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

Optimal. Leaf size=130 \[ \frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2 c d-b e}+2 \sqrt {c} \sqrt {e} x}{\sqrt {b e+2 c d}}\right )}{\sqrt {c} \sqrt {b e+2 c d}}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2 c d-b e}-2 \sqrt {c} \sqrt {e} x}{\sqrt {b e+2 c d}}\right )}{\sqrt {c} \sqrt {b e+2 c d}} \]

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Rubi [A]  time = 0.13, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1990, 1161, 618, 204} \begin {gather*} \frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2 c d-b e}+2 \sqrt {c} \sqrt {e} x}{\sqrt {b e+2 c d}}\right )}{\sqrt {c} \sqrt {b e+2 c d}}-\frac {e^{3/2} \tan ^{-1}\left (\frac {\sqrt {2 c d-b e}-2 \sqrt {c} \sqrt {e} x}{\sqrt {b e+2 c d}}\right )}{\sqrt {c} \sqrt {b e+2 c d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(b*x^2 + c*(d^2/e^2 + x^4)),x]

[Out]

-((e^(3/2)*ArcTan[(Sqrt[2*c*d - b*e] - 2*Sqrt[c]*Sqrt[e]*x)/Sqrt[2*c*d + b*e]])/(Sqrt[c]*Sqrt[2*c*d + b*e])) +
(e^(3/2)*ArcTan[(Sqrt[2*c*d - b*e] + 2*Sqrt[c]*Sqrt[e]*x)/Sqrt[2*c*d + b*e]])/(Sqrt[c]*Sqrt[2*c*d + b*e])

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 1990

Int[(u_)^(q_.)*(v_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*ExpandToSum[v, x]^p, x] /; FreeQ[{p, q}, x] &&
BinomialQ[u, x] && TrinomialQ[v, x] &&  !(BinomialMatchQ[u, x] && TrinomialMatchQ[v, x])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 248, normalized size = 1.91 \begin {gather*} \frac {e^{3/2} \left (\frac {\left (\sqrt {b^2 e^2-4 c^2 d^2}-b e+2 c d\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {e} x}{\sqrt {b e-\sqrt {b^2 e^2-4 c^2 d^2}}}\right )}{\sqrt {b e-\sqrt {b^2 e^2-4 c^2 d^2}}}+\frac {\left (\sqrt {b^2 e^2-4 c^2 d^2}+b e-2 c d\right ) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {e} x}{\sqrt {\sqrt {b^2 e^2-4 c^2 d^2}+b e}}\right )}{\sqrt {\sqrt {b^2 e^2-4 c^2 d^2}+b e}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2 e^2-4 c^2 d^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^(3/2)*(((2*c*d - b*e + Sqrt[-4*c^2*d^2 + b^2*e^2])*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[e]*x)/Sqrt[b*e - Sqrt[-4*c^
2*d^2 + b^2*e^2]]])/Sqrt[b*e - Sqrt[-4*c^2*d^2 + b^2*e^2]] + ((-2*c*d + b*e + Sqrt[-4*c^2*d^2 + b^2*e^2])*ArcT
an[(Sqrt[2]*Sqrt[c]*Sqrt[e]*x)/Sqrt[b*e + Sqrt[-4*c^2*d^2 + b^2*e^2]]])/Sqrt[b*e + Sqrt[-4*c^2*d^2 + b^2*e^2]]
))/(Sqrt[2]*Sqrt[c]*Sqrt[-4*c^2*d^2 + b^2*e^2])

