\(\int \frac {(a+b x^3)^2}{x^{11}} \, dx\) [233]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {a^2}{10 x^{10}}-\frac {2 a b}{7 x^7}-\frac {b^2}{4 x^4} \]

[Out]

-1/10*a^2/x^10-2/7*a*b/x^7-1/4*b^2/x^4

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} \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {a^2}{10 x^{10}}-\frac {2 a b}{7 x^7}-\frac {b^2}{4 x^4} \]

[In]

Int[(a + b*x^3)^2/x^11,x]

[Out]

-1/10*a^2/x^10 - (2*a*b)/(7*x^7) - b^2/(4*x^4)

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^{11}}+\frac {2 a b}{x^8}+\frac {b^2}{x^5}\right ) \, dx \\ & = -\frac {a^2}{10 x^{10}}-\frac {2 a b}{7 x^7}-\frac {b^2}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {a^2}{10 x^{10}}-\frac {2 a b}{7 x^7}-\frac {b^2}{4 x^4} \]

[In]

Integrate[(a + b*x^3)^2/x^11,x]

[Out]

-1/10*a^2/x^10 - (2*a*b)/(7*x^7) - b^2/(4*x^4)

Maple [A] (verified)

Time = 3.58 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(-\frac {a^{2}}{10 x^{10}}-\frac {2 a b}{7 x^{7}}-\frac {b^{2}}{4 x^{4}}\)	\(25\)
norman	\(\frac {-\frac {1}{4} b^{2} x^{6}-\frac {2}{7} a b \,x^{3}-\frac {1}{10} a^{2}}{x^{10}}\)	\(26\)
risch	\(\frac {-\frac {1}{4} b^{2} x^{6}-\frac {2}{7} a b \,x^{3}-\frac {1}{10} a^{2}}{x^{10}}\)	\(26\)
gosper	\(-\frac {35 b^{2} x^{6}+40 a b \,x^{3}+14 a^{2}}{140 x^{10}}\)	\(27\)
parallelrisch	\(\frac {-35 b^{2} x^{6}-40 a b \,x^{3}-14 a^{2}}{140 x^{10}}\)	\(27\)

[In]

int((b*x^3+a)^2/x^11,x,method=_RETURNVERBOSE)

[Out]

-1/10*a^2/x^10-2/7*a*b/x^7-1/4*b^2/x^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {35 \, b^{2} x^{6} + 40 \, a b x^{3} + 14 \, a^{2}}{140 \, x^{10}} \]

[In]

integrate((b*x^3+a)^2/x^11,x, algorithm="fricas")

[Out]

-1/140*(35*b^2*x^6 + 40*a*b*x^3 + 14*a^2)/x^10

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=\frac {- 14 a^{2} - 40 a b x^{3} - 35 b^{2} x^{6}}{140 x^{10}} \]

[In]

integrate((b*x**3+a)**2/x**11,x)

[Out]

(-14*a**2 - 40*a*b*x**3 - 35*b**2*x**6)/(140*x**10)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {35 \, b^{2} x^{6} + 40 \, a b x^{3} + 14 \, a^{2}}{140 \, x^{10}} \]

[In]

integrate((b*x^3+a)^2/x^11,x, algorithm="maxima")

[Out]

-1/140*(35*b^2*x^6 + 40*a*b*x^3 + 14*a^2)/x^10

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {35 \, b^{2} x^{6} + 40 \, a b x^{3} + 14 \, a^{2}}{140 \, x^{10}} \]

[In]

integrate((b*x^3+a)^2/x^11,x, algorithm="giac")

[Out]

-1/140*(35*b^2*x^6 + 40*a*b*x^3 + 14*a^2)/x^10

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^2}{x^{11}} \, dx=-\frac {\frac {a^2}{10}+\frac {2\,a\,b\,x^3}{7}+\frac {b^2\,x^6}{4}}{x^{10}} \]

[In]

int((a + b*x^3)^2/x^11,x)

[Out]

-(a^2/10 + (b^2*x^6)/4 + (2*a*b*x^3)/7)/x^10