\(\int \frac {x^7}{b x^2+c x^4} \, dx\) [177]

Optimal result

Integrand size = 17, antiderivative size = 40 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {b^2 \log \left (b+c x^2\right )}{2 c^3} \]

[Out]

-1/2*b*x^2/c^2+1/4*x^4/c+1/2*b^2*ln(c*x^2+b)/c^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1598, 272, 45} \[ \int \frac {x^7}{b x^2+c x^4} \, dx=\frac {b^2 \log \left (b+c x^2\right )}{2 c^3}-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c} \]

[In]

Int[x^7/(b*x^2 + c*x^4),x]

[Out]

-1/2*(b*x^2)/c^2 + x^4/(4*c) + (b^2*Log[b + c*x^2])/(2*c^3)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 {x^5}{b+c x^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^2}{b+c x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {b}{c^2}+\frac {x}{c}+\frac {b^2}{c^2 (b+c x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {b^2 \log \left (b+c x^2\right )}{2 c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=-\frac {b x^2}{2 c^2}+\frac {x^4}{4 c}+\frac {b^2 \log \left (b+c x^2\right )}{2 c^3} \]

[In]

Integrate[x^7/(b*x^2 + c*x^4),x]

[Out]

-1/2*(b*x^2)/c^2 + x^4/(4*c) + (b^2*Log[b + c*x^2])/(2*c^3)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85

method	result	size
parallelrisch	\(\frac {c^{2} x^{4}-2 b c \,x^{2}+2 b^{2} \ln \left (c \,x^{2}+b \right )}{4 c^{3}}\)	\(34\)
default	\(-\frac {-\frac {1}{2} c \,x^{4}+b \,x^{2}}{2 c^{2}}+\frac {b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{3}}\)	\(35\)
norman	\(\frac {\frac {x^{5}}{4 c}-\frac {b \,x^{3}}{2 c^{2}}}{x}+\frac {b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{3}}\)	\(40\)
risch	\(\frac {x^{4}}{4 c}-\frac {b \,x^{2}}{2 c^{2}}+\frac {b^{2}}{4 c^{3}}+\frac {b^{2} \ln \left (c \,x^{2}+b \right )}{2 c^{3}}\)	\(43\)

[In]

int(x^7/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*(c^2*x^4-2*b*c*x^2+2*b^2*ln(c*x^2+b))/c^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=\frac {c^{2} x^{4} - 2 \, b c x^{2} + 2 \, b^{2} \log \left (c x^{2} + b\right )}{4 \, c^{3}} \]

[In]

integrate(x^7/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/4*(c^2*x^4 - 2*b*c*x^2 + 2*b^2*log(c*x^2 + b))/c^3

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=\frac {b^{2} \log {\left (b + c x^{2} \right )}}{2 c^{3}} - \frac {b x^{2}}{2 c^{2}} + \frac {x^{4}}{4 c} \]

[In]

integrate(x**7/(c*x**4+b*x**2),x)

[Out]

b**2*log(b + c*x**2)/(2*c**3) - b*x**2/(2*c**2) + x**4/(4*c)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=\frac {b^{2} \log \left (c x^{2} + b\right )}{2 \, c^{3}} + \frac {c x^{4} - 2 \, b x^{2}}{4 \, c^{2}} \]

[In]

integrate(x^7/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/2*b^2*log(c*x^2 + b)/c^3 + 1/4*(c*x^4 - 2*b*x^2)/c^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=\frac {b^{2} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{3}} + \frac {c x^{4} - 2 \, b x^{2}}{4 \, c^{2}} \]

[In]

integrate(x^7/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/2*b^2*log(abs(c*x^2 + b))/c^3 + 1/4*(c*x^4 - 2*b*x^2)/c^2

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {x^7}{b x^2+c x^4} \, dx=\frac {2\,b^2\,\ln \left (c\,x^2+b\right )+c^2\,x^4-2\,b\,c\,x^2}{4\,c^3} \]

[In]

int(x^7/(b*x^2 + c*x^4),x)

[Out]

(2*b^2*log(b + c*x^2) + c^2*x^4 - 2*b*c*x^2)/(4*c^3)