3.986 \(\int \frac {e^{\tanh ^{-1}(a x)}}{(c-a^2 c x^2)^{7/2}} \, dx\)

Optimal. Leaf size=277 \[ \frac {3 \sqrt {1-a^2 x^2}}{16 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{32 a c^3 (a x+1)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{24 a c^3 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a c^3 \sqrt {c-a^2 c x^2}} \]

[Out]

1/24*(-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)+3/16*(-a^2*x^2+1)^(1/2)/a/c^3/(-a*x+1)/(-a^2*c*x^2+c)^(1/2)-1/32*(-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)+5/16*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 = 0.13, antiderivative size = 277, 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 = {6143, 6140, 44, 207} \[ \frac {3 \sqrt {1-a^2 x^2}}{16 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{8 a c^3 (a x+1) \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2}}{32 a c^3 (1-a x)^2 \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2}}{32 a c^3 (a x+1)^2 \sqrt {c-a^2 c x^2}}+\frac {\sqrt {1-a^2 x^2}}{24 a c^3 (1-a x)^3 \sqrt {c-a^2 c x^2}}+\frac {5 \sqrt {1-a^2 x^2} \tanh ^{-1}(a x)}{16 a c^3 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - a^2*x^2]/(24*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*Sqr
t[c - a^2*c*x^2]) + (3*Sqrt[1 - a^2*x^2])/(16*a*c^3*(1 - a*x)*Sqrt[c - a^2*c*x^2]) - 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])/(16*a*c^3*Sqrt[c - a^2*c*x^2])

Rule 44

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] && L
tQ[m + n + 2, 0])

Rule 207

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

Rule 6140

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 6143

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^{\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^{\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)^4 (1+a x)^3} \, dx}{c^3 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2} \int \left (\frac {1}{8 (-1+a x)^4}-\frac {3}{16 (-1+a x)^3}+\frac {3}{16 (-1+a x)^2}+\frac {1}{16 (1+a x)^3}+\frac {1}{8 (1+a x)^2}-\frac {5}{16 \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}}{24 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 {3 \sqrt {1-a^2 x^2}}{16 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\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}{16 c^3 \sqrt {c-a^2 c x^2}}\\ &=\frac {\sqrt {1-a^2 x^2}}{24 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 {3 \sqrt {1-a^2 x^2}}{16 a c^3 (1-a x) \sqrt {c-a^2 c x^2}}-\frac {\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)}{16 a c^3 \sqrt {c-a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 103, normalized size = 0.37 \[ \frac {\sqrt {1-a^2 x^2} \left (-15 a^4 x^4+15 a^3 x^3+25 a^2 x^2-25 a x+15 (a x-1)^3 (a x+1)^2 \tanh ^{-1}(a x)-8\right )}{48 a c^3 (a x-1)^3 (a x+1)^2 \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(-8 - 25*a*x + 25*a^2*x^2 + 15*a^3*x^3 - 15*a^4*x^4 + 15*(-1 + a*x)^3*(1 + a*x)^2*ArcTanh[a
*x]))/(48*a*c^3*(-1 + a*x)^3*(1 + a*x)^2*Sqrt[c - a^2*c*x^2])

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fricas [A]  time = 1.20, size = 567, normalized size = 2.05 \[ \left [\frac {15 \, {\left (a^{7} x^{7} - a^{6} x^{6} - 3 \, a^{5} x^{5} + 3 \, a^{4} x^{4} + 3 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - a x + 1\right )} \sqrt {c} \log \left (-\frac {a^{6} c x^{6} + 5 \, a^{4} c x^{4} - 5 \, a^{2} c x^{2} - 4 \, {\left (a^{3} x^{3} + a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} \sqrt {c} - c}{a^{6} x^{6} - 3 \, a^{4} x^{4} + 3 \, a^{2} x^{2} - 1}\right ) + 4 \, {\left (8 \, a^{5} x^{5} + 7 \, a^{4} x^{4} - 31 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 33 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{192 \, {\left (a^{8} c^{4} x^{7} - a^{7} c^{4} x^{6} - 3 \, a^{6} c^{4} x^{5} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\right )}}, \frac {15 \, {\left (a^{7} x^{7} - a^{6} x^{6} - 3 \, a^{5} x^{5} + 3 \, a^{4} x^{4} + 3 \, a^{3} x^{3} - 3 \, a^{2} x^{2} - a x + 1\right )} \sqrt {-c} \arctan \left (\frac {2 \, \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} a \sqrt {-c} x}{a^{4} c x^{4} - c}\right ) + 2 \, {\left (8 \, a^{5} x^{5} + 7 \, a^{4} x^{4} - 31 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 33 \, a x\right )} \sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1}}{96 \, {\left (a^{8} c^{4} x^{7} - a^{7} c^{4} x^{6} - 3 \, a^{6} c^{4} x^{5} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{4} c^{4} x^{3} - 3 \, a^{3} c^{4} x^{2} - a^{2} c^{4} x + a c^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

