3.23.44 \(\int \frac {1+a x^2}{(-1+a x^2) \sqrt {a x+\sqrt {b+a^2 x^2}}} \, dx\)

Optimal. Leaf size=167 \[ \text {RootSum}\left [\text {$\#$1}^8-4 \text {$\#$1}^4 a-2 \text {$\#$1}^4 b+b^2\& ,\frac {\text {$\#$1}^4 \left (-\log \left (\sqrt {\sqrt {a^2 x^2+b}+a x}-\text {$\#$1}\right )\right )-b \log \left (\sqrt {\sqrt {a^2 x^2+b}+a x}-\text {$\#$1}\right )}{-\text {$\#$1}^7+2 \text {$\#$1}^3 a+\text {$\#$1}^3 b}\& \right ]-\frac {b}{3 a \left (\sqrt {a^2 x^2+b}+a x\right )^{3/2}}+\frac {\sqrt {\sqrt {a^2 x^2+b}+a x}}{a} \]

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Rubi [A]  time = 1.26, antiderivative size = 319, normalized size of antiderivative = 1.91, number of steps used = 23, number of rules used = 10, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6725, 2117, 14, 2119, 1628, 828, 826, 1166, 208, 205} \begin {gather*} -\frac {b}{3 a \left (\sqrt {a^2 x^2+b}+a x\right )^{3/2}}+\frac {\sqrt {\sqrt {a^2 x^2+b}+a x}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b}+a x}}{\sqrt {\sqrt {a}-\sqrt {a+b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a}-\sqrt {a+b}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b}+a x}}{\sqrt {\sqrt {a+b}+\sqrt {a}}}\right )}{\sqrt {a} \sqrt {\sqrt {a+b}+\sqrt {a}}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b}+a x}}{\sqrt {\sqrt {a}-\sqrt {a+b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a}-\sqrt {a+b}}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2 x^2+b}+a x}}{\sqrt {\sqrt {a+b}+\sqrt {a}}}\right )}{\sqrt {a} \sqrt {\sqrt {a+b}+\sqrt {a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + a*x^2)/((-1 + a*x^2)*Sqrt[a*x + Sqrt[b + a^2*x^2]]),x]

[Out]

-1/3*b/(a*(a*x + Sqrt[b + a^2*x^2])^(3/2)) + Sqrt[a*x + Sqrt[b + a^2*x^2]]/a - (2*ArcTan[Sqrt[a*x + Sqrt[b + a
^2*x^2]]/Sqrt[Sqrt[a] - Sqrt[a + b]]])/(Sqrt[a]*Sqrt[Sqrt[a] - Sqrt[a + b]]) - (2*ArcTan[Sqrt[a*x + Sqrt[b + a
^2*x^2]]/Sqrt[Sqrt[a] + Sqrt[a + b]]])/(Sqrt[a]*Sqrt[Sqrt[a] + Sqrt[a + b]]) - (2*ArcTanh[Sqrt[a*x + Sqrt[b +
a^2*x^2]]/Sqrt[Sqrt[a] - Sqrt[a + b]]])/(Sqrt[a]*Sqrt[Sqrt[a] - Sqrt[a + b]]) - (2*ArcTanh[Sqrt[a*x + Sqrt[b +
a^2*x^2]]/Sqrt[Sqrt[a] + Sqrt[a + b]]])/(Sqrt[a]*Sqrt[Sqrt[a] + Sqrt[a + b]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 826

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

Rule 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 1166

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

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2117

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[((g + h*x^n)^p*(d^2 + a*f^2 - 2*d*x + x^2))/(d - x)^2, x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rule 2119

