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

Optimal. Leaf size=56 \[ -\frac{a^2 A}{2 x^2}-\frac{a^2 B}{x}+2 a A c \log (x)+2 a B c x+\frac{1}{2} A c^2 x^2+\frac{1}{3} B c^2 x^3 \]

[Out]

-(a^2*A)/(2*x^2) - (a^2*B)/x + 2*a*B*c*x + (A*c^2*x^2)/2 + (B*c^2*x^3)/3 + 2*a*A*c*Log[x]

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Rubi [A]  time = 0.025971, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {766} \[ -\frac{a^2 A}{2 x^2}-\frac{a^2 B}{x}+2 a A c \log (x)+2 a B c x+\frac{1}{2} A c^2 x^2+\frac{1}{3} B c^2 x^3 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^2)/x^3,x]

[Out]

-(a^2*A)/(2*x^2) - (a^2*B)/x + 2*a*B*c*x + (A*c^2*x^2)/2 + (B*c^2*x^3)/3 + 2*a*A*c*Log[x]

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+c x^2\right )^2}{x^3} \, dx &=\int \left (2 a B c+\frac{a^2 A}{x^3}+\frac{a^2 B}{x^2}+\frac{2 a A c}{x}+A c^2 x+B c^2 x^2\right ) \, dx\\ &=-\frac{a^2 A}{2 x^2}-\frac{a^2 B}{x}+2 a B c x+\frac{1}{2} A c^2 x^2+\frac{1}{3} B c^2 x^3+2 a A c \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0067204, size = 56, normalized size = 1. \[ -\frac{a^2 A}{2 x^2}-\frac{a^2 B}{x}+2 a A c \log (x)+2 a B c x+\frac{1}{2} A c^2 x^2+\frac{1}{3} B c^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*A)/(2*x^2) - (a^2*B)/x + 2*a*B*c*x + (A*c^2*x^2)/2 + (B*c^2*x^3)/3 + 2*a*A*c*Log[x]

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Maple [A]  time = 0.007, size = 51, normalized size = 0.9 \begin{align*} -{\frac{A{a}^{2}}{2\,{x}^{2}}}-{\frac{B{a}^{2}}{x}}+2\,aBcx+{\frac{A{c}^{2}{x}^{2}}{2}}+{\frac{B{c}^{2}{x}^{3}}{3}}+2\,aAc\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*A/x^2-a^2*B/x+2*a*B*c*x+1/2*A*c^2*x^2+1/3*B*c^2*x^3+2*a*A*c*ln(x)

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Maxima [A]  time = 1.07479, size = 68, normalized size = 1.21 \begin{align*} \frac{1}{3} \, B c^{2} x^{3} + \frac{1}{2} \, A c^{2} x^{2} + 2 \, B a c x + 2 \, A a c \log \left (x\right ) - \frac{2 \, B a^{2} x + A a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*c^2*x^3 + 1/2*A*c^2*x^2 + 2*B*a*c*x + 2*A*a*c*log(x) - 1/2*(2*B*a^2*x + A*a^2)/x^2

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Fricas [A]  time = 1.83406, size = 130, normalized size = 2.32 \begin{align*} \frac{2 \, B c^{2} x^{5} + 3 \, A c^{2} x^{4} + 12 \, B a c x^{3} + 12 \, A a c x^{2} \log \left (x\right ) - 6 \, B a^{2} x - 3 \, A a^{2}}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*B*c^2*x^5 + 3*A*c^2*x^4 + 12*B*a*c*x^3 + 12*A*a*c*x^2*log(x) - 6*B*a^2*x - 3*A*a^2)/x^2

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Sympy [A]  time = 0.519386, size = 56, normalized size = 1. \begin{align*} 2 A a c \log{\left (x \right )} + \frac{A c^{2} x^{2}}{2} + 2 B a c x + \frac{B c^{2} x^{3}}{3} - \frac{A a^{2} + 2 B a^{2} x}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a*c*log(x) + A*c**2*x**2/2 + 2*B*a*c*x + B*c**2*x**3/3 - (A*a**2 + 2*B*a**2*x)/(2*x**2)

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Giac [A]  time = 1.13916, size = 69, normalized size = 1.23 \begin{align*} \frac{1}{3} \, B c^{2} x^{3} + \frac{1}{2} \, A c^{2} x^{2} + 2 \, B a c x + 2 \, A a c \log \left ({\left | x \right |}\right ) - \frac{2 \, B a^{2} x + A a^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*c^2*x^3 + 1/2*A*c^2*x^2 + 2*B*a*c*x + 2*A*a*c*log(abs(x)) - 1/2*(2*B*a^2*x + A*a^2)/x^2