\(\int \frac {(b x^2+c x^4)^3}{x^{14}} \, dx\) [170]

Optimal result

Integrand size = 17, antiderivative size = 39 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {b^3}{7 x^7}-\frac {3 b^2 c}{5 x^5}-\frac {b c^2}{x^3}-\frac {c^3}{x} \]

[Out]

-1/7*b^3/x^7-3/5*b^2*c/x^5-b*c^2/x^3-c^3/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 276} \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {b^3}{7 x^7}-\frac {3 b^2 c}{5 x^5}-\frac {b c^2}{x^3}-\frac {c^3}{x} \]

[In]

Int[(b*x^2 + c*x^4)^3/x^14,x]

[Out]

-1/7*b^3/x^7 - (3*b^2*c)/(5*x^5) - (b*c^2)/x^3 - c^3/x

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]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (b+c x^2\right )^3}{x^8} \, dx \\ & = \int \left (\frac {b^3}{x^8}+\frac {3 b^2 c}{x^6}+\frac {3 b c^2}{x^4}+\frac {c^3}{x^2}\right ) \, dx \\ & = -\frac {b^3}{7 x^7}-\frac {3 b^2 c}{5 x^5}-\frac {b c^2}{x^3}-\frac {c^3}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {b^3}{7 x^7}-\frac {3 b^2 c}{5 x^5}-\frac {b c^2}{x^3}-\frac {c^3}{x} \]

[In]

Integrate[(b*x^2 + c*x^4)^3/x^14,x]

[Out]

-1/7*b^3/x^7 - (3*b^2*c)/(5*x^5) - (b*c^2)/x^3 - c^3/x

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {b^{3}}{7 x^{7}}-\frac {3 b^{2} c}{5 x^{5}}-\frac {b \,c^{2}}{x^{3}}-\frac {c^{3}}{x}\)	\(36\)
risch	\(\frac {-c^{3} x^{6}-b \,c^{2} x^{4}-\frac {3}{5} b^{2} c \,x^{2}-\frac {1}{7} b^{3}}{x^{7}}\)	\(37\)
gosper	\(-\frac {35 c^{3} x^{6}+35 b \,c^{2} x^{4}+21 b^{2} c \,x^{2}+5 b^{3}}{35 x^{7}}\)	\(38\)
parallelrisch	\(\frac {-35 c^{3} x^{6}-35 b \,c^{2} x^{4}-21 b^{2} c \,x^{2}-5 b^{3}}{35 x^{7}}\)	\(38\)
norman	\(\frac {-\frac {1}{7} b^{3} x^{6}-c^{3} x^{12}-b \,c^{2} x^{10}-\frac {3}{5} b^{2} c \,x^{8}}{x^{13}}\)	\(40\)

[In]

int((c*x^4+b*x^2)^3/x^14,x,method=_RETURNVERBOSE)

[Out]

-1/7*b^3/x^7-3/5*b^2*c/x^5-b*c^2/x^3-c^3/x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {35 \, c^{3} x^{6} + 35 \, b c^{2} x^{4} + 21 \, b^{2} c x^{2} + 5 \, b^{3}}{35 \, x^{7}} \]

[In]

integrate((c*x^4+b*x^2)^3/x^14,x, algorithm="fricas")

[Out]

-1/35*(35*c^3*x^6 + 35*b*c^2*x^4 + 21*b^2*c*x^2 + 5*b^3)/x^7

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=\frac {- 5 b^{3} - 21 b^{2} c x^{2} - 35 b c^{2} x^{4} - 35 c^{3} x^{6}}{35 x^{7}} \]

[In]

integrate((c*x**4+b*x**2)**3/x**14,x)

[Out]

(-5*b**3 - 21*b**2*c*x**2 - 35*b*c**2*x**4 - 35*c**3*x**6)/(35*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {35 \, c^{3} x^{6} + 35 \, b c^{2} x^{4} + 21 \, b^{2} c x^{2} + 5 \, b^{3}}{35 \, x^{7}} \]

[In]

integrate((c*x^4+b*x^2)^3/x^14,x, algorithm="maxima")

[Out]

-1/35*(35*c^3*x^6 + 35*b*c^2*x^4 + 21*b^2*c*x^2 + 5*b^3)/x^7

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {35 \, c^{3} x^{6} + 35 \, b c^{2} x^{4} + 21 \, b^{2} c x^{2} + 5 \, b^{3}}{35 \, x^{7}} \]

[In]

integrate((c*x^4+b*x^2)^3/x^14,x, algorithm="giac")

[Out]

-1/35*(35*c^3*x^6 + 35*b*c^2*x^4 + 21*b^2*c*x^2 + 5*b^3)/x^7

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^{14}} \, dx=-\frac {\frac {b^3}{7}+\frac {3\,b^2\,c\,x^2}{5}+b\,c^2\,x^4+c^3\,x^6}{x^7} \]

[In]

int((b*x^2 + c*x^4)^3/x^14,x)

[Out]

-(b^3/7 + c^3*x^6 + (3*b^2*c*x^2)/5 + b*c^2*x^4)/x^7