3.3.37 \(\int \frac {1}{x^4 (a+b x^2) (c+d x^2)} \, dx\)

Optimal. Leaf size=100 \[ \frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)}+\frac {a d+b c}{a^2 c^2 x}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)}-\frac {1}{3 a c x^3} \]

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Rubi [A]  time = 0.18, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {480, 583, 522, 205} \begin {gather*} \frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (b c-a d)}+\frac {a d+b c}{a^2 c^2 x}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)}-\frac {1}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)*(c + d*x^2)),x]

[Out]

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

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Mathematica [A]  time = 0.13, size = 101, normalized size = 1.01 \begin {gather*} -\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2} (a d-b c)}+\frac {a d+b c}{a^2 c^2 x}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{c^{5/2} (b c-a d)}-\frac {1}{3 a c x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)*(c + d*x^2)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^4 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^4*(a + b*x^2)*(c + d*x^2)),x]

[Out]

IntegrateAlgebraic[1/(x^4*(a + b*x^2)*(c + d*x^2)), x]

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fricas [A]  time = 1.16, size = 560, normalized size = 5.60 \begin {gather*} \left [-\frac {3 \, b^{2} c^{2} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 3 \, a^{2} d^{2} x^{3} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} + 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) + 2 \, a b c^{2} - 2 \, a^{2} c d - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{6 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{3}}, -\frac {6 \, a^{2} d^{2} x^{3} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) + 3 \, b^{2} c^{2} x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 2 \, a b c^{2} - 2 \, a^{2} c d - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{6 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{3}}, \frac {6 \, b^{2} c^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 3 \, a^{2} d^{2} x^{3} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{2} + 2 \, c x \sqrt {-\frac {d}{c}} - c}{d x^{2} + c}\right ) - 2 \, a b c^{2} + 2 \, a^{2} c d + 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{6 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{3}}, \frac {3 \, b^{2} c^{2} x^{3} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) - 3 \, a^{2} d^{2} x^{3} \sqrt {\frac {d}{c}} \arctan \left (x \sqrt {\frac {d}{c}}\right ) - a b c^{2} + a^{2} c d + 3 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{3 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.35, size = 98, normalized size = 0.98 \begin {gather*} \frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b c - a^{3} d\right )} \sqrt {a b}} - \frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b c^{3} - a c^{2} d\right )} \sqrt {c d}} + \frac {3 \, b c x^{2} + 3 \, a d x^{2} - a c}{3 \, a^{2} c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 98, normalized size = 0.98 \begin {gather*} -\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\left (a d -b c \right ) \sqrt {a b}\, a^{2}}+\frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{\left (a d -b c \right ) \sqrt {c d}\, c^{2}}+\frac {d}{a \,c^{2} x}+\frac {b}{a^{2} c x}-\frac {1}{3 a c \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)/(d*x^2+c),x)

[Out]

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

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maxima [A]  time = 2.41, size = 96, normalized size = 0.96 \begin {gather*} \frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (a^{2} b c - a^{3} d\right )} \sqrt {a b}} - \frac {d^{3} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{{\left (b c^{3} - a c^{2} d\right )} \sqrt {c d}} + \frac {3 \, {\left (b c + a d\right )} x^{2} - a c}{3 \, a^{2} c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.61, size = 367, normalized size = 3.67 \begin {gather*} \frac {\ln \left (a^{13}\,b^2\,d^5-a^8\,b^7\,c^5+c^5\,x\,{\left (-a^5\,b^5\right )}^{3/2}+a^{10}\,d^5\,x\,\sqrt {-a^5\,b^5}\right )\,\sqrt {-a^5\,b^5}}{2\,a^6\,d-2\,a^5\,b\,c}-\frac {\ln \left (a^8\,b^7\,c^5-a^{13}\,b^2\,d^5+c^5\,x\,{\left (-a^5\,b^5\right )}^{3/2}+a^{10}\,d^5\,x\,\sqrt {-a^5\,b^5}\right )\,\sqrt {-a^5\,b^5}}{2\,\left (a^6\,d-a^5\,b\,c\right )}-\frac {\frac {1}{3\,a\,c}-\frac {x^2\,\left (a\,d+b\,c\right )}{a^2\,c^2}}{x^3}-\frac {\ln \left (a^5\,c^8\,d^7-b^5\,c^{13}\,d^2+a^5\,x\,{\left (-c^5\,d^5\right )}^{3/2}+b^5\,c^{10}\,x\,\sqrt {-c^5\,d^5}\right )\,\sqrt {-c^5\,d^5}}{2\,\left (b\,c^6-a\,c^5\,d\right )}+\frac {\ln \left (b^5\,c^{13}\,d^2-a^5\,c^8\,d^7+a^5\,x\,{\left (-c^5\,d^5\right )}^{3/2}+b^5\,c^{10}\,x\,\sqrt {-c^5\,d^5}\right )\,\sqrt {-c^5\,d^5}}{2\,b\,c^6-2\,a\,c^5\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^2)*(c + d*x^2)),x)

[Out]

