\(\int \frac {(a+b x^3)^2}{x^{12}} \, dx\) [234]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {a^2}{11 x^{11}}-\frac {a b}{4 x^8}-\frac {b^2}{5 x^5} \]

[Out]

-1/11*a^2/x^11-1/4*a*b/x^8-1/5*b^2/x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {a^2}{11 x^{11}}-\frac {a b}{4 x^8}-\frac {b^2}{5 x^5} \]

[In]

Int[(a + b*x^3)^2/x^12,x]

[Out]

-1/11*a^2/x^11 - (a*b)/(4*x^8) - b^2/(5*x^5)

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a^2}{x^{12}}+\frac {2 a b}{x^9}+\frac {b^2}{x^6}\right ) \, dx \\ & = -\frac {a^2}{11 x^{11}}-\frac {a b}{4 x^8}-\frac {b^2}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {a^2}{11 x^{11}}-\frac {a b}{4 x^8}-\frac {b^2}{5 x^5} \]

[In]

Integrate[(a + b*x^3)^2/x^12,x]

[Out]

-1/11*a^2/x^11 - (a*b)/(4*x^8) - b^2/(5*x^5)

Maple [A] (verified)

Time = 3.69 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {a^{2}}{11 x^{11}}-\frac {a b}{4 x^{8}}-\frac {b^{2}}{5 x^{5}}\)	\(25\)
norman	\(\frac {-\frac {1}{5} b^{2} x^{6}-\frac {1}{4} a b \,x^{3}-\frac {1}{11} a^{2}}{x^{11}}\)	\(26\)
risch	\(\frac {-\frac {1}{5} b^{2} x^{6}-\frac {1}{4} a b \,x^{3}-\frac {1}{11} a^{2}}{x^{11}}\)	\(26\)
gosper	\(-\frac {44 b^{2} x^{6}+55 a b \,x^{3}+20 a^{2}}{220 x^{11}}\)	\(27\)
parallelrisch	\(\frac {-44 b^{2} x^{6}-55 a b \,x^{3}-20 a^{2}}{220 x^{11}}\)	\(27\)

[In]

int((b*x^3+a)^2/x^12,x,method=_RETURNVERBOSE)

[Out]

-1/11*a^2/x^11-1/4*a*b/x^8-1/5*b^2/x^5

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {44 \, b^{2} x^{6} + 55 \, a b x^{3} + 20 \, a^{2}}{220 \, x^{11}} \]

[In]

integrate((b*x^3+a)^2/x^12,x, algorithm="fricas")

[Out]

-1/220*(44*b^2*x^6 + 55*a*b*x^3 + 20*a^2)/x^11

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=\frac {- 20 a^{2} - 55 a b x^{3} - 44 b^{2} x^{6}}{220 x^{11}} \]

[In]

integrate((b*x**3+a)**2/x**12,x)

[Out]

(-20*a**2 - 55*a*b*x**3 - 44*b**2*x**6)/(220*x**11)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {44 \, b^{2} x^{6} + 55 \, a b x^{3} + 20 \, a^{2}}{220 \, x^{11}} \]

[In]

integrate((b*x^3+a)^2/x^12,x, algorithm="maxima")

[Out]

-1/220*(44*b^2*x^6 + 55*a*b*x^3 + 20*a^2)/x^11

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {44 \, b^{2} x^{6} + 55 \, a b x^{3} + 20 \, a^{2}}{220 \, x^{11}} \]

[In]

integrate((b*x^3+a)^2/x^12,x, algorithm="giac")

[Out]

-1/220*(44*b^2*x^6 + 55*a*b*x^3 + 20*a^2)/x^11

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{12}} \, dx=-\frac {\frac {a^2}{11}+\frac {a\,b\,x^3}{4}+\frac {b^2\,x^6}{5}}{x^{11}} \]

[In]

int((a + b*x^3)^2/x^12,x)

[Out]

-(a^2/11 + (b^2*x^6)/5 + (a*b*x^3)/4)/x^11