3.778 \(\int \frac {1}{x^7 (a+b x^4) (c+d x^4)} \, dx\)

Optimal. Leaf size=112 \[ \frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 a^{5/2} (b c-a d)}+\frac {a d+b c}{2 a^2 c^2 x^2}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 c^{5/2} (b c-a d)}-\frac {1}{6 a c x^6} \]

[Out]

-1/6/a/c/x^6+1/2*(a*d+b*c)/a^2/c^2/x^2+1/2*b^(5/2)*arctan(x^2*b^(1/2)/a^(1/2))/a^(5/2)/(-a*d+b*c)-1/2*d^(5/2)*
arctan(x^2*d^(1/2)/c^(1/2))/c^(5/2)/(-a*d+b*c)

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Rubi [A]  time = 0.22, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {465, 480, 583, 522, 205} \[ \frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 a^{5/2} (b c-a d)}+\frac {a d+b c}{2 a^2 c^2 x^2}-\frac {d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 c^{5/2} (b c-a d)}-\frac {1}{6 a c x^6} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^4)*(c + d*x^4)),x]

[Out]

-1/(6*a*c*x^6) + (b*c + a*d)/(2*a^2*c^2*x^2) + (b^(5/2)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*a^(5/2)*(b*c - a*d))
- (d^(5/2)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c]])/(2*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 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]

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

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Mathematica [A]  time = 0.26, size = 193, normalized size = 1.72 \[ \frac {\frac {3 b^{5/2} x^6 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{5/2}}+\frac {3 b^{5/2} x^6 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{a^{5/2}}-\frac {3 b^2 x^4}{a^2}+\frac {b}{a}-\frac {3 d^{5/2} x^6 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{5/2}}-\frac {3 d^{5/2} x^6 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{c^{5/2}}+\frac {3 d^2 x^4}{c^2}-\frac {d}{c}}{6 x^6 (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a + b*x^4)*(c + d*x^4)),x]

[Out]

(b/a - d/c - (3*b^2*x^4)/a^2 + (3*d^2*x^4)/c^2 + (3*b^(5/2)*x^6*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(5/
2) + (3*b^(5/2)*x^6*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(5/2) - (3*d^(5/2)*x^6*ArcTan[1 - (Sqrt[2]*d^(1
/4)*x)/c^(1/4)])/c^(5/2) - (3*d^(5/2)*x^6*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/c^(5/2))/(6*(-(b*c) + a*d)*
x^6)

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fricas [A]  time = 3.46, size = 592, normalized size = 5.29 \[ \left [-\frac {3 \, b^{2} c^{2} x^{6} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{4} - 2 \, a x^{2} \sqrt {-\frac {b}{a}} - a}{b x^{4} + a}\right ) + 3 \, a^{2} d^{2} x^{6} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{4} + 2 \, c x^{2} \sqrt {-\frac {d}{c}} - c}{d x^{4} + c}\right ) - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{4} + 2 \, a b c^{2} - 2 \, a^{2} c d}{12 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{6}}, \frac {6 \, a^{2} d^{2} x^{6} \sqrt {\frac {d}{c}} \arctan \left (\frac {c \sqrt {\frac {d}{c}}}{d x^{2}}\right ) - 3 \, b^{2} c^{2} x^{6} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{4} - 2 \, a x^{2} \sqrt {-\frac {b}{a}} - a}{b x^{4} + a}\right ) + 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{4} - 2 \, a b c^{2} + 2 \, a^{2} c d}{12 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{6}}, -\frac {6 \, b^{2} c^{2} x^{6} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b x^{2}}\right ) + 3 \, a^{2} d^{2} x^{6} \sqrt {-\frac {d}{c}} \log \left (\frac {d x^{4} + 2 \, c x^{2} \sqrt {-\frac {d}{c}} - c}{d x^{4} + c}\right ) - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{4} + 2 \, a b c^{2} - 2 \, a^{2} c d}{12 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{6}}, -\frac {3 \, b^{2} c^{2} x^{6} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b x^{2}}\right ) - 3 \, a^{2} d^{2} x^{6} \sqrt {\frac {d}{c}} \arctan \left (\frac {c \sqrt {\frac {d}{c}}}{d x^{2}}\right ) - 3 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{4} + a b c^{2} - a^{2} c d}{6 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")

[Out]