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.75, size = 232, normalized size = 1.78 \begin {gather*} \left [\frac {1}{2} \, e \sqrt {-\frac {e}{2 \, c^{2} d + b c e}} \log \left (\frac {c e^{2} x^{4} + c d^{2} - {\left (4 \, c d e + b e^{2}\right )} x^{2} + 2 \, {\left ({\left (2 \, c^{2} d e + b c e^{2}\right )} x^{3} - {\left (2 \, c^{2} d^{2} + b c d e\right )} x\right )} \sqrt {-\frac {e}{2 \, c^{2} d + b c e}}}{c e^{2} x^{4} + b e^{2} x^{2} + c d^{2}}\right ), e \sqrt {\frac {e}{2 \, c^{2} d + b c e}} \arctan \left (c x \sqrt {\frac {e}{2 \, c^{2} d + b c e}}\right ) + e \sqrt {\frac {e}{2 \, c^{2} d + b c e}} \arctan \left (\frac {{\left (c e x^{3} + {\left (c d + b e\right )} x\right )} \sqrt {\frac {e}{2 \, c^{2} d + b c e}}}{d}\right )\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*e*sqrt(-e/(2*c^2*d + b*c*e))*log((c*e^2*x^4 + c*d^2 - (4*c*d*e + b*e^2)*x^2 + 2*((2*c^2*d*e + b*c*e^2)*x^
3 - (2*c^2*d^2 + b*c*d*e)*x)*sqrt(-e/(2*c^2*d + b*c*e)))/(c*e^2*x^4 + b*e^2*x^2 + c*d^2)), e*sqrt(e/(2*c^2*d +
b*c*e))*arctan(c*x*sqrt(e/(2*c^2*d + b*c*e))) + e*sqrt(e/(2*c^2*d + b*c*e))*arctan((c*e*x^3 + (c*d + b*e)*x)*
sqrt(e/(2*c^2*d + b*c*e))/d)]

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giac [B]  time = 1.35, size = 2202, normalized size = 16.94

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(32*c^5*d^4*e^4 - 16*sqrt(2)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^4*d^4*e^2 - 16*b^2*c^
3*d^2*e^6 + 8*b*c^4*d^2*e^6 + 8*sqrt(2)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c^2*d^2*e^4 -
8*sqrt(2)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^3*d^2*e^4 + 4*sqrt(2)*sqrt(b*c*e^4 + sqrt(
-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^4*d^2*e^4 - 4*sqrt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4
*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^2*d^2*e^2 - 8*(4*c^2*d^2*e^2 - b^2*e^4)*c^3*d^2*e^2 + 2*b^4*c*e^8 - 2*b^3*c
^2*e^8 - sqrt(2)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^4*e^6 + 2*sqrt(2)*sqrt(b*c*e^4 + sqrt(
-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^3*c*e^6 - sqrt(2)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*
c^2*e^6 + sqrt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^3*e^4
- 2*sqrt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c*e^4 + sq
rt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^2*e^4 + 2*(4*c^2
*d^2*e^2 - b^2*e^4)*b^2*c*e^4 - 2*(4*c^2*d^2*e^2 - b^2*e^4)*b*c^2*e^4 - 2*(8*c^5*d^3*e^4 - 4*sqrt(2)*sqrt(-4*c
^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^3*d^3 - 2*b^2*c^3*d*e^6 + sqrt(2)
*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c*d*e^2 - 2*sqrt(2)*s
qrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^2*d*e^2 + sqrt(2)*sqrt(
-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 + sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^3*d*e^2 - 2*(4*c^2*d^2*e^2 -
b^2*e^4)*c^3*d*e^2)*e)*arctan(2*sqrt(1/2)*x/sqrt((b + sqrt(-4*c^2*d^2*e^(-2) + b^2))/c))/((16*c^5*d^5*e^2 - 8*
b^2*c^3*d^3*e^4 + 8*b*c^4*d^3*e^4 - 4*c^5*d^3*e^4 + b^4*c*d*e^6 - 2*b^3*c^2*d*e^6 + b^2*c^3*d*e^6)*abs(c)) + 1
/4*(32*c^5*d^4*e^4 + 16*sqrt(2)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^4*d^4*e^2 - 16*b^2*c^3*
d^2*e^6 + 8*b*c^4*d^2*e^6 - 8*sqrt(2)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c^2*d^2*e^4 + 8
*sqrt(2)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^3*d^2*e^4 - 4*sqrt(2)*sqrt(b*c*e^4 - sqrt(-4
*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^4*d^2*e^4 - 4*sqrt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c
^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^2*d^2*e^2 - 8*(4*c^2*d^2*e^2 - b^2*e^4)*c^3*d^2*e^2 + 2*b^4*c*e^8 - 2*b^3*c^2
*e^8 + sqrt(2)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^4*e^6 - 2*sqrt(2)*sqrt(b*c*e^4 - sqrt(-4
*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^3*c*e^6 + sqrt(2)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c^
2*e^6 + sqrt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^3*e^4 -
2*sqrt(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c*e^4 + sqrt
(2)*sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^2*e^4 + 2*(4*c^2*d
^2*e^2 - b^2*e^4)*b^2*c*e^4 - 2*(4*c^2*d^2*e^2 - b^2*e^4)*b*c^2*e^4 - 2*(8*c^5*d^3*e^4 - 4*sqrt(2)*sqrt(-4*c^2
*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^3*d^3 - 2*b^2*c^3*d*e^6 + sqrt(2)*s
qrt(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b^2*c*d*e^2 - 2*sqrt(2)*sqr
t(-4*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*b*c^2*d*e^2 + sqrt(2)*sqrt(-4
*c^2*d^2*e^2 + b^2*e^4)*sqrt(b*c*e^4 - sqrt(-4*c^2*d^2*e^2 + b^2*e^4)*c*e^2)*c^3*d*e^2 - 2*(4*c^2*d^2*e^2 - b^
2*e^4)*c^3*d*e^2)*e)*arctan(2*sqrt(1/2)*x/sqrt((b - sqrt(-4*c^2*d^2*e^(-2) + b^2))/c))/((16*c^5*d^5*e^2 - 8*b^
2*c^3*d^3*e^4 + 8*b*c^4*d^3*e^4 - 4*c^5*d^3*e^4 + b^4*c*d*e^6 - 2*b^3*c^2*d*e^6 + b^2*c^3*d*e^6)*abs(c))

