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

Optimal. Leaf size=291 \[ -\frac {(b c-a d) (5 a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(b c-a d) (5 a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} b^{9/4}}+\frac {x (b c-a d)^2}{4 a b^2 \left (a+b x^4\right )}+\frac {d^2 x}{b^2} \]

[Out]

d^2*x/b^2+1/4*(-a*d+b*c)^2*x/a/b^2/(b*x^4+a)+1/16*(-a*d+b*c)*(5*a*d+3*b*c)*arctan(-1+b^(1/4)*x*2^(1/2)/a^(1/4)
)/a^(7/4)/b^(9/4)*2^(1/2)+1/16*(-a*d+b*c)*(5*a*d+3*b*c)*arctan(1+b^(1/4)*x*2^(1/2)/a^(1/4))/a^(7/4)/b^(9/4)*2^
(1/2)-1/32*(-a*d+b*c)*(5*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^(9/4)*2^(1/2)
+1/32*(-a*d+b*c)*(5*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^(9/4)*2^(1/2)

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Rubi [A]  time = 0.38, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {390, 385, 211, 1165, 628, 1162, 617, 204} \[ -\frac {(b c-a d) (5 a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(b c-a d) (5 a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{16 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(b c-a d) (5 a d+3 b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} b^{9/4}}+\frac {x (b c-a d)^2}{4 a b^2 \left (a+b x^4\right )}+\frac {d^2 x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^4)^2/(a + b*x^4)^2,x]

[Out]

(d^2*x)/b^2 + ((b*c - a*d)^2*x)/(4*a*b^2*(a + b*x^4)) - ((b*c - a*d)*(3*b*c + 5*a*d)*ArcTan[1 - (Sqrt[2]*b^(1/
4)*x)/a^(1/4)])/(8*Sqrt[2]*a^(7/4)*b^(9/4)) + ((b*c - a*d)*(3*b*c + 5*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1
/4)])/(8*Sqrt[2]*a^(7/4)*b^(9/4)) - ((b*c - a*d)*(3*b*c + 5*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqr
t[b]*x^2])/(16*Sqrt[2]*a^(7/4)*b^(9/4)) + ((b*c - a*d)*(3*b*c + 5*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^(9/4))

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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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 {\left (c+d x^4\right )^2}{\left (a+b x^4\right )^2} \, dx &=\int \left (\frac {d^2}{b^2}+\frac {b^2 c^2-a^2 d^2+2 b d (b c-a d) x^4}{b^2 \left (a+b x^4\right )^2}\right ) \, dx\\ &=\frac {d^2 x}{b^2}+\frac {\int \frac {b^2 c^2-a^2 d^2+2 b d (b c-a d) x^4}{\left (a+b x^4\right )^2} \, dx}{b^2}\\ &=\frac {d^2 x}{b^2}+\frac {(b c-a d)^2 x}{4 a b^2 \left (a+b x^4\right )}+\frac {((b c-a d) (3 b c+5 a d)) \int \frac {1}{a+b x^4} \, dx}{4 a b^2}\\ &=\frac {d^2 x}{b^2}+\frac {(b c-a d)^2 x}{4 a b^2 \left (a+b x^4\right )}+\frac {((b c-a d) (3 b c+5 a d)) \int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx}{8 a^{3/2} b^2}+\frac {((b c-a d) (3 b c+5 a d)) \int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx}{8 a^{3/2} b^2}\\ &=\frac {d^2 x}{b^2}+\frac {(b c-a d)^2 x}{4 a b^2 \left (a+b x^4\right )}+\frac {((b c-a d) (3 b c+5 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^{5/2}}+\frac {((b c-a d) (3 b c+5 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^{5/2}}-\frac {((b c-a d) (3 b c+5 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^{9/4}}-\frac {((b c-a d) (3 b c+5 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^{9/4}}\\ &=\frac {d^2 x}{b^2}+\frac {(b c-a d)^2 x}{4 a b^2 \left (a+b x^4\right )}-\frac {(b c-a d) (3 b c+5 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^{9/4}}+\frac {(b c-a d) (3 b c+5 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^{9/4}}+\frac {((b c-a d) (3 b c+5 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^{9/4}}-\frac {((b c-a d) (3 b c+5 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^{9/4}}\\ &=\frac {d^2 x}{b^2}+\frac {(b c-a d)^2 x}{4 a b^2 \left (a+b x^4\right )}-\frac {(b c-a d) (3 b c+5 a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(b c-a d) (3 b c+5 a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(b c-a d) (3 b c+5 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^{9/4}}+\frac {(b c-a d) (3 b c+5 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^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 297, normalized size = 1.02 \[ \frac {\frac {\sqrt {2} \left (5 a^2 d^2-2 a b c d-3 b^2 c^2\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{a^{7/4}}+\frac {\sqrt {2} \left (-5 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{a^{7/4}}+\frac {2 \sqrt {2} \left (5 a^2 d^2-2 a b c d-3 b^2 c^2\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{7/4}}+\frac {2 \sqrt {2} \left (-5 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{a^{7/4}}+\frac {8 \sqrt [4]{b} x (b c-a d)^2}{a \left (a+b x^4\right )}+32 \sqrt [4]{b} d^2 x}{32 b^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^(1/4)*d^2*x + (8*b^(1/4)*(b*c - a*d)^2*x)/(a*(a + b*x^4)) + (2*Sqrt[2]*(-3*b^2*c^2 - 2*a*b*c*d + 5*a^2*d
^2)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(7/4) + (2*Sqrt[2]*(3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*ArcTan[1
+ (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(7/4) + (Sqrt[2]*(-3*b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*Log[Sqrt[a] - Sqrt[2]
*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/a^(7/4) + (Sqrt[2]*(3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*Log[Sqrt[a] + Sqrt[2
]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/a^(7/4))/(32*b^(9/4))

