3.840 \(\int \frac {x^m (A+B x)}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=81 \[ \frac {x^{m+1} (a B (m+1)+A b (2-m)) \, _2F_1\left (3,m+1;m+2;-\frac {b x}{a}\right )}{3 a^4 b (m+1)}+\frac {x^{m+1} (A b-a B)}{3 a b (a+b x)^3} \]

[Out]

1/3*(A*b-B*a)*x^(1+m)/a/b/(b*x+a)^3+1/3*(A*b*(2-m)+a*B*(1+m))*x^(1+m)*hypergeom([3, 1+m],[2+m],-b*x/a)/a^4/b/(
1+m)

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Rubi [A]  time = 0.03, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {27, 78, 64} \[ \frac {x^{m+1} (a B (m+1)+A b (2-m)) \, _2F_1\left (3,m+1;m+2;-\frac {b x}{a}\right )}{3 a^4 b (m+1)}+\frac {x^{m+1} (A b-a B)}{3 a b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

((A*b - a*B)*x^(1 + m))/(3*a*b*(a + b*x)^3) + ((A*b*(2 - m) + a*B*(1 + m))*x^(1 + m)*Hypergeometric2F1[3, 1 +
m, 2 + m, -((b*x)/a)])/(3*a^4*b*(1 + m))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rubi steps

\begin {align*} \int \frac {x^m (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {x^m (A+B x)}{(a+b x)^4} \, dx\\ &=\frac {(A b-a B) x^{1+m}}{3 a b (a+b x)^3}-\frac {(A b (-2+m)-a B (1+m)) \int \frac {x^m}{(a+b x)^3} \, dx}{3 a b}\\ &=\frac {(A b-a B) x^{1+m}}{3 a b (a+b x)^3}+\frac {(A b (2-m)+a B (1+m)) x^{1+m} \, _2F_1\left (3,1+m;2+m;-\frac {b x}{a}\right )}{3 a^4 b (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 71, normalized size = 0.88 \[ \frac {x^{m+1} \left (\frac {a^3 (A b-a B)}{(a+b x)^3}-\frac {(A b (m-2)-a B (m+1)) \, _2F_1\left (3,m+1;m+2;-\frac {b x}{a}\right )}{m+1}\right )}{3 a^4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^m*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(x^(1 + m)*((a^3*(A*b - a*B))/(a + b*x)^3 - ((A*b*(-2 + m) - a*B*(1 + m))*Hypergeometric2F1[3, 1 + m, 2 + m, -
((b*x)/a)])/(1 + m)))/(3*a^4*b)

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} x^{m}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

integral((B*x + A)*x^m/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

integrate((B*x + A)*x^m/(b^2*x^2 + 2*a*b*x + a^2)^2, x)

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maple [F]  time = 1.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) x^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

int(x^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x + A\right )} x^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

integrate((B*x + A)*x^m/(b^2*x^2 + 2*a*b*x + a^2)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

int((x^m*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \left (A + B x\right )}{\left (a + b x\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Integral(x**m*(A + B*x)/(a + b*x)**4, x)

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