
n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - S
qrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b - Sqrt[b^2 - 4*a*c]))/(2*c) + (-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - (
x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 -
(1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b + Sqrt[b^2 - 4
*a*c]))/(2*c) + (-b + Sqrt[b^2 - 4*a*c])^2/(2*c))))/(a*(-b^2 + 4*a*c)*(1 + m)) - (4*c*x*(d*x)^m*((1 - (x^n/(-1
/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m
)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b - Sqrt[b^2 - 4*a*c]))
/(2*c) + (-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - (x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n
)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sq
rt[b^2 - 4*a*c])/c + x^n))])/((b*(-b + Sqrt[b^2 - 4*a*c]))/(2*c) + (-b + Sqrt[b^2 - 4*a*c])^2/(2*c))))/((-b^2
+ 4*a*c)*(1 + m)) - (b^2*x*(d*x)^m*((1 - (x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hyperge
ometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 -
4*a*c])/c + x^n))])/((b*(-b - Sqrt[b^2 - 4*a*c]))/(2*c) + (-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - (x^n/(-1/2*
(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n
, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b + Sqrt[b^2 - 4*a*c]))/(2
*c) + (-b + Sqrt[b^2 - 4*a*c])^2/(2*c))))/(a*(-b^2 + 4*a*c)*(1 + m)*n) + (2*c*x*(d*x)^m*((1 - (x^n/(-1/2*(-b -
Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/
2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b - Sqrt[b^2 - 4*a*c]))/(2*c) +
(-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - (x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hyperg
eometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 -
4*a*c])/c + x^n))])/((b*(-b + Sqrt[b^2 - 4*a*c]))/(2*c) + (-b + Sqrt[b^2 - 4*a*c])^2/(2*c))))/((-b^2 + 4*a*c)
*(1 + m)*n) - (b^2*m*x*(d*x)^m*((1 - (x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeomet
ric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*
c])/c + x^n))])/((b*(-b - Sqrt[b^2 - 4*a*c]))/(2*c) + (-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - (x^n/(-1/2*(-b
+ Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1
/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b + Sqrt[b^2 - 4*a*c]))/(2*c)
+ (-b + Sqrt[b^2 - 4*a*c])^2/(2*c))))/(a*(-b^2 + 4*a*c)*(1 + m)*n) + (2*c*m*x*(d*x)^m*((1 - (x^n/(-1/2*(-b - S
qrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeometric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*
(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))])/((b*(-b - Sqrt[b^2 - 4*a*c]))/(2*c) + (
-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - (x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(-n^(-1) - m/n)*Hypergeo
metric2F1[-((1 + m)/n), -((1 + m)/n), 1 - (1 + m)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4
*a*c])/c + x^n))])/((b*(-b + Sqrt[b^2 - 4*a*c]))/(2*c) + (-b + Sqrt[b^2 - 4*a*c])^2/(2*c))))/((-b^2 + 4*a*c)*(
1 + m)*n)

________________________________________________________________________________________

fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d x\right )^{m}}{c^{2} x^{4 \, n} + b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2} + 2 \, {\left (b c x^{n} + a c\right )} x^{2 \, n}}, x\right ) \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{\left (b \,x^{n}+c \,x^{2 n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b c d^{m} x e^{\left (m \log \relax (x) + n \log \relax (x)\right )} + {\left (b^{2} d^{m} - 2 \, a c d^{m}\right )} x x^{m}}{a^{2} b^{2} n - 4 \, a^{3} c n + {\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} + {\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}} + \int -\frac {b c d^{m} {\left (m - n + 1\right )} e^{\left (m \log \relax (x) + n \log \relax (x)\right )} + {\left (b^{2} d^{m} {\left (m - n + 1\right )} - 2 \, a c d^{m} {\left (m - 2 \, n + 1\right )}\right )} x^{m}}{a^{2} b^{2} n - 4 \, a^{3} c n + {\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} + {\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}}\,{d x} \]

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________