### 3.1821 $$\int \frac{(a+b x)^5}{(a c+(b c+a d) x+b d x^2)^3} \, dx$$

Optimal. Leaf size=59 $\frac{2 b (b c-a d)}{d^3 (c+d x)}-\frac{(b c-a d)^2}{2 d^3 (c+d x)^2}+\frac{b^2 \log (c+d x)}{d^3}$

[Out]

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

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Rubi [A]  time = 0.0432568, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.069, Rules used = {626, 43} $\frac{2 b (b c-a d)}{d^3 (c+d x)}-\frac{(b c-a d)^2}{2 d^3 (c+d x)^2}+\frac{b^2 \log (c+d x)}{d^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^5/(a*c + (b*c + a*d)*x + b*d*x^2)^3,x]

[Out]

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

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

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

Mathematica [A]  time = 0.0236377, size = 48, normalized size = 0.81 $\frac{\frac{(b c-a d) (a d+3 b c+4 b d x)}{(c+d x)^2}+2 b^2 \log (c+d x)}{2 d^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x)^5/(a*c + (b*c + a*d)*x + b*d*x^2)^3,x]

[Out]

(((b*c - a*d)*(3*b*c + a*d + 4*b*d*x))/(c + d*x)^2 + 2*b^2*Log[c + d*x])/(2*d^3)

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Maple [A]  time = 0.046, size = 92, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}\ln \left ( dx+c \right ) }{{d}^{3}}}-{\frac{{a}^{2}}{2\,d \left ( dx+c \right ) ^{2}}}+{\frac{abc}{{d}^{2} \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{2}{c}^{2}}{2\,{d}^{3} \left ( dx+c \right ) ^{2}}}-2\,{\frac{ab}{{d}^{2} \left ( dx+c \right ) }}+2\,{\frac{{b}^{2}c}{{d}^{3} \left ( dx+c \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^5/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x)

[Out]

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

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Maxima [A]  time = 1.15721, size = 108, normalized size = 1.83 \begin{align*} \frac{3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2} + 4 \,{\left (b^{2} c d - a b d^{2}\right )} x}{2 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} + \frac{b^{2} \log \left (d x + c\right )}{d^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60057, size = 205, normalized size = 3.47 \begin{align*} \frac{3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2} + 4 \,{\left (b^{2} c d - a b d^{2}\right )} x + 2 \,{\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \left (d x + c\right )}{2 \,{\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.15639, size = 80, normalized size = 1.36 \begin{align*} \frac{b^{2} \log{\left (c + d x \right )}}{d^{3}} - \frac{a^{2} d^{2} + 2 a b c d - 3 b^{2} c^{2} + x \left (4 a b d^{2} - 4 b^{2} c d\right )}{2 c^{2} d^{3} + 4 c d^{4} x + 2 d^{5} x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**5/(a*c+(a*d+b*c)*x+b*d*x**2)**3,x)

[Out]

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

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Giac [A]  time = 1.23623, size = 93, normalized size = 1.58 \begin{align*} \frac{b^{2} \log \left ({\left | d x + c \right |}\right )}{d^{3}} + \frac{4 \,{\left (b^{2} c - a b d\right )} x + \frac{3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}}{d}}{2 \,{\left (d x + c\right )}^{2} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^5/(a*c+(a*d+b*c)*x+b*d*x^2)^3,x, algorithm="giac")

[Out]

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