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

Optimal. Leaf size=275 \[ -\frac {\left (3 \sqrt {c} d-\sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (3 \sqrt {c} d-\sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} e+3 \sqrt {c} d\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} e+3 \sqrt {c} d\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {x \left (d+e x^2\right )}{4 a \left (a+c x^4\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 275, 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 = {1179, 1168, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {\left (3 \sqrt {c} d-\sqrt {a} e\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (3 \sqrt {c} d-\sqrt {a} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{16 \sqrt {2} a^{7/4} c^{3/4}}-\frac {\left (\sqrt {a} e+3 \sqrt {c} d\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {\left (\sqrt {a} e+3 \sqrt {c} d\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{8 \sqrt {2} a^{7/4} c^{3/4}}+\frac {x \left (d+e x^2\right )}{4 a \left (a+c x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(a + c*x^4)^2,x]

[Out]

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

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 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]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)*(a + c*x^4)^(p + 1))/(
4*a*(p + 1)), x] + Dist[1/(4*a*(p + 1)), Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 267, normalized size = 0.97 \begin {gather*} \frac {\frac {\sqrt {2} \left (a^{3/4} e-3 \sqrt [4]{a} \sqrt {c} d\right ) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{c^{3/4}}+\frac {\sqrt {2} \left (3 \sqrt [4]{a} \sqrt {c} d-a^{3/4} e\right ) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{c} x+\sqrt {a}+\sqrt {c} x^2\right )}{c^{3/4}}-\frac {2 \sqrt {2} \sqrt [4]{a} \left (\sqrt {a} e+3 \sqrt {c} d\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{c^{3/4}}+\frac {2 \sqrt {2} \sqrt [4]{a} \left (\sqrt {a} e+3 \sqrt {c} d\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}+1\right )}{c^{3/4}}+\frac {8 a x \left (d+e x^2\right )}{a+c x^4}}{32 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d+e x^2}{\left (a+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d + e*x^2)/(a + c*x^4)^2,x]

[Out]

IntegrateAlgebraic[(d + e*x^2)/(a + c*x^4)^2, x]

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fricas [B]  time = 1.10, size = 873, normalized size = 3.17 \begin {gather*} \frac {4 \, e x^{3} - {\left (a c x^{4} + a^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} + 6 \, d e}{a^{3} c}} \log \left (-{\left (81 \, c^{2} d^{4} - a^{2} e^{4}\right )} x + {\left (a^{6} c^{2} e \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} + 27 \, a^{2} c^{2} d^{3} - 3 \, a^{3} c d e^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} + 6 \, d e}{a^{3} c}}\right ) + {\left (a c x^{4} + a^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} + 6 \, d e}{a^{3} c}} \log \left (-{\left (81 \, c^{2} d^{4} - a^{2} e^{4}\right )} x - {\left (a^{6} c^{2} e \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} + 27 \, a^{2} c^{2} d^{3} - 3 \, a^{3} c d e^{2}\right )} \sqrt {-\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} + 6 \, d e}{a^{3} c}}\right ) + {\left (a c x^{4} + a^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} - 6 \, d e}{a^{3} c}} \log \left (-{\left (81 \, c^{2} d^{4} - a^{2} e^{4}\right )} x + {\left (a^{6} c^{2} e \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} - 27 \, a^{2} c^{2} d^{3} + 3 \, a^{3} c d e^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} - 6 \, d e}{a^{3} c}}\right ) - {\left (a c x^{4} + a^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} - 6 \, d e}{a^{3} c}} \log \left (-{\left (81 \, c^{2} d^{4} - a^{2} e^{4}\right )} x - {\left (a^{6} c^{2} e \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} - 27 \, a^{2} c^{2} d^{3} + 3 \, a^{3} c d e^{2}\right )} \sqrt {\frac {a^{3} c \sqrt {-\frac {81 \, c^{2} d^{4} - 18 \, a c d^{2} e^{2} + a^{2} e^{4}}{a^{7} c^{3}}} - 6 \, d e}{a^{3} c}}\right ) + 4 \, d x}{16 \, {\left (a c x^{4} + a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*e*x^3 - (a*c*x^4 + a^2)*sqrt(-(a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) + 6*d*e)
/(a^3*c))*log(-(81*c^2*d^4 - a^2*e^4)*x + (a^6*c^2*e*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3))
+ 27*a^2*c^2*d^3 - 3*a^3*c*d*e^2)*sqrt(-(a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) + 6*d*
e)/(a^3*c))) + (a*c*x^4 + a^2)*sqrt(-(a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) + 6*d*e)/
(a^3*c))*log(-(81*c^2*d^4 - a^2*e^4)*x - (a^6*c^2*e*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) +
27*a^2*c^2*d^3 - 3*a^3*c*d*e^2)*sqrt(-(a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) + 6*d*e
)/(a^3*c))) + (a*c*x^4 + a^2)*sqrt((a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) - 6*d*e)/(a
^3*c))*log(-(81*c^2*d^4 - a^2*e^4)*x + (a^6*c^2*e*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) - 2
7*a^2*c^2*d^3 + 3*a^3*c*d*e^2)*sqrt((a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) - 6*d*e)/(
a^3*c))) - (a*c*x^4 + a^2)*sqrt((a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) - 6*d*e)/(a^3*
c))*log(-(81*c^2*d^4 - a^2*e^4)*x - (a^6*c^2*e*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) - 27*a
^2*c^2*d^3 + 3*a^3*c*d*e^2)*sqrt((a^3*c*sqrt(-(81*c^2*d^4 - 18*a*c*d^2*e^2 + a^2*e^4)/(a^7*c^3)) - 6*d*e)/(a^3
*c))) + 4*d*x)/(a*c*x^4 + a^2)

