3.239 \(\int \frac {1}{x^6 (a+b x^2) (c+d x^2)} \, dx\)

Optimal. Leaf size=134 \[ -\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} (b c-a d)}+\frac {a d+b c}{3 a^2 c^2 x^3}-\frac {a^2 d^2+a b c d+b^2 c^2}{a^3 c^3 x}+\frac {d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)}-\frac {1}{5 a c x^5} \]

[Out]

-1/5/a/c/x^5+1/3*(a*d+b*c)/a^2/c^2/x^3+(-a^2*d^2-a*b*c*d-b^2*c^2)/a^3/c^3/x-b^(7/2)*arctan(x*b^(1/2)/a^(1/2))/
a^(7/2)/(-a*d+b*c)+d^(7/2)*arctan(x*d^(1/2)/c^(1/2))/c^(7/2)/(-a*d+b*c)

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Rubi [A]  time = 0.23, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {480, 583, 522, 205} \[ -\frac {a^2 d^2+a b c d+b^2 c^2}{a^3 c^3 x}-\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} (b c-a d)}+\frac {a d+b c}{3 a^2 c^2 x^3}+\frac {d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)}-\frac {1}{5 a c x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^2)*(c + d*x^2)),x]

[Out]

-1/(5*a*c*x^5) + (b*c + a*d)/(3*a^2*c^2*x^3) - (b^2*c^2 + a*b*c*d + a^2*d^2)/(a^3*c^3*x) - (b^(7/2)*ArcTan[(Sq
rt[b]*x)/Sqrt[a]])/(a^(7/2)*(b*c - a*d)) + (d^(7/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(7/2)*(b*c - a*d))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
 + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
 IGtQ[n, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx &=-\frac {1}{5 a c x^5}+\frac {\int \frac {-5 (b c+a d)-5 b d x^2}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{5 a c}\\ &=-\frac {1}{5 a c x^5}+\frac {b c+a d}{3 a^2 c^2 x^3}-\frac {\int \frac {-15 \left (b^2 c^2+a b c d+a^2 d^2\right )-15 b d (b c+a d) x^2}{x^2 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{15 a^2 c^2}\\ &=-\frac {1}{5 a c x^5}+\frac {b c+a d}{3 a^2 c^2 x^3}-\frac {b^2 c^2+a b c d+a^2 d^2}{a^3 c^3 x}+\frac {\int \frac {-15 (b c+a d) \left (b^2 c^2+a^2 d^2\right )-15 b d \left (b^2 c^2+a b c d+a^2 d^2\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{15 a^3 c^3}\\ &=-\frac {1}{5 a c x^5}+\frac {b c+a d}{3 a^2 c^2 x^3}-\frac {b^2 c^2+a b c d+a^2 d^2}{a^3 c^3 x}-\frac {b^4 \int \frac {1}{a+b x^2} \, dx}{a^3 (b c-a d)}+\frac {d^4 \int \frac {1}{c+d x^2} \, dx}{c^3 (b c-a d)}\\ &=-\frac {1}{5 a c x^5}+\frac {b c+a d}{3 a^2 c^2 x^3}-\frac {b^2 c^2+a b c d+a^2 d^2}{a^3 c^3 x}-\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} (b c-a d)}+\frac {d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 135, normalized size = 1.01 \[ \frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2} (a d-b c)}+\frac {a d+b c}{3 a^2 c^2 x^3}+\frac {-a^2 d^2-a b c d-b^2 c^2}{a^3 c^3 x}+\frac {d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{7/2} (b c-a d)}-\frac {1}{5 a c x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^2)*(c + d*x^2)),x]

[Out]

-1/5*1/(a*c*x^5) + (b*c + a*d)/(3*a^2*c^2*x^3) + (-(b^2*c^2) - a*b*c*d - a^2*d^2)/(a^3*c^3*x) + (b^(7/2)*ArcTa
n[(Sqrt[b]*x)/Sqrt[a]])/(a^(7/2)*(-(b*c) + a*d)) + (d^(7/2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(c^(7/2)*(b*c - a*d))