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fricas [B]  time = 1.08, size = 1335, normalized size = 4.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a*b*d^2*x^5 - 4*(a*b^3*x^4 + a^2*b^2)*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3
*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d
^8)/(a^7*b^9))^(1/4)*arctan((a^5*b^7*x*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5
*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7
*b^9))^(3/4) - a^5*b^7*sqrt((a^4*b^4*sqrt(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c
^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a
^7*b^9)) + (9*b^4*c^4 + 12*a*b^3*c^3*d - 26*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + 25*a^4*d^4)*x^2)/(9*b^4*c^4 + 1
2*a*b^3*c^3*d - 26*a^2*b^2*c^2*d^2 - 20*a^3*b*c*d^3 + 25*a^4*d^4))*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b
^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7
*b*c*d^7 + 625*a^8*d^8)/(a^7*b^9))^(3/4))/(27*b^6*c^6 + 54*a*b^5*c^5*d - 99*a^2*b^4*c^4*d^2 - 172*a^3*b^3*c^3*
d^3 + 165*a^4*b^2*c^2*d^4 + 150*a^5*b*c*d^5 - 125*a^6*d^6)) - (a*b^3*x^4 + a^2*b^2)*(-(81*b^8*c^8 + 216*a*b^7*
c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c
^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*b^9))^(1/4)*log(a^2*b^2*(-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^
2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*
a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*b^9))^(1/4) - (3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*x) + (a*b^3*x^4 + a^2*b^2)*(
-(81*b^8*c^8 + 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^
3*c^3*d^5 - 900*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*b^9))^(1/4)*log(-a^2*b^2*(-(81*b^8*c^8
+ 216*a*b^7*c^7*d - 324*a^2*b^6*c^6*d^2 - 984*a^3*b^5*c^5*d^3 + 646*a^4*b^4*c^4*d^4 + 1640*a^5*b^3*c^3*d^5 - 9
00*a^6*b^2*c^2*d^6 - 1000*a^7*b*c*d^7 + 625*a^8*d^8)/(a^7*b^9))^(1/4) - (3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*x)
+ 4*(b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x)/(a*b^3*x^4 + a^2*b^2)

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giac [A]  time = 0.17, size = 376, normalized size = 1.29 \[ \frac {d^{2} x}{b^{2}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 2 \, \left (a b^{3}\right )^{\frac {1}{4}} a b c d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 2 \, \left (a b^{3}\right )^{\frac {1}{4}} a b c d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 2 \, \left (a b^{3}\right )^{\frac {1}{4}} a b c d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{3}} - \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c^{2} + 2 \, \left (a b^{3}\right )^{\frac {1}{4}} a b c d - 5 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} d^{2}\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{3}} + \frac {b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{4 \, {\left (b x^{4} + a\right )} a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*x/b^2 + 1/16*sqrt(2)*(3*(a*b^3)^(1/4)*b^2*c^2 + 2*(a*b^3)^(1/4)*a*b*c*d - 5*(a*b^3)^(1/4)*a^2*d^2)*arctan(
1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^2*b^3) + 1/16*sqrt(2)*(3*(a*b^3)^(1/4)*b^2*c^2 + 2*(a*
b^3)^(1/4)*a*b*c*d - 5*(a*b^3)^(1/4)*a^2*d^2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^2
*b^3) + 1/32*sqrt(2)*(3*(a*b^3)^(1/4)*b^2*c^2 + 2*(a*b^3)^(1/4)*a*b*c*d - 5*(a*b^3)^(1/4)*a^2*d^2)*log(x^2 + s
qrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^2*b^3) - 1/32*sqrt(2)*(3*(a*b^3)^(1/4)*b^2*c^2 + 2*(a*b^3)^(1/4)*a*b*c*d
- 5*(a*b^3)^(1/4)*a^2*d^2)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^2*b^3) + 1/4*(b^2*c^2*x - 2*a*b*c*d
*x + a^2*d^2*x)/((b*x^4 + a)*a*b^2)

