\(\int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx\) [177]

Optimal result

Integrand size = 54, antiderivative size = 32 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=a c x+\frac {1}{2} a d x^2+\frac {1}{3} b c x^3+\frac {1}{4} b d x^4 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, 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} \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=a c x+\frac {1}{2} a d x^2+\frac {1}{3} b c x^3+\frac {1}{4} b d x^4 \]

[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 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 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] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=a c x+\frac {1}{2} a d x^2+\frac {1}{3} b c x^3+\frac {1}{4} b d x^4 \]

[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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
default	\(a c x +\frac {1}{2} a d \,x^{2}+\frac {1}{3} b c \,x^{3}+\frac {1}{4} b d \,x^{4}\)	\(27\)
norman	\(a c x +\frac {1}{2} a d \,x^{2}+\frac {1}{3} b c \,x^{3}+\frac {1}{4} b d \,x^{4}\)	\(27\)
risch	\(a c x +\frac {1}{2} a d \,x^{2}+\frac {1}{3} b c \,x^{3}+\frac {1}{4} b d \,x^{4}\)	\(27\)
parallelrisch	\(a c x +\frac {1}{2} a d \,x^{2}+\frac {1}{3} b c \,x^{3}+\frac {1}{4} b d \,x^{4}\)	\(27\)
parts	\(a c x +\frac {1}{2} a d \,x^{2}+\frac {1}{3} b c \,x^{3}+\frac {1}{4} b d \,x^{4}\)	\(27\)
gosper	\(\frac {x \left (3 x^{3} b d +4 b c \,x^{2}+6 a d x +12 a c \right )}{12}\)	\(28\)

[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,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=\frac {1}{4} \, b d x^{4} + \frac {1}{3} \, b c x^{3} + \frac {1}{2} \, a d x^{2} + a c x \]

[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),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

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=a c x + \frac {a d x^{2}}{2} + \frac {b c x^{3}}{3} + \frac {b d x^{4}}{4} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=\frac {1}{4} \, b d x^{4} + \frac {1}{3} \, b c x^{3} + \frac {1}{2} \, a d x^{2} + a c x \]

[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),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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=\frac {1}{4} \, b d x^{4} + \frac {1}{3} \, b c x^{3} + \frac {1}{2} \, a d x^{2} + a c x \]

[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),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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {a^2 c+a^2 d x+2 a b c x^2+2 a b d x^3+b^2 c x^4+b^2 d x^5}{a+b x^2} \, dx=\frac {b\,d\,x^4}{4}+\frac {b\,c\,x^3}{3}+\frac {a\,d\,x^2}{2}+a\,c\,x \]

[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),x)

[Out]

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