3.233 \(\int \frac{e^{\tanh ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{8 \sqrt{2} a c^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}+\frac{\sqrt{1-a^2 x^2}}{2 a (c-a c x)^{5/2}} \]

[Out]

Sqrt[1 - a^2*x^2]/(2*a*(c - a*c*x)^(5/2)) - Sqrt[1 - a^2*x^2]/(8*a*c*(c - a*c*x)^(3/2)) - ArcTanh[(Sqrt[c]*Sqr
t[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])]/(8*Sqrt[2]*a*c^(5/2))

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Rubi [A]  time = 0.106377, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {6127, 663, 673, 661, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{8 \sqrt{2} a c^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}+\frac{\sqrt{1-a^2 x^2}}{2 a (c-a c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(c - a*c*x)^(5/2),x]

[Out]

Sqrt[1 - a^2*x^2]/(2*a*(c - a*c*x)^(5/2)) - Sqrt[1 - a^2*x^2]/(8*a*c*(c - a*c*x)^(3/2)) - ArcTanh[(Sqrt[c]*Sqr
t[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])]/(8*Sqrt[2]*a*c^(5/2))

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 673

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[(m + 2*p + 2)/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[m, 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^{\tanh ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx &=c \int \frac{\sqrt{1-a^2 x^2}}{(c-a c x)^{7/2}} \, dx\\ &=\frac{\sqrt{1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac{\int \frac{1}{(c-a c x)^{3/2} \sqrt{1-a^2 x^2}} \, dx}{4 c}\\ &=\frac{\sqrt{1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}-\frac{\int \frac{1}{\sqrt{c-a c x} \sqrt{1-a^2 x^2}} \, dx}{16 c^2}\\ &=\frac{\sqrt{1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}+\frac{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 )}{8 c}\\ &=\frac{\sqrt{1-a^2 x^2}}{2 a (c-a c x)^{5/2}}-\frac{\sqrt{1-a^2 x^2}}{8 a c (c-a c x)^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{8 \sqrt{2} a c^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0184426, size = 57, normalized size = 0.47 \[ \frac{(a x+1)^{3/2} \sqrt{c-a c x} \text{Hypergeometric2F1}\left (\frac{3}{2},3,\frac{5}{2},\frac{1}{2} (a x+1)\right )}{12 a c^3 \sqrt{1-a x}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a*x]/(c - a*c*x)^(5/2),x]

[Out]

((1 + a*x)^(3/2)*Sqrt[c - a*c*x]*Hypergeometric2F1[3/2, 3, 5/2, (1 + a*x)/2])/(12*a*c^3*Sqrt[1 - a*x])

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Maple [A]  time = 0.102, size = 156, normalized size = 1.3 \begin{align*}{\frac{1}{16\, \left ( ax-1 \right ) ^{3}a}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ( ax-1 \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( ax+1 \right ) }{\frac{1}{\sqrt{c}}}} \right ){x}^{2}{a}^{2}c-2\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ) xac-2\,xa\sqrt{c \left ( ax+1 \right ) }\sqrt{c}+\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c \left ( ax+1 \right ) }{\frac{1}{\sqrt{c}}}} \right ) c-6\,\sqrt{c \left ( ax+1 \right ) }\sqrt{c} \right ){c}^{-{\frac{7}{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)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x)

[Out]

1/16*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(7/2)*(2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x^2
*a^2*c-2*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x*a*c-2*x*a*(c*(a*x+1))^(1/2)*c^(1/2)+2^(1/2)*
arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*c-6*(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{a x + 1}{\sqrt{-a^{2} x^{2} + 1}{\left (-a c x + c\right )}^{\frac{5}{2}}}\,{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)/(-a*c*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(-a*c*x + c)^(5/2)), x)

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Fricas [A]  time = 2.02664, size = 695, normalized size = 5.7 \begin{align*} \left [\frac{\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 (a x + 3\right )}}{32 \,{\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}}, -\frac{\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 (a x + 3\right )}}{16 \,{\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/32*(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)*(a*x + 3))/
(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3), -1/16*(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)*(a*x + 3))/(a^4*c^3*x^3 - 3*a^3*c^3*x^2 + 3*a^2*c^3*x - a*c^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**(5/2),x)

[Out]

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

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Giac [A]  time = 1.30308, size = 103, normalized size = 0.84 \begin{align*} \frac{\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a c x + c}}{2 \, \sqrt{-c}}\right )}{\sqrt{-c} c} + \frac{2 \,{\left ({\left (a c x + c\right )}^{\frac{3}{2}} + 2 \, \sqrt{a c x + c} c\right )}}{{\left (a c x - c\right )}^{2} c}}{16 \, a{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^(5/2),x, algorithm="giac")

[Out]

1/16*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(sqrt(-c)*c) + 2*((a*c*x + c)^(3/2) + 2*sqrt(a*c*x
+ c)*c)/((a*c*x - c)^2*c))/(a*abs(c))