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maple [B]  time = 0.06, size = 475, normalized size = 1.63 \[ \frac {a \,d^{2} x}{4 \left (b \,x^{4}+a \right ) b^{2}}+\frac {c^{2} x}{4 \left (b \,x^{4}+a \right ) a}-\frac {c d x}{2 \left (b \,x^{4}+a \right ) b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c 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 )}{16 a b}+\frac {3 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, c^{2} \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^{2} \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^{2} \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 {d^{2} x}{b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{16 b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{2} \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{16 b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, d^{2} \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 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*d^2*x+1/4/b^2*a*x/(b*x^4+a)*d^2-1/2/b*x/(b*x^4+a)*c*d+1/4/a*x/(b*x^4+a)*c^2-5/16/b^2*(a/b)^(1/4)*2^(1/2)
*arctan(2^(1/2)/(a/b)^(1/4)*x-1)*d^2+1/8/b/a*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)*c*d+3/16/a^2*
(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x-1)*c^2-5/32/b^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)))*d^2+1/16/b/a*(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*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^2-5/16/b^2*(a/b)^(1/4)*2^(1/2)*arct
an(2^(1/2)/(a/b)^(1/4)*x+1)*d^2+1/8/b/a*(a/b)^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x+1)*c*d+3/16/a^2*(a/b)
^(1/4)*2^(1/2)*arctan(2^(1/2)/(a/b)^(1/4)*x+1)*c^2

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maxima [A]  time = 1.26, size = 319, normalized size = 1.10 \[ \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x}{4 \, {\left (a b^{3} x^{4} + a^{2} b^{2}\right )}} + \frac {d^{2} x}{b^{2}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, b^{2} c^{2} + 2 \, a b c d - 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (3 \, b^{2} c^{2} + 2 \, a b c d - 5 \, a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (3 \, b^{2} c^{2} + 2 \, a b c d - 5 \, a^{2} d^{2}\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, b^{2} c^{2} + 2 \, a b c d - 5 \, a^{2} d^{2}\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{32 \, a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x/(a*b^3*x^4 + a^2*b^2) + d^2*x/b^2 + 1/32*(2*sqrt(2)*(3*b^2*c^2 + 2*a*b*c
*d - 5*a^2*d^2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqr
t(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(3*b^2*c^2 + 2*a*b*c*d - 5*a^2*d^2)*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^2*c^2 + 2*a*b*c*d - 5*a
^2*d^2)*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^2*c^2 + 2*a*b*
c*d - 5*a^2*d^2)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/(a*b^2)

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mupad [B]  time = 0.30, size = 1254, normalized size = 4.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^4)^2/(a + b*x^4)^2,x)

[Out]