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maple [B]  time = 0.01, size = 582, normalized size = 4.48 \begin {gather*} \frac {\sqrt {2}\, b \,e^{4} \arctanh \left (\frac {\sqrt {2}\, c e x}{\sqrt {\left (-b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}\right )}{2 \sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\, \sqrt {\left (-b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}+\frac {\sqrt {2}\, b \,e^{4} \arctan \left (\frac {\sqrt {2}\, c e x}{\sqrt {\left (b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}\right )}{2 \sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\, \sqrt {\left (b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}-\frac {\sqrt {2}\, c d \,e^{3} \arctanh \left (\frac {\sqrt {2}\, c e x}{\sqrt {\left (-b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}\right )}{\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\, \sqrt {\left (-b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}-\frac {\sqrt {2}\, c d \,e^{3} \arctan \left (\frac {\sqrt {2}\, c e x}{\sqrt {\left (b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}\right )}{\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\, \sqrt {\left (b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}-\frac {\sqrt {2}\, e^{2} \arctanh \left (\frac {\sqrt {2}\, c e x}{\sqrt {\left (-b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}+\frac {\sqrt {2}\, e^{2} \arctan \left (\frac {\sqrt {2}\, c e x}{\sqrt {\left (b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}}\right )}{2 \sqrt {\left (b \,e^{2}+\sqrt {\left (b e -2 c d \right ) \left (b e +2 c d \right ) e^{2}}\right ) c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^4/((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2)*2^(1/2)/((-b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*arc
tanh(2^(1/2)/((-b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*c*e*x)*b-e^3*c/((b*e-2*c*d)*(b*e+2*c*d)*e^
2)^(1/2)*2^(1/2)/((-b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*arctanh(2^(1/2)/((-b*e^2+((b*e-2*c*d)*
(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*c*e*x)*d-1/2*e^2*2^(1/2)/((-b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/
2)*arctanh(2^(1/2)/((-b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*c*e*x)+1/2/((b*e-2*c*d)*(b*e+2*c*d)*
e^2)^(1/2)*2^(1/2)/((b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*b*e^4*arctan(2^(1/2)/((b*e^2+((b*e-2*
c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*c*e*x)-1/((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2)*2^(1/2)/((b*e^2+((b*e-2*c*d
)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*c*d*e^3*arctan(2^(1/2)/((b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2
)*c*e*x)+1/2*2^(1/2)/((b*e^2+((b*e-2*c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*e^2*arctan(2^(1/2)/((b*e^2+((b*e-2*
c*d)*(b*e+2*c*d)*e^2)^(1/2))*c)^(1/2)*c*e*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.13, size = 232, normalized size = 1.78 \begin {gather*} \frac {e^{3/2}\,\left (\mathrm {atan}\left (\frac {c\,\sqrt {e}\,x}{\sqrt {c\,\left (b\,e+2\,c\,d\right )}}\right )-\mathrm {atan}\left (\frac {\left (2\,d\,c^2+b\,e\,c\right )\,\left (x\,\left (\frac {\sqrt {e}\,\left (c\,d\,e^7-\frac {4\,c^3\,d^2\,e^7}{2\,d\,c^2+b\,e\,c}\right )}{d\,\sqrt {c\,\left (b\,e+2\,c\,d\right )}\,\left (b\,e-2\,c\,d\right )}+\frac {e^{3/2}\,\left (2\,c^2\,d\,e^6-b\,c\,e^7\right )}{c\,d\,\sqrt {2\,d\,c^2+b\,e\,c}\,\left (b\,e-2\,c\,d\right )}\right )+\frac {\sqrt {e}\,x^3\,\left (c\,e^8-\frac {2\,b\,c^2\,e^9}{2\,d\,c^2+b\,e\,c}\right )}{d\,\sqrt {c\,\left (b\,e+2\,c\,d\right )}\,\left (b\,e-2\,c\,d\right )}\right )}{c\,e^7}\right )\right )}{\sqrt {2\,d\,c^2+b\,e\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(3/2)*(atan((c*e^(1/2)*x)/(c*(b*e + 2*c*d))^(1/2)) - atan(((2*c^2*d + b*c*e)*(x*((e^(1/2)*(c*d*e^7 - (4*c^3
*d^2*e^7)/(2*c^2*d + b*c*e)))/(d*(c*(b*e + 2*c*d))^(1/2)*(b*e - 2*c*d)) + (e^(3/2)*(2*c^2*d*e^6 - b*c*e^7))/(c
*d*(2*c^2*d + b*c*e)^(1/2)*(b*e - 2*c*d))) + (e^(1/2)*x^3*(c*e^8 - (2*b*c^2*e^9)/(2*c^2*d + b*c*e)))/(d*(c*(b*
e + 2*c*d))^(1/2)*(b*e - 2*c*d))))/(c*e^7))))/(2*c^2*d + b*c*e)^(1/2)

