3.773 \(\int \frac {x^{13}}{(a+b x^4) (c+d x^4)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.27, 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, 479, 582, 522, 205} \[ -\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{5/2} (b c-a d)}-\frac {x^2 (a d+b c)}{2 b^2 d^2}+\frac {c^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{5/2} (b c-a d)}+\frac {x^6}{6 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^13/((a + b*x^4)*(c + d*x^4)),x]

[Out]

-((b*c + a*d)*x^2)/(2*b^2*d^2) + x^6/(6*b*d) - (a^(5/2)*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/(2*b^(5/2)*(b*c - a*d))
+ (c^(5/2)*ArcTan[(Sqrt[d]*x^2)/Sqrt[c]])/(2*d^(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 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[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 582

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

Rubi steps

\begin {align*} \int \frac {x^{13}}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^6}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )\\ &=\frac {x^6}{6 b d}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (3 a c+3 (b c+a d) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )}{6 b d}\\ &=-\frac {(b c+a d) x^2}{2 b^2 d^2}+\frac {x^6}{6 b d}+\frac {\operatorname {Subst}\left (\int \frac {3 a c (b c+a d)+3 \left (b^2 c^2+a d (b c+a d)\right ) x^2}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx,x,x^2\right )}{6 b^2 d^2}\\ &=-\frac {(b c+a d) x^2}{2 b^2 d^2}+\frac {x^6}{6 b d}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^2\right )}{2 b^2 (b c-a d)}+\frac {c^3 \operatorname {Subst}\left (\int \frac {1}{c+d x^2} \, dx,x,x^2\right )}{2 d^2 (b c-a d)}\\ &=-\frac {(b c+a d) x^2}{2 b^2 d^2}+\frac {x^6}{6 b d}-\frac {a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{5/2} (b c-a d)}+\frac {c^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{2 d^{5/2} (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 104, normalized size = 0.93 \[ \frac {1}{6} \left (\frac {3 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{b^{5/2} (a d-b c)}+\frac {x^2 \left (-3 a d-3 b c+b d x^4\right )}{b^2 d^2}+\frac {3 c^{5/2} \tan ^{-1}\left (\frac {\sqrt {d} x^2}{\sqrt {c}}\right )}{d^{5/2} (b c-a d)}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^13/((a + b*x^4)*(c + d*x^4)),x]

[Out]

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

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fricas [A]  time = 2.61, size = 576, normalized size = 5.14 \[ \left [\frac {2 \, {\left (b^{2} c d - a b d^{2}\right )} x^{6} - 3 \, a^{2} d^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} + 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) - 3 \, b^{2} c^{2} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} - 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right ) - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{12 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}}, \frac {2 \, {\left (b^{2} c d - a b d^{2}\right )} x^{6} - 6 \, a^{2} d^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) - 3 \, b^{2} c^{2} \sqrt {-\frac {c}{d}} \log \left (\frac {d x^{4} - 2 \, d x^{2} \sqrt {-\frac {c}{d}} - c}{d x^{4} + c}\right ) - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{12 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}}, \frac {2 \, {\left (b^{2} c d - a b d^{2}\right )} x^{6} + 6 \, b^{2} c^{2} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right ) - 3 \, a^{2} d^{2} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{4} + 2 \, b x^{2} \sqrt {-\frac {a}{b}} - a}{b x^{4} + a}\right ) - 6 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{12 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}}, \frac {{\left (b^{2} c d - a b d^{2}\right )} x^{6} - 3 \, a^{2} d^{2} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x^{2} \sqrt {\frac {a}{b}}}{a}\right ) + 3 \, b^{2} c^{2} \sqrt {\frac {c}{d}} \arctan \left (\frac {d x^{2} \sqrt {\frac {c}{d}}}{c}\right ) - 3 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{2}}{6 \, {\left (b^{3} c d^{2} - a b^{2} d^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 112, normalized size = 1.00 \[ -\frac {a^{3} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {a b}} + \frac {c^{3} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d^{2} - a d^{3}\right )} \sqrt {c d}} + \frac {b^{2} d^{2} x^{6} - 3 \, b^{2} c d x^{2} - 3 \, a b d^{2} x^{2}}{6 \, b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(b*x^4+a)/(d*x^4+c),x)

[Out]

