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

Optimal. Leaf size=55 \[ \frac {1}{2} A b^2 x^2+\frac {1}{4} c x^4 (A c+2 b B)+\frac {1}{3} b x^3 (2 A c+b B)+\frac {1}{5} B c^2 x^5 \]

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Rubi [A]  time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {765} \begin {gather*} \frac {1}{2} A b^2 x^2+\frac {1}{4} c x^4 (A c+2 b B)+\frac {1}{3} b x^3 (2 A c+b B)+\frac {1}{5} B c^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^2)/x,x]

[Out]

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

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 )^2}{x} \, dx &=\int \left (A b^2 x+b (b B+2 A c) x^2+c (2 b B+A c) x^3+B c^2 x^4\right ) \, dx\\ &=\frac {1}{2} A b^2 x^2+\frac {1}{3} b (b B+2 A c) x^3+\frac {1}{4} c (2 b B+A c) x^4+\frac {1}{5} B c^2 x^5\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 0.89 \begin {gather*} \frac {1}{60} x^2 \left (30 A b^2+15 c x^2 (A c+2 b B)+20 b x (2 A c+b B)+12 B c^2 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(30*A*b^2 + 20*b*(b*B + 2*A*c)*x + 15*c*(2*b*B + A*c)*x^2 + 12*B*c^2*x^3))/60

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IntegrateAlgebraic [A]  time = 0.03, size = 65, normalized size = 1.18 \begin {gather*} \frac {1}{2} A b^2 x^2+\frac {2}{3} A b c x^3+\frac {1}{4} A c^2 x^4+\frac {1}{3} b^2 B x^3+\frac {1}{2} b B c x^4+\frac {1}{5} B c^2 x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{2} \, A b^{2} x^{2} + \frac {1}{4} \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{2} + 2 \, A b c\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 53, normalized size = 0.96 \begin {gather*} \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{2} \, B b c x^{4} + \frac {1}{4} \, A c^{2} x^{4} + \frac {1}{3} \, B b^{2} x^{3} + \frac {2}{3} \, A b c x^{3} + \frac {1}{2} \, A b^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 52, normalized size = 0.95 \begin {gather*} \frac {B \,c^{2} x^{5}}{5}+\frac {A \,b^{2} x^{2}}{2}+\frac {\left (A \,c^{2}+2 b B c \right ) x^{4}}{4}+\frac {\left (2 A b c +b^{2} B \right ) x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 51, normalized size = 0.93 \begin {gather*} \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{2} \, A b^{2} x^{2} + \frac {1}{4} \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{2} + 2 \, A b c\right )} x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 51, normalized size = 0.93 \begin {gather*} x^3\,\left (\frac {B\,b^2}{3}+\frac {2\,A\,c\,b}{3}\right )+x^4\,\left (\frac {A\,c^2}{4}+\frac {B\,b\,c}{2}\right )+\frac {A\,b^2\,x^2}{2}+\frac {B\,c^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^2*(A + B*x))/x,x)

[Out]

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

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sympy [A]  time = 0.08, size = 54, normalized size = 0.98 \begin {gather*} \frac {A b^{2} x^{2}}{2} + \frac {B c^{2} x^{5}}{5} + x^{4} \left (\frac {A c^{2}}{4} + \frac {B b c}{2}\right ) + x^{3} \left (\frac {2 A b c}{3} + \frac {B b^{2}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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