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 + ((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))

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Rubi [A]  time = 0.377291, antiderivative size = 291, normalized size of antiderivative = 1., 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 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 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 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.176814, 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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Maple [B]  time = 0.009, size = 475, normalized size = 1.6 \begin{align*}{\frac{{d}^{2}x}{{b}^{2}}}+{\frac{ax{d}^{2}}{4\,{b}^{2} \left ( b{x}^{4}+a \right ) }}-{\frac{cxd}{2\,b \left ( b{x}^{4}+a \right ) }}+{\frac{x{c}^{2}}{4\,a \left ( b{x}^{4}+a \right ) }}-{\frac{5\,\sqrt{2}{d}^{2}}{16\,{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{\sqrt{2}cd}{8\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}{c}^{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-1 \right ) }-{\frac{5\,\sqrt{2}{d}^{2}}{32\,{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}cd}{16\,ab}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}{c}^{2}}{32\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{a}{b}}}x\sqrt{2}+\sqrt{{\frac{a}{b}}} \right ) ^{-1}} \right ) }-{\frac{5\,\sqrt{2}{d}^{2}}{16\,{b}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{\sqrt{2}cd}{8\,ab}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) }+{\frac{3\,\sqrt{2}{c}^{2}}{16\,{a}^{2}}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.92587, size = 2909, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [A]  time = 2.06516, size = 219, normalized size = 0.75 \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 1.12698, size = 508, normalized size = 1.75 \begin{align*} \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}} \end{align*}

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)