3.6.3 \(\int \frac {1}{x^3 (a+b x^4) (c+d x^4)} \, dx\)

Optimal. Leaf size=92 \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 a^{3/2} (b c-a d)}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 c^{3/2} (b c-a d)}-\frac {1}{2 a c x^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 92, 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 = {465, 480, 522, 205} \begin {gather*} -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 a^{3/2} (b c-a d)}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 c^{3/2} (b c-a d)}-\frac {1}{2 a c x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^4)*(c + d*x^4)),x]

[Out]

-1/(2*a*c*x^2) - (b^(3/2)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*a^(3/2)*(b*c - a*d)) + (d^(3/2)*ArcTan[(Sqrt[d]*x^
2)/Sqrt[c]])/(2*c^(3/2)*(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 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

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]

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 169, normalized size = 1.84 \begin {gather*} \frac {-\frac {b^{3/2} x^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{3/2}}-\frac {b^{3/2} x^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{a^{3/2}}+\frac {b}{a}+\frac {d^{3/2} x^2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{3/2}}+\frac {d^{3/2} x^2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{c^{3/2}}-\frac {d}{c}}{2 x^2 (a d-b c)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^4)*(c + d*x^4)),x]

[Out]

(b/a - d/c - (b^(3/2)*x^2*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(3/2) - (b^(3/2)*x^2*ArcTan[1 + (Sqrt[2]*
b^(1/4)*x)/a^(1/4)])/a^(3/2) + (d^(3/2)*x^2*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(3/2) + (d^(3/2)*x^2*Ar
cTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(3/2))/(2*(-(b*c) + a*d)*x^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^3 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^3*(a + b*x^4)*(c + d*x^4)),x]

[Out]

IntegrateAlgebraic[1/(x^3*(a + b*x^4)*(c + d*x^4)), x]

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fricas [A]  time = 0.88, size = 432, normalized size = 4.70 \begin {gather*} \left [-\frac {b c x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{4} + 2 \, a x^{2} \sqrt {-\frac {b}{a}} - a}{b x^{4} + a}\right ) + a d x^{2} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{4} - 2 \, c x^{2} \sqrt {-\frac {d}{c}} - c}{d x^{4} + c}\right ) + 2 \, b c - 2 \, a d}{4 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}, -\frac {2 \, a d x^{2} \sqrt {\frac {d}{c}} \arctan \left (\frac {c \sqrt {\frac {d}{c}}}{d x^{2}}\right ) + b c x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{4} + 2 \, a x^{2} \sqrt {-\frac {b}{a}} - a}{b x^{4} + a}\right ) + 2 \, b c - 2 \, a d}{4 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}, \frac {2 \, b c x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b x^{2}}\right ) - a d x^{2} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{4} - 2 \, c x^{2} \sqrt {-\frac {d}{c}} - c}{d x^{4} + c}\right ) - 2 \, b c + 2 \, a d}{4 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}, \frac {b c x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b x^{2}}\right ) - a d x^{2} \sqrt {\frac {d}{c}} \arctan \left (\frac {c \sqrt {\frac {d}{c}}}{d x^{2}}\right ) - b c + a d}{2 \, {\left (a b c^{2} - a^{2} c d\right )} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")

[Out]

[-1/4*(b*c*x^2*sqrt(-b/a)*log((b*x^4 + 2*a*x^2*sqrt(-b/a) - a)/(b*x^4 + a)) + a*d*x^2*sqrt(-d/c)*log((d*x^4 -
2*c*x^2*sqrt(-d/c) - c)/(d*x^4 + c)) + 2*b*c - 2*a*d)/((a*b*c^2 - a^2*c*d)*x^2), -1/4*(2*a*d*x^2*sqrt(d/c)*arc
tan(c*sqrt(d/c)/(d*x^2)) + b*c*x^2*sqrt(-b/a)*log((b*x^4 + 2*a*x^2*sqrt(-b/a) - a)/(b*x^4 + a)) + 2*b*c - 2*a*
d)/((a*b*c^2 - a^2*c*d)*x^2), 1/4*(2*b*c*x^2*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*x^2)) - a*d*x^2*sqrt(-d/c)*log((d
*x^4 - 2*c*x^2*sqrt(-d/c) - c)/(d*x^4 + c)) - 2*b*c + 2*a*d)/((a*b*c^2 - a^2*c*d)*x^2), 1/2*(b*c*x^2*sqrt(b/a)
*arctan(a*sqrt(b/a)/(b*x^2)) - a*d*x^2*sqrt(d/c)*arctan(c*sqrt(d/c)/(d*x^2)) - b*c + a*d)/((a*b*c^2 - a^2*c*d)
*x^2)]

