3.1032 \(\int \frac{(e x)^m}{(a+b x^n)^2 (c+d x^n)} \, dx\)

Optimal. Leaf size=175 \[ \frac{b (e x)^{m+1} (a d (m-2 n+1)-b c (m-n+1)) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^2 e (m+1) n (b c-a d)^2}+\frac{d^2 (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c e (m+1) (b c-a d)^2}+\frac{b (e x)^{m+1}}{a e n (b c-a d) \left (a+b x^n\right )} \]

[Out]

(b*(e*x)^(1 + m))/(a*(b*c - a*d)*e*n*(a + b*x^n)) + (b*(a*d*(1 + m - 2*n) - b*c*(1 + m - n))*(e*x)^(1 + m)*Hyp
ergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a^2*(b*c - a*d)^2*e*(1 + m)*n) + (d^2*(e*x)^(1 + m
)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(b*c - a*d)^2*e*(1 + m))

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Rubi [A]  time = 0.288443, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {504, 597, 364} \[ \frac{b (e x)^{m+1} (a d (m-2 n+1)-b c (m-n+1)) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^2 e (m+1) n (b c-a d)^2}+\frac{d^2 (e x)^{m+1} \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c e (m+1) (b c-a d)^2}+\frac{b (e x)^{m+1}}{a e n (b c-a d) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(e*x)^(1 + m))/(a*(b*c - a*d)*e*n*(a + b*x^n)) + (b*(a*d*(1 + m - 2*n) - b*c*(1 + m - n))*(e*x)^(1 + m)*Hyp
ergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((b*x^n)/a)])/(a^2*(b*c - a*d)^2*e*(1 + m)*n) + (d^2*(e*x)^(1 + m
)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(b*c - a*d)^2*e*(1 + m))

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]

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

Mathematica [A]  time = 0.185273, size = 141, normalized size = 0.81 \[ \frac{x (e x)^m \left (\frac{b^2 c-a b d}{a^2 n+a b n x^n}+\frac{b (a d (m-2 n+1)-b c (m-n+1)) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{b x^n}{a}\right )}{a^2 (m+1) n}+\frac{d^2 \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{d x^n}{c}\right )}{c m+c}\right )}{(b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.081, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m}}{ \left ( a+b{x}^{n} \right ) ^{2} \left ( c+d{x}^{n} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} e^{m} \int \frac{x^{m}}{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 e^{m} x x^{m}}{a^{2} b c n - a^{3} d n +{\left (a b^{2} c n - a^{2} b d n\right )} x^{n}} -{\left (b^{2} c e^{m}{\left (m - n + 1\right )} - a b d e^{m}{\left (m - 2 \, n + 1\right )}\right )} \int \frac{x^{m}}{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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*e^m*integrate(x^m/(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*e^
m*x*x^m/(a^2*b*c*n - a^3*d*n + (a*b^2*c*n - a^2*b*d*n)*x^n) - (b^2*c*e^m*(m - n + 1) - a*b*d*e^m*(m - 2*n + 1)
)*integrate(x^m/(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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (e x\right )^{m}}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*x)**m/((a + b*x**n)**2*(c + d*x**n)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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