3.7.30 \(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [630]

Optimal. Leaf size=74 \[ -\frac {2 (1+a x)}{5 a \left (c-a^2 c x^2\right )^{5/2}}-\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

-2/5*(a*x+1)/a/(-a^2*c*x^2+c)^(5/2)-1/5*x/c/(-a^2*c*x^2+c)^(3/2)-2/5*x/c^2/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {6302, 6276, 667, 198, 197} \begin {gather*} -\frac {2 x}{5 c^2 \sqrt {c-a^2 c x^2}}-\frac {x}{5 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {2 (a x+1)}{5 a \left (c-a^2 c x^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 667

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

Rule 6276

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

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 53, normalized size = 0.72 \begin {gather*} -\frac {2+a x-4 a^2 x^2+2 a^3 x^3}{5 a c^2 (-1+a x)^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(2 + a*x - 4*a^2*x^2 + 2*a^3*x^3)/(a*c^2*(-1 + a*x)^2*Sqrt[c - a^2*c*x^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(205\) vs. \(2(62)=124\).
time = 0.20, size = 206, normalized size = 2.78

method	result	size
gosper	\(-\frac {\left (a x +1\right )^{2} \left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right )}{5 a \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}\)	\(47\)
trager	\(\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \sqrt {-a^{2} c \,x^{2}+c}}{5 c^{3} \left (a x -1\right )^{3} a \left (a x +1\right )}\)	\(57\)
default	\(\frac {x}{3 c \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{2} \sqrt {-a^{2} c \,x^{2}+c}}+\frac {\frac {2}{5 a c \left (x -\frac {1}{a}\right ) \left (-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {8 a \left (-\frac {-2 a^{2} c \left (x -\frac {1}{a}\right )-2 a c}{6 a^{2} c^{2} \left (-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c \right )^{\frac {3}{2}}}-\frac {-2 a^{2} c \left (x -\frac {1}{a}\right )-2 a c}{3 a^{2} c^{3} \sqrt {-a^{2} c \left (x -\frac {1}{a}\right )^{2}-2 \left (x -\frac {1}{a}\right ) a c}}\right )}{5}}{a}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x-1)*(a*x+1)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*x/c/(-a^2*c*x^2+c)^(3/2)+2/3*x/c^2/(-a^2*c*x^2+c)^(1/2)+2/a*(1/5/a/c/(x-1/a)/(-a^2*c*(x-1/a)^2-2*(x-1/a)*a
*c)^(3/2)-4/5*a*(-1/6*(-2*a^2*c*(x-1/a)-2*a*c)/a^2/c^2/(-a^2*c*(x-1/a)^2-2*(x-1/a)*a*c)^(3/2)-1/3/a^2/c^3*(-2*
a^2*c*(x-1/a)-2*a*c)/(-a^2*c*(x-1/a)^2-2*(x-1/a)*a*c)^(1/2)))

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Maxima [A]
time = 0.29, size = 80, normalized size = 1.08 \begin {gather*} \frac {2}{5 \, {\left ({\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x - {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a c\right )}} - \frac {2 \, x}{5 \, \sqrt {-a^{2} c x^{2} + c} c^{2}} - \frac {x}{5 \, {\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/((-a^2*c*x^2 + c)^(3/2)*a^2*c*x - (-a^2*c*x^2 + c)^(3/2)*a*c) - 2/5*x/(sqrt(-a^2*c*x^2 + c)*c^2) - 1/5*x/(
(-a^2*c*x^2 + c)^(3/2)*c)

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Fricas [A]
time = 0.40, size = 75, normalized size = 1.01 \begin {gather*} \frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt {-a^{2} c x^{2} + c}}{5 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2*a^3*x^3 - 4*a^2*x^2 + a*x + 2)*sqrt(-a^2*c*x^2 + c)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a x + 1}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \left (a x - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.38, size = 56, normalized size = 0.76 \begin {gather*} \frac {\sqrt {c-a^2\,c\,x^2}\,\left (2\,a^3\,x^3-4\,a^2\,x^2+a\,x+2\right )}{5\,a\,c^3\,{\left (a\,x-1\right )}^3\,\left (a\,x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c - a^2*c*x^2)^(1/2)*(a*x - 4*a^2*x^2 + 2*a^3*x^3 + 2))/(5*a*c^3*(a*x - 1)^3*(a*x + 1))

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