3.2164 \(\int \frac{(A+B x) (a c+b c x)^m}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)
Optimal. Leaf size=64 \[ -\frac{c^5 (A b-a B) (a c+b c x)^{m-5}}{b^2 (5-m)}-\frac{B c^4 (a c+b c x)^{m-4}}{b^2 (4-m)} \]
[Out]
-(((A*b - a*B)*c^5*(a*c + b*c*x)^(-5 + m))/(b^2*(5 - m))) - (B*c^4*(a*c + b*c*x)^(-4 + m))/(b^2*(4 - m))
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Rubi [A] time = 0.0489636, antiderivative size = 64, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used =
{27, 21, 43} \[ -\frac{c^5 (A b-a B) (a c+b c x)^{m-5}}{b^2 (5-m)}-\frac{B c^4 (a c+b c x)^{m-4}}{b^2 (4-m)} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*(a*c + b*c*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
[Out]
-(((A*b - a*B)*c^5*(a*c + b*c*x)^(-5 + m))/(b^2*(5 - m))) - (B*c^4*(a*c + b*c*x)^(-4 + m))/(b^2*(4 - 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 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{(A+B x) (a c+b c x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{(A+B x) (a c+b c x)^m}{(a+b x)^6} \, dx\\ &=c^6 \int (A+B x) (a c+b c x)^{-6+m} \, dx\\ &=c^6 \int \left (\frac{(A b-a B) (a c+b c x)^{-6+m}}{b}+\frac{B (a c+b c x)^{-5+m}}{b c}\right ) \, dx\\ &=-\frac{(A b-a B) c^5 (a c+b c x)^{-5+m}}{b^2 (5-m)}-\frac{B c^4 (a c+b c x)^{-4+m}}{b^2 (4-m)}\\ \end{align*}
Mathematica [A] time = 0.0387008, size = 48, normalized size = 0.75 \[ \frac{(c (a+b x))^m (-a B+A b (m-4)+b B (m-5) x)}{b^2 (m-5) (m-4) (a+b x)^5} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*(a*c + b*c*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
[Out]
((c*(a + b*x))^m*(-(a*B) + A*b*(-4 + m) + b*B*(-5 + m)*x))/(b^2*(-5 + m)*(-4 + m)*(a + b*x)^5)
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Maple [A] time = 0.003, size = 73, normalized size = 1.1 \begin{align*}{\frac{ \left ( Bbmx+Abm-5\,bBx-4\,Ab-aB \right ) \left ( bcx+ac \right ) ^{m}}{ \left ( bx+a \right ) \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{2}{b}^{2} \left ({m}^{2}-9\,m+20 \right ) }} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^3,x)
[Out]
(B*b*m*x+A*b*m-5*B*b*x-4*A*b-B*a)*(b*c*x+a*c)^m/(b*x+a)/(b^2*x^2+2*a*b*x+a^2)^2/b^2/(m^2-9*m+20)
