3.1.98 \(\int \frac {A+B x^2}{a-\sqrt {a} x^2+x^4} \, dx\)

Optimal. Leaf size=160 \[ -\frac {\left (A-\sqrt {a} B\right ) \log \left (-\sqrt {3} \sqrt [4]{a} x+\sqrt {a}+x^2\right )}{4 \sqrt {3} a^{3/4}}+\frac {\left (A-\sqrt {a} B\right ) \log \left (\sqrt {3} \sqrt [4]{a} x+\sqrt {a}+x^2\right )}{4 \sqrt {3} a^{3/4}}-\frac {\left (\sqrt {a} B+A\right ) \tan ^{-1}\left (\sqrt {3}-\frac {2 x}{\sqrt [4]{a}}\right )}{2 a^{3/4}}+\frac {\left (\sqrt {a} B+A\right ) \tan ^{-1}\left (\frac {2 x}{\sqrt [4]{a}}+\sqrt {3}\right )}{2 a^{3/4}} \]

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Rubi [A]  time = 0.12, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1169, 634, 617, 204, 628} \begin {gather*} -\frac {\left (A-\sqrt {a} B\right ) \log \left (-\sqrt {3} \sqrt [4]{a} x+\sqrt {a}+x^2\right )}{4 \sqrt {3} a^{3/4}}+\frac {\left (A-\sqrt {a} B\right ) \log \left (\sqrt {3} \sqrt [4]{a} x+\sqrt {a}+x^2\right )}{4 \sqrt {3} a^{3/4}}-\frac {\left (\sqrt {a} B+A\right ) \tan ^{-1}\left (\sqrt {3}-\frac {2 x}{\sqrt [4]{a}}\right )}{2 a^{3/4}}+\frac {\left (\sqrt {a} B+A\right ) \tan ^{-1}\left (\frac {2 x}{\sqrt [4]{a}}+\sqrt {3}\right )}{2 a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(a - Sqrt[a]*x^2 + x^4),x]

[Out]

-((A + Sqrt[a]*B)*ArcTan[Sqrt[3] - (2*x)/a^(1/4)])/(2*a^(3/4)) + ((A + Sqrt[a]*B)*ArcTan[Sqrt[3] + (2*x)/a^(1/
4)])/(2*a^(3/4)) - ((A - Sqrt[a]*B)*Log[Sqrt[a] - Sqrt[3]*a^(1/4)*x + x^2])/(4*Sqrt[3]*a^(3/4)) + ((A - Sqrt[a
]*B)*Log[Sqrt[a] + Sqrt[3]*a^(1/4)*x + x^2])/(4*Sqrt[3]*a^(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 634

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

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rubi steps

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

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Mathematica [C]  time = 0.13, size = 138, normalized size = 0.86 \begin {gather*} \frac {\sqrt [4]{-1} \left (\frac {\left (\left (\sqrt {3}-i\right ) \sqrt {a} B-2 i A\right ) \tan ^{-1}\left (\frac {(1+i) x}{\sqrt {\sqrt {3}-i} \sqrt [4]{a}}\right )}{\sqrt {\sqrt {3}-i}}-\frac {\left (\left (\sqrt {3}+i\right ) \sqrt {a} B+2 i A\right ) \tanh ^{-1}\left (\frac {(1+i) x}{\sqrt {\sqrt {3}+i} \sqrt [4]{a}}\right )}{\sqrt {\sqrt {3}+i}}\right )}{\sqrt {6} a^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(a - Sqrt[a]*x^2 + x^4),x]

[Out]

((-1)^(1/4)*((((-2*I)*A + (-I + Sqrt[3])*Sqrt[a]*B)*ArcTan[((1 + I)*x)/(Sqrt[-I + Sqrt[3]]*a^(1/4))])/Sqrt[-I
+ Sqrt[3]] - (((2*I)*A + (I + Sqrt[3])*Sqrt[a]*B)*ArcTanh[((1 + I)*x)/(Sqrt[I + Sqrt[3]]*a^(1/4))])/Sqrt[I + S
qrt[3]]))/(Sqrt[6]*a^(3/4))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^2}{a-\sqrt {a} x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x^2)/(a - Sqrt[a]*x^2 + x^4),x]

[Out]

IntegrateAlgebraic[(A + B*x^2)/(a - Sqrt[a]*x^2 + x^4), x]

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fricas [B]  time = 1.68, size = 1141, normalized size = 7.13

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(a+x^4-x^2*a^(1/2)),x, algorithm="fricas")

[Out]

