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

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

[Out]

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

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Rubi [A]  time = 0.365765, 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} \[ \frac{(b c-a d) (3 a d+5 b c) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt{c}+\sqrt{d} x^2\right )}{16 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (3 a d+5 b c) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt{c}+\sqrt{d} x^2\right )}{16 \sqrt{2} c^{7/4} d^{9/4}}+\frac{(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}-\frac{(b c-a d) (3 a d+5 b c) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{8 \sqrt{2} c^{7/4} d^{9/4}}+\frac{x (b c-a d)^2}{4 c d^2 \left (c+d x^4\right )}+\frac{b^2 x}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^2/(c + d*x^4)^2,x]

[Out]

(b^2*x)/d^2 + ((b*c - a*d)^2*x)/(4*c*d^2*(c + d*x^4)) + ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/
4)*x)/c^(1/4)])/(8*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*b*c + 3*a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1
/4)])/(8*Sqrt[2]*c^(7/4)*d^(9/4)) + ((b*c - a*d)*(5*b*c + 3*a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqr
t[d]*x^2])/(16*Sqrt[2]*c^(7/4)*d^(9/4)) - ((b*c - a*d)*(5*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^(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 (a+b x^4\right )^2}{\left (c+d x^4\right )^2} \, dx &=\int \left (\frac{b^2}{d^2}-\frac{b^2 c^2-a^2 d^2+2 b d (b c-a d) x^4}{d^2 \left (c+d x^4\right )^2}\right ) \, dx\\ &=\frac{b^2 x}{d^2}-\frac{\int \frac{b^2 c^2-a^2 d^2+2 b d (b c-a d) x^4}{\left (c+d x^4\right )^2} \, dx}{d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{4 c d^2 \left (c+d x^4\right )}-\frac{((b c-a d) (5 b c+3 a d)) \int \frac{1}{c+d x^4} \, dx}{4 c d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{4 c d^2 \left (c+d x^4\right )}-\frac{((b c-a d) (5 b c+3 a d)) \int \frac{\sqrt{c}-\sqrt{d} x^2}{c+d x^4} \, dx}{8 c^{3/2} d^2}-\frac{((b c-a d) (5 b c+3 a d)) \int \frac{\sqrt{c}+\sqrt{d} x^2}{c+d x^4} \, dx}{8 c^{3/2} d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{4 c d^2 \left (c+d x^4\right )}-\frac{((b c-a d) (5 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^{5/2}}-\frac{((b c-a d) (5 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^{5/2}}+\frac{((b c-a d) (5 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^{9/4}}+\frac{((b c-a d) (5 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^{9/4}}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{4 c d^2 \left (c+d x^4\right )}+\frac{(b c-a d) (5 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^{9/4}}-\frac{(b c-a d) (5 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^{9/4}}-\frac{((b c-a d) (5 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^{9/4}}+\frac{((b c-a d) (5 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^{9/4}}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{4 c d^2 \left (c+d x^4\right )}+\frac{(b c-a d) (5 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^{9/4}}-\frac{(b c-a d) (5 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^{9/4}}+\frac{(b c-a d) (5 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^{9/4}}-\frac{(b c-a d) (5 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^{9/4}}\\ \end{align*}

