3.1305 \(\int \frac {e^{\frac {1}{2} \tanh ^{-1}(a x)} x}{(c-a^2 c x^2)^{9/8}} \, dx\)

Optimal. Leaf size=133 \[ \frac {8 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \, _2F_1\left (\frac {5}{8},\frac {7}{8};\frac {13}{8};\frac {1}{2} (1-a x)\right )}{15 a^2 c \sqrt [8]{c-a^2 c x^2}}+\frac {4 \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{3 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

[Out]

4/3*(a*x+1)^(1/8)*(-a^2*x^2+1)^(1/8)/a^2/c/(-a*x+1)^(3/8)/(-a^2*c*x^2+c)^(1/8)+8/15*2^(1/8)*(-a*x+1)^(5/8)*(-a
^2*x^2+1)^(1/8)*hypergeom([5/8, 7/8],[13/8],-1/2*a*x+1/2)/a^2/c/(-a^2*c*x^2+c)^(1/8)

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Rubi [A]  time = 0.17, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6153, 6150, 78, 69} \[ \frac {8 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \, _2F_1\left (\frac {5}{8},\frac {7}{8};\frac {13}{8};\frac {1}{2} (1-a x)\right )}{15 a^2 c \sqrt [8]{c-a^2 c x^2}}+\frac {4 \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{3 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(1 + a*x)^(1/8)*(1 - a^2*x^2)^(1/8))/(3*a^2*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8)) + (8*2^(1/8)*(1 - a*x)
^(5/8)*(1 - a^2*x^2)^(1/8)*Hypergeometric2F1[5/8, 7/8, 13/8, (1 - a*x)/2])/(15*a^2*c*(c - a^2*c*x^2)^(1/8))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 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 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 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]

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 {x}{(1-a x)^{11/8} (1+a x)^{7/8}} \, dx}{c \sqrt [8]{c-a^2 c x^2}}\\ &=\frac {4 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{3 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}-\frac {\left (2 \sqrt [8]{1-a^2 x^2}\right ) \int \frac {1}{(1-a x)^{3/8} (1+a x)^{7/8}} \, dx}{3 a c \sqrt [8]{c-a^2 c x^2}}\\ &=\frac {4 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{3 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac {8 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \, _2F_1\left (\frac {5}{8},\frac {7}{8};\frac {13}{8};\frac {1}{2} (1-a x)\right )}{15 a^2 c \sqrt [8]{c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 93, normalized size = 0.70 \[ -\frac {4 \sqrt [8]{1-a^2 x^2} \left (2 \sqrt [8]{2} (a x-1) \, _2F_1\left (\frac {5}{8},\frac {7}{8};\frac {13}{8};\frac {1}{2} (1-a x)\right )-5 \sqrt [8]{a x+1}\right )}{15 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 - a^2*x^2)^(1/8)*(-5*(1 + a*x)^(1/8) + 2*2^(1/8)*(-1 + a*x)*Hypergeometric2F1[5/8, 7/8, 13/8, (1 - a*x)
/2]))/(15*a^2*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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

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maple [F]  time = 0.28, 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 \[ \int \frac {x \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {9}{8}}}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}}{{\left (c-a^2\,c\,x^2\right )}^{9/8}} \,d x \]

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)^(9/8),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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