\(\int \frac {(a+b x^4)^3}{x^2} \, dx\) [640]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=-\frac {a^3}{x}+a^2 b x^3+\frac {3}{7} a b^2 x^7+\frac {b^3 x^{11}}{11} \]

[Out]

-a^3/x+a^2*b*x^3+3/7*a*b^2*x^7+1/11*b^3*x^11

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, 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} \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=-\frac {a^3}{x}+a^2 b x^3+\frac {3}{7} a b^2 x^7+\frac {b^3 x^{11}}{11} \]

[In]

Int[(a + b*x^4)^3/x^2,x]

[Out]

-(a^3/x) + a^2*b*x^3 + (3*a*b^2*x^7)/7 + (b^3*x^11)/11

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^3}{x^2}+3 a^2 b x^2+3 a b^2 x^6+b^3 x^{10}\right ) \, dx \\ & = -\frac {a^3}{x}+a^2 b x^3+\frac {3}{7} a b^2 x^7+\frac {b^3 x^{11}}{11} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=-\frac {a^3}{x}+a^2 b x^3+\frac {3}{7} a b^2 x^7+\frac {b^3 x^{11}}{11} \]

[In]

Integrate[(a + b*x^4)^3/x^2,x]

[Out]

-(a^3/x) + a^2*b*x^3 + (3*a*b^2*x^7)/7 + (b^3*x^11)/11

Maple [A] (verified)

Time = 3.90 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {a^{3}}{x}+a^{2} b \,x^{3}+\frac {3 a \,b^{2} x^{7}}{7}+\frac {b^{3} x^{11}}{11}\)	\(35\)
risch	\(-\frac {a^{3}}{x}+a^{2} b \,x^{3}+\frac {3 a \,b^{2} x^{7}}{7}+\frac {b^{3} x^{11}}{11}\)	\(35\)
norman	\(\frac {\frac {1}{11} b^{3} x^{12}+\frac {3}{7} a \,b^{2} x^{8}+a^{2} b \,x^{4}-a^{3}}{x}\)	\(36\)
gosper	\(-\frac {-7 b^{3} x^{12}-33 a \,b^{2} x^{8}-77 a^{2} b \,x^{4}+77 a^{3}}{77 x}\)	\(38\)
parallelrisch	\(\frac {7 b^{3} x^{12}+33 a \,b^{2} x^{8}+77 a^{2} b \,x^{4}-77 a^{3}}{77 x}\)	\(38\)

[In]

int((b*x^4+a)^3/x^2,x,method=_RETURNVERBOSE)

[Out]

-a^3/x+a^2*b*x^3+3/7*a*b^2*x^7+1/11*b^3*x^11

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=\frac {7 \, b^{3} x^{12} + 33 \, a b^{2} x^{8} + 77 \, a^{2} b x^{4} - 77 \, a^{3}}{77 \, x} \]

[In]

integrate((b*x^4+a)^3/x^2,x, algorithm="fricas")

[Out]

1/77*(7*b^3*x^12 + 33*a*b^2*x^8 + 77*a^2*b*x^4 - 77*a^3)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=- \frac {a^{3}}{x} + a^{2} b x^{3} + \frac {3 a b^{2} x^{7}}{7} + \frac {b^{3} x^{11}}{11} \]

[In]

integrate((b*x**4+a)**3/x**2,x)

[Out]

-a**3/x + a**2*b*x**3 + 3*a*b**2*x**7/7 + b**3*x**11/11

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=\frac {1}{11} \, b^{3} x^{11} + \frac {3}{7} \, a b^{2} x^{7} + a^{2} b x^{3} - \frac {a^{3}}{x} \]

[In]

integrate((b*x^4+a)^3/x^2,x, algorithm="maxima")

[Out]

1/11*b^3*x^11 + 3/7*a*b^2*x^7 + a^2*b*x^3 - a^3/x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=\frac {1}{11} \, b^{3} x^{11} + \frac {3}{7} \, a b^{2} x^{7} + a^{2} b x^{3} - \frac {a^{3}}{x} \]

[In]

integrate((b*x^4+a)^3/x^2,x, algorithm="giac")

[Out]

1/11*b^3*x^11 + 3/7*a*b^2*x^7 + a^2*b*x^3 - a^3/x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^4\right )^3}{x^2} \, dx=\frac {b^3\,x^{11}}{11}-\frac {a^3}{x}+a^2\,b\,x^3+\frac {3\,a\,b^2\,x^7}{7} \]

[In]

int((a + b*x^4)^3/x^2,x)

[Out]

(b^3*x^11)/11 - a^3/x + a^2*b*x^3 + (3*a*b^2*x^7)/7