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

Optimal. Leaf size=122 \[ \frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{4 \sqrt{2} a c^{3/2}}-\frac{3 \sqrt{1-a^2 x^2}}{4 a (c-a c x)^{3/2}} \]

[Out]

(-3*Sqrt[1 - a^2*x^2])/(4*a*(c - a*c*x)^(3/2)) + (c^2*(1 - a^2*x^2)^(3/2))/(2*a*(c - a*c*x)^(7/2)) + (3*ArcTan
h[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])])/(4*Sqrt[2]*a*c^(3/2))

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Rubi [A]  time = 0.104767, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6127, 663, 661, 208} \[ \frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{4 \sqrt{2} a c^{3/2}}-\frac{3 \sqrt{1-a^2 x^2}}{4 a (c-a c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[1 - a^2*x^2])/(4*a*(c - a*c*x)^(3/2)) + (c^2*(1 - a^2*x^2)^(3/2))/(2*a*(c - a*c*x)^(7/2)) + (3*ArcTan
h[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])])/(4*Sqrt[2]*a*c^(3/2))

Rule 6127

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^n, Int[(c + d*x)^(p - n)*(1 -
a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]

Rule 663

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rule 661

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{e^{3 \tanh ^{-1}(a x)}}{(c-a c x)^{3/2}} \, dx &=c^3 \int \frac{\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{9/2}} \, dx\\ &=\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}-\frac{1}{4} (3 c) \int \frac{\sqrt{1-a^2 x^2}}{(c-a c x)^{5/2}} \, dx\\ &=-\frac{3 \sqrt{1-a^2 x^2}}{4 a (c-a c x)^{3/2}}+\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}+\frac{3 \int \frac{1}{\sqrt{c-a c x} \sqrt{1-a^2 x^2}} \, dx}{8 c}\\ &=-\frac{3 \sqrt{1-a^2 x^2}}{4 a (c-a c x)^{3/2}}+\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}-\frac{1}{4} (3 a) \operatorname{Subst}\left (\int \frac{1}{-2 a^2 c+a^2 c^2 x^2} \, dx,x,\frac{\sqrt{1-a^2 x^2}}{\sqrt{c-a c x}}\right )\\ &=-\frac{3 \sqrt{1-a^2 x^2}}{4 a (c-a c x)^{3/2}}+\frac{c^2 \left (1-a^2 x^2\right )^{3/2}}{2 a (c-a c x)^{7/2}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{4 \sqrt{2} a c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.075281, size = 91, normalized size = 0.75 \[ \frac{\sqrt{c-a c x} \left (10 a^2 x^2+8 a x+3 (a x-1)^2 \sqrt{2 a x+2} \tanh ^{-1}\left (\frac{\sqrt{a x+1}}{\sqrt{2}}\right )-2\right )}{8 a c^2 (a x-1)^2 \sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c - a*c*x]*(-2 + 8*a*x + 10*a^2*x^2 + 3*(-1 + a*x)^2*Sqrt[2 + 2*a*x]*ArcTanh[Sqrt[1 + a*x]/Sqrt[2]]))/(8
*a*c^2*(-1 + a*x)^2*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.109, size = 158, normalized size = 1.3 \begin{align*} -{\frac{1}{8\, \left ( ax-1 \right ) ^{3}a}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ( ax-1 \right ) } \left ( 3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ){x}^{2}{a}^{2}c-6\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ) xac+10\,xa\sqrt{c \left ( ax+1 \right ) }\sqrt{c}+3\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ) c-2\,\sqrt{c \left ( ax+1 \right ) }\sqrt{c} \right ){c}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{c \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(5/2)*(3*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x
^2*a^2*c-6*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x*a*c+10*x*a*(c*(a*x+1))^(1/2)*c^(1/2)+3*2^(
1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*c-2*(c*(a*x+1))^(1/2)*c^(1/2))/(a*x-1)^3/(c*(a*x+1))^(1/2)
/a

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}{\left (-a c x + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.34228, size = 703, normalized size = 5.76 \begin{align*} \left [\frac{3 \, \sqrt{2}{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt{c} \log \left (-\frac{a^{2} c x^{2} + 2 \, a c x - 2 \, \sqrt{2} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{c} - 3 \, c}{a^{2} x^{2} - 2 \, a x + 1}\right ) - 4 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (5 \, a x - 1\right )}}{16 \,{\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}}, \frac{3 \, \sqrt{2}{\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{2} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c} \sqrt{-c}}{a^{2} c x^{2} - c}\right ) - 2 \, \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}{\left (5 \, a x - 1\right )}}{8 \,{\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt(c)*log(-(a^2*c*x^2 + 2*a*c*x - 2*sqrt(2)*sqrt(-a^2*x^2
+ 1)*sqrt(-a*c*x + c)*sqrt(c) - 3*c)/(a^2*x^2 - 2*a*x + 1)) - 4*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*(5*a*x -
1))/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2), 1/8*(3*sqrt(2)*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*sqrt
(-c)*arctan(sqrt(2)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)*sqrt(-c)/(a^2*c*x^2 - c)) - 2*sqrt(-a^2*x^2 + 1)*sqrt(
-a*c*x + c)*(5*a*x - 1))/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a x + 1\right )^{3}}{\left (- c \left (a x - 1\right )\right )^{\frac{3}{2}} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)/(-a*c*x+c)**(3/2),x)

[Out]

Integral((a*x + 1)**3/((-c*(a*x - 1))**(3/2)*(-(a*x - 1)*(a*x + 1))**(3/2)), x)

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Giac [A]  time = 1.37443, size = 99, normalized size = 0.81 \begin{align*} -\frac{\frac{3 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a c x + c}}{2 \, \sqrt{-c}}\right )}{\sqrt{-c}} - \frac{2 \,{\left (5 \,{\left (a c x + c\right )}^{\frac{3}{2}} - 6 \, \sqrt{a c x + c} c\right )}}{{\left (a c x - c\right )}^{2}}}{8 \, a{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(3/2),x, algorithm="giac")

[Out]

-1/8*(3*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/sqrt(-c) - 2*(5*(a*c*x + c)^(3/2) - 6*sqrt(a*c*x
+ c)*c)/(a*c*x - c)^2)/(a*abs(c))