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

Optimal. Leaf size=75 \[ \frac{d^2 x (3 b c-2 a d)}{b^3}-\frac{(b c-a d)^3}{b^4 (a+b x)}+\frac{3 d (b c-a d)^2 \log (a+b x)}{b^4}+\frac{d^3 x^2}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0614668, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{d^2 x (3 b c-2 a d)}{b^3}-\frac{(b c-a d)^3}{b^4 (a+b x)}+\frac{3 d (b c-a d)^2 \log (a+b x)}{b^4}+\frac{d^3 x^2}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3/(a + b*x)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0501473, size = 72, normalized size = 0.96 \[ \frac{2 b d^2 x (3 b c-2 a d)-\frac{2 (b c-a d)^3}{a+b x}+6 d (b c-a d)^2 \log (a+b x)+b^2 d^3 x^2}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3/(a + b*x)^2,x]

[Out]

(2*b*d^2*(3*b*c - 2*a*d)*x + b^2*d^3*x^2 - (2*(b*c - a*d)^3)/(a + b*x) + 6*d*(b*c - a*d)^2*Log[a + b*x])/(2*b^
4)

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Maple [B]  time = 0.007, size = 149, normalized size = 2. \begin{align*}{\frac{{d}^{3}{x}^{2}}{2\,{b}^{2}}}-2\,{\frac{a{d}^{3}x}{{b}^{3}}}+3\,{\frac{{d}^{2}xc}{{b}^{2}}}+{\frac{{a}^{3}{d}^{3}}{{b}^{4} \left ( bx+a \right ) }}-3\,{\frac{{a}^{2}c{d}^{2}}{{b}^{3} \left ( bx+a \right ) }}+3\,{\frac{a{c}^{2}d}{{b}^{2} \left ( bx+a \right ) }}-{\frac{{c}^{3}}{b \left ( bx+a \right ) }}+3\,{\frac{{d}^{3}\ln \left ( bx+a \right ){a}^{2}}{{b}^{4}}}-6\,{\frac{{d}^{2}\ln \left ( bx+a \right ) ac}{{b}^{3}}}+3\,{\frac{d\ln \left ( bx+a \right ){c}^{2}}{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05798, size = 159, normalized size = 2.12 \begin{align*} -\frac{b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}}{b^{5} x + a b^{4}} + \frac{b d^{3} x^{2} + 2 \,{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} x}{2 \, b^{3}} + \frac{3 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \log \left (b x + a\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.19866, size = 354, normalized size = 4.72 \begin{align*} \frac{b^{3} d^{3} x^{3} - 2 \, b^{3} c^{3} + 6 \, a b^{2} c^{2} d - 6 \, a^{2} b c d^{2} + 2 \, a^{3} d^{3} + 3 \,{\left (2 \, b^{3} c d^{2} - a b^{2} d^{3}\right )} x^{2} + 2 \,{\left (3 \, a b^{2} c d^{2} - 2 \, a^{2} b d^{3}\right )} x + 6 \,{\left (a b^{2} c^{2} d - 2 \, a^{2} b c d^{2} + a^{3} d^{3} +{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{5} x + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.851266, size = 100, normalized size = 1.33 \begin{align*} \frac{a^{3} d^{3} - 3 a^{2} b c d^{2} + 3 a b^{2} c^{2} d - b^{3} c^{3}}{a b^{4} + b^{5} x} + \frac{d^{3} x^{2}}{2 b^{2}} - \frac{x \left (2 a d^{3} - 3 b c d^{2}\right )}{b^{3}} + \frac{3 d \left (a d - b c\right )^{2} \log{\left (a + b x \right )}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15804, size = 225, normalized size = 3. \begin{align*} \frac{{\left (d^{3} + \frac{6 \,{\left (b^{2} c d^{2} - a b d^{3}\right )}}{{\left (b x + a\right )} b}\right )}{\left (b x + a\right )}^{2}}{2 \, b^{4}} - \frac{3 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \log \left (\frac{{\left | b x + a \right |}}{{\left (b x + a\right )}^{2}{\left | b \right |}}\right )}{b^{4}} - \frac{\frac{b^{5} c^{3}}{b x + a} - \frac{3 \, a b^{4} c^{2} d}{b x + a} + \frac{3 \, a^{2} b^{3} c d^{2}}{b x + a} - \frac{a^{3} b^{2} d^{3}}{b x + a}}{b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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