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

Optimal. Leaf size=245 \[ -\frac {(a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}-\frac {(a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {x (b c-a d)}{4 a b \left (a+b x^4\right )} \]

[Out]

1/4*(-a*d+b*c)*x/a/b/(b*x^4+a)+1/16*(a*d+3*b*c)*arctan(-1+b^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/4)/b^(5/4)*2^(1/2)+1
/16*(a*d+3*b*c)*arctan(1+b^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/4)/b^(5/4)*2^(1/2)-1/32*(a*d+3*b*c)*ln(-a^(1/4)*b^(1/
4)*x*2^(1/2)+a^(1/2)+x^2*b^(1/2))/a^(7/4)/b^(5/4)*2^(1/2)+1/32*(a*d+3*b*c)*ln(a^(1/4)*b^(1/4)*x*2^(1/2)+a^(1/2
)+x^2*b^(1/2))/a^(7/4)/b^(5/4)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {385, 211, 1165, 628, 1162, 617, 204} \[ -\frac {(a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}-\frac {(a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {x (b c-a d)}{4 a b \left (a+b x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^4)/(a + b*x^4)^2,x]

[Out]

((b*c - a*d)*x)/(4*a*b*(a + b*x^4)) - ((3*b*c + a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/
4)*b^(5/4)) + ((3*b*c + a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*b^(5/4)) - ((3*b*c +
a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(16*Sqrt[2]*a^(7/4)*b^(5/4)) + ((3*b*c + a*d)*Log
[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(16*Sqrt[2]*a^(7/4)*b^(5/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {c+d x^4}{\left (a+b x^4\right )^2} \, dx &=\frac {(b c-a d) x}{4 a b \left (a+b x^4\right )}+\frac {(3 b c+a d) \int \frac {1}{a+b x^4} \, dx}{4 a b}\\ &=\frac {(b c-a d) x}{4 a b \left (a+b x^4\right )}+\frac {(3 b c+a d) \int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx}{8 a^{3/2} b}+\frac {(3 b c+a d) \int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx}{8 a^{3/2} b}\\ &=\frac {(b c-a d) x}{4 a b \left (a+b x^4\right )}+\frac {(3 b c+a d) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{16 a^{3/2} b^{3/2}}+\frac {(3 b c+a d) \int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx}{16 a^{3/2} b^{3/2}}-\frac {(3 b c+a d) \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx}{16 \sqrt {2} a^{7/4} b^{5/4}}-\frac {(3 b c+a d) \int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx}{16 \sqrt {2} a^{7/4} b^{5/4}}\\ &=\frac {(b c-a d) x}{4 a b \left (a+b x^4\right )}-\frac {(3 b c+a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(3 b c+a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(3 b c+a d) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}-\frac {(3 b c+a d) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}\\ &=\frac {(b c-a d) x}{4 a b \left (a+b x^4\right )}-\frac {(3 b c+a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(3 b c+a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}-\frac {(3 b c+a d) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(3 b c+a d) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{5/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 212, normalized size = 0.87 \[ \frac {-\frac {8 a^{3/4} \sqrt [4]{b} x (a d-b c)}{a+b x^4}-\sqrt {2} (a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )+\sqrt {2} (a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )-2 \sqrt {2} (a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \sqrt {2} (a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{32 a^{7/4} b^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^(3/4)*b^(1/4)*(-(b*c) + a*d)*x)/(a + b*x^4) - 2*Sqrt[2]*(3*b*c + a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^
(1/4)] + 2*Sqrt[2]*(3*b*c + a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] - Sqrt[2]*(3*b*c + a*d)*Log[Sqrt[a] -
Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + Sqrt[2]*(3*b*c + a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sq
rt[b]*x^2])/(32*a^(7/4)*b^(5/4))