[1/192*(15*(a^7*x^7 - a^6*x^6 - 3*a^5*x^5 + 3*a^4*x^4 + 3*a^3*x^3 - 3*a^2*x^2 - 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*(8*a^5*x^5 + 7*a^4*x^4 - 31*a^3*x^3 - 9*a^2*x^2 + 33*a*x)*sqrt(-a^2*c*x^2
+ c)*sqrt(-a^2*x^2 + 1))/(a^8*c^4*x^7 - a^7*c^4*x^6 - 3*a^6*c^4*x^5 + 3*a^5*c^4*x^4 + 3*a^4*c^4*x^3 - 3*a^3*c
^4*x^2 - a^2*c^4*x + a*c^4), 1/96*(15*(a^7*x^7 - a^6*x^6 - 3*a^5*x^5 + 3*a^4*x^4 + 3*a^3*x^3 - 3*a^2*x^2 - 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*(8*a^5*x^5 +
7*a^4*x^4 - 31*a^3*x^3 - 9*a^2*x^2 + 33*a*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1))/(a^8*c^4*x^7 - a^7*c^4*
x^6 - 3*a^6*c^4*x^5 + 3*a^5*c^4*x^4 + 3*a^4*c^4*x^3 - 3*a^3*c^4*x^2 - a^2*c^4*x + a*c^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 238, normalized size = 0.86 \[ \frac {\sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (15 \ln \left (a x -1\right ) x^{5} a^{5}-15 \ln \left (a x +1\right ) x^{5} a^{5}-15 \ln \left (a x -1\right ) x^{4} a^{4}+15 \ln \left (a x +1\right ) x^{4} a^{4}+30 x^{4} a^{4}-30 \ln \left (a x -1\right ) x^{3} a^{3}+30 a^{3} x^{3} \ln \left (a x +1\right )-30 x^{3} a^{3}+30 \ln \left (a x -1\right ) x^{2} a^{2}-30 \ln \left (a x +1\right ) x^{2} a^{2}-50 a^{2} x^{2}+15 \ln \left (a x -1\right ) x a -15 a x \ln \left (a x +1\right )+50 a x -15 \ln \left (a x -1\right )+15 \ln \left (a x +1\right )+16\right )}{96 \left (a^{2} x^{2}-1\right ) c^{4} a \left (a x -1\right )^{3} \left (a x +1\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(-a^2*x^2+1)^(1/2)*(-c*(a^2*x^2-1))^(1/2)*(15*ln(a*x-1)*x^5*a^5-15*ln(a*x+1)*x^5*a^5-15*ln(a*x-1)*x^4*a^4
+15*ln(a*x+1)*x^4*a^4+30*x^4*a^4-30*ln(a*x-1)*x^3*a^3+30*a^3*x^3*ln(a*x+1)-30*x^3*a^3+30*ln(a*x-1)*x^2*a^2-30*
ln(a*x+1)*x^2*a^2-50*a^2*x^2+15*ln(a*x-1)*x*a-15*a*x*ln(a*x+1)+50*a*x-15*ln(a*x-1)+15*ln(a*x+1)+16)/(a^2*x^2-1
)/c^4/a/(a*x-1)^3/(a*x+1)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {7}{2}} \sqrt {-a^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/(-a^2*c*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a x + 1}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*x + 1)/(sqrt(-(a*x - 1)*(a*x + 1))*(-c*(a*x - 1)*(a*x + 1))**(7/2)), x)

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