1/2*sqrt(1/6)*sqrt(-(4*A*B*a + 3*sqrt(1/3)*a^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) + (B^2*a + A^2)*sqrt(a
))/a^2)*log(2*(B^6*a^3 - A^6)*x + 3*sqrt(1/6)*(A*B^4*a^3 - A^5*a - sqrt(1/3)*(2*B^3*a^4 + A^2*B*a^3)*sqrt(-(B^
4*a^2 - 2*A^2*B^2*a + A^4)/a^3) - (A^2*B^3*a^2 - A^4*B*a - sqrt(1/3)*(A*B^2*a^3 - A^3*a^2)*sqrt(-(B^4*a^2 - 2*
A^2*B^2*a + A^4)/a^3))*sqrt(a))*sqrt(-(4*A*B*a + 3*sqrt(1/3)*a^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) + (B
^2*a + A^2)*sqrt(a))/a^2)) - 1/2*sqrt(1/6)*sqrt(-(4*A*B*a + 3*sqrt(1/3)*a^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4
)/a^3) + (B^2*a + A^2)*sqrt(a))/a^2)*log(2*(B^6*a^3 - A^6)*x - 3*sqrt(1/6)*(A*B^4*a^3 - A^5*a - sqrt(1/3)*(2*B
^3*a^4 + A^2*B*a^3)*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) - (A^2*B^3*a^2 - A^4*B*a - sqrt(1/3)*(A*B^2*a^3 -
A^3*a^2)*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3))*sqrt(a))*sqrt(-(4*A*B*a + 3*sqrt(1/3)*a^2*sqrt(-(B^4*a^2 -
2*A^2*B^2*a + A^4)/a^3) + (B^2*a + A^2)*sqrt(a))/a^2)) + 1/2*sqrt(1/6)*sqrt(-(4*A*B*a - 3*sqrt(1/3)*a^2*sqrt(
-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) + (B^2*a + A^2)*sqrt(a))/a^2)*log(2*(B^6*a^3 - A^6)*x + 3*sqrt(1/6)*(A*B^4
*a^3 - A^5*a + sqrt(1/3)*(2*B^3*a^4 + A^2*B*a^3)*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) - (A^2*B^3*a^2 - A^4
*B*a + sqrt(1/3)*(A*B^2*a^3 - A^3*a^2)*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3))*sqrt(a))*sqrt(-(4*A*B*a - 3*s
qrt(1/3)*a^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) + (B^2*a + A^2)*sqrt(a))/a^2)) - 1/2*sqrt(1/6)*sqrt(-(4*
A*B*a - 3*sqrt(1/3)*a^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) + (B^2*a + A^2)*sqrt(a))/a^2)*log(2*(B^6*a^3
- A^6)*x - 3*sqrt(1/6)*(A*B^4*a^3 - A^5*a + sqrt(1/3)*(2*B^3*a^4 + A^2*B*a^3)*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A
^4)/a^3) - (A^2*B^3*a^2 - A^4*B*a + sqrt(1/3)*(A*B^2*a^3 - A^3*a^2)*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3))*
sqrt(a))*sqrt(-(4*A*B*a - 3*sqrt(1/3)*a^2*sqrt(-(B^4*a^2 - 2*A^2*B^2*a + A^4)/a^3) + (B^2*a + A^2)*sqrt(a))/a^
2))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(a+x^4-x^2*a^(1/2)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.04, size = 198, normalized size = 1.24 \begin {gather*} \frac {B \arctan \left (\frac {2 x +\sqrt {3}\, a^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2 a^{\frac {1}{4}}}-\frac {B \arctan \left (\frac {-2 x +\sqrt {3}\, a^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2 a^{\frac {1}{4}}}-\frac {\sqrt {3}\, B \ln \left (x^{2}+\sqrt {3}\, a^{\frac {1}{4}} x +\sqrt {a}\right )}{12 a^{\frac {1}{4}}}+\frac {\sqrt {3}\, B \ln \left (-x^{2}+\sqrt {3}\, a^{\frac {1}{4}} x -\sqrt {a}\right )}{12 a^{\frac {1}{4}}}+\frac {A \arctan \left (\frac {2 x +\sqrt {3}\, a^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2 a^{\frac {3}{4}}}-\frac {A \arctan \left (\frac {-2 x +\sqrt {3}\, a^{\frac {1}{4}}}{a^{\frac {1}{4}}}\right )}{2 a^{\frac {3}{4}}}+\frac {\sqrt {3}\, A \ln \left (x^{2}+\sqrt {3}\, a^{\frac {1}{4}} x +\sqrt {a}\right )}{12 a^{\frac {3}{4}}}-\frac {\sqrt {3}\, A \ln \left (-x^{2}+\sqrt {3}\, a^{\frac {1}{4}} x -\sqrt {a}\right )}{12 a^{\frac {3}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/(a+x^4-a^(1/2)*x^2),x)

