3.177 \(\int \frac{a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx\)

Optimal. Leaf size=32 \[ a c x+\frac{1}{2} a d x^2+\frac{1}{3} b c x^3+\frac{1}{4} b d x^4 \]

[Out]

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

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Rubi [A]  time = 0.0340129, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 54, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.019, Rules used = {1586} \[ a c x+\frac{1}{2} a d x^2+\frac{1}{3} b c x^3+\frac{1}{4} b d x^4 \]

Antiderivative was successfully verified.

[In]

Int[(a^2*c + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 + b^2*c*x^4 + b^2*d*x^5)/(a + b*x^2),x]

[Out]

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps

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

Mathematica [A]  time = 0.0018921, size = 32, normalized size = 1. \[ a c x+\frac{1}{2} a d x^2+\frac{1}{3} b c x^3+\frac{1}{4} b d x^4 \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2*c + a^2*d*x + 2*a*b*c*x^2 + 2*a*b*d*x^3 + b^2*c*x^4 + b^2*d*x^5)/(a + b*x^2),x]

[Out]

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

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Maple [A]  time = 0., size = 27, normalized size = 0.8 \begin{align*} acx+{\frac{ad{x}^{2}}{2}}+{\frac{bc{x}^{3}}{3}}+{\frac{bd{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02178, size = 35, normalized size = 1.09 \begin{align*} \frac{1}{4} \, b d x^{4} + \frac{1}{3} \, b c x^{3} + \frac{1}{2} \, a d x^{2} + a c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.25276, size = 66, normalized size = 2.06 \begin{align*} \frac{1}{4} \, b d x^{4} + \frac{1}{3} \, b c x^{3} + \frac{1}{2} \, a d x^{2} + a c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.080349, size = 29, normalized size = 0.91 \begin{align*} a c x + \frac{a d x^{2}}{2} + \frac{b c x^{3}}{3} + \frac{b d x^{4}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*c*x + a*d*x**2/2 + b*c*x**3/3 + b*d*x**4/4

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Giac [A]  time = 1.13044, size = 35, normalized size = 1.09 \begin{align*} \frac{1}{4} \, b d x^{4} + \frac{1}{3} \, b c x^{3} + \frac{1}{2} \, a d x^{2} + a c x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*d*x^5+b^2*c*x^4+2*a*b*d*x^3+2*a*b*c*x^2+a^2*d*x+a^2*c)/(b*x^2+a),x, algorithm="giac")

[Out]

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