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giac [A]  time = 0.44, size = 273, normalized size = 0.99 \begin {gather*} \frac {x^{3} e + d x}{4 \, {\left (c x^{4} + a\right )} a} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d + \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d + \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{c}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{c}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} c^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d - \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c^{3}} - \frac {\sqrt {2} {\left (3 \, \left (a c^{3}\right )^{\frac {1}{4}} c^{2} d - \left (a c^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{c}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{c}}\right )}{32 \, a^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 303, normalized size = 1.10 \begin {gather*} \frac {e \,x^{3}}{4 \left (c \,x^{4}+a \right ) a}+\frac {d x}{4 \left (c \,x^{4}+a \right ) a}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} a c}+\frac {\sqrt {2}\, e \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{16 \left (\frac {a}{c}\right )^{\frac {1}{4}} a c}+\frac {\sqrt {2}\, e \ln \left (\frac {x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{32 \left (\frac {a}{c}\right )^{\frac {1}{4}} a c}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}-1\right )}{16 a^{2}}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{c}\right )^{\frac {1}{4}}}+1\right )}{16 a^{2}}+\frac {3 \left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, d \ln \left (\frac {x^{2}+\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}{x^{2}-\left (\frac {a}{c}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {a}{c}}}\right )}{32 a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.27, size = 253, normalized size = 0.92 \begin {gather*} \frac {e x^{3} + d x}{4 \, {\left (a c x^{4} + a^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, \sqrt {c} d + \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} {\left (3 \, \sqrt {c} d + \sqrt {a} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {c} x - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}} \sqrt {c}} + \frac {\sqrt {2} {\left (3 \, \sqrt {c} d - \sqrt {a} e\right )} \log \left (\sqrt {c} x^{2} + \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (3 \, \sqrt {c} d - \sqrt {a} e\right )} \log \left (\sqrt {c} x^{2} - \sqrt {2} a^{\frac {1}{4}} c^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} c^{\frac {3}{4}}}}{32 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.40, size = 637, normalized size = 2.32 \begin {gather*} \frac {\frac {e\,x^3}{4\,a}+\frac {d\,x}{4\,a}}{c\,x^4+a}-2\,\mathrm {atanh}\left (\frac {c^2\,e^2\,x\,\sqrt {\frac {e^2\,\sqrt {-a^7\,c^3}}{256\,a^6\,c^3}-\frac {9\,d^2\,\sqrt {-a^7\,c^3}}{256\,a^7\,c^2}-\frac {3\,d\,e}{128\,a^3\,c}}}{2\,\left (\frac {c\,e^3}{32\,a}-\frac {9\,c^2\,d^2\,e}{32\,a^2}-\frac {27\,c\,d^3\,\sqrt {-a^7\,c^3}}{32\,a^6}+\frac {3\,d\,e^2\,\sqrt {-a^7\,c^3}}{32\,a^5}\right )}-\frac {9\,c^3\,d^2\,x\,\sqrt {\frac {e^2\,\sqrt {-a^7\,c^3}}{256\,a^6\,c^3}-\frac {9\,d^2\,\sqrt {-a^7\,c^3}}{256\,a^7\,c^2}-\frac {3\,d\,e}{128\,a^3\,c}}}{2\,\left (\frac {c\,e^3}{32}-\frac {9\,c^2\,d^2\,e}{32\,a}-\frac {27\,c\,d^3\,\sqrt {-a^7\,c^3}}{32\,a^5}+\frac {3\,d\,e^2\,\sqrt {-a^7\,c^3}}{32\,a^4}\right )}\right )\,\sqrt {-\frac {9\,c\,d^2\,\sqrt {-a^7\,c^3}-a\,e^2\,\sqrt {-a^7\,c^3}+6\,a^4\,c^2\,d\,e}{256\,a^7\,c^3}}-2\,\mathrm {atanh}\left (\frac {c^2\,e^2\,x\,\sqrt {\frac {9\,d^2\,\sqrt {-a^7\,c^3}}{256\,a^7\,c^2}-\frac {3\,d\,e}{128\,a^3\,c}-\frac {e^2\,\sqrt {-a^7\,c^3}}{256\,a^6\,c^3}}}{2\,\left (\frac {c\,e^3}{32\,a}-\frac {9\,c^2\,d^2\,e}{32\,a^2}+\frac {27\,c\,d^3\,\sqrt {-a^7\,c^3}}{32\,a^6}-\frac {3\,d\,e^2\,\sqrt {-a^7\,c^3}}{32\,a^5}\right )}-\frac {9\,c^3\,d^2\,x\,\sqrt {\frac {9\,d^2\,\sqrt {-a^7\,c^3}}{256\,a^7\,c^2}-\frac {3\,d\,e}{128\,a^3\,c}-\frac {e^2\,\sqrt {-a^7\,c^3}}{256\,a^6\,c^3}}}{2\,\left (\frac {c\,e^3}{32}-\frac {9\,c^2\,d^2\,e}{32\,a}+\frac {27\,c\,d^3\,\sqrt {-a^7\,c^3}}{32\,a^5}-\frac {3\,d\,e^2\,\sqrt {-a^7\,c^3}}{32\,a^4}\right )}\right )\,\sqrt {-\frac {a\,e^2\,\sqrt {-a^7\,c^3}-9\,c\,d^2\,\sqrt {-a^7\,c^3}+6\,a^4\,c^2\,d\,e}{256\,a^7\,c^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^2)/(a + c*x^4)^2,x)