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giac [A]  time = 0.21, size = 80, normalized size = 0.87 \begin {gather*} -\frac {b^{2} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (a b c - a^{2} d\right )} \sqrt {a b}} + \frac {d^{2} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c^{2} - a c d\right )} \sqrt {c d}} - \frac {1}{2 \, a c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 81, normalized size = 0.88 \begin {gather*} \frac {b^{2} \arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 \left (a d -b c \right ) \sqrt {a b}\, a}-\frac {d^{2} \arctan \left (\frac {d \,x^{2}}{\sqrt {c d}}\right )}{2 \left (a d -b c \right ) \sqrt {c d}\, c}-\frac {1}{2 a c \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a)/(d*x^4+c),x)

[Out]

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

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maxima [A]  time = 1.20, size = 80, normalized size = 0.87 \begin {gather*} -\frac {b^{2} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (a b c - a^{2} d\right )} \sqrt {a b}} + \frac {d^{2} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c^{2} - a c d\right )} \sqrt {c d}} - \frac {1}{2 \, a c x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.35, size = 354, normalized size = 3.85 \begin {gather*} \frac {\ln \left (c^3\,x^2\,{\left (-a^3\,b^3\right )}^{3/2}-a^8\,b\,d^3+a^5\,b^4\,c^3+a^6\,d^3\,x^2\,\sqrt {-a^3\,b^3}\right )\,\sqrt {-a^3\,b^3}}{4\,a^4\,d-4\,a^3\,b\,c}-\frac {\ln \left (c^3\,x^2\,{\left (-a^3\,b^3\right )}^{3/2}+a^8\,b\,d^3-a^5\,b^4\,c^3+a^6\,d^3\,x^2\,\sqrt {-a^3\,b^3}\right )\,\sqrt {-a^3\,b^3}}{4\,\left (a^4\,d-a^3\,b\,c\right )}-\frac {1}{2\,a\,c\,x^2}-\frac {\ln \left (a^3\,x^2\,{\left (-c^3\,d^3\right )}^{3/2}+b^3\,c^8\,d-a^3\,c^5\,d^4+b^3\,c^6\,x^2\,\sqrt {-c^3\,d^3}\right )\,\sqrt {-c^3\,d^3}}{4\,\left (b\,c^4-a\,c^3\,d\right )}+\frac {\ln \left (a^3\,x^2\,{\left (-c^3\,d^3\right )}^{3/2}-b^3\,c^8\,d+a^3\,c^5\,d^4+b^3\,c^6\,x^2\,\sqrt {-c^3\,d^3}\right )\,\sqrt {-c^3\,d^3}}{4\,b\,c^4-4\,a\,c^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^4)*(c + d*x^4)),x)

[Out]

(log(c^3*x^2*(-a^3*b^3)^(3/2) - a^8*b*d^3 + a^5*b^4*c^3 + a^6*d^3*x^2*(-a^3*b^3)^(1/2))*(-a^3*b^3)^(1/2))/(4*a
^4*d - 4*a^3*b*c) - (log(c^3*x^2*(-a^3*b^3)^(3/2) + a^8*b*d^3 - a^5*b^4*c^3 + a^6*d^3*x^2*(-a^3*b^3)^(1/2))*(-
a^3*b^3)^(1/2))/(4*(a^4*d - a^3*b*c)) - 1/(2*a*c*x^2) - (log(a^3*x^2*(-c^3*d^3)^(3/2) + b^3*c^8*d - a^3*c^5*d^
4 + b^3*c^6*x^2*(-c^3*d^3)^(1/2))*(-c^3*d^3)^(1/2))/(4*(b*c^4 - a*c^3*d)) + (log(a^3*x^2*(-c^3*d^3)^(3/2) - b^
3*c^8*d + a^3*c^5*d^4 + b^3*c^6*x^2*(-c^3*d^3)^(1/2))*(-c^3*d^3)^(1/2))/(4*b*c^4 - 4*a*c^3*d)

