\(\int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{(a+b x^2)^2} \, dx\) [178]

Optimal result

Integrand size = 54, antiderivative size = 12 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=c x+\frac {d x^2}{2} \]

[Out]

c*x+1/2*d*x^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {1600} \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=c x+\frac {d x^2}{2} \]

[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)^2,x]

[Out]

c*x + (d*x^2)/2

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=c x+\frac {d x^2}{2} \]

[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)^2,x]

[Out]

c*x + (d*x^2)/2

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
gosper	\(\frac {x \left (d x +2 c \right )}{2}\)	\(11\)
default	\(c x +\frac {1}{2} d \,x^{2}\)	\(11\)
risch	\(c x +\frac {1}{2} d \,x^{2}\)	\(11\)
parallelrisch	\(c x +\frac {1}{2} d \,x^{2}\)	\(11\)
parts	\(c x +\frac {1}{2} d \,x^{2}\)	\(11\)
norman	\(\frac {a c x +b c \,x^{3}-\frac {a^{2} d}{2 b}+\frac {b d \,x^{4}}{2}}{b \,x^{2}+a}\)	\(38\)

[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)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*x*(d*x+2*c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=\frac {1}{2} \, d x^{2} + c x \]

[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)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

1/2*d*x^2 + c*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=c x + \frac {d x^{2}}{2} \]

[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)**2,x)

[Out]

c*x + d*x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=\frac {1}{2} \, d x^{2} + c x \]

[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)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

1/2*d*x^2 + c*x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=\frac {1}{2} \, d x^{2} + c x \]

[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)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

1/2*d*x^2 + c*x

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{\left (a+b x^2\right )^2} \, dx=\frac {d\,x^2}{2}+c\,x \]

[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)/(a + b*x^2)^2,x)

[Out]

c*x + (d*x^2)/2