3.69 \(\int \frac{(a+b x^2) (A+B x+C x^2+D x^3)}{x^4} \, dx\)

Optimal. Leaf size=54 \[ -\frac{a C+A b}{x}-\frac{a A}{3 x^3}+\log (x) (a D+b B)-\frac{a B}{2 x^2}+b C x+\frac{1}{2} b D x^2 \]

[Out]

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

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Rubi [A]  time = 0.0487032, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {1802} \[ -\frac{a C+A b}{x}-\frac{a A}{3 x^3}+\log (x) (a D+b B)-\frac{a B}{2 x^2}+b C x+\frac{1}{2} b D x^2 \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x^4,x]

[Out]

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

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx &=\int \left (b C+\frac{a A}{x^4}+\frac{a B}{x^3}+\frac{A b+a C}{x^2}+\frac{b B+a D}{x}+b D x\right ) \, dx\\ &=-\frac{a A}{3 x^3}-\frac{a B}{2 x^2}-\frac{A b+a C}{x}+b C x+\frac{1}{2} b D x^2+(b B+a D) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0174944, size = 55, normalized size = 1.02 \[ \frac{-a C-A b}{x}-\frac{a A}{3 x^3}+\log (x) (a D+b B)-\frac{a B}{2 x^2}+b C x+\frac{1}{2} b D x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(A + B*x + C*x^2 + D*x^3))/x^4,x]

[Out]

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

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Maple [A]  time = 0.005, size = 51, normalized size = 0.9 \begin{align*}{\frac{bD{x}^{2}}{2}}+bCx+B\ln \left ( x \right ) b+D\ln \left ( x \right ) a-{\frac{Aa}{3\,{x}^{3}}}-{\frac{Ba}{2\,{x}^{2}}}-{\frac{Ab}{x}}-{\frac{aC}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x)

[Out]

1/2*b*D*x^2+b*C*x+B*ln(x)*b+D*ln(x)*a-1/3*a*A/x^3-1/2*a*B/x^2-1/x*A*b-1/x*a*C

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Maxima [A]  time = 1.00688, size = 66, normalized size = 1.22 \begin{align*} \frac{1}{2} \, D b x^{2} + C b x +{\left (D a + B b\right )} \log \left (x\right ) - \frac{3 \, B a x + 6 \,{\left (C a + A b\right )} x^{2} + 2 \, A a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x, algorithm="maxima")

[Out]

1/2*D*b*x^2 + C*b*x + (D*a + B*b)*log(x) - 1/6*(3*B*a*x + 6*(C*a + A*b)*x^2 + 2*A*a)/x^3

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [A]  time = 0.80509, size = 53, normalized size = 0.98 \begin{align*} C b x + \frac{D b x^{2}}{2} + \left (B b + D a\right ) \log{\left (x \right )} - \frac{2 A a + 3 B a x + x^{2} \left (6 A b + 6 C a\right )}{6 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(D*x**3+C*x**2+B*x+A)/x**4,x)

[Out]

C*b*x + D*b*x**2/2 + (B*b + D*a)*log(x) - (2*A*a + 3*B*a*x + x**2*(6*A*b + 6*C*a))/(6*x**3)

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Giac [A]  time = 1.19355, size = 68, normalized size = 1.26 \begin{align*} \frac{1}{2} \, D b x^{2} + C b x +{\left (D a + B b\right )} \log \left ({\left | x \right |}\right ) - \frac{3 \, B a x + 6 \,{\left (C a + A b\right )} x^{2} + 2 \, A a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A)/x^4,x, algorithm="giac")

[Out]

1/2*D*b*x^2 + C*b*x + (D*a + B*b)*log(abs(x)) - 1/6*(3*B*a*x + 6*(C*a + A*b)*x^2 + 2*A*a)/x^3