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

Optimal result

Integrand size = 52, antiderivative size = 25 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {1600} \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {b^2 x^5}{5} \]

[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

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 22, normalized size of antiderivative = 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	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
parallelrisch	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
parts	\(a^{2} x +\frac {2}{3} a b \,x^{3}+\frac {1}{5} b^{2} x^{5}\)	\(22\)
gosper	\(\frac {x \left (3 b^{2} x^{4}+10 a b \,x^{2}+15 a^{2}\right )}{15}\)	\(25\)

[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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=\frac {1}{5} \, b^{2} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{c+d x} \, dx=a^2\,x+\frac {2\,a\,b\,x^3}{3}+\frac {b^2\,x^5}{5} \]

[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