3.1300 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/4}} \, dx\)

Optimal. Leaf size=106 \[ \frac{\sqrt [4]{1-a^2 x^2}}{a c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}-\frac{\sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} a c \sqrt [4]{c-a^2 c x^2}} \]

[Out]

(1 - a^2*x^2)^(1/4)/(a*c*Sqrt[1 - a*x]*(c - a^2*c*x^2)^(1/4)) - ((1 - a^2*x^2)^(1/4)*ArcTanh[Sqrt[1 - a*x]/Sqr
t[2]])/(Sqrt[2]*a*c*(c - a^2*c*x^2)^(1/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0978393, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {6143, 6140, 51, 63, 206} \[ \frac{\sqrt [4]{1-a^2 x^2}}{a c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}-\frac{\sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} a c \sqrt [4]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(ArcTanh[a*x]/2)/(c - a^2*c*x^2)^(5/4),x]

[Out]

(1 - a^2*x^2)^(1/4)/(a*c*Sqrt[1 - a*x]*(c - a^2*c*x^2)^(1/4)) - ((1 - a^2*x^2)^(1/4)*ArcTanh[Sqrt[1 - a*x]/Sqr
t[2]])/(Sqrt[2]*a*c*(c - a^2*c*x^2)^(1/4))

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])

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0166965, size = 64, normalized size = 0.6 \[ \frac{\sqrt [4]{1-a^2 x^2} \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{1}{2} (1-a x)\right )}{a c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(ArcTanh[a*x]/2)/(c - a^2*c*x^2)^(5/4),x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.213, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

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)^(5/4),x)

[Out]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*c*x^2+c)^(5/4),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

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)^(5/4),x, algorithm="maxima")

[Out]

integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(5/4), x)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)^(5/4),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**(5/4),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

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)^(5/4),x, algorithm="giac")

[Out]

integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(5/4), x)