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

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)/(c + d*x^4)^2,x]

[Out]

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

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Mathematica [A]  time = 0.20, size = 212, normalized size = 0.87 \[ \frac {-\frac {8 c^{3/4} \sqrt [4]{d} x (b c-a d)}{c+d x^4}-\sqrt {2} (3 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )+\sqrt {2} (3 a d+b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )-2 \sqrt {2} (3 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+2 \sqrt {2} (3 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{32 c^{7/4} d^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.19, size = 711, normalized size = 2.90 \[ \frac {4 \, {\left (c d^{2} x^{4} + c^{2} d\right )} \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {c^{5} d^{4} x \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {3}{4}} - c^{5} d^{4} \sqrt {\frac {c^{4} d^{2} \sqrt {-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}} + {\left (b^{2} c^{2} + 6 \, a b c d + 9 \, a^{2} d^{2}\right )} x^{2}}{b^{2} c^{2} + 6 \, a b c d + 9 \, a^{2} d^{2}}} \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {3}{4}}}{b^{3} c^{3} + 9 \, a b^{2} c^{2} d + 27 \, a^{2} b c d^{2} + 27 \, a^{3} d^{3}}\right ) + {\left (c d^{2} x^{4} + c^{2} d\right )} \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {1}{4}} \log \left (c^{2} d \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {1}{4}} + {\left (b c + 3 \, a d\right )} x\right ) - {\left (c d^{2} x^{4} + c^{2} d\right )} \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {1}{4}} \log \left (-c^{2} d \left (-\frac {b^{4} c^{4} + 12 \, a b^{3} c^{3} d + 54 \, a^{2} b^{2} c^{2} d^{2} + 108 \, a^{3} b c d^{3} + 81 \, a^{4} d^{4}}{c^{7} d^{5}}\right )^{\frac {1}{4}} + {\left (b c + 3 \, a d\right )} x\right ) - 4 \, {\left (b c - a d\right )} x}{16 \, {\left (c d^{2} x^{4} + c^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.16, size = 236, normalized size = 0.96 \[ -\frac {{\left (b c - a d\right )} x}{4 \, {\left (c d^{2} x^{4} + c^{2} d\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (b c + 3 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (b c + 3 \, a d\right )} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (b c + 3 \, a d\right )} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}}{32 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.52, size = 740, normalized size = 3.02 \[ \frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}-\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}+\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}+\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}}{\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}-\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}-\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}+\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}}\right )\,\left (3\,a\,d+b\,c\right )}{8\,{\left (-c\right )}^{7/4}\,d^{5/4}}+\frac {x\,\left (a\,d-b\,c\right )}{4\,c\,d\,\left (d\,x^4+c\right )}+\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}-\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}+\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}+\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )\,1{}\mathrm {i}}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}}{\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}-\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}-\frac {\left (\frac {x\,\left (9\,a^2\,d^3+6\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{4\,c^2}+\frac {\left (3\,a\,d+b\,c\right )\,\left (12\,a\,d^3+4\,b\,c\,d^2\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}\right )\,\left (3\,a\,d+b\,c\right )}{16\,{\left (-c\right )}^{7/4}\,d^{5/4}}}\right )\,\left (3\,a\,d+b\,c\right )\,1{}\mathrm {i}}{8\,{\left (-c\right )}^{7/4}\,d^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)/(c + d*x^4)^2,x)

[Out]

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

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sympy [A]  time = 0.85, size = 112, normalized size = 0.46 \[ \frac {x \left (a d - b c\right )}{4 c^{2} d + 4 c d^{2} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} c^{7} d^{5} + 81 a^{4} d^{4} + 108 a^{3} b c d^{3} + 54 a^{2} b^{2} c^{2} d^{2} + 12 a b^{3} c^{3} d + b^{4} c^{4}, \left (t \mapsto t \log {\left (\frac {16 t c^{2} d}{3 a d + b c} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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