________________________________________________________________________________________

fricas [B]  time = 0.88, size = 711, normalized size = 2.90 \[ \frac {4 \, {\left (a b^{2} x^{4} + a^{2} b\right )} \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {a^{5} b^{4} x \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {3}{4}} - a^{5} b^{4} \sqrt {\frac {a^{4} b^{2} \sqrt {-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}} + {\left (9 \, b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2}}{9 \, b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}}} \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {3}{4}}}{27 \, b^{3} c^{3} + 27 \, a b^{2} c^{2} d + 9 \, a^{2} b c d^{2} + a^{3} d^{3}}\right ) + {\left (a b^{2} x^{4} + a^{2} b\right )} \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (a^{2} b \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + {\left (3 \, b c + a d\right )} x\right ) - {\left (a b^{2} x^{4} + a^{2} b\right )} \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (-a^{2} b \left (-\frac {81 \, b^{4} c^{4} + 108 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 12 \, a^{3} b c d^{3} + a^{4} d^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + {\left (3 \, b c + a d\right )} x\right ) + 4 \, {\left (b c - a d\right )} x}{16 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(a*b^2*x^4 + a^2*b)*(-(81*b^4*c^4 + 108*a*b^3*c^3*d + 54*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 + a^4*d^4)/(
a^7*b^5))^(1/4)*arctan(-(a^5*b^4*x*(-(81*b^4*c^4 + 108*a*b^3*c^3*d + 54*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 + a^4
*d^4)/(a^7*b^5))^(3/4) - a^5*b^4*sqrt((a^4*b^2*sqrt(-(81*b^4*c^4 + 108*a*b^3*c^3*d + 54*a^2*b^2*c^2*d^2 + 12*a
^3*b*c*d^3 + a^4*d^4)/(a^7*b^5)) + (9*b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2)/(9*b^2*c^2 + 6*a*b*c*d + a^2*d^2))*(
-(81*b^4*c^4 + 108*a*b^3*c^3*d + 54*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 + a^4*d^4)/(a^7*b^5))^(3/4))/(27*b^3*c^3
+ 27*a*b^2*c^2*d + 9*a^2*b*c*d^2 + a^3*d^3)) + (a*b^2*x^4 + a^2*b)*(-(81*b^4*c^4 + 108*a*b^3*c^3*d + 54*a^2*b^
2*c^2*d^2 + 12*a^3*b*c*d^3 + a^4*d^4)/(a^7*b^5))^(1/4)*log(a^2*b*(-(81*b^4*c^4 + 108*a*b^3*c^3*d + 54*a^2*b^2*
c^2*d^2 + 12*a^3*b*c*d^3 + a^4*d^4)/(a^7*b^5))^(1/4) + (3*b*c + a*d)*x) - (a*b^2*x^4 + a^2*b)*(-(81*b^4*c^4 +
108*a*b^3*c^3*d + 54*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 + a^4*d^4)/(a^7*b^5))^(1/4)*log(-a^2*b*(-(81*b^4*c^4 + 1
08*a*b^3*c^3*d + 54*a^2*b^2*c^2*d^2 + 12*a^3*b*c*d^3 + a^4*d^4)/(a^7*b^5))^(1/4) + (3*b*c + a*d)*x) + 4*(b*c -
a*d)*x)/(a*b^2*x^4 + a^2*b)

________________________________________________________________________________________

giac [A]  time = 0.17, size = 266, normalized size = 1.09 \[ \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b c + \left (a b^{3}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{2}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b c + \left (a b^{3}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{2}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b c + \left (a b^{3}\right )^{\frac {1}{4}} a d\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{2}} - \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b c + \left (a b^{3}\right )^{\frac {1}{4}} a d\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{2}} + \frac {b c x - a d x}{4 \, {\left (b x^{4} + a\right )} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(2)*(3*(a*b^3)^(1/4)*b*c + (a*b^3)^(1/4)*a*d)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1
/4))/(a^2*b^2) + 1/16*sqrt(2)*(3*(a*b^3)^(1/4)*b*c + (a*b^3)^(1/4)*a*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b
)^(1/4))/(a/b)^(1/4))/(a^2*b^2) + 1/32*sqrt(2)*(3*(a*b^3)^(1/4)*b*c + (a*b^3)^(1/4)*a*d)*log(x^2 + sqrt(2)*x*(
a/b)^(1/4) + sqrt(a/b))/(a^2*b^2) - 1/32*sqrt(2)*(3*(a*b^3)^(1/4)*b*c + (a*b^3)^(1/4)*a*d)*log(x^2 - sqrt(2)*x
*(a/b)^(1/4) + sqrt(a/b))/(a^2*b^2) + 1/4*(b*c*x - a*d*x)/((b*x^4 + a)*a*b)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 295, normalized size = 1.20 \[ \frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{16 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{16 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{32 a b}+\frac {3 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{16 a^{2}}+\frac {3 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{16 a^{2}}+\frac {3 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c \ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{b}}}\right )}{32 a^{2}}-\frac {\left (a d -b c \right ) x}{4 \left (b \,x^{4}+a \right ) a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*d-b*c)/a/b*x/(b*x^4+a)+1/16/a/b*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)*d+3/16/a^2*(a/b)^(
1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)*c+1/32/a/b*(a/b)^(1/4)*2^(1/2)*ln((x^2+(a/b)^(1/4)*2^(1/2)*x+(a/b
)^(1/2))/(x^2-(a/b)^(1/4)*2^(1/2)*x+(a/b)^(1/2)))*d+3/32/a^2*(a/b)^(1/4)*2^(1/2)*ln((x^2+(a/b)^(1/4)*2^(1/2)*x
+(a/b)^(1/2))/(x^2-(a/b)^(1/4)*2^(1/2)*x+(a/b)^(1/2)))*c+1/16/a/b*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/
4)*x+1)*d+3/16/a^2*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x+1)*c

________________________________________________________________________________________

maxima [A]  time = 1.35, size = 236, normalized size = 0.96 \[ \frac {{\left (b c - a d\right )} x}{4 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, b c + a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (3 \, b c + a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (3 \, b c + a d\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, b c + a d\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{32 \, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*c - a*d)*x/(a*b^2*x^4 + a^2*b) + 1/32*(2*sqrt(2)*(3*b*c + a*d)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2
)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(3*b*c + a*d)*arctan(1/2
*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + sqrt
(2)*(3*b*c + a*d)*log(sqrt(b)*x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) - sqrt(2)*(3*b*c +
a*d)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/(a*b)

