3.14.7 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^{9/8}} \, dx\) [1307]

Optimal. Leaf size=73 \[ -\frac {2\ 2^{5/8} \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2} F_1\left (\frac {1}{8};\frac {11}{8},1;\frac {9}{8};\frac {1}{2} (1+a x),1+a x\right )}{c \sqrt [8]{c-a^2 c x^2}} \]

[Out]

-2*2^(5/8)*(a*x+1)^(1/8)*(-a^2*x^2+1)^(1/8)*AppellF1(1/8,11/8,1,9/8,1/2*a*x+1/2,a*x+1)/c/(-a^2*c*x^2+c)^(1/8)

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Rubi [A]
time = 0.16, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {6288, 6285, 141} \begin {gather*} -\frac {2\ 2^{5/8} \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2} F_1\left (\frac {1}{8};\frac {11}{8},1;\frac {9}{8};\frac {1}{2} (a x+1),a x+1\right )}{c \sqrt [8]{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)^(9/8)),x]

[Out]

(-2*2^(5/8)*(1 + a*x)^(1/8)*(1 - a^2*x^2)^(1/8)*AppellF1[1/8, 11/8, 1, 9/8, (1 + a*x)/2, 1 + a*x])/(c*(c - a^2
*c*x^2)^(1/8))

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

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 )^{9/8}} \, dx &=\frac {\sqrt [8]{1-a^2 x^2} \int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)}}{x \left (1-a^2 x^2\right )^{9/8}} \, dx}{c \sqrt [8]{c-a^2 c x^2}}\\ &=\frac {\sqrt [8]{1-a^2 x^2} \int \frac {1}{x (1-a x)^{11/8} (1+a x)^{7/8}} \, dx}{c \sqrt [8]{c-a^2 c x^2}}\\ &=-\frac {2\ 2^{5/8} \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2} F_1\left (\frac {1}{8};\frac {11}{8},1;\frac {9}{8};\frac {1}{2} (1+a x),1+a x\right )}{c \sqrt [8]{c-a^2 c x^2}}\\ \end {align*}

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Mathematica [F]
time = 10.40, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{9/8}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[E^(ArcTanh[a*x]/2)/(x*(c - a^2*c*x^2)^(9/8)),x]

[Out]

Integrate[E^(ArcTanh[a*x]/2)/(x*(c - a^2*c*x^2)^(9/8)), x]

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}}{x \left (-a^{2} c \,x^{2}+c \right )^{\frac {9}{8}}}\, dx\]

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)^(9/8),x)

[Out]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/x/(-a^2*c*x^2+c)^(9/8),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)^(9/8),x, algorithm="maxima")

[Out]

integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/((-a^2*c*x^2 + c)^(9/8)*x), 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)^(9/8),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)**(9/8),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)^(9/8),x, algorithm="giac")

[Out]

integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/((-a^2*c*x^2 + c)^(9/8)*x), 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}}}}{x\,{\left (c-a^2\,c\,x^2\right )}^{9/8}} \,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)/(x*(c - a^2*c*x^2)^(9/8)),x)

[Out]

int(((a*x + 1)/(1 - a^2*x^2)^(1/2))^(1/2)/(x*(c - a^2*c*x^2)^(9/8)), x)

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