3.13.78 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\) [1278]

Optimal. Leaf size=275 \[ \frac {\sqrt {1-a^2 x^2}}{32 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{16 a c^3 (1+a x)^4 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{12 a c^3 (1+a x)^3 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{32 a c^3 \sqrt {c-a^2 c x^2}} \]

[Out]

1/32*(-a^2*x^2+1)^(1/2)/a/c^3/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/16*(-a^2*x^2+1)^(1/2)/a/c^3/(a*x+1)^4/(-a^2*c*x^
2+c)^(1/2)-1/12*(-a^2*x^2+1)^(1/2)/a/c^3/(a*x+1)^3/(-a^2*c*x^2+c)^(1/2)-3/32*(-a^2*x^2+1)^(1/2)/a/c^3/(a*x+1)^
2/(-a^2*c*x^2+c)^(1/2)-1/8*(-a^2*x^2+1)^(1/2)/a/c^3/(a*x+1)/(-a^2*c*x^2+c)^(1/2)+5/32*arctanh(a*x)*(-a^2*x^2+1
)^(1/2)/a/c^3/(-a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 275, 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}}{32 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (a x+1) \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (a x+1)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{12 a c^3 (a x+1)^3 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{16 a c^3 (a x+1)^4 \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{32 a c^3 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(7/2)),x]

[Out]

Sqrt[1 - a^2*x^2]/(32*a*c^3*(1 - a*x)*Sqrt[c - a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(16*a*c^3*(1 + a*x)^4*Sqrt[c -
a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(12*a*c^3*(1 + a*x)^3*Sqrt[c - a^2*c*x^2]) - (3*Sqrt[1 - a^2*x^2])/(32*a*c^3*(
1 + a*x)^2*Sqrt[c - a^2*c*x^2]) - Sqrt[1 - a^2*x^2]/(8*a*c^3*(1 + a*x)*Sqrt[c - a^2*c*x^2]) + (5*Sqrt[1 - a^2*
x^2]*ArcTanh[a*x])/(32*a*c^3*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 )^{7/2}} \, dx &=\frac {\sqrt {1-a^2 x^2} \int \frac {e^{-3 \tanh ^{-1}(a x)}}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^3 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \frac {1}{(1-a x)^2 (1+a x)^5} \, dx}{c^3 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{32 (-1+a x)^2}+\frac {1}{4 (1+a x)^5}+\frac {1}{4 (1+a x)^4}+\frac {3}{16 (1+a x)^3}+\frac {1}{8 (1+a x)^2}-\frac {5}{32 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{32 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{16 a c^3 (1+a x)^4 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{12 a c^3 (1+a x)^3 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (1+a x) \sqrt {c-a^2 c x^2}}-\frac {\left (5 \sqrt {1-a^2 x^2}\right ) \int \frac {1}{-1+a^2 x^2} \, dx}{32 c^3 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{32 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{16 a c^3 (1+a x)^4 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{12 a c^3 (1+a x)^3 \sqrt {c-a^2 c x^2}}-\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (1+a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (1+a x) \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{32 a c^3 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 101, normalized size = 0.37 \begin {gather*} \frac {\sqrt {1-a^2 x^2} \left (32+15 a x-35 a^2 x^2-45 a^3 x^3-15 a^4 x^4+15 (-1+a x) (1+a x)^4 \tanh ^{-1}(a x)\right )}{96 a c^3 (-1+a x) (1+a x)^4 \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2)^(7/2)),x]

[Out]

(Sqrt[1 - a^2*x^2]*(32 + 15*a*x - 35*a^2*x^2 - 45*a^3*x^3 - 15*a^4*x^4 + 15*(-1 + a*x)*(1 + a*x)^4*ArcTanh[a*x
]))/(96*a*c^3*(-1 + a*x)*(1 + a*x)^4*Sqrt[c - a^2*c*x^2])

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Maple [A]
time = 0.06, size = 238, normalized size = 0.87

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x +1\right ) a^{5} x^{5}-15 \ln \left (a x -1\right ) a^{5} x^{5}+45 \ln \left (a x +1\right ) a^{4} x^{4}-45 \ln \left (a x -1\right ) a^{4} x^{4}-30 a^{4} x^{4}+30 \ln \left (a x +1\right ) a^{3} x^{3}-30 \ln \left (a x -1\right ) a^{3} x^{3}-90 a^{3} x^{3}-30 \ln \left (a x +1\right ) a^{2} x^{2}+30 \ln \left (a x -1\right ) a^{2} x^{2}-70 a^{2} x^{2}-45 \ln \left (a x +1\right ) a x +45 \ln \left (a x -1\right ) a x +30 a x -15 \ln \left (a x +1\right )+15 \ln \left (a x -1\right )+64\right )}{192 \left (a^{2} x^{2}-1\right ) c^{4} a \left (a x +1\right )^{4} \left (a x -1\right )}\)	\(238\)

