3.1303 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)} x^3}{(c-a^2 c x^2)^{9/8}} \, dx\)

Optimal. Leaf size=200 \[ \frac{64 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \text{Hypergeometric2F1}\left (\frac{5}{8},\frac{7}{8},\frac{13}{8},\frac{1}{2} (1-a x)\right )}{105 a^4 c \sqrt [8]{c-a^2 c x^2}}-\frac{4 x^2 \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{7 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac{8 (6-a x) \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{21 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

[Out]

(-4*x^2*(1 + a*x)^(1/8)*(1 - a^2*x^2)^(1/8))/(7*a^2*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8)) + (8*(6 - a*x)*(1
+ a*x)^(1/8)*(1 - a^2*x^2)^(1/8))/(21*a^4*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8)) + (64*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])/(105*a^4*c*(c - a^2*c*x^2)^(1/8))

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Rubi [A]  time = 0.25395, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {6153, 6150, 100, 146, 69} \[ \frac{64 \sqrt [8]{2} (1-a x)^{5/8} \sqrt [8]{1-a^2 x^2} \, _2F_1\left (\frac{5}{8},\frac{7}{8};\frac{13}{8};\frac{1}{2} (1-a x)\right )}{105 a^4 c \sqrt [8]{c-a^2 c x^2}}-\frac{4 x^2 \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{7 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac{8 (6-a x) \sqrt [8]{a x+1} \sqrt [8]{1-a^2 x^2}}{21 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^(ArcTanh[a*x]/2)*x^3)/(c - a^2*c*x^2)^(9/8),x]

[Out]

(-4*x^2*(1 + a*x)^(1/8)*(1 - a^2*x^2)^(1/8))/(7*a^2*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8)) + (8*(6 - a*x)*(1
+ a*x)^(1/8)*(1 - a^2*x^2)^(1/8))/(21*a^4*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8)) + (64*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])/(105*a^4*c*(c - a^2*c*x^2)^(1/8))

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 100

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

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

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]))

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)} x^3}{\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^3}{\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^3}{(1-a x)^{11/8} (1+a x)^{7/8}} \, dx}{c \sqrt [8]{c-a^2 c x^2}}\\ &=-\frac{4 x^2 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{7 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}-\frac{\left (4 \sqrt [8]{1-a^2 x^2}\right ) \int \frac{x \left (-2-\frac{a x}{2}\right )}{(1-a x)^{11/8} (1+a x)^{7/8}} \, dx}{7 a^2 c \sqrt [8]{c-a^2 c x^2}}\\ &=-\frac{4 x^2 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{7 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac{8 (6-a x) \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{21 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}-\frac{\left (16 \sqrt [8]{1-a^2 x^2}\right ) \int \frac{1}{(1-a x)^{3/8} (1+a x)^{7/8}} \, dx}{21 a^3 c \sqrt [8]{c-a^2 c x^2}}\\ &=-\frac{4 x^2 \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{7 a^2 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac{8 (6-a x) \sqrt [8]{1+a x} \sqrt [8]{1-a^2 x^2}}{21 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}}+\frac{64 \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 )}{105 a^4 c \sqrt [8]{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0694418, size = 107, normalized size = 0.54 \[ -\frac{4 \sqrt [8]{1-a^2 x^2} \left (16 \sqrt [8]{2} (a x-1) \text{Hypergeometric2F1}\left (\frac{5}{8},\frac{7}{8},\frac{13}{8},\frac{1}{2} (1-a x)\right )+5 \sqrt [8]{a x+1} \left (3 a^2 x^2+2 a x-12\right )\right )}{105 a^4 c (1-a x)^{3/8} \sqrt [8]{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(ArcTanh[a*x]/2)*x^3)/(c - a^2*c*x^2)^(9/8),x]

[Out]

(-4*(1 - a^2*x^2)^(1/8)*(5*(1 + a*x)^(1/8)*(-12 + 2*a*x + 3*a^2*x^2) + 16*2^(1/8)*(-1 + a*x)*Hypergeometric2F1
[5/8, 7/8, 13/8, (1 - a*x)/2]))/(105*a^4*c*(1 - a*x)^(3/8)*(c - a^2*c*x^2)^(1/8))

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Maple [F]  time = 0.215, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3}\sqrt{{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{-{\frac{9}{8}}}}\, 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^3/(-a^2*c*x^2+c)^(9/8),x)

[Out]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)*x^3/(-a^2*c*x^2+c)^(9/8),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{9}{8}}}\,{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^3/(-a^2*c*x^2+c)^(9/8),x, algorithm="maxima")

[Out]

integrate(x^3*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(9/8), 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^3/(-a^2*c*x^2+c)^(9/8),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**3/(-a**2*c*x**2+c)**(9/8),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} c x^{2} + c\right )}^{\frac{9}{8}}}\,{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^3/(-a^2*c*x^2+c)^(9/8),x, algorithm="giac")

[Out]

integrate(x^3*sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*c*x^2 + c)^(9/8), x)