3.13.100 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/4}} \, dx\) [1300]

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]

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

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Rubi [A]
time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 = {6278, 6275, 53, 65, 212} \begin {gather*} \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 {gather*}

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 53

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 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 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 6275

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 6278

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 \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} \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 64, normalized size = 0.60 \begin {gather*} \frac {\sqrt [4]{1-a^2 x^2} \, _2F_1\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}} \end {gather*}

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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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)/(-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)

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{4}}}\, dx \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)/(-a**2*c*x**2+c)**(5/4),x)

[Out]

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

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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)/(-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)

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

[Out]

int(((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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