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+a x^2}{\left (-1+a x^2\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}} \, dx &=\int \left (\frac {1}{\sqrt {a x+\sqrt {b+a^2 x^2}}}+\frac {2}{\left (-1+a x^2\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}}\right ) \, dx\\ &=2 \int \frac {1}{\left (-1+a x^2\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}} \, dx+\int \frac {1}{\sqrt {a x+\sqrt {b+a^2 x^2}}} \, dx\\ &=2 \int \left (-\frac {1}{2 \left (1-\sqrt {a} x\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}}-\frac {1}{2 \left (1+\sqrt {a} x\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}}\right ) \, dx+\frac {\operatorname {Subst}\left (\int \frac {b+x^2}{x^{5/2}} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )}{2 a}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {b}{x^{5/2}}+\frac {1}{\sqrt {x}}\right ) \, dx,x,a x+\sqrt {b+a^2 x^2}\right )}{2 a}-\int \frac {1}{\left (1-\sqrt {a} x\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}} \, dx-\int \frac {1}{\left (1+\sqrt {a} x\right ) \sqrt {a x+\sqrt {b+a^2 x^2}}} \, dx\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-\operatorname {Subst}\left (\int \frac {b+x^2}{x^{3/2} \left (\sqrt {a} b+2 a x-\sqrt {a} x^2\right )} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )-\operatorname {Subst}\left (\int \frac {b+x^2}{x^{3/2} \left (-\sqrt {a} b+2 a x+\sqrt {a} x^2\right )} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-\operatorname {Subst}\left (\int \left (-\frac {1}{\sqrt {a} x^{3/2}}+\frac {2 \left (b+\sqrt {a} x\right )}{x^{3/2} \left (\sqrt {a} b+2 a x-\sqrt {a} x^2\right )}\right ) \, dx,x,a x+\sqrt {b+a^2 x^2}\right )-\operatorname {Subst}\left (\int \left (\frac {1}{\sqrt {a} x^{3/2}}+\frac {2 \left (b-\sqrt {a} x\right )}{x^{3/2} \left (-\sqrt {a} b+2 a x+\sqrt {a} x^2\right )}\right ) \, dx,x,a x+\sqrt {b+a^2 x^2}\right )\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-2 \operatorname {Subst}\left (\int \frac {b+\sqrt {a} x}{x^{3/2} \left (\sqrt {a} b+2 a x-\sqrt {a} x^2\right )} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )-2 \operatorname {Subst}\left (\int \frac {b-\sqrt {a} x}{x^{3/2} \left (-\sqrt {a} b+2 a x+\sqrt {a} x^2\right )} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-\frac {2 \operatorname {Subst}\left (\int \frac {-a b+\sqrt {a} b x}{\sqrt {x} \left (\sqrt {a} b+2 a x-\sqrt {a} x^2\right )} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )}{\sqrt {a} b}+\frac {2 \operatorname {Subst}\left (\int \frac {-a b-\sqrt {a} b x}{\sqrt {x} \left (-\sqrt {a} b+2 a x+\sqrt {a} x^2\right )} \, dx,x,a x+\sqrt {b+a^2 x^2}\right )}{\sqrt {a} b}\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-\frac {4 \operatorname {Subst}\left (\int \frac {-a b+\sqrt {a} b x^2}{\sqrt {a} b+2 a x^2-\sqrt {a} x^4} \, dx,x,\sqrt {a x+\sqrt {b+a^2 x^2}}\right )}{\sqrt {a} b}+\frac {4 \operatorname {Subst}\left (\int \frac {-a b-\sqrt {a} b x^2}{-\sqrt {a} b+2 a x^2+\sqrt {a} x^4} \, dx,x,\sqrt {a x+\sqrt {b+a^2 x^2}}\right )}{\sqrt {a} b}\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-2 \operatorname {Subst}\left (\int \frac {1}{a-\sqrt {a} \sqrt {a+b}-\sqrt {a} x^2} \, dx,x,\sqrt {a x+\sqrt {b+a^2 x^2}}\right )-2 \operatorname {Subst}\left (\int \frac {1}{a+\sqrt {a} \sqrt {a+b}-\sqrt {a} x^2} \, dx,x,\sqrt {a x+\sqrt {b+a^2 x^2}}\right )-2 \operatorname {Subst}\left (\int \frac {1}{a-\sqrt {a} \sqrt {a+b}+\sqrt {a} x^2} \, dx,x,\sqrt {a x+\sqrt {b+a^2 x^2}}\right )-2 \operatorname {Subst}\left (\int \frac {1}{a+\sqrt {a} \sqrt {a+b}+\sqrt {a} x^2} \, dx,x,\sqrt {a x+\sqrt {b+a^2 x^2}}\right )\\ &=-\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{\sqrt {\sqrt {a}-\sqrt {a+b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a}-\sqrt {a+b}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{\sqrt {\sqrt {a}+\sqrt {a+b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a}+\sqrt {a+b}}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{\sqrt {\sqrt {a}-\sqrt {a+b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a}-\sqrt {a+b}}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{\sqrt {\sqrt {a}+\sqrt {a+b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a}+\sqrt {a+b}}}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 166, normalized size = 0.99 \begin {gather*} \frac {-a \text {RootSum}\left [\text {$\#$1}^8 \left (-b^2\right )+4 \text {$\#$1}^4 a+2 \text {$\#$1}^4 b-1\&,\frac {\text {$\#$1}^4 b \log \left (\frac {1}{\sqrt {\sqrt {a^2 x^2+b}+a x}}-\text {$\#$1}\right )+\log \left (\frac {1}{\sqrt {\sqrt {a^2 x^2+b}+a x}}-\text {$\#$1}\right )}{\text {$\#$1}^5 b^2-2 \text {$\#$1} a-\text {$\#$1} b}\&\right ]-\frac {b}{3 \left (\sqrt {a^2 x^2+b}+a x\right )^{3/2}}+\sqrt {\sqrt {a^2 x^2+b}+a x}}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + a*x^2)/((-1 + a*x^2)*Sqrt[a*x + Sqrt[b + a^2*x^2]]),x]

