3.1788 \(\int \frac{(a c+(b c+a d) x+b d x^2)^3}{(a+b x)^2} \, dx\)

Optimal. Leaf size=38 \[ \frac{b (c+d x)^5}{5 d^2}-\frac{(c+d x)^4 (b c-a d)}{4 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0228092, antiderivative size = 38, 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} \[ \frac{b (c+d x)^5}{5 d^2}-\frac{(c+d x)^4 (b c-a d)}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0077657, size = 67, normalized size = 1.76 \[ \frac{1}{2} c^2 x^2 (3 a d+b c)+\frac{1}{4} d^2 x^4 (a d+3 b c)+c d x^3 (a d+b c)+a c^3 x+\frac{1}{5} b d^3 x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.041, size = 94, normalized size = 2.5 \begin{align*}{\frac{{d}^{3}b{x}^{5}}{5}}+{\frac{ \left ( 2\,c{d}^{2}b+{d}^{2} \left ( ad+bc \right ) \right ){x}^{4}}{4}}+{\frac{ \left ({c}^{2}bd+2\,cd \left ( ad+bc \right ) +ac{d}^{2} \right ){x}^{3}}{3}}+{\frac{ \left ({c}^{2} \left ( ad+bc \right ) +2\,a{c}^{2}d \right ){x}^{2}}{2}}+a{c}^{3}x \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)^3/(b*x+a)^2,x)

[Out]

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

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Maxima [B]  time = 1.01124, size = 93, normalized size = 2.45 \begin{align*} \frac{1}{5} \, b d^{3} x^{5} + a c^{3} x + \frac{1}{4} \,{\left (3 \, b c d^{2} + a d^{3}\right )} x^{4} +{\left (b c^{2} d + a c d^{2}\right )} x^{3} + \frac{1}{2} \,{\left (b c^{3} + 3 \, a c^{2} d\right )} x^{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)^3/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.58697, size = 150, normalized size = 3.95 \begin{align*} \frac{1}{5} \, b d^{3} x^{5} + a c^{3} x + \frac{1}{4} \,{\left (3 \, b c d^{2} + a d^{3}\right )} x^{4} +{\left (b c^{2} d + a c d^{2}\right )} x^{3} + \frac{1}{2} \,{\left (b c^{3} + 3 \, a c^{2} d\right )} x^{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)^3/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.240515, size = 73, normalized size = 1.92 \begin{align*} a c^{3} x + \frac{b d^{3} x^{5}}{5} + x^{4} \left (\frac{a d^{3}}{4} + \frac{3 b c d^{2}}{4}\right ) + x^{3} \left (a c d^{2} + b c^{2} d\right ) + x^{2} \left (\frac{3 a c^{2} d}{2} + \frac{b c^{3}}{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)**3/(b*x+a)**2,x)

[Out]

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

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Giac [B]  time = 1.16105, size = 209, normalized size = 5.5 \begin{align*} \frac{{\left (\frac{10 \, b^{3} c^{3}}{{\left (b x + a\right )}^{3}} + \frac{20 \, b^{2} c^{2} d}{{\left (b x + a\right )}^{2}} - \frac{30 \, a b^{2} c^{2} d}{{\left (b x + a\right )}^{3}} + \frac{15 \, b c d^{2}}{b x + a} - \frac{40 \, a b c d^{2}}{{\left (b x + a\right )}^{2}} + \frac{30 \, a^{2} b c d^{2}}{{\left (b x + a\right )}^{3}} - \frac{15 \, a d^{3}}{b x + a} + \frac{20 \, a^{2} d^{3}}{{\left (b x + a\right )}^{2}} - \frac{10 \, a^{3} d^{3}}{{\left (b x + a\right )}^{3}} + 4 \, d^{3}\right )}{\left (b x + a\right )}^{5}}{20 \, b^{4}} \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)^3/(b*x+a)^2,x, algorithm="giac")

[Out]

1/20*(10*b^3*c^3/(b*x + a)^3 + 20*b^2*c^2*d/(b*x + a)^2 - 30*a*b^2*c^2*d/(b*x + a)^3 + 15*b*c*d^2/(b*x + a) -
40*a*b*c*d^2/(b*x + a)^2 + 30*a^2*b*c*d^2/(b*x + a)^3 - 15*a*d^3/(b*x + a) + 20*a^2*d^3/(b*x + a)^2 - 10*a^3*d
^3/(b*x + a)^3 + 4*d^3)*(b*x + a)^5/b^4