\(\int \frac {x^7 (A+B x^2)}{b x^2+c x^4} \, dx\) [44]

Optimal result

Integrand size = 24, antiderivative size = 75 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {b (b B-A c) x^2}{2 c^3}-\frac {(b B-A c) x^4}{4 c^2}+\frac {B x^6}{6 c}-\frac {b^2 (b B-A c) \log \left (b+c x^2\right )}{2 c^4} \]

[Out]

1/2*b*(-A*c+B*b)*x^2/c^3-1/4*(-A*c+B*b)*x^4/c^2+1/6*B*x^6/c-1/2*b^2*(-A*c+B*b)*ln(c*x^2+b)/c^4

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1598, 457, 78} \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {b^2 (b B-A c) \log \left (b+c x^2\right )}{2 c^4}+\frac {b x^2 (b B-A c)}{2 c^3}-\frac {x^4 (b B-A c)}{4 c^2}+\frac {B x^6}{6 c} \]

[In]

Int[(x^7*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(b*(b*B - A*c)*x^2)/(2*c^3) - ((b*B - A*c)*x^4)/(4*c^2) + (B*x^6)/(6*c) - (b^2*(b*B - A*c)*Log[b + c*x^2])/(2*
c^4)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && 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 \left (A+B x^2\right )}{b+c x^2} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {x^2 (A+B x)}{b+c x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {b (b B-A c)}{c^3}+\frac {(-b B+A c) x}{c^2}+\frac {B x^2}{c}-\frac {b^2 (b B-A c)}{c^3 (b+c x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {b (b B-A c) x^2}{2 c^3}-\frac {(b B-A c) x^4}{4 c^2}+\frac {B x^6}{6 c}-\frac {b^2 (b B-A c) \log \left (b+c x^2\right )}{2 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {c x^2 \left (6 b^2 B-3 b c \left (2 A+B x^2\right )+c^2 x^2 \left (3 A+2 B x^2\right )\right )+6 b^2 (-b B+A c) \log \left (b+c x^2\right )}{12 c^4} \]

[In]

Integrate[(x^7*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(c*x^2*(6*b^2*B - 3*b*c*(2*A + B*x^2) + c^2*x^2*(3*A + 2*B*x^2)) + 6*b^2*(-(b*B) + A*c)*Log[b + c*x^2])/(12*c^
4)

Maple [A] (verified)

Time = 1.75 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97

method	result	size
norman	\(\frac {\frac {B \,x^{7}}{6 c}+\frac {\left (A c -B b \right ) x^{5}}{4 c^{2}}-\frac {b \left (A c -B b \right ) x^{3}}{2 c^{3}}}{x}+\frac {b^{2} \left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{4}}\)	\(73\)
default	\(-\frac {-\frac {1}{3} B \,c^{2} x^{6}-\frac {1}{2} A \,c^{2} x^{4}+\frac {1}{2} x^{4} B b c +A b c \,x^{2}-b^{2} B \,x^{2}}{2 c^{3}}+\frac {b^{2} \left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{4}}\)	\(74\)
parallelrisch	\(\frac {2 B \,c^{3} x^{6}+3 A \,c^{3} x^{4}-3 B b \,c^{2} x^{4}-6 A b \,c^{2} x^{2}+6 B \,b^{2} c \,x^{2}+6 A \ln \left (c \,x^{2}+b \right ) b^{2} c -6 B \ln \left (c \,x^{2}+b \right ) b^{3}}{12 c^{4}}\)	\(84\)
risch	\(\frac {B \,x^{6}}{6 c}+\frac {A \,x^{4}}{4 c}-\frac {x^{4} B b}{4 c^{2}}-\frac {A b \,x^{2}}{2 c^{2}}+\frac {b^{2} B \,x^{2}}{2 c^{3}}+\frac {b^{2} \ln \left (c \,x^{2}+b \right ) A}{2 c^{3}}-\frac {b^{3} \ln \left (c \,x^{2}+b \right ) B}{2 c^{4}}\)	\(86\)

[In]

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

[Out]

(1/6*B*x^7/c+1/4/c^2*(A*c-B*b)*x^5-1/2*b/c^3*(A*c-B*b)*x^3)/x+1/2*b^2*(A*c-B*b)/c^4*ln(c*x^2+b)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 \, B c^{3} x^{6} - 3 \, {\left (B b c^{2} - A c^{3}\right )} x^{4} + 6 \, {\left (B b^{2} c - A b c^{2}\right )} x^{2} - 6 \, {\left (B b^{3} - A b^{2} c\right )} \log \left (c x^{2} + b\right )}{12 \, c^{4}} \]

[In]

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

[Out]

1/12*(2*B*c^3*x^6 - 3*(B*b*c^2 - A*c^3)*x^4 + 6*(B*b^2*c - A*b*c^2)*x^2 - 6*(B*b^3 - A*b^2*c)*log(c*x^2 + b))/
c^4

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.93 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {B x^{6}}{6 c} - \frac {b^{2} \left (- A c + B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{4}} + x^{4} \left (\frac {A}{4 c} - \frac {B b}{4 c^{2}}\right ) + x^{2} \left (- \frac {A b}{2 c^{2}} + \frac {B b^{2}}{2 c^{3}}\right ) \]

[In]

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

[Out]

B*x**6/(6*c) - b**2*(-A*c + B*b)*log(b + c*x**2)/(2*c**4) + x**4*(A/(4*c) - B*b/(4*c**2)) + x**2*(-A*b/(2*c**2
) + B*b**2/(2*c**3))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 \, B c^{2} x^{6} - 3 \, {\left (B b c - A c^{2}\right )} x^{4} + 6 \, {\left (B b^{2} - A b c\right )} x^{2}}{12 \, c^{3}} - \frac {{\left (B b^{3} - A b^{2} c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{4}} \]

[In]

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

[Out]

1/12*(2*B*c^2*x^6 - 3*(B*b*c - A*c^2)*x^4 + 6*(B*b^2 - A*b*c)*x^2)/c^3 - 1/2*(B*b^3 - A*b^2*c)*log(c*x^2 + b)/
c^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 \, B c^{2} x^{6} - 3 \, B b c x^{4} + 3 \, A c^{2} x^{4} + 6 \, B b^{2} x^{2} - 6 \, A b c x^{2}}{12 \, c^{3}} - \frac {{\left (B b^{3} - A b^{2} c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} \]

[In]

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

[Out]

1/12*(2*B*c^2*x^6 - 3*B*b*c*x^4 + 3*A*c^2*x^4 + 6*B*b^2*x^2 - 6*A*b*c*x^2)/c^3 - 1/2*(B*b^3 - A*b^2*c)*log(abs
(c*x^2 + b))/c^4

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {x^7 \left (A+B x^2\right )}{b x^2+c x^4} \, dx=x^4\,\left (\frac {A}{4\,c}-\frac {B\,b}{4\,c^2}\right )+\frac {B\,x^6}{6\,c}-\frac {\ln \left (c\,x^2+b\right )\,\left (B\,b^3-A\,b^2\,c\right )}{2\,c^4}-\frac {b\,x^2\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{2\,c} \]

[In]

int((x^7*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

x^4*(A/(4*c) - (B*b)/(4*c^2)) + (B*x^6)/(6*c) - (log(b + c*x^2)*(B*b^3 - A*b^2*c))/(2*c^4) - (b*x^2*(A/c - (B*
b)/c^2))/(2*c)