[Out]

(-1/3*b/(a*x + Sqrt[b + a^2*x^2])^(3/2) + Sqrt[a*x + Sqrt[b + a^2*x^2]] - a*RootSum[-1 + 4*a*#1^4 + 2*b*#1^4 -
b^2*#1^8 & , (Log[1/Sqrt[a*x + Sqrt[b + a^2*x^2]] - #1] + b*Log[1/Sqrt[a*x + Sqrt[b + a^2*x^2]] - #1]*#1^4)/(
-2*a*#1 - b*#1 + b^2*#1^5) & ])/a

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IntegrateAlgebraic [A]  time = 0.25, size = 167, normalized size = 1.00 \begin {gather*} -\frac {b}{3 a \left (a x+\sqrt {b+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b+a^2 x^2}}}{a}+\text {RootSum}\left [b^2-4 a \text {$\#$1}^4-2 b \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-b \log \left (\sqrt {a x+\sqrt {b+a^2 x^2}}-\text {$\#$1}\right )-\log \left (\sqrt {a x+\sqrt {b+a^2 x^2}}-\text {$\#$1}\right ) \text {$\#$1}^4}{2 a \text {$\#$1}^3+b \text {$\#$1}^3-\text {$\#$1}^7}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + a*x^2)/((-1 + a*x^2)*Sqrt[a*x + Sqrt[b + a^2*x^2]]),x]

[Out]

-1/3*b/(a*(a*x + Sqrt[b + a^2*x^2])^(3/2)) + Sqrt[a*x + Sqrt[b + a^2*x^2]]/a + RootSum[b^2 - 4*a*#1^4 - 2*b*#1
^4 + #1^8 & , (-(b*Log[Sqrt[a*x + Sqrt[b + a^2*x^2]] - #1]) - Log[Sqrt[a*x + Sqrt[b + a^2*x^2]] - #1]*#1^4)/(2
*a*#1^3 + b*#1^3 - #1^7) & ]

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fricas [B]  time = 0.62, size = 1097, normalized size = 6.57