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fricas [A]  time = 0.58, size = 669, normalized size = 4.99 \[ \left [-\frac {15 \, b^{3} c^{3} x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 15 \, a^{3} d^{3} x^{5} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} - 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) + 6 \, a^{2} b c^{3} - 6 \, a^{3} c^{2} d + 30 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x^{4} - 10 \, {\left (a b^{2} c^{3} - a^{3} c d^{2}\right )} x^{2}}{30 \, {\left (a^{3} b c^{4} - a^{4} c^{3} d\right )} x^{5}}, \frac {30 \, a^{3} d^{3} x^{5} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) - 15 \, b^{3} c^{3} x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 6 \, a^{2} b c^{3} + 6 \, a^{3} c^{2} d - 30 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x^{4} + 10 \, {\left (a b^{2} c^{3} - a^{3} c d^{2}\right )} x^{2}}{30 \, {\left (a^{3} b c^{4} - a^{4} c^{3} d\right )} x^{5}}, -\frac {30 \, b^{3} c^{3} x^{5} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 15 \, a^{3} d^{3} x^{5} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} - 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) + 6 \, a^{2} b c^{3} - 6 \, a^{3} c^{2} d + 30 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x^{4} - 10 \, {\left (a b^{2} c^{3} - a^{3} c d^{2}\right )} x^{2}}{30 \, {\left (a^{3} b c^{4} - a^{4} c^{3} d\right )} x^{5}}, -\frac {15 \, b^{3} c^{3} x^{5} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 15 \, a^{3} d^{3} x^{5} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) + 3 \, a^{2} b c^{3} - 3 \, a^{3} c^{2} d + 15 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x^{4} - 5 \, {\left (a b^{2} c^{3} - a^{3} c d^{2}\right )} x^{2}}{15 \, {\left (a^{3} b c^{4} - a^{4} c^{3} d\right )} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a)/(d*x^2+c),x, algorithm="fricas")

[Out]

[-1/30*(15*b^3*c^3*x^5*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 15*a^3*d^3*x^5*sqrt(-d/c)*
log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 6*a^2*b*c^3 - 6*a^3*c^2*d + 30*(b^3*c^3 - a^3*d^3)*x^4 - 10*
(a*b^2*c^3 - a^3*c*d^2)*x^2)/((a^3*b*c^4 - a^4*c^3*d)*x^5), 1/30*(30*a^3*d^3*x^5*sqrt(d/c)*arctan(x*sqrt(d/c))
 - 15*b^3*c^3*x^5*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 6*a^2*b*c^3 + 6*a^3*c^2*d - 30*
(b^3*c^3 - a^3*d^3)*x^4 + 10*(a*b^2*c^3 - a^3*c*d^2)*x^2)/((a^3*b*c^4 - a^4*c^3*d)*x^5), -1/30*(30*b^3*c^3*x^5
*sqrt(b/a)*arctan(x*sqrt(b/a)) + 15*a^3*d^3*x^5*sqrt(-d/c)*log((d*x^2 - 2*c*x*sqrt(-d/c) - c)/(d*x^2 + c)) + 6
*a^2*b*c^3 - 6*a^3*c^2*d + 30*(b^3*c^3 - a^3*d^3)*x^4 - 10*(a*b^2*c^3 - a^3*c*d^2)*x^2)/((a^3*b*c^4 - a^4*c^3*
d)*x^5), -1/15*(15*b^3*c^3*x^5*sqrt(b/a)*arctan(x*sqrt(b/a)) - 15*a^3*d^3*x^5*sqrt(d/c)*arctan(x*sqrt(d/c)) +
3*a^2*b*c^3 - 3*a^3*c^2*d + 15*(b^3*c^3 - a^3*d^3)*x^4 - 5*(a*b^2*c^3 - a^3*c*d^2)*x^2)/((a^3*b*c^4 - a^4*c^3*
d)*x^5)]

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giac [A]  time = 0.33, size = 139, normalized size = 1.04 \[ -\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{3} b c - a^{4} d\right )} \sqrt {a b}} + \frac {d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b c^{4} - a c^{3} d\right )} \sqrt {c d}} - \frac {15 \, b^{2} c^{2} x^{4} + 15 \, a b c d x^{4} + 15 \, a^{2} d^{2} x^{4} - 5 \, a b c^{2} x^{2} - 5 \, a^{2} c d x^{2} + 3 \, a^{2} c^{2}}{15 \, a^{3} c^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a)/(d*x^2+c),x, algorithm="giac")

[Out]

-b^4*arctan(b*x/sqrt(a*b))/((a^3*b*c - a^4*d)*sqrt(a*b)) + d^4*arctan(d*x/sqrt(c*d))/((b*c^4 - a*c^3*d)*sqrt(c
*d)) - 1/15*(15*b^2*c^2*x^4 + 15*a*b*c*d*x^4 + 15*a^2*d^2*x^4 - 5*a*b*c^2*x^2 - 5*a^2*c*d*x^2 + 3*a^2*c^2)/(a^
3*c^3*x^5)

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maple [A]  time = 0.01, size = 141, normalized size = 1.05 \[ \frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right ) \sqrt {a b}\, a^{3}}-\frac {d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\left (a d -b c \right ) \sqrt {c d}\, c^{3}}-\frac {d^{2}}{a \,c^{3} x}-\frac {b d}{a^{2} c^{2} x}-\frac {b^{2}}{a^{3} c x}+\frac {d}{3 a \,c^{2} x^{3}}+\frac {b}{3 a^{2} c \,x^{3}}-\frac {1}{5 a c \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^2+a)/(d*x^2+c),x)

