3.13.59 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^2} \, dx\) [1259]

Optimal. Leaf size=94 \[ -\frac {\sqrt {1-a^2 x^2}}{5 a c^2 (1+a x)^3}-\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (1+a x)^2}-\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (1+a x)} \]

[Out]

-1/5*(-a^2*x^2+1)^(1/2)/a/c^2/(a*x+1)^3-2/15*(-a^2*x^2+1)^(1/2)/a/c^2/(a*x+1)^2-2/15*(-a^2*x^2+1)^(1/2)/a/c^2/
(a*x+1)

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Rubi [A]
time = 0.05, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6274, 669, 673, 665} \begin {gather*} -\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (a x+1)}-\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (a x+1)^2}-\frac {\sqrt {1-a^2 x^2}}{5 a c^2 (a x+1)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*Sqrt[1 - a^2*x^2]/(a*c^2*(1 + a*x)^3) - (2*Sqrt[1 - a^2*x^2])/(15*a*c^2*(1 + a*x)^2) - (2*Sqrt[1 - a^2*x^
2])/(15*a*c^2*(1 + a*x))

Rule 665

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*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 669

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

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[Simplify[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, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p +
2], 0]

Rule 6274

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[(1 - a^2*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] && ILtQ[(n - 1)/2, 0] &&
!IntegerQ[p - n/2]

Rubi steps

\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac {\int \frac {(1-a x)^3}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^2}\\ &=\frac {\int \frac {1}{(1+a x)^3 \sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{5 a c^2 (1+a x)^3}+\frac {2 \int \frac {1}{(1+a x)^2 \sqrt {1-a^2 x^2}} \, dx}{5 c^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{5 a c^2 (1+a x)^3}-\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (1+a x)^2}+\frac {2 \int \frac {1}{(1+a x) \sqrt {1-a^2 x^2}} \, dx}{15 c^2}\\ &=-\frac {\sqrt {1-a^2 x^2}}{5 a c^2 (1+a x)^3}-\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (1+a x)^2}-\frac {2 \sqrt {1-a^2 x^2}}{15 a c^2 (1+a x)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.46 \begin {gather*} -\frac {\sqrt {1-a x} \left (7+6 a x+2 a^2 x^2\right )}{15 a c^2 (1+a x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[1 - a*x]*(7 + 6*a*x + 2*a^2*x^2))/(a*c^2*(1 + a*x)^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 3 vs. order 2.
time = 0.06, size = 1031, normalized size = 10.97

method	result	size
gosper	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{2} x^{2}+6 a x +7\right )}{15 \left (a x +1\right )^{3} c^{2} a}\)	\(42\)
trager	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{2} x^{2}+6 a x +7\right )}{15 \left (a x +1\right )^{3} c^{2} a}\)	\(42\)
default	\(\text {Expression too large to display}\)	\(1031\)

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^2*c*x^2+c)^2,x,method=_RETURNVERBOSE)

[Out]

1/c^2*(5/64/a*(1/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+a*(-1/4*(-2*a^2*(x+1/a)+2*a)/a^2*(-a^2*(x+1/a)^2+2*a*(x+
1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2))))+1/4/a^4*(-1/3/a/(x+1/a)
^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-1/3*a*(-1/a/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-2*a*(1/a/(x+1/a
)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+3*a*(1/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+a*(-1/4*(-2*a^2*(x+1/a)+2*a
)/a^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/
2)))))))+1/8/a^2*(1/a/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+3*a*(1/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)
+a*(-1/4*(-2*a^2*(x+1/a)+2*a)/a^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^
2*(x+1/a)^2+2*a*(x+1/a))^(1/2)))))-1/20/a^6/(x+1/a)^5*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)-5/64/a*(1/3*(-a^2*(x-
1/a)^2-2*(x-1/a)*a)^(3/2)-a*(-1/4*(-2*a^2*(x-1/a)-2*a)/a^2*(-a^2*(x-1/a)^2-2*(x-1/a)*a)^(1/2)+1/2/(a^2)^(1/2)*
arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*(x-1/a)*a)^(1/2))))+3/16/a^3*(-1/a/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a
))^(5/2)-2*a*(1/a/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+3*a*(1/3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+a*(
-1/4*(-2*a^2*(x+1/a)+2*a)/a^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+1/2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x
+1/a)^2+2*a*(x+1/a))^(1/2))))))+1/32/a^2*(-1/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*(x-1/a)*a)^(5/2)-3*a*(1/3*(-a^2*(x-
1/a)^2-2*(x-1/a)*a)^(3/2)-a*(-1/4*(-2*a^2*(x-1/a)-2*a)/a^2*(-a^2*(x-1/a)^2-2*(x-1/a)*a)^(1/2)+1/2/(a^2)^(1/2)*
arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*(x-1/a)*a)^(1/2))))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 89, normalized size = 0.95 \begin {gather*} -\frac {7 \, a^{3} x^{3} + 21 \, a^{2} x^{2} + 21 \, a x + {\left (2 \, a^{2} x^{2} + 6 \, a x + 7\right )} \sqrt {-a^{2} x^{2} + 1} + 7}{15 \, {\left (a^{4} c^{2} x^{3} + 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x + a c^{2}\right )}} \end {gather*}

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^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/15*(7*a^3*x^3 + 21*a^2*x^2 + 21*a*x + (2*a^2*x^2 + 6*a*x + 7)*sqrt(-a^2*x^2 + 1) + 7)/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} + 3 a^{2} x^{2} \sqrt {- a^{2} x^{2} + 1} + 3 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx}{c^{2}} \end {gather*}

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**2*c*x**2+c)**2,x)

[Out]

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

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Giac [A]
time = 0.42, size = 145, normalized size = 1.54 \begin {gather*} \frac {2 \, {\left (\frac {20 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} + \frac {40 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {30 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} + \frac {15 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} + 7\right )}}{15 \, c^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )}^{5} {\left | a \right |}} \end {gather*}

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^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

2/15*(20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 40*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 30*(sqrt(-
a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) + 15*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) + 7)/(c^2*((sqrt(-a^2*x^
2 + 1)*abs(a) + a)/(a^2*x) + 1)^5*abs(a))

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Mupad [B]
time = 0.06, size = 125, normalized size = 1.33 \begin {gather*} -\frac {\sqrt {1-a^2\,x^2}\,\left (\frac {a^3}{5\,c^2\,{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )}^3}-\frac {2\,a^3}{15\,c^2\,\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )}+\frac {2\,a^4}{15\,c^2\,{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )}^2\,\sqrt {-a^2}}\right )}{a^3\,\sqrt {-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((1 - a^2*x^2)^(1/2)*(a^3/(5*c^2*(x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)^3) - (2*a^3)/(15*c^2*(x*(-a^2)^(1/2) + (-a
^2)^(1/2)/a)) + (2*a^4)/(15*c^2*(x*(-a^2)^(1/2) + (-a^2)^(1/2)/a)^2*(-a^2)^(1/2))))/(a^3*(-a^2)^(1/2))

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