3.19.87 \(\int \frac {(A+B x) (d+e x)^m}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [1887]

Optimal. Leaf size=126 \[ -\frac {(A b-a B) (d+e x)^{1+m}}{3 b (b d-a e) (a+b x)^3}-\frac {e^2 (b (3 B d-A e (2-m))-a B e (1+m)) (d+e x)^{1+m} \, _2F_1\left (3,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{3 b (b d-a e)^4 (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 125, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {27, 79, 70} \begin {gather*} -\frac {e^2 (d+e x)^{m+1} (-a B e (m+1)-A b e (2-m)+3 b B d) \, _2F_1\left (3,m+1;m+2;\frac {b (d+e x)}{b d-a e}\right )}{3 b (m+1) (b d-a e)^4}-\frac {(A b-a B) (d+e x)^{m+1}}{3 b (a+b x)^3 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((A*b - a*B)*(d + e*x)^(1 + m))/(b*(b*d - a*e)*(a + b*x)^3) - (e^2*(3*b*B*d - A*b*e*(2 - m) - a*B*e*(1 +
m))*(d + e*x)^(1 + m)*Hypergeometric2F1[3, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/(3*b*(b*d - a*e)^4*(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 70

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

Rule 79

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] || L
tQ[p, n]))))

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 109, normalized size = 0.87 \begin {gather*} \frac {(d+e x)^{1+m} \left (\frac {-A b+a B}{(a+b x)^3}-\frac {e^2 (3 b B d+A b e (-2+m)-a B e (1+m)) \, _2F_1\left (3,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^3 (1+m)}\right )}{3 b (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*((-(A*b) + a*B)/(a + b*x)^3 - (e^2*(3*b*B*d + A*b*e*(-2 + m) - a*B*e*(1 + m))*Hypergeometri
c2F1[3, 1 + m, 2 + m, (b*(d + e*x))/(b*d - a*e)])/((b*d - a*e)^3*(1 + m))))/(3*b*(b*d - a*e))

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Maple [F]
time = 0.65, size = 0, normalized size = 0.00 \[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((B*x+A)*(e*x+d)^m/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integral((B*x + A)*(x*e + d)^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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (d + e x\right )^{m}}{\left (a + b x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((B*x + A)*(x*e + d)^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 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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