3.12.71 \(\int \frac {e^{3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{5/2}} \, dx\) [1171]

Optimal. Leaf size=185 \[ \frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \]

[Out]

1/6*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*x+1)^3/(-a^2*c*x^2+c)^(1/2)+1/8*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*x+1)^2/(-a^2*c*x
^2+c)^(1/2)+1/8*(-a^2*x^2+1)^(1/2)/a/c^2/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)+1/8*arctanh(a*x)*(-a^2*x^2+1)^(1/2)/a/c
^2/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6278, 6275, 46, 213} \begin {gather*} \frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3*ArcTanh[a*x])/(c - a^2*c*x^2)^(5/2),x]

[Out]

Sqrt[1 - a^2*x^2]/(6*a*c^2*(1 - a*x)^3*Sqrt[c - a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 - a*x)^2*Sqrt[c -
a^2*c*x^2]) + Sqrt[1 - a^2*x^2]/(8*a*c^2*(1 - a*x)*Sqrt[c - a^2*c*x^2]) + (Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(8*
a*c^2*Sqrt[c - a^2*c*x^2])

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 6275

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

Rule 6278

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

Rubi steps

\begin {align*} \int \frac {e^{3 \tanh ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{3 \tanh ^{-1}(a x)}}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {1}{(1-a x)^4 (1+a x)} \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{2 (-1+a x)^4}-\frac {1}{4 (-1+a x)^3}+\frac {1}{8 (-1+a x)^2}-\frac {1}{8 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} \int \frac {1}{-1+a^2 x^2} \, dx}{8 c^2 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{6 a c^2 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{8 a c^2 (1-a x) \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{8 a c^2 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 73, normalized size = 0.39 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (-10+9 a x-3 a^2 x^2+3 (-1+a x)^3 \tanh ^{-1}(a x)\right )}{24 a c^2 (-1+a x)^3 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcTanh[a*x])/(c - a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[1 - a^2*x^2]*(-10 + 9*a*x - 3*a^2*x^2 + 3*(-1 + a*x)^3*ArcTanh[a*x]))/(24*a*c^2*(-1 + a*x)^3*Sqrt[c - a^
2*c*x^2])

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Maple [A]
time = 0.06, size = 159, normalized size = 0.86

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 \ln \left (a x +1\right ) a^{3} x^{3}-3 \ln \left (a x -1\right ) a^{3} x^{3}-9 \ln \left (a x +1\right ) a^{2} x^{2}+9 \ln \left (a x -1\right ) a^{2} x^{2}-6 a^{2} x^{2}+9 \ln \left (a x +1\right ) a x -9 \ln \left (a x -1\right ) a x +18 a x -3 \ln \left (a x +1\right )+3 \ln \left (a x -1\right )-20\right )}{48 \left (a^{2} x^{2}-1\right ) c^{3} a \left (a x -1\right )^{3}}\)	\(159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/48*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(3*ln(a*x+1)*a^3*x^3-3*ln(a*x-1)*a^3*x^3-9*ln(a*x+1)*a^2*x^2+9
*ln(a*x-1)*a^2*x^2-6*a^2*x^2+9*ln(a*x+1)*a*x-9*ln(a*x-1)*a*x+18*a*x-3*ln(a*x+1)+3*ln(a*x-1)-20)/(a^2*x^2-1)/c^
3/a/(a*x-1)^3

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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((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 459, normalized size = 2.48 \begin {gather*} \left [\frac {3 \, {\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) + 4 \, {\left (10 \, a^{3} x^{3} - 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{96 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}}, \frac {3 \, {\left (a^{5} x^{5} - 3 \, a^{4} x^{4} + 2 \, a^{3} x^{3} + 2 \, a^{2} x^{2} - 3 \, a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) + 2 \, {\left (10 \, a^{3} x^{3} - 27 \, a^{2} x^{2} + 21 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{48 \, {\left (a^{6} c^{3} x^{5} - 3 \, a^{5} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{3} + 2 \, a^{3} c^{3} x^{2} - 3 \, a^{2} c^{3} x + a c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[1/96*(3*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 + 2*a^2*x^2 - 3*a*x + 1)*sqrt(c)*log(-(a^6*c*x^6 + 5*a^4*c*x^4 - 5*a
^2*c*x^2 - 4*(a^3*x^3 + a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*sqrt(c) - c)/(a^6*x^6 - 3*a^4*x^4 + 3*a^2
*x^2 - 1)) + 4*(10*a^3*x^3 - 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 - 3*a^
5*c^3*x^4 + 2*a^4*c^3*x^3 + 2*a^3*c^3*x^2 - 3*a^2*c^3*x + a*c^3), 1/48*(3*(a^5*x^5 - 3*a^4*x^4 + 2*a^3*x^3 + 2
*a^2*x^2 - 3*a*x + 1)*sqrt(-c)*arctan(2*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*a*sqrt(-c)*x/(a^4*c*x^4 - c))
+ 2*(10*a^3*x^3 - 27*a^2*x^2 + 21*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^6*c^3*x^5 - 3*a^5*c^3*x^4 +
2*a^4*c^3*x^3 + 2*a^3*c^3*x^2 - 3*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 {\left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + 1)**3/((-(a*x - 1)*(a*x + 1))**(3/2)*(-c*(a*x - 1)*(a*x + 1))**(5/2)), 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((a*x+1)^3/(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x+1\right )}^3}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,{\left (1-a^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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