________________________________________________________________________________________

mupad [B]  time = 1.53, size = 740, normalized size = 3.02 \[ \frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}-\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}+\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}+\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}}{\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}-\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}-\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}+\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}}\right )\,\left (a\,d+3\,b\,c\right )}{8\,{\left (-a\right )}^{7/4}\,b^{5/4}}-\frac {x\,\left (a\,d-b\,c\right )}{4\,a\,b\,\left (b\,x^4+a\right )}+\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}-\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}+\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}+\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}}{\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}-\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}-\frac {\left (\frac {x\,\left (a^2\,b\,d^2+6\,a\,b^2\,c\,d+9\,b^3\,c^2\right )}{4\,a^2}+\frac {\left (a\,d+3\,b\,c\right )\,\left (12\,c\,b^3+4\,a\,d\,b^2\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}\right )\,\left (a\,d+3\,b\,c\right )}{16\,{\left (-a\right )}^{7/4}\,b^{5/4}}}\right )\,\left (a\,d+3\,b\,c\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{7/4}\,b^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^4)/(a + b*x^4)^2,x)

[Out]

(atan(((((x*(9*b^3*c^2 + a^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2) - ((a*d + 3*b*c)*(12*b^3*c + 4*a*b^2*d))/(16*(-a)^(
7/4)*b^(5/4)))*(a*d + 3*b*c)*1i)/(16*(-a)^(7/4)*b^(5/4)) + (((x*(9*b^3*c^2 + a^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2)
+ ((a*d + 3*b*c)*(12*b^3*c + 4*a*b^2*d))/(16*(-a)^(7/4)*b^(5/4)))*(a*d + 3*b*c)*1i)/(16*(-a)^(7/4)*b^(5/4)))/
((((x*(9*b^3*c^2 + a^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2) - ((a*d + 3*b*c)*(12*b^3*c + 4*a*b^2*d))/(16*(-a)^(7/4)*b
^(5/4)))*(a*d + 3*b*c))/(16*(-a)^(7/4)*b^(5/4)) - (((x*(9*b^3*c^2 + a^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2) + ((a*d
+ 3*b*c)*(12*b^3*c + 4*a*b^2*d))/(16*(-a)^(7/4)*b^(5/4)))*(a*d + 3*b*c))/(16*(-a)^(7/4)*b^(5/4))))*(a*d + 3*b*
c)*1i)/(8*(-a)^(7/4)*b^(5/4)) + (atan(((((x*(9*b^3*c^2 + a^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2) - ((a*d + 3*b*c)*(1
2*b^3*c + 4*a*b^2*d)*1i)/(16*(-a)^(7/4)*b^(5/4)))*(a*d + 3*b*c))/(16*(-a)^(7/4)*b^(5/4)) + (((x*(9*b^3*c^2 + a
^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2) + ((a*d + 3*b*c)*(12*b^3*c + 4*a*b^2*d)*1i)/(16*(-a)^(7/4)*b^(5/4)))*(a*d + 3
*b*c))/(16*(-a)^(7/4)*b^(5/4)))/((((x*(9*b^3*c^2 + a^2*b*d^2 + 6*a*b^2*c*d))/(4*a^2) - ((a*d + 3*b*c)*(12*b^3*
c + 4*a*b^2*d)*1i)/(16*(-a)^(7/4)*b^(5/4)))*(a*d + 3*b*c)*1i)/(16*(-a)^(7/4)*b^(5/4)) - (((x*(9*b^3*c^2 + a^2*
b*d^2 + 6*a*b^2*c*d))/(4*a^2) + ((a*d + 3*b*c)*(12*b^3*c + 4*a*b^2*d)*1i)/(16*(-a)^(7/4)*b^(5/4)))*(a*d + 3*b*
c)*1i)/(16*(-a)^(7/4)*b^(5/4))))*(a*d + 3*b*c))/(8*(-a)^(7/4)*b^(5/4)) - (x*(a*d - b*c))/(4*a*b*(a + b*x^4))

________________________________________________________________________________________

sympy [A]  time = 0.97, size = 112, normalized size = 0.46 \[ \frac {x \left (- a d + b c\right )}{4 a^{2} b + 4 a b^{2} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{7} b^{5} + a^{4} d^{4} + 12 a^{3} b c d^{3} + 54 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{3} d + 81 b^{4} c^{4}, \left (t \mapsto t \log {\left (\frac {16 t a^{2} b}{a d + 3 b c} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(-a*d + b*c)/(4*a**2*b + 4*a*b**2*x**4) + RootSum(65536*_t**4*a**7*b**5 + a**4*d**4 + 12*a**3*b*c*d**3 + 54*
a**2*b**2*c**2*d**2 + 108*a*b**3*c**3*d + 81*b**4*c**4, Lambda(_t, _t*log(16*_t*a**2*b/(a*d + 3*b*c) + x)))

________________________________________________________________________________________