3.34 \(\int \frac{(A+B x) (b x+c x^2)^3}{x^3} \, dx\)

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

[Out]

-((b*B - A*c)*(b + c*x)^4)/(4*c^2) + (B*(b + c*x)^5)/(5*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0200986, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {765} \[ \frac{B (b+c x)^5}{5 c^2}-\frac{(b+c x)^4 (b B-A c)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^3,x]

[Out]

-((b*B - A*c)*(b + c*x)^4)/(4*c^2) + (B*(b + c*x)^5)/(5*c^2)

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.0088554, size = 67, normalized size = 1.76 \[ \frac{1}{2} b^2 x^2 (3 A c+b B)+A b^3 x+\frac{1}{4} c^2 x^4 (A c+3 b B)+b c x^3 (A c+b B)+\frac{1}{5} B c^3 x^5 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^3,x]

[Out]

A*b^3*x + (b^2*(b*B + 3*A*c)*x^2)/2 + b*c*(b*B + A*c)*x^3 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^5)/5

________________________________________________________________________________________

Maple [B]  time = 0.002, size = 73, normalized size = 1.9 \begin{align*}{\frac{B{c}^{3}{x}^{5}}{5}}+{\frac{ \left ( A{c}^{3}+3\,Bb{c}^{2} \right ){x}^{4}}{4}}+{\frac{ \left ( 3\,Ab{c}^{2}+3\,B{b}^{2}c \right ){x}^{3}}{3}}+{\frac{ \left ( 3\,A{b}^{2}c+{b}^{3}B \right ){x}^{2}}{2}}+A{b}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^3,x)

[Out]

1/5*B*c^3*x^5+1/4*(A*c^3+3*B*b*c^2)*x^4+1/3*(3*A*b*c^2+3*B*b^2*c)*x^3+1/2*(3*A*b^2*c+B*b^3)*x^2+A*b^3*x

________________________________________________________________________________________

Maxima [B]  time = 1.15159, size = 93, normalized size = 2.45 \begin{align*} \frac{1}{5} \, B c^{3} x^{5} + A b^{3} x + \frac{1}{4} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} +{\left (B b^{2} c + A b c^{2}\right )} x^{3} + \frac{1}{2} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*c^3*x^5 + A*b^3*x + 1/4*(3*B*b*c^2 + A*c^3)*x^4 + (B*b^2*c + A*b*c^2)*x^3 + 1/2*(B*b^3 + 3*A*b^2*c)*x^2

________________________________________________________________________________________

Fricas [B]  time = 1.76434, size = 150, normalized size = 3.95 \begin{align*} \frac{1}{5} \, B c^{3} x^{5} + A b^{3} x + \frac{1}{4} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} +{\left (B b^{2} c + A b c^{2}\right )} x^{3} + \frac{1}{2} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*c^3*x^5 + A*b^3*x + 1/4*(3*B*b*c^2 + A*c^3)*x^4 + (B*b^2*c + A*b*c^2)*x^3 + 1/2*(B*b^3 + 3*A*b^2*c)*x^2

________________________________________________________________________________________

Sympy [B]  time = 0.077004, size = 73, normalized size = 1.92 \begin{align*} A b^{3} x + \frac{B c^{3} x^{5}}{5} + x^{4} \left (\frac{A c^{3}}{4} + \frac{3 B b c^{2}}{4}\right ) + x^{3} \left (A b c^{2} + B b^{2} c\right ) + x^{2} \left (\frac{3 A b^{2} c}{2} + \frac{B b^{3}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**3,x)

[Out]

A*b**3*x + B*c**3*x**5/5 + x**4*(A*c**3/4 + 3*B*b*c**2/4) + x**3*(A*b*c**2 + B*b**2*c) + x**2*(3*A*b**2*c/2 +
B*b**3/2)

________________________________________________________________________________________

Giac [B]  time = 1.21094, size = 97, normalized size = 2.55 \begin{align*} \frac{1}{5} \, B c^{3} x^{5} + \frac{3}{4} \, B b c^{2} x^{4} + \frac{1}{4} \, A c^{3} x^{4} + B b^{2} c x^{3} + A b c^{2} x^{3} + \frac{1}{2} \, B b^{3} x^{2} + \frac{3}{2} \, A b^{2} c x^{2} + A b^{3} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*B*c^3*x^5 + 3/4*B*b*c^2*x^4 + 1/4*A*c^3*x^4 + B*b^2*c*x^3 + A*b*c^2*x^3 + 1/2*B*b^3*x^2 + 3/2*A*b^2*c*x^2
+ A*b^3*x