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

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

[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))

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Rubi [A]  time = 0.116336, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1169, 634, 617, 204, 628} \[ -\frac{\left (A-\sqrt{a} B\right ) \log \left (-\sqrt{3} \sqrt [4]{a} x+\sqrt{a}+x^2\right )}{4 \sqrt{3} a^{3/4}}+\frac{\left (A-\sqrt{a} B\right ) \log \left (\sqrt{3} \sqrt [4]{a} x+\sqrt{a}+x^2\right )}{4 \sqrt{3} a^{3/4}}-\frac{\left (\sqrt{a} B+A\right ) \tan ^{-1}\left (\sqrt{3}-\frac{2 x}{\sqrt [4]{a}}\right )}{2 a^{3/4}}+\frac{\left (\sqrt{a} B+A\right ) \tan ^{-1}\left (\frac{2 x}{\sqrt [4]{a}}+\sqrt{3}\right )}{2 a^{3/4}} \]

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 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]

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 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])

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]

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*}

Mathematica [C]  time = 0.136933, size = 138, normalized size = 0.86 \[ \frac{\sqrt [4]{-1} \left (\frac{\left (\left (\sqrt{3}-i\right ) \sqrt{a} B-2 i A\right ) \tan ^{-1}\left (\frac{(1+i) x}{\sqrt{\sqrt{3}-i} \sqrt [4]{a}}\right )}{\sqrt{\sqrt{3}-i}}-\frac{\left (\left (\sqrt{3}+i\right ) \sqrt{a} B+2 i A\right ) \tanh ^{-1}\left (\frac{(1+i) x}{\sqrt{\sqrt{3}+i} \sqrt [4]{a}}\right )}{\sqrt{\sqrt{3}+i}}\right )}{\sqrt{6} a^{3/4}} \]

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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Maple [A]  time = 0.097, size = 194, normalized size = 1.2 \begin{align*}{\frac{A\sqrt{3}}{12}\ln \left ({x}^{2}+\sqrt [4]{a}x\sqrt{3}+\sqrt{a} \right ){a}^{-{\frac{3}{4}}}}-{\frac{B\sqrt{3}}{12}\ln \left ({x}^{2}+\sqrt [4]{a}x\sqrt{3}+\sqrt{a} \right ){\frac{1}{\sqrt [4]{a}}}}+{\frac{A}{2}\arctan \left ({ \left ( 2\,x+\sqrt{3}\sqrt [4]{a} \right ){\frac{1}{\sqrt [4]{a}}}} \right ){a}^{-{\frac{3}{4}}}}+{\frac{B}{2}\arctan \left ({ \left ( 2\,x+\sqrt{3}\sqrt [4]{a} \right ){\frac{1}{\sqrt [4]{a}}}} \right ){\frac{1}{\sqrt [4]{a}}}}-{\frac{A\sqrt{3}}{12}\ln \left ({x}^{2}-\sqrt [4]{a}x\sqrt{3}+\sqrt{a} \right ){a}^{-{\frac{3}{4}}}}+{\frac{B\sqrt{3}}{12}\ln \left ({x}^{2}-\sqrt [4]{a}x\sqrt{3}+\sqrt{a} \right ){\frac{1}{\sqrt [4]{a}}}}+{\frac{A}{2}\arctan \left ({ \left ( 2\,x-\sqrt{3}\sqrt [4]{a} \right ){\frac{1}{\sqrt [4]{a}}}} \right ){a}^{-{\frac{3}{4}}}}+{\frac{B}{2}\arctan \left ({ \left ( 2\,x-\sqrt{3}\sqrt [4]{a} \right ){\frac{1}{\sqrt [4]{a}}}} \right ){\frac{1}{\sqrt [4]{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/a^(3/4)*ln(x^2+a^(1/4)*x*3^(1/2)+a^(1/2))*A*3^(1/2)-1/12/a^(1/4)*ln(x^2+a^(1/4)*x*3^(1/2)+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-a^(1/4)*x*3^(1/2)+a^(1/2))*A*3^(1/2)+1/12/a^(1/4)*ln(x^2-a^(1/4)*x*3^(1/2)+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., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{2} + A}{x^{4} - \sqrt{a} x^{2} + a}\,{d x} \end{align*}

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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Fricas [B]  time = 2.4112, size = 2531, normalized size = 15.82 \begin{align*} \text{result too large to display} \end{align*}

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

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

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