Mathematica [A]  time = 0.172948, size = 298, normalized size = 1.02 \[ \frac{\frac{\sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \log \left (-\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt{c}+\sqrt{d} x^2\right )}{c^{7/4}}-\frac{\sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \log \left (\sqrt{2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt{c}+\sqrt{d} x^2\right )}{c^{7/4}}+\frac{2 \sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{c^{7/4}}-\frac{2 \sqrt{2} \left (-3 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{c^{7/4}}+\frac{8 \sqrt [4]{d} x (b c-a d)^2}{c \left (c+d x^4\right )}+32 b^2 \sqrt [4]{d} x}{32 d^{9/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.01, size = 475, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}x}{{d}^{2}}}+{\frac{{a}^{2}x}{4\,c \left ( d{x}^{4}+c \right ) }}-{\frac{xab}{2\,d \left ( d{x}^{4}+c \right ) }}+{\frac{cx{b}^{2}}{4\,{d}^{2} \left ( d{x}^{4}+c \right ) }}+{\frac{3\,\sqrt{2}{a}^{2}}{16\,{c}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{\sqrt{2}ab}{8\,cd}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }-{\frac{5\,\sqrt{2}{b}^{2}}{16\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}{a}^{2}}{32\,{c}^{2}}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}ab}{16\,cd}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) }-{\frac{5\,\sqrt{2}{b}^{2}}{32\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}{a}^{2}}{16\,{c}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }+{\frac{\sqrt{2}ab}{8\,cd}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }-{\frac{5\,\sqrt{2}{b}^{2}}{16\,{d}^{2}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x/d^2+1/4/c*x/(d*x^4+c)*a^2-1/2/d*x/(d*x^4+c)*a*b+1/4/d^2*c*x/(d*x^4+c)*b^2+3/16/c^2*(c/d)^(1/4)*2^(1/2)*a
rctan(2^(1/2)/(c/d)^(1/4)*x-1)*a^2+1/8/d/c*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x-1)*a*b-5/16/d^2*(c
/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x-1)*b^2+3/32/c^2*(c/d)^(1/4)*2^(1/2)*ln((x^2+(c/d)^(1/4)*x*2^(1/
2)+(c/d)^(1/2))/(x^2-(c/d)^(1/4)*x*2^(1/2)+(c/d)^(1/2)))*a^2+1/16/d/c*(c/d)^(1/4)*2^(1/2)*ln((x^2+(c/d)^(1/4)*
x*2^(1/2)+(c/d)^(1/2))/(x^2-(c/d)^(1/4)*x*2^(1/2)+(c/d)^(1/2)))*a*b-5/32/d^2*(c/d)^(1/4)*2^(1/2)*ln((x^2+(c/d)
^(1/4)*x*2^(1/2)+(c/d)^(1/2))/(x^2-(c/d)^(1/4)*x*2^(1/2)+(c/d)^(1/2)))*b^2+3/16/c^2*(c/d)^(1/4)*2^(1/2)*arctan
(2^(1/2)/(c/d)^(1/4)*x+1)*a^2+1/8/d/c*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x+1)*a*b-5/16/d^2*(c/d)^(
1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x+1)*b^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((b*x^4+a)^2/(d*x^4+c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.57329, 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((b*x^4+a)^2/(d*x^4+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.74498, size = 219, normalized size = 0.75 \begin{align*} \frac{b^{2} x}{d^{2}} + \frac{x \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{4 c^{2} d^{2} + 4 c d^{3} x^{4}} + \operatorname{RootSum}{\left (65536 t^{4} c^{7} d^{9} + 81 a^{8} d^{8} + 216 a^{7} b c d^{7} - 324 a^{6} b^{2} c^{2} d^{6} - 984 a^{5} b^{3} c^{3} d^{5} + 646 a^{4} b^{4} c^{4} d^{4} + 1640 a^{3} b^{5} c^{5} d^{3} - 900 a^{2} b^{6} c^{6} d^{2} - 1000 a b^{7} c^{7} d + 625 b^{8} c^{8}, \left ( t \mapsto t \log{\left (\frac{16 t c^{2} d^{2}}{3 a^{2} d^{2} + 2 a b c d - 5 b^{2} c^{2}} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*x/d**2 + x*(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(4*c**2*d**2 + 4*c*d**3*x**4) + RootSum(65536*_t**4*c**7*d
**9 + 81*a**8*d**8 + 216*a**7*b*c*d**7 - 324*a**6*b**2*c**2*d**6 - 984*a**5*b**3*c**3*d**5 + 646*a**4*b**4*c**
4*d**4 + 1640*a**3*b**5*c**5*d**3 - 900*a**2*b**6*c**6*d**2 - 1000*a*b**7*c**7*d + 625*b**8*c**8, Lambda(_t, _
t*log(16*_t*c**2*d**2/(3*a**2*d**2 + 2*a*b*c*d - 5*b**2*c**2) + x)))

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Giac [A]  time = 1.13484, size = 508, normalized size = 1.75 \begin{align*} \frac{b^{2} x}{d^{2}} - \frac{\sqrt{2}{\left (5 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\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^{3}} - \frac{\sqrt{2}{\left (5 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\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^{3}} - \frac{\sqrt{2}{\left (5 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\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^{3}} + \frac{\sqrt{2}{\left (5 \, \left (c d^{3}\right )^{\frac{1}{4}} b^{2} c^{2} - 2 \, \left (c d^{3}\right )^{\frac{1}{4}} a b c d - 3 \, \left (c d^{3}\right )^{\frac{1}{4}} a^{2} d^{2}\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^{3}} + \frac{b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{4 \,{\left (d x^{4} + c\right )} c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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