(log(a^13*b^2*d^5 - a^8*b^7*c^5 + c^5*x*(-a^5*b^5)^(3/2) + a^10*d^5*x*(-a^5*b^5)^(1/2))*(-a^5*b^5)^(1/2))/(2*a
^6*d - 2*a^5*b*c) - (log(a^8*b^7*c^5 - a^13*b^2*d^5 + c^5*x*(-a^5*b^5)^(3/2) + a^10*d^5*x*(-a^5*b^5)^(1/2))*(-
a^5*b^5)^(1/2))/(2*(a^6*d - a^5*b*c)) - (1/(3*a*c) - (x^2*(a*d + b*c))/(a^2*c^2))/x^3 - (log(a^5*c^8*d^7 - b^5
*c^13*d^2 + a^5*x*(-c^5*d^5)^(3/2) + b^5*c^10*x*(-c^5*d^5)^(1/2))*(-c^5*d^5)^(1/2))/(2*(b*c^6 - a*c^5*d)) + (l
og(b^5*c^13*d^2 - a^5*c^8*d^7 + a^5*x*(-c^5*d^5)^(3/2) + b^5*c^10*x*(-c^5*d^5)^(1/2))*(-c^5*d^5)^(1/2))/(2*b*c
^6 - 2*a*c^5*d)

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sympy [B]  time = 40.89, size = 1353, normalized size = 13.53

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)/(d*x**2+c),x)

[Out]

sqrt(-b**5/a**5)*log(x + (-a**10*c**5*d**5*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 + 2*a**9*b*c**6*d**4*(-b**5/a**5
)**(3/2)/(a*d - b*c)**3 - a**8*b**2*c**7*d**3*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 - a**8*d**8*sqrt(-b**5/a**5)/
(a*d - b*c) - a**7*b**3*c**8*d**2*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 + 2*a**6*b**4*c**9*d*(-b**5/a**5)**(3/2)/
(a*d - b*c)**3 - a**5*b**5*c**10*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 - b**8*c**8*sqrt(-b**5/a**5)/(a*d - b*c))/
(a**4*b**3*d**7 + a**3*b**4*c*d**6 + a**2*b**5*c**2*d**5 + a*b**6*c**3*d**4 + b**7*c**4*d**3))/(2*(a*d - b*c))
- sqrt(-b**5/a**5)*log(x + (a**10*c**5*d**5*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 - 2*a**9*b*c**6*d**4*(-b**5/a*
*5)**(3/2)/(a*d - b*c)**3 + a**8*b**2*c**7*d**3*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 + a**8*d**8*sqrt(-b**5/a**5
)/(a*d - b*c) + a**7*b**3*c**8*d**2*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 - 2*a**6*b**4*c**9*d*(-b**5/a**5)**(3/2
)/(a*d - b*c)**3 + a**5*b**5*c**10*(-b**5/a**5)**(3/2)/(a*d - b*c)**3 + b**8*c**8*sqrt(-b**5/a**5)/(a*d - b*c)
)/(a**4*b**3*d**7 + a**3*b**4*c*d**6 + a**2*b**5*c**2*d**5 + a*b**6*c**3*d**4 + b**7*c**4*d**3))/(2*(a*d - b*c
)) + sqrt(-d**5/c**5)*log(x + (-a**10*c**5*d**5*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 + 2*a**9*b*c**6*d**4*(-d**5
/c**5)**(3/2)/(a*d - b*c)**3 - a**8*b**2*c**7*d**3*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 - a**8*d**8*sqrt(-d**5/c
**5)/(a*d - b*c) - a**7*b**3*c**8*d**2*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 + 2*a**6*b**4*c**9*d*(-d**5/c**5)**(
3/2)/(a*d - b*c)**3 - a**5*b**5*c**10*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 - b**8*c**8*sqrt(-d**5/c**5)/(a*d - b
*c))/(a**4*b**3*d**7 + a**3*b**4*c*d**6 + a**2*b**5*c**2*d**5 + a*b**6*c**3*d**4 + b**7*c**4*d**3))/(2*(a*d -
b*c)) - sqrt(-d**5/c**5)*log(x + (a**10*c**5*d**5*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 - 2*a**9*b*c**6*d**4*(-d*
*5/c**5)**(3/2)/(a*d - b*c)**3 + a**8*b**2*c**7*d**3*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 + a**8*d**8*sqrt(-d**5
/c**5)/(a*d - b*c) + a**7*b**3*c**8*d**2*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 - 2*a**6*b**4*c**9*d*(-d**5/c**5)*
*(3/2)/(a*d - b*c)**3 + a**5*b**5*c**10*(-d**5/c**5)**(3/2)/(a*d - b*c)**3 + b**8*c**8*sqrt(-d**5/c**5)/(a*d -
b*c))/(a**4*b**3*d**7 + a**3*b**4*c*d**6 + a**2*b**5*c**2*d**5 + a*b**6*c**3*d**4 + b**7*c**4*d**3))/(2*(a*d
- b*c)) + (-a*c + x**2*(3*a*d + 3*b*c))/(3*a**2*c**2*x**3)

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