3.1299 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^{5/4}} \, dx\)

Optimal. Leaf size=105 \[ \frac{\sqrt [4]{1-a^2 x^2}}{a^2 c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac{\sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} a^2 c \sqrt [4]{c-a^2 c x^2}} \]

[Out]

(1 - a^2*x^2)^(1/4)/(a^2*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]/S
qrt[2]])/(Sqrt[2]*a^2*c*(c - a^2*c*x^2)^(1/4))

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Rubi [A]  time = 0.162256, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6153, 6150, 78, 63, 206} \[ \frac{\sqrt [4]{1-a^2 x^2}}{a^2 c \sqrt{1-a x} \sqrt [4]{c-a^2 c x^2}}+\frac{\sqrt [4]{1-a^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} a^2 c \sqrt [4]{c-a^2 c x^2}} \]

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)/(a^2*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]/S
qrt[2]])/(Sqrt[2]*a^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 78

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

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)} 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{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}}{a^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}{\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}}{a^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 )}{a^2 c \sqrt [4]{c-a^2 c x^2}}\\ &=\frac{\sqrt [4]{1-a^2 x^2}}{a^2 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^2 c \sqrt [4]{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0531292, size = 74, normalized size = 0.7 \[ \frac{\sqrt [4]{1-a^2 x^2} \left (\frac{1}{a^2 \sqrt{1-a x}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{1-a x}}{\sqrt{2}}\right )}{\sqrt{2} a^2}\right )}{c \sqrt [4]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

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

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Maple [F]  time = 0.214, size = 0, normalized size = 0. \begin{align*} \int{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{x \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)*x/(-a^2*c*x^2+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(x*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(5/4), 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{x \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)*x/(-a^2*c*x^2+c)^(5/4),x, algorithm="giac")

[Out]

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