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(12*a*b*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a - b)/(a^2*b^2))^(1/4)*arctan((a^3*b^2*sqrt((a + b)/(a^3
*b^4)) + a^2)*sqrt(a*x + (2*a^3*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a^2 + a*b)*sqrt(-(2*a^2*b^2*sqrt((a + b)/(a^3*
b^4)) - 2*a - b)/(a^2*b^2)) + sqrt(a^2*x^2 + b))*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a - b)/(a^2*b^2))^(3
/4) - (a^3*b^2*sqrt((a + b)/(a^3*b^4)) + a^2)*sqrt(a*x + sqrt(a^2*x^2 + b))*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4
)) - 2*a - b)/(a^2*b^2))^(3/4)) - 12*a*b*((2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2))^(1/4)*arcta
n(((a^3*b^2*sqrt((a + b)/(a^3*b^4)) - a^2)*sqrt(a*x - (2*a^3*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a^2 - a*b)*sqrt((
2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2)) + sqrt(a^2*x^2 + b))*sqrt((2*a^2*b^2*sqrt((a + b)/(a^3
*b^4)) + 2*a + b)/(a^2*b^2)) - (a^3*b^2*sqrt((a + b)/(a^3*b^4)) - a^2)*sqrt(a*x + sqrt(a^2*x^2 + b))*sqrt((2*a
^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2)))*((2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2)
)^(1/4)) + 3*a*b*((2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2))^(1/4)*log(8*(a^2*b^2*sqrt((a + b)/(
a^3*b^4)) - a)*((2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2))^(1/4) + 8*sqrt(a*x + sqrt(a^2*x^2 + b
))) - 3*a*b*((2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2))^(1/4)*log(-8*(a^2*b^2*sqrt((a + b)/(a^3*
b^4)) - a)*((2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) + 2*a + b)/(a^2*b^2))^(1/4) + 8*sqrt(a*x + sqrt(a^2*x^2 + b)))
- 3*a*b*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a - b)/(a^2*b^2))^(1/4)*log(8*(a^2*b^2*sqrt((a + b)/(a^3*b^4)
) + a)*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a - b)/(a^2*b^2))^(1/4) + 8*sqrt(a*x + sqrt(a^2*x^2 + b))) + 3
*a*b*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a - b)/(a^2*b^2))^(1/4)*log(-8*(a^2*b^2*sqrt((a + b)/(a^3*b^4))
+ a)*(-(2*a^2*b^2*sqrt((a + b)/(a^3*b^4)) - 2*a - b)/(a^2*b^2))^(1/4) + 8*sqrt(a*x + sqrt(a^2*x^2 + b))) - 2*(
a^2*x^2 - sqrt(a^2*x^2 + b)*a*x - b)*sqrt(a*x + sqrt(a^2*x^2 + b)))/(a*b)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + 1}{{\left (a x^{2} - 1\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^2 + 1)/((a*x^2 - 1)*sqrt(a*x + sqrt(a^2*x^2 + b))), x)

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {a \,x^{2}+1}{\left (a \,x^{2}-1\right ) \sqrt {a x +\sqrt {a^{2} x^{2}+b}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2+1)/(a*x^2-1)/(a*x+(a^2*x^2+b)^(1/2))^(1/2),x)

[Out]

int((a*x^2+1)/(a*x^2-1)/(a*x+(a^2*x^2+b)^(1/2))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + 1}{{\left (a x^{2} - 1\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^2 + 1)/((a*x^2 - 1)*sqrt(a*x + sqrt(a^2*x^2 + b))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a\,x^2+1}{\sqrt {\sqrt {a^2\,x^2+b}+a\,x}\,\left (a\,x^2-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + 1)/(((b + a^2*x^2)^(1/2) + a*x)^(1/2)*(a*x^2 - 1)),x)

[Out]

int((a*x^2 + 1)/(((b + a^2*x^2)^(1/2) + a*x)^(1/2)*(a*x^2 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x^{2} + 1}{\sqrt {a x + \sqrt {a^{2} x^{2} + b}} \left (a x^{2} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**2+1)/(a*x**2-1)/(a*x+(a**2*x**2+b)**(1/2))**(1/2),x)

[Out]

Integral((a*x**2 + 1)/(sqrt(a*x + sqrt(a**2*x**2 + b))*(a*x**2 - 1)), x)

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