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)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/192*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(15*ln(a*x+1)*a^5*x^5-15*ln(a*x-1)*a^5*x^5+45*ln(a*x+1)*a^4*x
^4-45*ln(a*x-1)*a^4*x^4-30*a^4*x^4+30*ln(a*x+1)*a^3*x^3-30*ln(a*x-1)*a^3*x^3-90*a^3*x^3-30*ln(a*x+1)*a^2*x^2+3
0*ln(a*x-1)*a^2*x^2-70*a^2*x^2-45*ln(a*x+1)*a*x+45*ln(a*x-1)*a*x+30*a*x-15*ln(a*x+1)+15*ln(a*x-1)+64)/(a^2*x^2
-1)/c^4/a/(a*x+1)^4/(a*x-1)

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Maxima [A]
time = 0.29, size = 122, normalized size = 0.44 \begin {gather*} -\frac {15 \, a^{4} x^{4} + 45 \, a^{3} x^{3} + 35 \, a^{2} x^{2} - 15 \, a x - 32}{96 \, {\left (a^{6} c^{\frac {7}{2}} x^{5} + 3 \, a^{5} c^{\frac {7}{2}} x^{4} + 2 \, a^{4} c^{\frac {7}{2}} x^{3} - 2 \, a^{3} c^{\frac {7}{2}} x^{2} - 3 \, a^{2} c^{\frac {7}{2}} x - a c^{\frac {7}{2}}\right )}} + \frac {5 \, \log \left (a x + 1\right )}{64 \, a c^{\frac {7}{2}}} - \frac {5 \, \log \left (a x - 1\right )}{64 \, a c^{\frac {7}{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)^(7/2),x, algorithm="maxima")

[Out]

-1/96*(15*a^4*x^4 + 45*a^3*x^3 + 35*a^2*x^2 - 15*a*x - 32)/(a^6*c^(7/2)*x^5 + 3*a^5*c^(7/2)*x^4 + 2*a^4*c^(7/2
)*x^3 - 2*a^3*c^(7/2)*x^2 - 3*a^2*c^(7/2)*x - a*c^(7/2)) + 5/64*log(a*x + 1)/(a*c^(7/2)) - 5/64*log(a*x - 1)/(
a*c^(7/2))

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Fricas [A]
time = 0.36, size = 559, normalized size = 2.03 \begin {gather*} \left [\frac {15 \, {\left (a^{7} x^{7} + 3 \, a^{6} x^{6} + a^{5} x^{5} - 5 \, a^{4} x^{4} - 5 \, a^{3} x^{3} + 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 (32 \, a^{5} x^{5} + 81 \, a^{4} x^{4} + 19 \, a^{3} x^{3} - 99 \, a^{2} x^{2} - 81 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{384 \, {\left (a^{8} c^{4} x^{7} + 3 \, a^{7} c^{4} x^{6} + a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} - 5 \, a^{4} c^{4} x^{3} + a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x + a c^{4}\right )}}, \frac {15 \, {\left (a^{7} x^{7} + 3 \, a^{6} x^{6} + a^{5} x^{5} - 5 \, a^{4} x^{4} - 5 \, a^{3} x^{3} + 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 (32 \, a^{5} x^{5} + 81 \, a^{4} x^{4} + 19 \, a^{3} x^{3} - 99 \, a^{2} x^{2} - 81 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{192 \, {\left (a^{8} c^{4} x^{7} + 3 \, a^{7} c^{4} x^{6} + a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} - 5 \, a^{4} c^{4} x^{3} + a^{3} c^{4} x^{2} + 3 \, a^{2} c^{4} x + a c^{4}\right )}}\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)^(7/2),x, algorithm="fricas")

[Out]

[1/384*(15*(a^7*x^7 + 3*a^6*x^6 + a^5*x^5 - 5*a^4*x^4 - 5*a^3*x^3 + 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*(32*a^5*x^5 + 81*a^4*x^4 + 19*a^3*x^3 - 99*a^2*x^2 - 81*a*x)*sqrt(-a^2*c*
x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^8*c^4*x^7 + 3*a^7*c^4*x^6 + a^6*c^4*x^5 - 5*a^5*c^4*x^4 - 5*a^4*c^4*x^3 + a^3*
c^4*x^2 + 3*a^2*c^4*x + a*c^4), 1/192*(15*(a^7*x^7 + 3*a^6*x^6 + a^5*x^5 - 5*a^4*x^4 - 5*a^3*x^3 + 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*(32*a^5*
x^5 + 81*a^4*x^4 + 19*a^3*x^3 - 99*a^2*x^2 - 81*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^8*c^4*x^7 + 3
*a^7*c^4*x^6 + a^6*c^4*x^5 - 5*a^5*c^4*x^4 - 5*a^4*c^4*x^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*} \int \frac {\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 {7}{2}} \left (a x + 1\right )^{3}}\, dx \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)**(7/2),x)

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)/((-c*(a*x - 1)*(a*x + 1))**(7/2)*(a*x + 1)**3), 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)^3*(-a^2*x^2+1)^(3/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="giac")

[Out]

integrate((-a^2*x^2 + 1)^(3/2)/((-a^2*c*x^2 + c)^(7/2)*(a*x + 1)^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (c-a^2\,c\,x^2\right )}^{7/2}\,{\left (a\,x+1\right )}^3} \,d x \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)^(7/2)*(a*x + 1)^3),x)

[Out]

int((1 - a^2*x^2)^(3/2)/((c - a^2*c*x^2)^(7/2)*(a*x + 1)^3), x)

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