[Out]

1/12/a^(3/4)*ln(x^2+3^(1/2)*a^(1/4)*x+a^(1/2))*A*3^(1/2)-1/12/a^(1/4)*ln(x^2+3^(1/2)*a^(1/4)*x+a^(1/2))*B*3^(1
/2)+1/2/a^(3/4)*arctan((2*x+3^(1/2)*a^(1/4))/a^(1/4))*A+1/2/a^(1/4)*arctan((2*x+3^(1/2)*a^(1/4))/a^(1/4))*B-1/
12/a^(3/4)*ln(-x^2+3^(1/2)*a^(1/4)*x-a^(1/2))*A*3^(1/2)+1/12/a^(1/4)*ln(-x^2+3^(1/2)*a^(1/4)*x-a^(1/2))*B*3^(1
/2)-1/2/a^(3/4)*arctan((-2*x+3^(1/2)*a^(1/4))/a^(1/4))*A-1/2/a^(1/4)*arctan((-2*x+3^(1/2)*a^(1/4))/a^(1/4))*B

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {B x^{2} + A}{x^{4} - \sqrt {a} x^{2} + a}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/(a+x^4-x^2*a^(1/2)),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)/(x^4 - sqrt(a)*x^2 + a), x)

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mupad [B]  time = 4.99, size = 1155, normalized size = 7.22 \begin {gather*} -2\,\mathrm {atanh}\left (\frac {6\,A^2\,x\,\sqrt {\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {A^2}{24\,a^{3/2}}-\frac {A\,B}{6\,a}}}{2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}+\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}-\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}}-\frac {6\,B^2\,a\,x\,\sqrt {\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {A^2}{24\,a^{3/2}}-\frac {A\,B}{6\,a}}}{2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}+\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}-\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}}-\frac {2\,A^2\,x\,\sqrt {-27\,a^3}\,\sqrt {\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {A^2}{24\,a^{3/2}}-\frac {A\,B}{6\,a}}}{3\,a^{3/2}\,\left (2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}+\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}-\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}\right )}+\frac {2\,B^2\,x\,\sqrt {-27\,a^3}\,\sqrt {\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {A^2}{24\,a^{3/2}}-\frac {A\,B}{6\,a}}}{3\,\sqrt {a}\,\left (2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}+\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}-\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}\right )}\right )\,\sqrt {\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {A^2}{24\,a^{3/2}}-\frac {A\,B}{6\,a}}-2\,\mathrm {atanh}\left (\frac {6\,A^2\,x\,\sqrt {\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2}{24\,a^{3/2}}-\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {A\,B}{6\,a}}}{2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}-\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}+\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}}-\frac {6\,B^2\,a\,x\,\sqrt {\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2}{24\,a^{3/2}}-\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {A\,B}{6\,a}}}{2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}-\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}+\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}}+\frac {2\,A^2\,x\,\sqrt {-27\,a^3}\,\sqrt {\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2}{24\,a^{3/2}}-\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {A\,B}{6\,a}}}{3\,a^{3/2}\,\left (2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}-\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}+\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}\right )}-\frac {2\,B^2\,x\,\sqrt {-27\,a^3}\,\sqrt {\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2}{24\,a^{3/2}}-\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {A\,B}{6\,a}}}{3\,\sqrt {a}\,\left (2\,A^2\,B-2\,B^3\,a+\frac {A^3}{\sqrt {a}}-A\,B^2\,\sqrt {a}-\frac {A^3\,\sqrt {-27\,a^3}}{3\,a^2}+\frac {A\,B^2\,\sqrt {-27\,a^3}}{3\,a}\right )}\right )\,\sqrt {\frac {A^2\,\sqrt {-27\,a^3}}{72\,a^3}-\frac {B^2}{24\,\sqrt {a}}-\frac {A^2}{24\,a^{3/2}}-\frac {B^2\,\sqrt {-27\,a^3}}{72\,a^2}-\frac {A\,B}{6\,a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(a + x^4 - a^(1/2)*x^2),x)

[Out]

