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

Optimal. Leaf size=160 \[ -\frac{15 \sqrt{c-a c x}}{32 a c^5 \sqrt{1-a^2 x^2}}+\frac{5}{16 a c^4 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}+\frac{1}{4 a c^3 \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{32 \sqrt{2} a c^{9/2}} \]

[Out]

1/(4*a*c^3*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2]) + 5/(16*a*c^4*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2]) - (15*Sqrt[c
- a*c*x])/(32*a*c^5*Sqrt[1 - a^2*x^2]) + (15*ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])])/(
32*Sqrt[2]*a*c^(9/2))

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Rubi [A]  time = 0.129878, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6127, 673, 667, 661, 208} \[ -\frac{15 \sqrt{c-a c x}}{32 a c^5 \sqrt{1-a^2 x^2}}+\frac{5}{16 a c^4 \sqrt{1-a^2 x^2} \sqrt{c-a c x}}+\frac{1}{4 a c^3 \sqrt{1-a^2 x^2} (c-a c x)^{3/2}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{32 \sqrt{2} a c^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^(9/2)),x]

[Out]

1/(4*a*c^3*(c - a*c*x)^(3/2)*Sqrt[1 - a^2*x^2]) + 5/(16*a*c^4*Sqrt[c - a*c*x]*Sqrt[1 - a^2*x^2]) - (15*Sqrt[c
- a*c*x])/(32*a*c^5*Sqrt[1 - a^2*x^2]) + (15*ArcTanh[(Sqrt[c]*Sqrt[1 - a^2*x^2])/(Sqrt[2]*Sqrt[c - a*c*x])])/(
32*Sqrt[2]*a*c^(9/2))

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 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 667

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(d*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*e*(p + 1)), x] + Dist[(d*(m + 2*p + 2))/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x
] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && 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)^{9/2}} \, dx &=\frac{\int \frac{1}{(c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac{1}{4 a c^3 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}+\frac{5 \int \frac{1}{\sqrt{c-a c x} \left (1-a^2 x^2\right )^{3/2}} \, dx}{8 c^4}\\ &=\frac{1}{4 a c^3 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}+\frac{5}{16 a c^4 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}+\frac{15 \int \frac{\sqrt{c-a c x}}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{32 c^5}\\ &=\frac{1}{4 a c^3 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}+\frac{5}{16 a c^4 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}-\frac{15 \sqrt{c-a c x}}{32 a c^5 \sqrt{1-a^2 x^2}}+\frac{15 \int \frac{1}{\sqrt{c-a c x} \sqrt{1-a^2 x^2}} \, dx}{64 c^4}\\ &=\frac{1}{4 a c^3 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}+\frac{5}{16 a c^4 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}-\frac{15 \sqrt{c-a c x}}{32 a c^5 \sqrt{1-a^2 x^2}}-\frac{(15 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 )}{32 c^3}\\ &=\frac{1}{4 a c^3 (c-a c x)^{3/2} \sqrt{1-a^2 x^2}}+\frac{5}{16 a c^4 \sqrt{c-a c x} \sqrt{1-a^2 x^2}}-\frac{15 \sqrt{c-a c x}}{32 a c^5 \sqrt{1-a^2 x^2}}+\frac{15 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-a^2 x^2}}{\sqrt{2} \sqrt{c-a c x}}\right )}{32 \sqrt{2} a c^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0299716, size = 57, normalized size = 0.36 \[ -\frac{(1-a x)^{3/2} \text{Hypergeometric2F1}\left (-\frac{1}{2},3,\frac{1}{2},\frac{1}{2} (a x+1)\right )}{4 a c^3 \sqrt{a x+1} (c-a c x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a*c*x)^(9/2)),x]

[Out]

-((1 - a*x)^(3/2)*Hypergeometric2F1[-1/2, 3, 1/2, (1 + a*x)/2])/(4*a*c^3*Sqrt[1 + a*x]*(c - a*c*x)^(3/2))

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Maple [A]  time = 0.119, size = 173, normalized size = 1.1 \begin{align*} -{\frac{1}{64\, \left ( ax-1 \right ) ^{3} \left ( ax+1 \right ) a}\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ( ax-1 \right ) } \left ( 15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}{x}^{2}{a}^{2}\sqrt{c \left ( ax+1 \right ) }-30\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ) xa\sqrt{c \left ( ax+1 \right ) }-30\,{x}^{2}{a}^{2}\sqrt{c}+15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( ax+1 \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sqrt{2}\sqrt{c \left ( ax+1 \right ) }+40\,xa\sqrt{c}+6\,\sqrt{c} \right ){c}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(-a^2*x^2+1)^(1/2)*(-c*(a*x-1))^(1/2)/c^(11/2)*(15*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2
)*x^2*a^2*(c*(a*x+1))^(1/2)-30*2^(1/2)*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*x*a*(c*(a*x+1))^(1/2)-30
*x^2*a^2*c^(1/2)+15*arctanh(1/2*(c*(a*x+1))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*(c*(a*x+1))^(1/2)+40*x*a*c^(1/2)+6*
c^(1/2))/(a*x-1)^3/(a*x+1)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63633, size = 747, normalized size = 4.67 \begin{align*} \left [\frac{15 \, \sqrt{2}{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, 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 \,{\left (15 \, a^{2} x^{2} - 20 \, a x - 3\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{128 \,{\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\right )}}, \frac{15 \, \sqrt{2}{\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, 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 \,{\left (15 \, a^{2} x^{2} - 20 \, a x - 3\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-a c x + c}}{64 \,{\left (a^{5} c^{5} x^{4} - 2 \, a^{4} c^{5} x^{3} + 2 \, a^{2} c^{5} x - a c^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),x, algorithm="fricas")

[Out]

[1/128*(15*sqrt(2)*(a^4*x^4 - 2*a^3*x^3 + 2*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*(15*a^2*x^2 - 20*a*x - 3)*sqrt(-a^2*x^2 + 1
)*sqrt(-a*c*x + c))/(a^5*c^5*x^4 - 2*a^4*c^5*x^3 + 2*a^2*c^5*x - a*c^5), 1/64*(15*sqrt(2)*(a^4*x^4 - 2*a^3*x^3
+ 2*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*(15*a^
2*x^2 - 20*a*x - 3)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c))/(a^5*c^5*x^4 - 2*a^4*c^5*x^3 + 2*a^2*c^5*x - a*c^5)]

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

[Out]

Timed out

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Giac [A]  time = 1.30077, size = 134, normalized size = 0.84 \begin{align*} -\frac{{\left (\frac{15 \, \sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{a c x + c}}{2 \, \sqrt{-c}}\right )}{a \sqrt{-c} c^{3}} + \frac{16}{\sqrt{a c x + c} a c^{3}} + \frac{2 \,{\left (7 \,{\left (a c x + c\right )}^{\frac{3}{2}} - 18 \, \sqrt{a c x + c} c\right )}}{{\left (a c x - c\right )}^{2} a c^{3}}\right )}{\left | c \right |}}{64 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)^3*(-a^2*x^2+1)^(3/2)/(-a*c*x+c)^(9/2),x, algorithm="giac")

[Out]

-1/64*(15*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(a*c*x + c)/sqrt(-c))/(a*sqrt(-c)*c^3) + 16/(sqrt(a*c*x + c)*a*c^3) +
2*(7*(a*c*x + c)^(3/2) - 18*sqrt(a*c*x + c)*c)/((a*c*x - c)^2*a*c^3))*abs(c)/c^2