3.2.58 \(\int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx\) [158]

Optimal. Leaf size=25 \[ a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[Out]

a^2*x+2/3*a*b*x^3+1/5*b^2*x^5

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Rubi [A]
time = 0.03, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {1600} \begin {gather*} a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \end {gather*}

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)/(c + d*x),x]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

Rule 1600

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

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Mathematica [A]
time = 0.00, size = 25, normalized size = 1.00 \begin {gather*} a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \end {gather*}

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)/(c + d*x),x]

[Out]

a^2*x + (2*a*b*x^3)/3 + (b^2*x^5)/5

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Maple [A]
time = 0.19, size = 22, normalized size = 0.88

method	result	size
default	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
norman	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
risch	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
gosper	\(\frac {x \left (3 b^{2} x^{4}+10 a b \,x^{2}+15 a^{2}\right )}{15}\)	\(25\)

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)/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

a^2*x+2/3*a*b*x^3+1/5*b^2*x^5

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Maxima [A]
time = 0.27, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \end {gather*}

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)/(d*x+c),x, algorithm="maxima")

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

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Fricas [A]
time = 0.37, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \end {gather*}

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)/(d*x+c),x, algorithm="fricas")

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

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Sympy [A]
time = 0.01, size = 22, normalized size = 0.88 \begin {gather*} a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5} \end {gather*}

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)/(d*x+c),x)

[Out]

a**2*x + 2*a*b*x**3/3 + b**2*x**5/5

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Giac [A]
time = 4.24, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \end {gather*}

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)/(d*x+c),x, algorithm="giac")

[Out]

1/5*b^2*x^5 + 2/3*a*b*x^3 + a^2*x

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Mupad [B]
time = 0.04, size = 21, normalized size = 0.84 \begin {gather*} a^2\,x+\frac {2\,a\,b\,x^3}{3}+\frac {b^2\,x^5}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + (b^2*x^5)/5 + (2*a*b*x^3)/3

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