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sympy [A]  time = 0.79, size = 160, normalized size = 1.23 \begin {gather*} - \frac {\sqrt {- \frac {e^{3}}{c \left (b e + 2 c d\right )}} \log {\left (- \frac {d}{e} + x^{2} + \frac {x \left (- b e \sqrt {- \frac {e^{3}}{c \left (b e + 2 c d\right )}} - 2 c d \sqrt {- \frac {e^{3}}{c \left (b e + 2 c d\right )}}\right )}{e^{2}} \right )}}{2} + \frac {\sqrt {- \frac {e^{3}}{c \left (b e + 2 c d\right )}} \log {\left (- \frac {d}{e} + x^{2} + \frac {x \left (b e \sqrt {- \frac {e^{3}}{c \left (b e + 2 c d\right )}} + 2 c d \sqrt {- \frac {e^{3}}{c \left (b e + 2 c d\right )}}\right )}{e^{2}} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-e**3/(c*(b*e + 2*c*d)))*log(-d/e + x**2 + x*(-b*e*sqrt(-e**3/(c*(b*e + 2*c*d))) - 2*c*d*sqrt(-e**3/(c*(
b*e + 2*c*d))))/e**2)/2 + sqrt(-e**3/(c*(b*e + 2*c*d)))*log(-d/e + x**2 + x*(b*e*sqrt(-e**3/(c*(b*e + 2*c*d)))
+ 2*c*d*sqrt(-e**3/(c*(b*e + 2*c*d))))/e**2)/2

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