[-1/12*(3*b^2*c^2*x^6*sqrt(-b/a)*log((b*x^4 - 2*a*x^2*sqrt(-b/a) - a)/(b*x^4 + a)) + 3*a^2*d^2*x^6*sqrt(-d/c)*
log((d*x^4 + 2*c*x^2*sqrt(-d/c) - c)/(d*x^4 + c)) - 6*(b^2*c^2 - a^2*d^2)*x^4 + 2*a*b*c^2 - 2*a^2*c*d)/((a^2*b
*c^3 - a^3*c^2*d)*x^6), 1/12*(6*a^2*d^2*x^6*sqrt(d/c)*arctan(c*sqrt(d/c)/(d*x^2)) - 3*b^2*c^2*x^6*sqrt(-b/a)*l
og((b*x^4 - 2*a*x^2*sqrt(-b/a) - a)/(b*x^4 + a)) + 6*(b^2*c^2 - a^2*d^2)*x^4 - 2*a*b*c^2 + 2*a^2*c*d)/((a^2*b*
c^3 - a^3*c^2*d)*x^6), -1/12*(6*b^2*c^2*x^6*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*x^2)) + 3*a^2*d^2*x^6*sqrt(-d/c)*l
og((d*x^4 + 2*c*x^2*sqrt(-d/c) - c)/(d*x^4 + c)) - 6*(b^2*c^2 - a^2*d^2)*x^4 + 2*a*b*c^2 - 2*a^2*c*d)/((a^2*b*
c^3 - a^3*c^2*d)*x^6), -1/6*(3*b^2*c^2*x^6*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*x^2)) - 3*a^2*d^2*x^6*sqrt(d/c)*arc
tan(c*sqrt(d/c)/(d*x^2)) - 3*(b^2*c^2 - a^2*d^2)*x^4 + a*b*c^2 - a^2*c*d)/((a^2*b*c^3 - a^3*c^2*d)*x^6)]

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giac [A]  time = 0.18, size = 103, normalized size = 0.92 \[ \frac {b^{3} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {a b}} - \frac {d^{3} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c^{3} - a c^{2} d\right )} \sqrt {c d}} + \frac {3 \, b c x^{4} + 3 \, a d x^{4} - a c}{6 \, a^{2} c^{2} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")

[Out]

1/2*b^3*arctan(b*x^2/sqrt(a*b))/((a^2*b*c - a^3*d)*sqrt(a*b)) - 1/2*d^3*arctan(d*x^2/sqrt(c*d))/((b*c^3 - a*c^
2*d)*sqrt(c*d)) + 1/6*(3*b*c*x^4 + 3*a*d*x^4 - a*c)/(a^2*c^2*x^6)

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maple [A]  time = 0.06, size = 105, normalized size = 0.94 \[ -\frac {b^{3} \arctan \left (\frac {b \,x^{2}}{\sqrt {a b}}\right )}{2 \left (a d -b c \right ) \sqrt {a b}\, a^{2}}+\frac {d^{3} \arctan \left (\frac {d \,x^{2}}{\sqrt {c d}}\right )}{2 \left (a d -b c \right ) \sqrt {c d}\, c^{2}}+\frac {d}{2 a \,c^{2} x^{2}}+\frac {b}{2 a^{2} c \,x^{2}}-\frac {1}{6 a c \,x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^4+a)/(d*x^4+c),x)

[Out]

1/2*d^3/c^2/(a*d-b*c)/(c*d)^(1/2)*arctan(1/(c*d)^(1/2)*d*x^2)-1/2*b^3/a^2/(a*d-b*c)/(a*b)^(1/2)*arctan(1/(a*b)
^(1/2)*b*x^2)-1/6/a/c/x^6+1/2/a/c^2/x^2*d+1/2/a^2/c/x^2*b

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maxima [A]  time = 1.51, size = 101, normalized size = 0.90 \[ \frac {b^{3} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (a^{2} b c - a^{3} d\right )} \sqrt {a b}} - \frac {d^{3} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c^{3} - a c^{2} d\right )} \sqrt {c d}} + \frac {3 \, {\left (b c + a d\right )} x^{4} - a c}{6 \, a^{2} c^{2} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")

[Out]

1/2*b^3*arctan(b*x^2/sqrt(a*b))/((a^2*b*c - a^3*d)*sqrt(a*b)) - 1/2*d^3*arctan(d*x^2/sqrt(c*d))/((b*c^3 - a*c^
2*d)*sqrt(c*d)) + 1/6*(3*(b*c + a*d)*x^4 - a*c)/(a^2*c^2*x^6)

