3.38 \(\int \frac {(e x)^m (A+B x^n)}{(c+d x^n)^3} \, dx\)

Optimal. Leaf size=112 \[ \frac {(e x)^{m+1} (B c (m+1)-A d (m-2 n+1)) \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{2 c^3 d e (m+1) n}-\frac {(e x)^{m+1} (B c-A d)}{2 c d e n \left (c+d x^n\right )^2} \]

[Out]

-1/2*(-A*d+B*c)*(e*x)^(1+m)/c/d/e/n/(c+d*x^n)^2+1/2*(B*c*(1+m)-A*d*(1+m-2*n))*(e*x)^(1+m)*hypergeom([2, (1+m)/
n],[(1+m+n)/n],-d*x^n/c)/c^3/d/e/(1+m)/n

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Rubi [A]  time = 0.06, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 364} \[ \frac {(e x)^{m+1} (B c (m+1)-A d (m-2 n+1)) \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )}{2 c^3 d e (m+1) n}-\frac {(e x)^{m+1} (B c-A d)}{2 c d e n \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(A + B*x^n))/(c + d*x^n)^3,x]

[Out]

-((B*c - A*d)*(e*x)^(1 + m))/(2*c*d*e*n*(c + d*x^n)^2) + ((B*c*(1 + m) - A*d*(1 + m - 2*n))*(e*x)^(1 + m)*Hype
rgeometric2F1[2, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(2*c^3*d*e*(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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (A+B x^n\right )}{\left (c+d x^n\right )^3} \, dx &=-\frac {(B c-A d) (e x)^{1+m}}{2 c d e n \left (c+d x^n\right )^2}+\frac {(B c (1+m)-A d (1+m-2 n)) \int \frac {(e x)^m}{\left (c+d x^n\right )^2} \, dx}{2 c d n}\\ &=-\frac {(B c-A d) (e x)^{1+m}}{2 c d e n \left (c+d x^n\right )^2}+\frac {(B c (1+m)-A d (1+m-2 n)) (e x)^{1+m} \, _2F_1\left (2,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {d x^n}{c}\right )}{2 c^3 d e (1+m) n}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 83, normalized size = 0.74 \[ \frac {x (e x)^m \left ((A d-B c) \, _2F_1\left (3,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )+B c \, _2F_1\left (2,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {d x^n}{c}\right )\right )}{c^3 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(A + B*x^n))/(c + d*x^n)^3,x]

[Out]

(x*(e*x)^m*(B*c*Hypergeometric2F1[2, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)] + (-(B*c) + A*d)*Hypergeometric2F
1[3, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)]))/(c^3*d*(1 + m))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{d^{3} x^{3 \, n} + 3 \, c d^{2} x^{2 \, n} + 3 \, c^{2} d x^{n} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(A+B*x^n)/(c+d*x^n)^3,x, algorithm="fricas")

[Out]

integral((B*x^n + A)*(e*x)^m/(d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{n} + A\right )} \left (e x\right )^{m}}{{\left (d x^{n} + c\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(A+B*x^n)/(c+d*x^n)^3,x, algorithm="giac")

[Out]

integrate((B*x^n + A)*(e*x)^m/(d*x^n + c)^3, x)

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maple [F]  time = 0.68, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{n}+A \right ) \left (e x \right )^{m}}{\left (d \,x^{n}+c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x^n+A)/(d*x^n+c)^3,x)

[Out]

int((e*x)^m*(B*x^n+A)/(d*x^n+c)^3,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(A+B*x^n)/(c+d*x^n)^3,x, algorithm="maxima")

[Out]

-((m^2 - m*(n - 2) - n + 1)*B*c*e^m - (m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*A*d*e^m)*integrate(1/2*x^m/(c^2*d^
2*n^2*x^n + c^3*d*n^2), x) + 1/2*((B*c^2*e^m*(m - n + 1) - A*c*d*e^m*(m - 3*n + 1))*x*x^m - (A*d^2*e^m*(m - 2*
n + 1) - B*c*d*e^m*(m + 1))*x*e^(m*log(x) + n*log(x)))/(c^2*d^3*n^2*x^(2*n) + 2*c^3*d^2*n^2*x^n + c^4*d*n^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )}{{\left (c+d\,x^n\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^m*(A + B*x^n))/(c + d*x^n)^3,x)

[Out]

int(((e*x)^m*(A + B*x^n))/(c + d*x^n)^3, x)

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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((e*x)**m*(A+B*x**n)/(c+d*x**n)**3,x)

[Out]

Timed out

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