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Maxima [B] time = 1.21029, size = 292, normalized size = 4.56 \begin{align*} \frac{{\left (b c^{m}{\left (m - 5\right )} x - a c^{m}\right )}{\left (b x + a\right )}^{m} B}{{\left (m^{2} - 9 \, m + 20\right )} b^{7} x^{5} + 5 \,{\left (m^{2} - 9 \, m + 20\right )} a b^{6} x^{4} + 10 \,{\left (m^{2} - 9 \, m + 20\right )} a^{2} b^{5} x^{3} + 10 \,{\left (m^{2} - 9 \, m + 20\right )} a^{3} b^{4} x^{2} + 5 \,{\left (m^{2} - 9 \, m + 20\right )} a^{4} b^{3} x +{\left (m^{2} - 9 \, m + 20\right )} a^{5} b^{2}} + \frac{{\left (b x + a\right )}^{m} A c^{m}}{b^{6}{\left (m - 5\right )} x^{5} + 5 \, a b^{5}{\left (m - 5\right )} x^{4} + 10 \, a^{2} b^{4}{\left (m - 5\right )} x^{3} + 10 \, a^{3} b^{3}{\left (m - 5\right )} x^{2} + 5 \, a^{4} b^{2}{\left (m - 5\right )} x + a^{5} b{\left (m - 5\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
[Out]
(b*c^m*(m - 5)*x - a*c^m)*(b*x + a)^m*B/((m^2 - 9*m + 20)*b^7*x^5 + 5*(m^2 - 9*m + 20)*a*b^6*x^4 + 10*(m^2 - 9
*m + 20)*a^2*b^5*x^3 + 10*(m^2 - 9*m + 20)*a^3*b^4*x^2 + 5*(m^2 - 9*m + 20)*a^4*b^3*x + (m^2 - 9*m + 20)*a^5*b
^2) + (b*x + a)^m*A*c^m/(b^6*(m - 5)*x^5 + 5*a*b^5*(m - 5)*x^4 + 10*a^2*b^4*(m - 5)*x^3 + 10*a^3*b^3*(m - 5)*x
^2 + 5*a^4*b^2*(m - 5)*x + a^5*b*(m - 5))
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Fricas [B] time = 1.58882, size = 439, normalized size = 6.86 \begin{align*} \frac{{\left (A b m - B a - 4 \, A b +{\left (B b m - 5 \, B b\right )} x\right )}{\left (b c x + a c\right )}^{m}}{a^{5} b^{2} m^{2} - 9 \, a^{5} b^{2} m + 20 \, a^{5} b^{2} +{\left (b^{7} m^{2} - 9 \, b^{7} m + 20 \, b^{7}\right )} x^{5} + 5 \,{\left (a b^{6} m^{2} - 9 \, a b^{6} m + 20 \, a b^{6}\right )} x^{4} + 10 \,{\left (a^{2} b^{5} m^{2} - 9 \, a^{2} b^{5} m + 20 \, a^{2} b^{5}\right )} x^{3} + 10 \,{\left (a^{3} b^{4} m^{2} - 9 \, a^{3} b^{4} m + 20 \, a^{3} b^{4}\right )} x^{2} + 5 \,{\left (a^{4} b^{3} m^{2} - 9 \, a^{4} b^{3} m + 20 \, a^{4} b^{3}\right )} x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
[Out]
(A*b*m - B*a - 4*A*b + (B*b*m - 5*B*b)*x)*(b*c*x + a*c)^m/(a^5*b^2*m^2 - 9*a^5*b^2*m + 20*a^5*b^2 + (b^7*m^2 -
9*b^7*m + 20*b^7)*x^5 + 5*(a*b^6*m^2 - 9*a*b^6*m + 20*a*b^6)*x^4 + 10*(a^2*b^5*m^2 - 9*a^2*b^5*m + 20*a^2*b^5
)*x^3 + 10*(a^3*b^4*m^2 - 9*a^3*b^4*m + 20*a^3*b^4)*x^2 + 5*(a^4*b^3*m^2 - 9*a^4*b^3*m + 20*a^4*b^3)*x)
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Sympy [A] time = 4.61326, size = 1367, normalized size = 21.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(b*c*x+a*c)**m/(b**2*x**2+2*a*b*x+a**2)**3,x)
[Out]
Piecewise(((a*c)**m*(A*x + B*x**2/2)/a**6, Eq(b, 0)), (-3*A*a*b*c**4/(5*a**2*b**2 + 5*a*b**3*x) + 2*A*b**2*c**
