3.1034 \(\int \frac {x}{(a+b x^n)^2 (c+d x^n)} \, dx\)

Optimal. Leaf size=143 \[ \frac {b x^2 (2 a d (1-n)-b c (2-n)) \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {b x^n}{a}\right )}{2 a^2 n (b c-a d)^2}+\frac {d^2 x^2 \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {d x^n}{c}\right )}{2 c (b c-a d)^2}+\frac {b x^2}{a n (b c-a d) \left (a+b x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {504, 597, 364} \[ \frac {b x^2 (2 a d (1-n)-b c (2-n)) \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {b x^n}{a}\right )}{2 a^2 n (b c-a d)^2}+\frac {d^2 x^2 \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {d x^n}{c}\right )}{2 c (b c-a d)^2}+\frac {b x^2}{a n (b c-a d) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/((a + b*x^n)^2*(c + d*x^n)),x]

[Out]

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

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 504

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && In
tBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 597

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, n, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 134, normalized size = 0.94 \[ \frac {x^2 \left (a \left (a d^2 n \left (a+b x^n\right ) \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {d x^n}{c}\right )+2 b c (b c-a d)\right )+b c \left (a+b x^n\right ) (b c (n-2)-2 a d (n-1)) \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {b x^n}{a}\right )\right )}{2 a^2 c n (b c-a d)^2 \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{b^{2} d x^{3 \, n} + a^{2} c + {\left (b^{2} c + 2 \, a b d\right )} x^{2 \, n} + {\left (2 \, a b c + a^{2} d\right )} x^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/(b^2*d*x^(3*n) + a^2*c + (b^2*c + 2*a*b*d)*x^(2*n) + (2*a*b*c + a^2*d)*x^n), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.97, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (b \,x^{n}+a \right )^{2} \left (d \,x^{n}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\left (a+b\,x^n\right )}^2\,\left (c+d\,x^n\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a + b*x^n)^2*(c + d*x^n)),x)

[Out]

int(x/((a + b*x^n)^2*(c + d*x^n)), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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