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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.961283, antiderivative size = 328, normalized size of antiderivative = 1., 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 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])

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])

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 = 2.28882, size = 1511, normalized size = 4.61 \[ \text{result too large to display} \]

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 - 4*a*c)*(-b + Sqrt[b^2 - 4*a*c])*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*(1 + m + n)*(b^2 - 2*a*
c + b*c*x^n)) + 2*b^2*c*Sqrt[b^2 - 4*a*c]*(1 + m + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hyperge
ometric2F1[1, (1 + m)/n, (1 + m + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hyperge
ometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 4*a*c^2*Sqrt[b^2 - 4*a*c]*(1 +
m + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (2*c*x^
n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x
^n)/(b + Sqrt[b^2 - 4*a*c])]) + 2*b^2*c*Sqrt[b^2 - 4*a*c]*m*(1 + m + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2
- 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2
- 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 4*a*c^2*Sqrt[
b^2 - 4*a*c]*m*(1 + m + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1
+ m + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1
+ m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 2*b^2*c*Sqrt[b^2 - 4*a*c]*n*(1 + m + n)*(a + x^n*(b + c*x^
n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c
])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c
])]) + 8*a*c^2*Sqrt[b^2 - 4*a*c]*n*(1 + m + n)*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric
2F1[1, (1 + m)/n, (1 + m + n)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric
2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) + 2*b*c^2*Sqrt[b^2 - 4*a*c]*(1 + m)*x^n*
(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m + n)/n, 2 + (1 + m)/n, (2*c*x^n)/
(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m + n)/n, 2 + (1 + m)/n, (-2*c*
x^n)/(b + Sqrt[b^2 - 4*a*c])]) + 2*b*c^2*Sqrt[b^2 - 4*a*c]*m*(1 + m)*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^
2 - 4*a*c])*Hypergeometric2F1[1, (1 + m + n)/n, 2 + (1 + m)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqr
t[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 + m + n)/n, 2 + (1 + m)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) - 2*b*
c^2*Sqrt[b^2 - 4*a*c]*(1 + m)*n*x^n*(a + x^n*(b + c*x^n))*(-((b + Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, (1 +
m + n)/n, 2 + (1 + m)/n, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]) + (b - Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1,
(1 + m + n)/n, 2 + (1 + m)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])))/(a*(b^2 - 4*a*c)^2*(-b + Sqrt[b^2 - 4*a*c
])*(b + Sqrt[b^2 - 4*a*c])*(1 + m)*n*(1 + m + n)*(a + x^n*(b + c*x^n))))

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Maple [F]  time = 0.036, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx \right ) ^{m}}{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \frac{b c d^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\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 \left (x\right ) + n \log \left (x\right )\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} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \end{align*}

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)