[Out]

((e*x^3)/(4*a) + (d*x)/(4*a))/(a + c*x^4) - 2*atanh((c^2*e^2*x*((e^2*(-a^7*c^3)^(1/2))/(256*a^6*c^3) - (9*d^2*
(-a^7*c^3)^(1/2))/(256*a^7*c^2) - (3*d*e)/(128*a^3*c))^(1/2))/(2*((c*e^3)/(32*a) - (9*c^2*d^2*e)/(32*a^2) - (2
7*c*d^3*(-a^7*c^3)^(1/2))/(32*a^6) + (3*d*e^2*(-a^7*c^3)^(1/2))/(32*a^5))) - (9*c^3*d^2*x*((e^2*(-a^7*c^3)^(1/
2))/(256*a^6*c^3) - (9*d^2*(-a^7*c^3)^(1/2))/(256*a^7*c^2) - (3*d*e)/(128*a^3*c))^(1/2))/(2*((c*e^3)/32 - (9*c
^2*d^2*e)/(32*a) - (27*c*d^3*(-a^7*c^3)^(1/2))/(32*a^5) + (3*d*e^2*(-a^7*c^3)^(1/2))/(32*a^4))))*(-(9*c*d^2*(-
a^7*c^3)^(1/2) - a*e^2*(-a^7*c^3)^(1/2) + 6*a^4*c^2*d*e)/(256*a^7*c^3))^(1/2) - 2*atanh((c^2*e^2*x*((9*d^2*(-a
^7*c^3)^(1/2))/(256*a^7*c^2) - (3*d*e)/(128*a^3*c) - (e^2*(-a^7*c^3)^(1/2))/(256*a^6*c^3))^(1/2))/(2*((c*e^3)/
(32*a) - (9*c^2*d^2*e)/(32*a^2) + (27*c*d^3*(-a^7*c^3)^(1/2))/(32*a^6) - (3*d*e^2*(-a^7*c^3)^(1/2))/(32*a^5)))
- (9*c^3*d^2*x*((9*d^2*(-a^7*c^3)^(1/2))/(256*a^7*c^2) - (3*d*e)/(128*a^3*c) - (e^2*(-a^7*c^3)^(1/2))/(256*a^
6*c^3))^(1/2))/(2*((c*e^3)/32 - (9*c^2*d^2*e)/(32*a) + (27*c*d^3*(-a^7*c^3)^(1/2))/(32*a^5) - (3*d*e^2*(-a^7*c
^3)^(1/2))/(32*a^4))))*(-(a*e^2*(-a^7*c^3)^(1/2) - 9*c*d^2*(-a^7*c^3)^(1/2) + 6*a^4*c^2*d*e)/(256*a^7*c^3))^(1
/2)

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sympy [A]  time = 1.03, size = 136, normalized size = 0.49 \begin {gather*} \operatorname {RootSum} {\left (65536 t^{4} a^{7} c^{3} + 3072 t^{2} a^{4} c^{2} d e + a^{2} e^{4} + 18 a c d^{2} e^{2} + 81 c^{2} d^{4}, \left (t \mapsto t \log {\left (x + \frac {4096 t^{3} a^{6} c^{2} e + 144 t a^{3} c d e^{2} - 432 t a^{2} c^{2} d^{3}}{a^{2} e^{4} - 81 c^{2} d^{4}} \right )} \right )\right )} + \frac {d x + e x^{3}}{4 a^{2} + 4 a c x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(65536*_t**4*a**7*c**3 + 3072*_t**2*a**4*c**2*d*e + a**2*e**4 + 18*a*c*d**2*e**2 + 81*c**2*d**4, Lambda
(_t, _t*log(x + (4096*_t**3*a**6*c**2*e + 144*_t*a**3*c*d*e**2 - 432*_t*a**2*c**2*d**3)/(a**2*e**4 - 81*c**2*d
**4)))) + (d*x + e*x**3)/(4*a**2 + 4*a*c*x**4)

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