4*x/(5*a**2*b**2 + 5*a*b**3*x) + 5*B*a**2*c**4*log(a/b + x)/(5*a**2*b**2 + 5*a*b**3*x) + 3*B*a**2*c**4/(5*a**2
*b**2 + 5*a*b**3*x) + 5*B*a*b*c**4*x*log(a/b + x)/(5*a**2*b**2 + 5*a*b**3*x) - 2*B*a*b*c**4*x/(5*a**2*b**2 + 5
*a*b**3*x), Eq(m, 4)), (A*c**5*log(a/b + x)/b - B*a*c**5*log(a/b + x)/b**2 + B*c**5*x/b, Eq(m, 5)), (A*b*m*(a*
c + b*c*x)**m/(a**5*b**2*m**2 - 9*a**5*b**2*m + 20*a**5*b**2 + 5*a**4*b**3*m**2*x - 45*a**4*b**3*m*x + 100*a**
4*b**3*x + 10*a**3*b**4*m**2*x**2 - 90*a**3*b**4*m*x**2 + 200*a**3*b**4*x**2 + 10*a**2*b**5*m**2*x**3 - 90*a**
2*b**5*m*x**3 + 200*a**2*b**5*x**3 + 5*a*b**6*m**2*x**4 - 45*a*b**6*m*x**4 + 100*a*b**6*x**4 + b**7*m**2*x**5
- 9*b**7*m*x**5 + 20*b**7*x**5) - 4*A*b*(a*c + b*c*x)**m/(a**5*b**2*m**2 - 9*a**5*b**2*m + 20*a**5*b**2 + 5*a*
*4*b**3*m**2*x - 45*a**4*b**3*m*x + 100*a**4*b**3*x + 10*a**3*b**4*m**2*x**2 - 90*a**3*b**4*m*x**2 + 200*a**3*
b**4*x**2 + 10*a**2*b**5*m**2*x**3 - 90*a**2*b**5*m*x**3 + 200*a**2*b**5*x**3 + 5*a*b**6*m**2*x**4 - 45*a*b**6
*m*x**4 + 100*a*b**6*x**4 + b**7*m**2*x**5 - 9*b**7*m*x**5 + 20*b**7*x**5) - B*a*(a*c + b*c*x)**m/(a**5*b**2*m
**2 - 9*a**5*b**2*m + 20*a**5*b**2 + 5*a**4*b**3*m**2*x - 45*a**4*b**3*m*x + 100*a**4*b**3*x + 10*a**3*b**4*m*
*2*x**2 - 90*a**3*b**4*m*x**2 + 200*a**3*b**4*x**2 + 10*a**2*b**5*m**2*x**3 - 90*a**2*b**5*m*x**3 + 200*a**2*b
**5*x**3 + 5*a*b**6*m**2*x**4 - 45*a*b**6*m*x**4 + 100*a*b**6*x**4 + b**7*m**2*x**5 - 9*b**7*m*x**5 + 20*b**7*
x**5) + B*b*m*x*(a*c + b*c*x)**m/(a**5*b**2*m**2 - 9*a**5*b**2*m + 20*a**5*b**2 + 5*a**4*b**3*m**2*x - 45*a**4
*b**3*m*x + 100*a**4*b**3*x + 10*a**3*b**4*m**2*x**2 - 90*a**3*b**4*m*x**2 + 200*a**3*b**4*x**2 + 10*a**2*b**5
*m**2*x**3 - 90*a**2*b**5*m*x**3 + 200*a**2*b**5*x**3 + 5*a*b**6*m**2*x**4 - 45*a*b**6*m*x**4 + 100*a*b**6*x**
4 + b**7*m**2*x**5 - 9*b**7*m*x**5 + 20*b**7*x**5) - 5*B*b*x*(a*c + b*c*x)**m/(a**5*b**2*m**2 - 9*a**5*b**2*m
+ 20*a**5*b**2 + 5*a**4*b**3*m**2*x - 45*a**4*b**3*m*x + 100*a**4*b**3*x + 10*a**3*b**4*m**2*x**2 - 90*a**3*b*
*4*m*x**2 + 200*a**3*b**4*x**2 + 10*a**2*b**5*m**2*x**3 - 90*a**2*b**5*m*x**3 + 200*a**2*b**5*x**3 + 5*a*b**6*
m**2*x**4 - 45*a*b**6*m*x**4 + 100*a*b**6*x**4 + b**7*m**2*x**5 - 9*b**7*m*x**5 + 20*b**7*x**5), True))
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (b c x + a c\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(b*c*x+a*c)^m/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
[Out]
integrate((B*x + A)*(b*c*x + a*c)^m/(b^2*x^2 + 2*a*b*x + a^2)^3, x)