3.2171 \(\int \frac{(a+b x) (a c+b c x)^m}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=27 \[ -\frac{c^4 (a c+b c x)^{m-4}}{b (4-m)} \]

[Out]

-((c^4*(a*c + b*c*x)^(-4 + m))/(b*(4 - m)))

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Rubi [A]  time = 0.0175843, antiderivative size = 27, 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 = {21, 27, 32} \[ -\frac{c^4 (a c+b c x)^{m-4}}{b (4-m)} \]

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]

-((c^4*(a*c + b*c*x)^(-4 + m))/(b*(4 - m)))

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 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 &=\frac{\int \frac{(a c+b c x)^{1+m}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx}{c}\\ &=\frac{\int \frac{(a c+b c x)^{1+m}}{(a+b x)^6} \, dx}{c}\\ &=c^5 \int (a c+b c x)^{-5+m} \, dx\\ &=-\frac{c^4 (a c+b c x)^{-4+m}}{b (4-m)}\\ \end{align*}

Mathematica [A]  time = 0.0128758, size = 25, normalized size = 0.93 \[ \frac{(c (a+b x))^m}{b (m-4) (a+b x)^4} \]

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/(b*(-4 + m)*(a + b*x)^4)

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Maple [A]  time = 0.003, size = 45, normalized size = 1.7 \begin{align*}{\frac{ \left ( bcx+ac \right ) ^{m}}{ \left ( bx+a \right ) ^{2} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) b \left ( -4+m \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*c*x+a*c)^m/(b*x+a)^2/(b^2*x^2+2*a*b*x+a^2)/b/(-4+m)

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Maxima [B]  time = 1.13534, size = 292, normalized size = 10.81 \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 = 1.63775, size = 201, normalized size = 7.44 \begin{align*} \frac{{\left (b c x + a c\right )}^{m}}{a^{4} b m - 4 \, a^{4} b +{\left (b^{5} m - 4 \, b^{5}\right )} x^{4} + 4 \,{\left (a b^{4} m - 4 \, a b^{4}\right )} x^{3} + 6 \,{\left (a^{2} b^{3} m - 4 \, a^{2} b^{3}\right )} x^{2} + 4 \,{\left (a^{3} b^{2} m - 4 \, a^{3} b^{2}\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]

(b*c*x + a*c)^m/(a^4*b*m - 4*a^4*b + (b^5*m - 4*b^5)*x^4 + 4*(a*b^4*m - 4*a*b^4)*x^3 + 6*(a^2*b^3*m - 4*a^2*b^
3)*x^2 + 4*(a^3*b^2*m - 4*a^3*b^2)*x)

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Sympy [A]  time = 3.5866, size = 136, normalized size = 5.04 \begin{align*} \begin{cases} \frac{c^{4} x}{a} & \text{for}\: b = 0 \wedge m = 4 \\\frac{x \left (a c\right )^{m}}{a^{5}} & \text{for}\: b = 0 \\\frac{c^{4} \log{\left (\frac{a}{b} + x \right )}}{b} & \text{for}\: m = 4 \\\frac{\left (a c + b c x\right )^{m}}{a^{4} b m - 4 a^{4} b + 4 a^{3} b^{2} m x - 16 a^{3} b^{2} x + 6 a^{2} b^{3} m x^{2} - 24 a^{2} b^{3} x^{2} + 4 a b^{4} m x^{3} - 16 a b^{4} x^{3} + b^{5} m x^{4} - 4 b^{5} x^{4}} & \text{otherwise} \end{cases} \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((c**4*x/a, Eq(b, 0) & Eq(m, 4)), (x*(a*c)**m/a**5, Eq(b, 0)), (c**4*log(a/b + x)/b, Eq(m, 4)), ((a*c
+ b*c*x)**m/(a**4*b*m - 4*a**4*b + 4*a**3*b**2*m*x - 16*a**3*b**2*x + 6*a**2*b**3*m*x**2 - 24*a**2*b**3*x**2
+ 4*a*b**4*m*x**3 - 16*a*b**4*x**3 + b**5*m*x**4 - 4*b**5*x**4), 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)