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sympy [B]  time = 84.89, size = 1103, normalized size = 11.99

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**4+a)/(d*x**4+c),x)

[Out]

-sqrt(-b**3/a**3)*log(x**2 + (-a**7*c**3*d**4*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 + 2*a**6*b*c**4*d**3*(-b**3/a
**3)**(3/2)/(a*d - b*c)**3 - 2*a**5*b**2*c**5*d**2*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 - a**5*d**5*sqrt(-b**3/a
**3)/(a*d - b*c) + 2*a**4*b**3*c**6*d*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 - a**3*b**4*c**7*(-b**3/a**3)**(3/2)/
(a*d - b*c)**3 - b**5*c**5*sqrt(-b**3/a**3)/(a*d - b*c))/(a**2*b**2*d**4 + a*b**3*c*d**3 + b**4*c**2*d**2))/(4
*(a*d - b*c)) + sqrt(-b**3/a**3)*log(x**2 + (a**7*c**3*d**4*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 - 2*a**6*b*c**4
*d**3*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 + 2*a**5*b**2*c**5*d**2*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 + a**5*d**
5*sqrt(-b**3/a**3)/(a*d - b*c) - 2*a**4*b**3*c**6*d*(-b**3/a**3)**(3/2)/(a*d - b*c)**3 + a**3*b**4*c**7*(-b**3
/a**3)**(3/2)/(a*d - b*c)**3 + b**5*c**5*sqrt(-b**3/a**3)/(a*d - b*c))/(a**2*b**2*d**4 + a*b**3*c*d**3 + b**4*
c**2*d**2))/(4*(a*d - b*c)) - sqrt(-d**3/c**3)*log(x**2 + (-a**7*c**3*d**4*(-d**3/c**3)**(3/2)/(a*d - b*c)**3
+ 2*a**6*b*c**4*d**3*(-d**3/c**3)**(3/2)/(a*d - b*c)**3 - 2*a**5*b**2*c**5*d**2*(-d**3/c**3)**(3/2)/(a*d - b*c
)**3 - a**5*d**5*sqrt(-d**3/c**3)/(a*d - b*c) + 2*a**4*b**3*c**6*d*(-d**3/c**3)**(3/2)/(a*d - b*c)**3 - a**3*b
**4*c**7*(-d**3/c**3)**(3/2)/(a*d - b*c)**3 - b**5*c**5*sqrt(-d**3/c**3)/(a*d - b*c))/(a**2*b**2*d**4 + a*b**3
*c*d**3 + b**4*c**2*d**2))/(4*(a*d - b*c)) + sqrt(-d**3/c**3)*log(x**2 + (a**7*c**3*d**4*(-d**3/c**3)**(3/2)/(
a*d - b*c)**3 - 2*a**6*b*c**4*d**3*(-d**3/c**3)**(3/2)/(a*d - b*c)**3 + 2*a**5*b**2*c**5*d**2*(-d**3/c**3)**(3
/2)/(a*d - b*c)**3 + a**5*d**5*sqrt(-d**3/c**3)/(a*d - b*c) - 2*a**4*b**3*c**6*d*(-d**3/c**3)**(3/2)/(a*d - b*
c)**3 + a**3*b**4*c**7*(-d**3/c**3)**(3/2)/(a*d - b*c)**3 + b**5*c**5*sqrt(-d**3/c**3)/(a*d - b*c))/(a**2*b**2
*d**4 + a*b**3*c*d**3 + b**4*c**2*d**2))/(4*(a*d - b*c)) - 1/(2*a*c*x**2)

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