- 2*atanh((6*A^2*x*((B^2*(-27*a^3)^(1/2))/(72*a^2) - B^2/(24*a^(1/2)) - (A^2*(-27*a^3)^(1/2))/(72*a^3) - A^2/(
24*a^(3/2)) - (A*B)/(6*a))^(1/2))/(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(1/2) + (A^3*(-27*a^3)^(1/2))/(3*
a^2) - (A*B^2*(-27*a^3)^(1/2))/(3*a)) - (6*B^2*a*x*((B^2*(-27*a^3)^(1/2))/(72*a^2) - B^2/(24*a^(1/2)) - (A^2*(
-27*a^3)^(1/2))/(72*a^3) - A^2/(24*a^(3/2)) - (A*B)/(6*a))^(1/2))/(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(
1/2) + (A^3*(-27*a^3)^(1/2))/(3*a^2) - (A*B^2*(-27*a^3)^(1/2))/(3*a)) - (2*A^2*x*(-27*a^3)^(1/2)*((B^2*(-27*a^
3)^(1/2))/(72*a^2) - B^2/(24*a^(1/2)) - (A^2*(-27*a^3)^(1/2))/(72*a^3) - A^2/(24*a^(3/2)) - (A*B)/(6*a))^(1/2)
)/(3*a^(3/2)*(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(1/2) + (A^3*(-27*a^3)^(1/2))/(3*a^2) - (A*B^2*(-27*a^
3)^(1/2))/(3*a))) + (2*B^2*x*(-27*a^3)^(1/2)*((B^2*(-27*a^3)^(1/2))/(72*a^2) - B^2/(24*a^(1/2)) - (A^2*(-27*a^
3)^(1/2))/(72*a^3) - A^2/(24*a^(3/2)) - (A*B)/(6*a))^(1/2))/(3*a^(1/2)*(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^
2*a^(1/2) + (A^3*(-27*a^3)^(1/2))/(3*a^2) - (A*B^2*(-27*a^3)^(1/2))/(3*a))))*((B^2*(-27*a^3)^(1/2))/(72*a^2) -
B^2/(24*a^(1/2)) - (A^2*(-27*a^3)^(1/2))/(72*a^3) - A^2/(24*a^(3/2)) - (A*B)/(6*a))^(1/2) - 2*atanh((6*A^2*x*
((A^2*(-27*a^3)^(1/2))/(72*a^3) - B^2/(24*a^(1/2)) - A^2/(24*a^(3/2)) - (B^2*(-27*a^3)^(1/2))/(72*a^2) - (A*B)
/(6*a))^(1/2))/(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(1/2) - (A^3*(-27*a^3)^(1/2))/(3*a^2) + (A*B^2*(-27*
a^3)^(1/2))/(3*a)) - (6*B^2*a*x*((A^2*(-27*a^3)^(1/2))/(72*a^3) - B^2/(24*a^(1/2)) - A^2/(24*a^(3/2)) - (B^2*(
-27*a^3)^(1/2))/(72*a^2) - (A*B)/(6*a))^(1/2))/(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(1/2) - (A^3*(-27*a^
3)^(1/2))/(3*a^2) + (A*B^2*(-27*a^3)^(1/2))/(3*a)) + (2*A^2*x*(-27*a^3)^(1/2)*((A^2*(-27*a^3)^(1/2))/(72*a^3)
- B^2/(24*a^(1/2)) - A^2/(24*a^(3/2)) - (B^2*(-27*a^3)^(1/2))/(72*a^2) - (A*B)/(6*a))^(1/2))/(3*a^(3/2)*(2*A^2
*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(1/2) - (A^3*(-27*a^3)^(1/2))/(3*a^2) + (A*B^2*(-27*a^3)^(1/2))/(3*a))) -
(2*B^2*x*(-27*a^3)^(1/2)*((A^2*(-27*a^3)^(1/2))/(72*a^3) - B^2/(24*a^(1/2)) - A^2/(24*a^(3/2)) - (B^2*(-27*a^
3)^(1/2))/(72*a^2) - (A*B)/(6*a))^(1/2))/(3*a^(1/2)*(2*A^2*B - 2*B^3*a + A^3/a^(1/2) - A*B^2*a^(1/2) - (A^3*(-
27*a^3)^(1/2))/(3*a^2) + (A*B^2*(-27*a^3)^(1/2))/(3*a))))*((A^2*(-27*a^3)^(1/2))/(72*a^3) - B^2/(24*a^(1/2)) -
A^2/(24*a^(3/2)) - (B^2*(-27*a^3)^(1/2))/(72*a^2) - (A*B)/(6*a))^(1/2)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/(a+x**4-x**2*a**(1/2)),x)

[Out]

Exception raised: PolynomialError

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