3.1291 \(\int e^{\frac {1}{2} \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2} \, dx\)

Optimal. Leaf size=429 \[ -\frac {(a x+1)^{3/4} (1-a x)^{5/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {3 (a x+1)^{3/4} \sqrt [4]{1-a x} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}} \]

[Out]

3/4*(-a*x+1)^(1/4)*(a*x+1)^(3/4)*(-a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-1/2*(-a*x+1)^(5/4)*(a*x+1)^(3/4)*(-
a^2*c*x^2+c)^(1/2)/a/(-a^2*x^2+1)^(1/2)-3/8*arctan(-1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))*(-a^2*c*x^2+c)^(1/
2)/a*2^(1/2)/(-a^2*x^2+1)^(1/2)-3/8*arctan(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4))*(-a^2*c*x^2+c)^(1/2)/a*2^(1
/2)/(-a^2*x^2+1)^(1/2)+3/16*ln(1-(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^(1/2))*(-a^2*c*x^
2+c)^(1/2)/a*2^(1/2)/(-a^2*x^2+1)^(1/2)-3/16*ln(1+(-a*x+1)^(1/4)*2^(1/2)/(a*x+1)^(1/4)+(-a*x+1)^(1/2)/(a*x+1)^
(1/2))*(-a^2*c*x^2+c)^(1/2)/a*2^(1/2)/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {6143, 6140, 50, 63, 240, 211, 1165, 628, 1162, 617, 204} \[ -\frac {(a x+1)^{3/4} (1-a x)^{5/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {3 (a x+1)^{3/4} \sqrt [4]{1-a x} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \log \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{a x+1}}+1\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(ArcTanh[a*x]/2)*Sqrt[c - a^2*c*x^2],x]

[Out]

(3*(1 - a*x)^(1/4)*(1 + a*x)^(3/4)*Sqrt[c - a^2*c*x^2])/(4*a*Sqrt[1 - a^2*x^2]) - ((1 - a*x)^(5/4)*(1 + a*x)^(
3/4)*Sqrt[c - a^2*c*x^2])/(2*a*Sqrt[1 - a^2*x^2]) + (3*Sqrt[c - a^2*c*x^2]*ArcTan[1 - (Sqrt[2]*(1 - a*x)^(1/4)
)/(1 + a*x)^(1/4)])/(4*Sqrt[2]*a*Sqrt[1 - a^2*x^2]) - (3*Sqrt[c - a^2*c*x^2]*ArcTan[1 + (Sqrt[2]*(1 - a*x)^(1/
4))/(1 + a*x)^(1/4)])/(4*Sqrt[2]*a*Sqrt[1 - a^2*x^2]) + (3*Sqrt[c - a^2*c*x^2]*Log[1 + Sqrt[1 - a*x]/Sqrt[1 +
a*x] - (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a*Sqrt[1 - a^2*x^2]) - (3*Sqrt[c - a^2*c*x^2]*Lo
g[1 + Sqrt[1 - a*x]/Sqrt[1 + a*x] + (Sqrt[2]*(1 - a*x)^(1/4))/(1 + a*x)^(1/4)])/(8*Sqrt[2]*a*Sqrt[1 - a^2*x^2]
)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

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 6140

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a*x)^(p - n/2)*
(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])

Rule 6143

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d*x^2)^Frac
Part[p])/(1 - a^2*x^2)^FracPart[p], Int[(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x
] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0])

Rubi steps

\begin {align*} \int e^{\frac {1}{2} \tanh ^{-1}(a x)} \sqrt {c-a^2 c x^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int e^{\frac {1}{2} \tanh ^{-1}(a x)} \sqrt {1-a^2 x^2} \, dx}{\sqrt {1-a^2 x^2}}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \sqrt [4]{1-a x} (1+a x)^{3/4} \, dx}{\sqrt {1-a^2 x^2}}\\ &=-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \int \frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}} \, dx}{4 \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \int \frac {1}{(1-a x)^{3/4} \sqrt [4]{1+a x}} \, dx}{8 \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-a x}\right )}{2 a \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{2 a \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 a \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 a \sqrt {1-a^2 x^2}}+\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}+\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {\sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}}+\frac {\left (3 \sqrt {c-a^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}}\\ &=\frac {3 \sqrt [4]{1-a x} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-a^2 x^2}}-\frac {(1-a x)^{5/4} (1+a x)^{3/4} \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{4 \sqrt {2} a \sqrt {1-a^2 x^2}}+\frac {3 \sqrt {c-a^2 c x^2} \log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}-\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}-\frac {3 \sqrt {c-a^2 c x^2} \log \left (1+\frac {\sqrt {1-a x}}{\sqrt {1+a x}}+\frac {\sqrt {2} \sqrt [4]{1-a x}}{\sqrt [4]{1+a x}}\right )}{8 \sqrt {2} a \sqrt {1-a^2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 71, normalized size = 0.17 \[ -\frac {4\ 2^{3/4} (1-a x)^{5/4} \sqrt {c-a^2 c x^2} \, _2F_1\left (-\frac {3}{4},\frac {5}{4};\frac {9}{4};\frac {1}{2} (1-a x)\right )}{5 a \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(ArcTanh[a*x]/2)*Sqrt[c - a^2*c*x^2],x]

[Out]

(-4*2^(3/4)*(1 - a*x)^(5/4)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[-3/4, 5/4, 9/4, (1 - a*x)/2])/(5*a*Sqrt[1 -
a^2*x^2])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} c x^{2} + c} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, \sqrt {-a^{2} c \,x^{2}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a^{2} c x^{2} + c} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {c-a^2\,c\,x^2}\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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