3.11.32 \(\int \frac {(e x)^m}{(a+b x^n)^2 (c+d x^n)} \, dx\) [1032]

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

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 175, normalized size of antiderivative = 1.00, 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 = {518, 611, 371} \begin {gather*} \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 )} \end {gather*}

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 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 518

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] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 611

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 {(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*}

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Mathematica [A]
time = 0.18, size = 141, normalized size = 0.81 \begin {gather*} \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 (1+m-2 n)-b c (1+m-n)) \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {b x^n}{a}\right )}{a^2 (1+m) n}+\frac {d^2 \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{c+c m}\right )}{(b c-a d)^2} \end {gather*}

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 = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m}}{\left (a +b \,x^{n}\right )^{2} \left (c +d \,x^{n}\right )}\, dx\]

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 = 0.00, 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((e*x)^m/(a+b*x^n)^2/(c+d*x^n),x, algorithm="maxima")

[Out]

d^2*integrate(e^(m*log(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*x*e^(m*log(x) + m)/(a^2*b*c*n - a^3*d*n + (a*b^2*c*n - a^2*b*d*n)*x^n) - ((m*e^m - (n - 1)*e^m)*b^2*c -
(m*e^m - (2*n - 1)*e^m)*a*b*d)*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 = 0.00, 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((e*x)^m/(a+b*x^n)^2/(c+d*x^n),x, algorithm="fricas")

[Out]

integral((x*e)^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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^m}{{\left (a+b\,x^n\right )}^2\,\left (c+d\,x^n\right )} \,d x \end {gather*}

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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