3.1301 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^{5/4}} \, dx\)

Optimal. Leaf size=144 \[ \frac{\sqrt [4]{1-a^2 x^2}}{c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}-\frac{2 \sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a x}\right )}{c \sqrt [4]{c-a^2 c x^2}}+\frac{\sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} c \sqrt [4]{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.235062, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {6153, 6150, 85, 156, 63, 208, 206} \[ \frac{\sqrt [4]{1-a^2 x^2}}{c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}-\frac{2 \sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a x}\right )}{c \sqrt [4]{c-a^2 c x^2}}+\frac{\sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} c \sqrt [4]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

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

Rule 6150

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

Rule 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 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 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)}}{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)}}{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}{x (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}}{c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac{\sqrt [4]{1-a^2 x^2} \int \frac{2 a+a^2 x}{x \sqrt{1-a x} (1+a x)} \, dx}{2 a c \sqrt [4]{c-a^2 c x^2}}\\ &=\frac{\sqrt [4]{1-a^2 x^2}}{c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac{\sqrt [4]{1-a^2 x^2} \int \frac{1}{x \sqrt{1-a x}} \, dx}{c \sqrt [4]{c-a^2 c x^2}}-\frac{\left (a \sqrt [4]{1-a^2 x^2}\right ) \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}}{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 )}{c \sqrt [4]{c-a^2 c x^2}}-\frac{\left (2 \sqrt [4]{1-a^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a}-\frac{x^2}{a}} \, 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}}{c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}-\frac{2 \sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a x}\right )}{c \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} c \sqrt [4]{c-a^2 c x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0316578, size = 79, normalized size = 0.55 \[ -\frac{\sqrt [4]{1-a^2 x^2} \left (\text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{1}{2} (1-a x)\right )-2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},1-a x\right )\right )}{c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

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Maple [F]  time = 0.211, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x}\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)/x/(-a^2*c*x^2+c)^(5/4),x)

[Out]

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

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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}} x}\,{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)/x/(-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), x)

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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)/x/(-a^2*c*x^2+c)^(5/4),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

Timed out

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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}} x}\,{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)/x/(-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), x)