[Out]

1/a^3*b^4/(a*d-b*c)/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)-1/c^3*d^4/(a*d-b*c)/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)
*d*x)-1/5/a/c/x^5+1/3/a/c^2/x^3*d+1/3/a^2/c/x^3*b-1/a/c^3/x*d^2-1/a^2/c^2/x*b*d-1/a^3/c/x*b^2

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maxima [A]  time = 2.45, size = 131, normalized size = 0.98 \[ -\frac {b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{3} b c - a^{4} d\right )} \sqrt {a b}} + \frac {d^{4} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b c^{4} - a c^{3} d\right )} \sqrt {c d}} - \frac {15 \, {\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} x^{4} + 3 \, a^{2} c^{2} - 5 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2}}{15 \, a^{3} c^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a)/(d*x^2+c),x, algorithm="maxima")

[Out]

-b^4*arctan(b*x/sqrt(a*b))/((a^3*b*c - a^4*d)*sqrt(a*b)) + d^4*arctan(d*x/sqrt(c*d))/((b*c^4 - a*c^3*d)*sqrt(c
*d)) - 1/15*(15*(b^2*c^2 + a*b*c*d + a^2*d^2)*x^4 + 3*a^2*c^2 - 5*(a*b*c^2 + a^2*c*d)*x^2)/(a^3*c^3*x^5)

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mupad [B]  time = 0.63, size = 397, normalized size = 2.96 \[ \frac {\ln \left (a^{11}\,b^{10}\,c^7-a^{18}\,b^3\,d^7+c^7\,x\,{\left (-a^7\,b^7\right )}^{3/2}+a^{14}\,d^7\,x\,\sqrt {-a^7\,b^7}\right )\,\sqrt {-a^7\,b^7}}{2\,a^8\,d-2\,a^7\,b\,c}-\frac {\ln \left (a^{18}\,b^3\,d^7-a^{11}\,b^{10}\,c^7+c^7\,x\,{\left (-a^7\,b^7\right )}^{3/2}+a^{14}\,d^7\,x\,\sqrt {-a^7\,b^7}\right )\,\sqrt {-a^7\,b^7}}{2\,\left (a^8\,d-a^7\,b\,c\right )}-\frac {\frac {1}{5\,a\,c}-\frac {x^2\,\left (a\,d+b\,c\right )}{3\,a^2\,c^2}+\frac {x^4\,\left (a^2\,d^2+a\,b\,c\,d+b^2\,c^2\right )}{a^3\,c^3}}{x^5}-\frac {\ln \left (b^7\,c^{18}\,d^3-a^7\,c^{11}\,d^{10}+a^7\,x\,{\left (-c^7\,d^7\right )}^{3/2}+b^7\,c^{14}\,x\,\sqrt {-c^7\,d^7}\right )\,\sqrt {-c^7\,d^7}}{2\,\left (b\,c^8-a\,c^7\,d\right )}+\frac {\ln \left (a^7\,c^{11}\,d^{10}-b^7\,c^{18}\,d^3+a^7\,x\,{\left (-c^7\,d^7\right )}^{3/2}+b^7\,c^{14}\,x\,\sqrt {-c^7\,d^7}\right )\,\sqrt {-c^7\,d^7}}{2\,b\,c^8-2\,a\,c^7\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^2)*(c + d*x^2)),x)

[Out]

(log(a^11*b^10*c^7 - a^18*b^3*d^7 + c^7*x*(-a^7*b^7)^(3/2) + a^14*d^7*x*(-a^7*b^7)^(1/2))*(-a^7*b^7)^(1/2))/(2
*a^8*d - 2*a^7*b*c) - (log(a^18*b^3*d^7 - a^11*b^10*c^7 + c^7*x*(-a^7*b^7)^(3/2) + a^14*d^7*x*(-a^7*b^7)^(1/2)
)*(-a^7*b^7)^(1/2))/(2*(a^8*d - a^7*b*c)) - (1/(5*a*c) - (x^2*(a*d + b*c))/(3*a^2*c^2) + (x^4*(a^2*d^2 + b^2*c
^2 + a*b*c*d))/(a^3*c^3))/x^5 - (log(b^7*c^18*d^3 - a^7*c^11*d^10 + a^7*x*(-c^7*d^7)^(3/2) + b^7*c^14*x*(-c^7*
d^7)^(1/2))*(-c^7*d^7)^(1/2))/(2*(b*c^8 - a*c^7*d)) + (log(a^7*c^11*d^10 - b^7*c^18*d^3 + a^7*x*(-c^7*d^7)^(3/
2) + b^7*c^14*x*(-c^7*d^7)^(1/2))*(-c^7*d^7)^(1/2))/(2*b*c^8 - 2*a*c^7*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**2+a)/(d*x**2+c),x)

[Out]

Timed out

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