3.13.99 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^{5/4}} \, dx\) [1299]

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]

1/2*(-a^2*x^2+1)^(1/4)*arctanh(1/2*(-a*x+1)^(1/2)*2^(1/2))/a^2/c/(-a^2*c*x^2+c)^(1/4)*2^(1/2)+(-a^2*x^2+1)^(1/
4)/a^2/c/(-a^2*c*x^2+c)^(1/4)/(-a*x+1)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 105, normalized size of antiderivative = 1.00, 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 = {6288, 6285, 79, 65, 212} \begin {gather*} \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 {gather*}

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

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

Rule 6285

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 6288

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]

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} \text {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*}

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Mathematica [A]
time = 0.04, size = 74, normalized size = 0.70 \begin {gather*} \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}} \end {gather*}

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 = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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