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

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maxima [A]  time = 1.32, size = 100, normalized size = 0.89 \[ -\frac {a^{3} \arctan \left (\frac {b x^{2}}{\sqrt {a b}}\right )}{2 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {a b}} + \frac {c^{3} \arctan \left (\frac {d x^{2}}{\sqrt {c d}}\right )}{2 \, {\left (b c d^{2} - a d^{3}\right )} \sqrt {c d}} + \frac {b d x^{6} - 3 \, {\left (b c + a d\right )} x^{2}}{6 \, b^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.70, size = 532, normalized size = 4.75 \[ \frac {\ln \left (d^{10}\,{\left (-a^5\,b^5\right )}^{5/2}+b^{20}\,c^{10}\,\sqrt {-a^5\,b^5}-a^2\,b^{23}\,c^{10}\,x^2-a^{12}\,b^{13}\,d^{10}\,x^2+2\,b^{10}\,c^5\,d^5\,{\left (-a^5\,b^5\right )}^{3/2}+2\,a^7\,b^{18}\,c^5\,d^5\,x^2\right )\,\sqrt {-a^5\,b^5}}{4\,b^6\,c-4\,a\,b^5\,d}-\frac {\ln \left (d^{10}\,{\left (-a^5\,b^5\right )}^{5/2}+b^{20}\,c^{10}\,\sqrt {-a^5\,b^5}+a^2\,b^{23}\,c^{10}\,x^2+a^{12}\,b^{13}\,d^{10}\,x^2+2\,b^{10}\,c^5\,d^5\,{\left (-a^5\,b^5\right )}^{3/2}-2\,a^7\,b^{18}\,c^5\,d^5\,x^2\right )\,\sqrt {-a^5\,b^5}}{4\,\left (b^6\,c-a\,b^5\,d\right )}-\frac {\ln \left (b^{10}\,{\left (-c^5\,d^5\right )}^{5/2}+a^{10}\,d^{20}\,\sqrt {-c^5\,d^5}+a^{10}\,c^2\,d^{23}\,x^2+b^{10}\,c^{12}\,d^{13}\,x^2+2\,a^5\,b^5\,d^{10}\,{\left (-c^5\,d^5\right )}^{3/2}-2\,a^5\,b^5\,c^7\,d^{18}\,x^2\right )\,\sqrt {-c^5\,d^5}}{4\,\left (a\,d^6-b\,c\,d^5\right )}+\frac {\ln \left (b^{10}\,{\left (-c^5\,d^5\right )}^{5/2}+a^{10}\,d^{20}\,\sqrt {-c^5\,d^5}-a^{10}\,c^2\,d^{23}\,x^2-b^{10}\,c^{12}\,d^{13}\,x^2+2\,a^5\,b^5\,d^{10}\,{\left (-c^5\,d^5\right )}^{3/2}+2\,a^5\,b^5\,c^7\,d^{18}\,x^2\right )\,\sqrt {-c^5\,d^5}}{4\,a\,d^6-4\,b\,c\,d^5}+\frac {x^6}{6\,b\,d}-\frac {x^2\,\left (a\,d+b\,c\right )}{2\,b^2\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/((a + b*x^4)*(c + d*x^4)),x)

[Out]

(log(d^10*(-a^5*b^5)^(5/2) + b^20*c^10*(-a^5*b^5)^(1/2) - a^2*b^23*c^10*x^2 - a^12*b^13*d^10*x^2 + 2*b^10*c^5*
d^5*(-a^5*b^5)^(3/2) + 2*a^7*b^18*c^5*d^5*x^2)*(-a^5*b^5)^(1/2))/(4*b^6*c - 4*a*b^5*d) - (log(d^10*(-a^5*b^5)^
(5/2) + b^20*c^10*(-a^5*b^5)^(1/2) + a^2*b^23*c^10*x^2 + a^12*b^13*d^10*x^2 + 2*b^10*c^5*d^5*(-a^5*b^5)^(3/2)
- 2*a^7*b^18*c^5*d^5*x^2)*(-a^5*b^5)^(1/2))/(4*(b^6*c - a*b^5*d)) - (log(b^10*(-c^5*d^5)^(5/2) + a^10*d^20*(-c
^5*d^5)^(1/2) + a^10*c^2*d^23*x^2 + b^10*c^12*d^13*x^2 + 2*a^5*b^5*d^10*(-c^5*d^5)^(3/2) - 2*a^5*b^5*c^7*d^18*
x^2)*(-c^5*d^5)^(1/2))/(4*(a*d^6 - b*c*d^5)) + (log(b^10*(-c^5*d^5)^(5/2) + a^10*d^20*(-c^5*d^5)^(1/2) - a^10*
c^2*d^23*x^2 - b^10*c^12*d^13*x^2 + 2*a^5*b^5*d^10*(-c^5*d^5)^(3/2) + 2*a^5*b^5*c^7*d^18*x^2)*(-c^5*d^5)^(1/2)
)/(4*a*d^6 - 4*b*c*d^5) + x^6/(6*b*d) - (x^2*(a*d + b*c))/(2*b^2*d^2)

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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(x**13/(b*x**4+a)/(d*x**4+c),x)

[Out]

Timed out

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