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mupad [B]  time = 5.53, size = 535, normalized size = 4.78 \[ \frac {\ln \left (c^{10}\,{\left (-a^5\,b^5\right )}^{5/2}+a^{20}\,d^{10}\,\sqrt {-a^5\,b^5}-a^{12}\,b^{13}\,c^{10}\,x^2-a^{22}\,b^3\,d^{10}\,x^2+2\,a^{10}\,c^5\,d^5\,{\left (-a^5\,b^5\right )}^{3/2}+2\,a^{17}\,b^8\,c^5\,d^5\,x^2\right )\,\sqrt {-a^5\,b^5}}{4\,a^6\,d-4\,a^5\,b\,c}-\frac {\ln \left (c^{10}\,{\left (-a^5\,b^5\right )}^{5/2}+a^{20}\,d^{10}\,\sqrt {-a^5\,b^5}+a^{12}\,b^{13}\,c^{10}\,x^2+a^{22}\,b^3\,d^{10}\,x^2+2\,a^{10}\,c^5\,d^5\,{\left (-a^5\,b^5\right )}^{3/2}-2\,a^{17}\,b^8\,c^5\,d^5\,x^2\right )\,\sqrt {-a^5\,b^5}}{4\,\left (a^6\,d-a^5\,b\,c\right )}-\frac {\frac {1}{6\,a\,c}-\frac {x^4\,\left (a\,d+b\,c\right )}{2\,a^2\,c^2}}{x^6}-\frac {\ln \left (a^{10}\,{\left (-c^5\,d^5\right )}^{5/2}+b^{10}\,c^{20}\,\sqrt {-c^5\,d^5}+a^{10}\,c^{12}\,d^{13}\,x^2+b^{10}\,c^{22}\,d^3\,x^2+2\,a^5\,b^5\,c^{10}\,{\left (-c^5\,d^5\right )}^{3/2}-2\,a^5\,b^5\,c^{17}\,d^8\,x^2\right )\,\sqrt {-c^5\,d^5}}{4\,\left (b\,c^6-a\,c^5\,d\right )}+\frac {\ln \left (a^{10}\,{\left (-c^5\,d^5\right )}^{5/2}+b^{10}\,c^{20}\,\sqrt {-c^5\,d^5}-a^{10}\,c^{12}\,d^{13}\,x^2-b^{10}\,c^{22}\,d^3\,x^2+2\,a^5\,b^5\,c^{10}\,{\left (-c^5\,d^5\right )}^{3/2}+2\,a^5\,b^5\,c^{17}\,d^8\,x^2\right )\,\sqrt {-c^5\,d^5}}{4\,b\,c^6-4\,a\,c^5\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a + b*x^4)*(c + d*x^4)),x)

[Out]

(log(c^10*(-a^5*b^5)^(5/2) + a^20*d^10*(-a^5*b^5)^(1/2) - a^12*b^13*c^10*x^2 - a^22*b^3*d^10*x^2 + 2*a^10*c^5*
d^5*(-a^5*b^5)^(3/2) + 2*a^17*b^8*c^5*d^5*x^2)*(-a^5*b^5)^(1/2))/(4*a^6*d - 4*a^5*b*c) - (log(c^10*(-a^5*b^5)^
(5/2) + a^20*d^10*(-a^5*b^5)^(1/2) + a^12*b^13*c^10*x^2 + a^22*b^3*d^10*x^2 + 2*a^10*c^5*d^5*(-a^5*b^5)^(3/2)
- 2*a^17*b^8*c^5*d^5*x^2)*(-a^5*b^5)^(1/2))/(4*(a^6*d - a^5*b*c)) - (1/(6*a*c) - (x^4*(a*d + b*c))/(2*a^2*c^2)
)/x^6 - (log(a^10*(-c^5*d^5)^(5/2) + b^10*c^20*(-c^5*d^5)^(1/2) + a^10*c^12*d^13*x^2 + b^10*c^22*d^3*x^2 + 2*a
^5*b^5*c^10*(-c^5*d^5)^(3/2) - 2*a^5*b^5*c^17*d^8*x^2)*(-c^5*d^5)^(1/2))/(4*(b*c^6 - a*c^5*d)) + (log(a^10*(-c
^5*d^5)^(5/2) + b^10*c^20*(-c^5*d^5)^(1/2) - a^10*c^12*d^13*x^2 - b^10*c^22*d^3*x^2 + 2*a^5*b^5*c^10*(-c^5*d^5
)^(3/2) + 2*a^5*b^5*c^17*d^8*x^2)*(-c^5*d^5)^(1/2))/(4*b*c^6 - 4*a*c^5*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**7/(b*x**4+a)/(d*x**4+c),x)

[Out]

Timed out

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