(d^2*x)/b^2 + (x*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(4*a*(a*b^2 + b^3*x^4)) + (atan((((a*d - b*c)*(5*a*d + 3*b*c
)*((x*(25*a^4*d^4 + 9*b^4*c^4 - 26*a^2*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*b*c*d^3))/(4*a^2*b) - ((a*d - b*c
)*(5*a*d + 3*b*c)*(12*b^3*c^2 - 20*a^2*b*d^2 + 8*a*b^2*c*d))/(16*(-a)^(7/4)*b^(9/4)))*1i)/(16*(-a)^(7/4)*b^(9/
4)) + ((a*d - b*c)*(5*a*d + 3*b*c)*((x*(25*a^4*d^4 + 9*b^4*c^4 - 26*a^2*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*
b*c*d^3))/(4*a^2*b) + ((a*d - b*c)*(5*a*d + 3*b*c)*(12*b^3*c^2 - 20*a^2*b*d^2 + 8*a*b^2*c*d))/(16*(-a)^(7/4)*b
^(9/4)))*1i)/(16*(-a)^(7/4)*b^(9/4)))/(((a*d - b*c)*(5*a*d + 3*b*c)*((x*(25*a^4*d^4 + 9*b^4*c^4 - 26*a^2*b^2*c
^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*b*c*d^3))/(4*a^2*b) - ((a*d - b*c)*(5*a*d + 3*b*c)*(12*b^3*c^2 - 20*a^2*b*d^2
+ 8*a*b^2*c*d))/(16*(-a)^(7/4)*b^(9/4))))/(16*(-a)^(7/4)*b^(9/4)) - ((a*d - b*c)*(5*a*d + 3*b*c)*((x*(25*a^4*
d^4 + 9*b^4*c^4 - 26*a^2*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*b*c*d^3))/(4*a^2*b) + ((a*d - b*c)*(5*a*d + 3*b
*c)*(12*b^3*c^2 - 20*a^2*b*d^2 + 8*a*b^2*c*d))/(16*(-a)^(7/4)*b^(9/4))))/(16*(-a)^(7/4)*b^(9/4))))*(a*d - b*c)
*(5*a*d + 3*b*c)*1i)/(8*(-a)^(7/4)*b^(9/4)) + (atan((((a*d - b*c)*(5*a*d + 3*b*c)*((x*(25*a^4*d^4 + 9*b^4*c^4
- 26*a^2*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*b*c*d^3))/(4*a^2*b) - ((a*d - b*c)*(5*a*d + 3*b*c)*(12*b^3*c^2
- 20*a^2*b*d^2 + 8*a*b^2*c*d)*1i)/(16*(-a)^(7/4)*b^(9/4))))/(16*(-a)^(7/4)*b^(9/4)) + ((a*d - b*c)*(5*a*d + 3*
b*c)*((x*(25*a^4*d^4 + 9*b^4*c^4 - 26*a^2*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*b*c*d^3))/(4*a^2*b) + ((a*d -
b*c)*(5*a*d + 3*b*c)*(12*b^3*c^2 - 20*a^2*b*d^2 + 8*a*b^2*c*d)*1i)/(16*(-a)^(7/4)*b^(9/4))))/(16*(-a)^(7/4)*b^
(9/4)))/(((a*d - b*c)*(5*a*d + 3*b*c)*((x*(25*a^4*d^4 + 9*b^4*c^4 - 26*a^2*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a
^3*b*c*d^3))/(4*a^2*b) - ((a*d - b*c)*(5*a*d + 3*b*c)*(12*b^3*c^2 - 20*a^2*b*d^2 + 8*a*b^2*c*d)*1i)/(16*(-a)^(
7/4)*b^(9/4)))*1i)/(16*(-a)^(7/4)*b^(9/4)) - ((a*d - b*c)*(5*a*d + 3*b*c)*((x*(25*a^4*d^4 + 9*b^4*c^4 - 26*a^2
*b^2*c^2*d^2 + 12*a*b^3*c^3*d - 20*a^3*b*c*d^3))/(4*a^2*b) + ((a*d - b*c)*(5*a*d + 3*b*c)*(12*b^3*c^2 - 20*a^2
*b*d^2 + 8*a*b^2*c*d)*1i)/(16*(-a)^(7/4)*b^(9/4)))*1i)/(16*(-a)^(7/4)*b^(9/4))))*(a*d - b*c)*(5*a*d + 3*b*c))/
(8*(-a)^(7/4)*b^(9/4))

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sympy [A]  time = 2.17, size = 219, normalized size = 0.75 \[ \frac {x \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{4 a^{2} b^{2} + 4 a b^{3} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{7} b^{9} + 625 a^{8} d^{8} - 1000 a^{7} b c d^{7} - 900 a^{6} b^{2} c^{2} d^{6} + 1640 a^{5} b^{3} c^{3} d^{5} + 646 a^{4} b^{4} c^{4} d^{4} - 984 a^{3} b^{5} c^{5} d^{3} - 324 a^{2} b^{6} c^{6} d^{2} + 216 a b^{7} c^{7} d + 81 b^{8} c^{8}, \left (t \mapsto t \log {\left (- \frac {16 t a^{2} b^{2}}{5 a^{2} d^{2} - 2 a b c d - 3 b^{2} c^{2}} + x \right )} \right )\right )} + \frac {d^{2} x}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(4*a**2*b**2 + 4*a*b**3*x**4) + RootSum(65536*_t**4*a**7*b**9 + 625*a**8
*d**8 - 1000*a**7*b*c*d**7 - 900*a**6*b**2*c**2*d**6 + 1640*a**5*b**3*c**3*d**5 + 646*a**4*b**4*c**4*d**4 - 98
4*a**3*b**5*c**5*d**3 - 324*a**2*b**6*c**6*d**2 + 216*a*b**7*c**7*d + 81*b**8*c**8, Lambda(_t, _t*log(-16*_t*a
**2*b**2/(5*a**2*d**2 - 2*a*b*c*d - 3*b**2*c**2) + x))) + d**2*x/b**2

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