\(\int \frac {(a+b x^4)^2}{x^4} \, dx\) [630]

Optimal result

Integrand size = 13, antiderivative size = 26 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=-\frac {a^2}{3 x^3}+2 a b x+\frac {b^2 x^5}{5} \]

[Out]

-1/3*a^2/x^3+2*a*b*x+1/5*b^2*x^5

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, 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} \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=-\frac {a^2}{3 x^3}+2 a b x+\frac {b^2 x^5}{5} \]

[In]

Int[(a + b*x^4)^2/x^4,x]

[Out]

-1/3*a^2/x^3 + 2*a*b*x + (b^2*x^5)/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 (2 a b+\frac {a^2}{x^4}+b^2 x^4\right ) \, dx \\ & = -\frac {a^2}{3 x^3}+2 a b x+\frac {b^2 x^5}{5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=-\frac {a^2}{3 x^3}+2 a b x+\frac {b^2 x^5}{5} \]

[In]

Integrate[(a + b*x^4)^2/x^4,x]

[Out]

-1/3*a^2/x^3 + 2*a*b*x + (b^2*x^5)/5

Maple [A] (verified)

Time = 3.94 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {a^{2}}{3 x^{3}}+2 a b x +\frac {b^{2} x^{5}}{5}\)	\(23\)
risch	\(-\frac {a^{2}}{3 x^{3}}+2 a b x +\frac {b^{2} x^{5}}{5}\)	\(23\)
norman	\(\frac {\frac {1}{5} b^{2} x^{8}+2 a b \,x^{4}-\frac {1}{3} a^{2}}{x^{3}}\)	\(26\)
gosper	\(-\frac {-3 b^{2} x^{8}-30 a b \,x^{4}+5 a^{2}}{15 x^{3}}\)	\(27\)
parallelrisch	\(\frac {3 b^{2} x^{8}+30 a b \,x^{4}-5 a^{2}}{15 x^{3}}\)	\(27\)

[In]

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

[Out]

-1/3*a^2/x^3+2*a*b*x+1/5*b^2*x^5

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=\frac {3 \, b^{2} x^{8} + 30 \, a b x^{4} - 5 \, a^{2}}{15 \, x^{3}} \]

[In]

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

[Out]

1/15*(3*b^2*x^8 + 30*a*b*x^4 - 5*a^2)/x^3

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=- \frac {a^{2}}{3 x^{3}} + 2 a b x + \frac {b^{2} x^{5}}{5} \]

[In]

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

[Out]

-a**2/(3*x**3) + 2*a*b*x + b**2*x**5/5

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=\frac {1}{5} \, b^{2} x^{5} + 2 \, a b x - \frac {a^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

1/5*b^2*x^5 + 2*a*b*x - 1/3*a^2/x^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=\frac {1}{5} \, b^{2} x^{5} + 2 \, a b x - \frac {a^{2}}{3 \, x^{3}} \]

[In]

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

[Out]

1/5*b^2*x^5 + 2*a*b*x - 1/3*a^2/x^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^2}{x^4} \, dx=\frac {b^2\,x^5}{5}-\frac {a^2}{3\,x^3}+2\,a\,b\,x \]

[In]

int((a + b*x^4)^2/x^4,x)

[Out]

(b^2*x^5)/5 - a^2/(3*x^3) + 2*a*b*x