3.3.4 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [204]

Optimal. Leaf size=97 \[ \frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6262, 673, 665} \begin {gather*} \frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*(c - a*c*x)^4),x]

[Out]

Sqrt[1 - a^2*x^2]/(5*a*c^4*(1 - a*x)^3) + (2*Sqrt[1 - a^2*x^2])/(15*a*c^4*(1 - a*x)^2) + (2*Sqrt[1 - a^2*x^2])
/(15*a*c^4*(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 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 6262

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]

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{(c-a c x)^4} \, dx &=\frac {\int \frac {1}{(c-a c x)^3 \sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3}+\frac {2 \int \frac {1}{(c-a c x)^2 \sqrt {1-a^2 x^2}} \, dx}{5 c^2}\\ &=\frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac {2 \int \frac {1}{(c-a c x) \sqrt {1-a^2 x^2}} \, dx}{15 c^3}\\ &=\frac {\sqrt {1-a^2 x^2}}{5 a c^4 (1-a x)^3}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)^2}+\frac {2 \sqrt {1-a^2 x^2}}{15 a c^4 (1-a x)}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 43, normalized size = 0.44 \begin {gather*} \frac {\sqrt {1+a x} \left (7-6 a x+2 a^2 x^2\right )}{15 a c^4 (1-a x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^ArcTanh[a*x]*(c - a*c*x)^4),x]

[Out]

(Sqrt[1 + a*x]*(7 - 6*a*x + 2*a^2*x^2))/(15*a*c^4*(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 = 1.10, size = 394, normalized size = 4.06

method	result	size
gosper	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{2} x^{2}-6 a x +7\right )}{15 a \left (a x -1\right )^{3} c^{4}}\)	\(42\)
trager	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (2 a^{2} x^{2}-6 a x +7\right )}{15 a \left (a x -1\right )^{3} c^{4}}\)	\(42\)
default	\(\frac {\frac {\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{a \left (x -\frac {1}{a}\right )^{2}}+a \left (\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )}{8 a^{2}}-\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{12 a^{4} \left (x -\frac {1}{a}\right )^{3}}+\frac {\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{16 a}+\frac {\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{5 a \left (x -\frac {1}{a}\right )^{4}}-\frac {\left (-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {1}{a}\right )^{3}}}{2 a^{4}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}-\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}}{16 a}}{c^{4}}\)	\(394\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

1/c^4*(1/8/a^2*(1/a/(x-1/a)^2*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)+a*((-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-a/(a^2)
^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2))))-1/12/a^4/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*(x-1/
a))^(3/2)+1/16/a*((-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)+a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x+1/a)^2+2*a*(x
+1/a))^(1/2)))+1/2/a^4*(1/5/a/(x-1/a)^4*(-a^2*(x-1/a)^2-2*a*(x-1/a))^(3/2)-1/15/(x-1/a)^3*(-a^2*(x-1/a)^2-2*a*
(x-1/a))^(3/2))-1/16/a*((-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-a/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x-1/a)^2-
2*a*(x-1/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)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,x, algorithm="maxima")

[Out]

integrate(sqrt(-a^2*x^2 + 1)/((a*c*x - c)^4*(a*x + 1)), x)

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Fricas [A]
time = 0.38, size = 91, normalized size = 0.94 \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^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x - a c^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{5} x^{5} - 3 a^{4} x^{4} + 2 a^{3} x^{3} + 2 a^{2} x^{2} - 3 a x + 1}\, dx}{c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a**2*x**2+1)**(1/2)/(-a*c*x+c)**4,x)

[Out]

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

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Giac [A]
time = 0.45, size = 145, normalized size = 1.49 \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^{4} {\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)*(-a^2*x^2+1)^(1/2)/(-a*c*x+c)^4,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^4*((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 = 127, normalized size = 1.31 \begin {gather*} \frac {\sqrt {1-a^2\,x^2}\,\left (\frac {2\,a^3}{15\,c^4\,\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}-\frac {a^3}{5\,c^4\,{\left (x\,\sqrt {-a^2}-\frac {\sqrt {-a^2}}{a}\right )}^3}+\frac {2\,a^4}{15\,c^4\,{\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)^(1/2)/((c - a*c*x)^4*(a*x + 1)),x)

[Out]

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