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

Optimal. Leaf size=223 \[ \frac {(b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}+\frac {(b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} c^{3/4} d^{5/4}}+\frac {b x}{d} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 223, 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 = {388, 211, 1165, 628, 1162, 617, 204} \[ \frac {(b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}+\frac {(b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} c^{3/4} d^{5/4}}+\frac {b x}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)/(c + d*x^4),x]

[Out]

(b*x)/d + ((b*c - a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*c^(3/4)*d^(5/4)) - ((b*c - a*d)*Arc
Tan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*c^(3/4)*d^(5/4)) + ((b*c - a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)
*d^(1/4)*x + Sqrt[d]*x^2])/(4*Sqrt[2]*c^(3/4)*d^(5/4)) - ((b*c - a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x
+ Sqrt[d]*x^2])/(4*Sqrt[2]*c^(3/4)*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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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}{c+d x^4} \, dx &=\frac {b x}{d}-\frac {(b c-a d) \int \frac {1}{c+d x^4} \, dx}{d}\\ &=\frac {b x}{d}-\frac {(b c-a d) \int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx}{2 \sqrt {c} d}-\frac {(b c-a d) \int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx}{2 \sqrt {c} d}\\ &=\frac {b x}{d}-\frac {(b c-a d) \int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx}{4 \sqrt {c} d^{3/2}}-\frac {(b c-a d) \int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx}{4 \sqrt {c} d^{3/2}}+\frac {(b c-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}{4 \sqrt {2} c^{3/4} d^{5/4}}+\frac {(b c-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}{4 \sqrt {2} c^{3/4} d^{5/4}}\\ &=\frac {b x}{d}+\frac {(b c-a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-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 )}{2 \sqrt {2} c^{3/4} d^{5/4}}+\frac {(b c-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 )}{2 \sqrt {2} c^{3/4} d^{5/4}}\\ &=\frac {b x}{d}+\frac {(b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{3/4} d^{5/4}}+\frac {(b c-a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}-\frac {(b c-a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{4 \sqrt {2} c^{3/4} d^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 196, normalized size = 0.88 \[ \frac {\sqrt {2} (b c-a d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )-\sqrt {2} (b c-a d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )+2 \sqrt {2} (b c-a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )-2 \sqrt {2} (b c-a d) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )+8 b c^{3/4} \sqrt [4]{d} x}{8 c^{3/4} d^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)/(c + d*x^4),x]

[Out]

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

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fricas [B]  time = 1.32, size = 639, normalized size = 2.87 \[ \frac {4 \, d \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {1}{4}} \arctan \left (\frac {c^{2} d^{4} x \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {3}{4}} - c^{2} d^{4} \sqrt {\frac {c^{2} d^{2} \sqrt {-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2}}{b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}} \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {3}{4}}}{b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}}\right ) + d \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {1}{4}} \log \left (c d \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {1}{4}} - {\left (b c - a d\right )} x\right ) - d \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {1}{4}} \log \left (-c d \left (-\frac {b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}}{c^{3} d^{5}}\right )^{\frac {1}{4}} - {\left (b c - a d\right )} x\right ) + 4 \, b x}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*d*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(1/4)*arctan((c^
2*d^4*x*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(3/4) - c^2*d^4*s
qrt((c^2*d^2*sqrt(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5)) + (b^2*c
^2 - 2*a*b*c*d + a^2*d^2)*x^2)/(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*(-(b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2
- 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(3/4))/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)) + d*(-(b^4*
c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(1/4)*log(c*d*(-(b^4*c^4 - 4*a*b
^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(1/4) - (b*c - a*d)*x) - d*(-(b^4*c^4 - 4*a
*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(1/4)*log(-c*d*(-(b^4*c^4 - 4*a*b^3*c^3*d
+ 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)/(c^3*d^5))^(1/4) - (b*c - a*d)*x) + 4*b*x)/d

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giac [A]  time = 0.16, size = 245, normalized size = 1.10 \[ \frac {b x}{d} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c - \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{4 \, c d^{2}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c - \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{4 \, c d^{2}} - \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c - \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{8 \, c d^{2}} + \frac {\sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c - \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{8 \, c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/d - 1/4*sqrt(2)*((c*d^3)^(1/4)*b*c - (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*d^2) - 1/4*sqrt(2)*((c*d^3)^(1/4)*b*c - (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*d^2) - 1/8*sqrt(2)*((c*d^3)^(1/4)*b*c - (c*d^3)^(1/4)*a*d)*log(x^2 + sqrt(2)*x*(c/d)^
(1/4) + sqrt(c/d))/(c*d^2) + 1/8*sqrt(2)*((c*d^3)^(1/4)*b*c - (c*d^3)^(1/4)*a*d)*log(x^2 - sqrt(2)*x*(c/d)^(1/
4) + sqrt(c/d))/(c*d^2)

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maple [A]  time = 0.05, size = 266, normalized size = 1.19 \[ \frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{4 c}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{4 c}+\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a \ln \left (\frac {x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}{x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}\right )}{8 c}+\frac {b x}{d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{4 d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{4 d}-\frac {\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b \ln \left (\frac {x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}{x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.28, size = 212, normalized size = 0.95 \[ \frac {b x}{d} - \frac {\frac {2 \, \sqrt {2} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (b c - a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (b c - a d\right )} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (b c - a d\right )} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/d - 1/8*(2*sqrt(2)*(b*c - a*d)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x + sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqr
t(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + 2*sqrt(2)*(b*c - 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))) + sqrt(2)*(b*c - a*d)*log(sqrt(d)*x^2 + sqr
t(2)*c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(3/4)*d^(1/4)) - sqrt(2)*(b*c - a*d)*log(sqrt(d)*x^2 - sqrt(2)*c^(1/4)*d^
(1/4)*x + sqrt(c))/(c^(3/4)*d^(1/4)))/d

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mupad [B]  time = 1.48, size = 720, normalized size = 3.23 \[ \frac {b\,x}{d}-\frac {\mathrm {atan}\left (\frac {\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )-\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}+\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )+\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}}{\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )-\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}-\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )+\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}}\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{2\,{\left (-c\right )}^{3/4}\,d^{5/4}}-\frac {\mathrm {atan}\left (\frac {\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )-\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}+\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )+\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}}{\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )-\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}-\frac {\left (x\,\left (4\,a^2\,d^3-8\,a\,b\,c\,d^2+4\,b^2\,c^2\,d\right )+\frac {\left (16\,b\,c^2\,d^2-16\,a\,c\,d^3\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}\right )\,\left (a\,d-b\,c\right )\,1{}\mathrm {i}}{4\,{\left (-c\right )}^{3/4}\,d^{5/4}}}\right )\,\left (a\,d-b\,c\right )}{2\,{\left (-c\right )}^{3/4}\,d^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)/(c + d*x^4),x)

[Out]

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

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sympy [A]  time = 0.66, size = 87, normalized size = 0.39 \[ \frac {b x}{d} + \operatorname {RootSum} {\left (256 t^{4} c^{3} d^{5} + a^{4} d^{4} - 4 a^{3} b c d^{3} + 6 a^{2} b^{2} c^{2} d^{2} - 4 a b^{3} c^{3} d + b^{4} c^{4}, \left (t \mapsto t \log {\left (\frac {4 t c d}{a d - b c} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x/d + RootSum(256*_t**4*c**3*d**5 + a**4*d**4 - 4*a**3*b*c*d**3 + 6*a**2*b**2*c**2*d**2 - 4*a*b**3*c**3*d +
b**4*c**4, Lambda(_t, _t*log(4*_t*c*d/(a*d - b*c) + x)))

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