### 3.1767 $$\int \frac{a c+(b c+a d) x+b d x^2}{(a+b x)^5} \, dx$$

Optimal. Leaf size=38 $-\frac{b c-a d}{3 b^2 (a+b x)^3}-\frac{d}{2 b^2 (a+b x)^2}$

[Out]

-(b*c - a*d)/(3*b^2*(a + b*x)^3) - d/(2*b^2*(a + b*x)^2)

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Rubi [A]  time = 0.0242644, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.074, Rules used = {24, 43} $-\frac{b c-a d}{3 b^2 (a+b x)^3}-\frac{d}{2 b^2 (a+b x)^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)/(a + b*x)^5,x]

[Out]

-(b*c - a*d)/(3*b^2*(a + b*x)^3) - d/(2*b^2*(a + b*x)^2)

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

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

Mathematica [A]  time = 0.0087894, size = 27, normalized size = 0.71 $-\frac{a d+2 b c+3 b d x}{6 b^2 (a+b x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)/(a + b*x)^5,x]

[Out]

-(2*b*c + a*d + 3*b*d*x)/(6*b^2*(a + b*x)^3)

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Maple [A]  time = 0.044, size = 35, normalized size = 0.9 \begin{align*} -{\frac{d}{2\,{b}^{2} \left ( bx+a \right ) ^{2}}}-{\frac{-ad+bc}{3\,{b}^{2} \left ( bx+a \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a*c+(a*d+b*c)*x+b*d*x^2)/(b*x+a)^5,x)

[Out]

-1/2*d/b^2/(b*x+a)^2-1/3*(-a*d+b*c)/b^2/(b*x+a)^3

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Maxima [A]  time = 1.08868, size = 68, normalized size = 1.79 \begin{align*} -\frac{3 \, b d x + 2 \, b c + a d}{6 \,{\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)/(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/6*(3*b*d*x + 2*b*c + a*d)/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2)

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Fricas [A]  time = 1.78756, size = 105, normalized size = 2.76 \begin{align*} -\frac{3 \, b d x + 2 \, b c + a d}{6 \,{\left (b^{5} x^{3} + 3 \, a b^{4} x^{2} + 3 \, a^{2} b^{3} x + a^{3} b^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)/(b*x+a)^5,x, algorithm="fricas")

[Out]

-1/6*(3*b*d*x + 2*b*c + a*d)/(b^5*x^3 + 3*a*b^4*x^2 + 3*a^2*b^3*x + a^3*b^2)

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Sympy [A]  time = 0.57405, size = 53, normalized size = 1.39 \begin{align*} - \frac{a d + 2 b c + 3 b d x}{6 a^{3} b^{2} + 18 a^{2} b^{3} x + 18 a b^{4} x^{2} + 6 b^{5} x^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x**2)/(b*x+a)**5,x)

[Out]

-(a*d + 2*b*c + 3*b*d*x)/(6*a**3*b**2 + 18*a**2*b**3*x + 18*a*b**4*x**2 + 6*b**5*x**3)

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Giac [A]  time = 1.14854, size = 55, normalized size = 1.45 \begin{align*} -\frac{c}{3 \,{\left (b x + a\right )}^{3} b} - \frac{d}{2 \,{\left (b x + a\right )}^{2} b^{2}} + \frac{a d}{3 \,{\left (b x + a\right )}^{3} b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)/(b*x+a)^5,x, algorithm="giac")

[Out]

-1/3*c/((b*x + a)^3*b) - 1/2*d/((b*x + a)^2*b^2) + 1/3*a*d/((b*x + a)^3*b^2)