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

Optimal. Leaf size=361 \[ \frac {3 \left (1-a^2 x^2\right )^{7/2}}{2 a^8 x^7 (1-a x) \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}-\frac {5 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 x^7 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}-\frac {11 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (1-a x)^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {\left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {\left (1-a^2 x^2\right )^{7/2}}{24 a^8 x^7 (1-a x)^3 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {51 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{32 a^8 x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}-\frac {19 \left (1-a^2 x^2\right )^{7/2} \log (a x+1)}{32 a^8 x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {\left (1-a^2 x^2\right )^{7/2}}{a^7 x^6 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \]

[Out]

(-a^2*x^2+1)^(7/2)/a^7/(c-c/a^2/x^2)^(7/2)/x^6+1/24*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)/x^7/(-a*x+1)^3-
11/32*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)/x^7/(-a*x+1)^2+3/2*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)
/x^7/(-a*x+1)+1/32*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a^2/x^2)^(7/2)/x^7/(a*x+1)^2-5/16*(-a^2*x^2+1)^(7/2)/a^8/(c-c/a
^2/x^2)^(7/2)/x^7/(a*x+1)+51/32*(-a^2*x^2+1)^(7/2)*ln(-a*x+1)/a^8/(c-c/a^2/x^2)^(7/2)/x^7-19/32*(-a^2*x^2+1)^(
7/2)*ln(a*x+1)/a^8/(c-c/a^2/x^2)^(7/2)/x^7

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Rubi [A]  time = 0.30, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6160, 6150, 88} \[ \frac {3 \left (1-a^2 x^2\right )^{7/2}}{2 a^8 x^7 (1-a x) \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}-\frac {5 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 x^7 (a x+1) \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}-\frac {11 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (1-a x)^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {\left (1-a^2 x^2\right )^{7/2}}{32 a^8 x^7 (a x+1)^2 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {\left (1-a^2 x^2\right )^{7/2}}{24 a^8 x^7 (1-a x)^3 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {\left (1-a^2 x^2\right )^{7/2}}{a^7 x^6 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}+\frac {51 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{32 a^8 x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}}-\frac {19 \left (1-a^2 x^2\right )^{7/2} \log (a x+1)}{32 a^8 x^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - a^2*x^2)^(7/2)/(a^7*(c - c/(a^2*x^2))^(7/2)*x^6) + (1 - a^2*x^2)^(7/2)/(24*a^8*(c - c/(a^2*x^2))^(7/2)*x^
7*(1 - a*x)^3) - (11*(1 - a^2*x^2)^(7/2))/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 - a*x)^2) + (3*(1 - a^2*x^2)^
(7/2))/(2*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 - a*x)) + (1 - a^2*x^2)^(7/2)/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^7
*(1 + a*x)^2) - (5*(1 - a^2*x^2)^(7/2))/(16*a^8*(c - c/(a^2*x^2))^(7/2)*x^7*(1 + a*x)) + (51*(1 - a^2*x^2)^(7/
2)*Log[1 - a*x])/(32*a^8*(c - c/(a^2*x^2))^(7/2)*x^7) - (19*(1 - a^2*x^2)^(7/2)*Log[1 + a*x])/(32*a^8*(c - c/(
a^2*x^2))^(7/2)*x^7)

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

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

Rule 6160

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

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2}} \, dx &=\frac {\left (1-a^2 x^2\right )^{7/2} \int \frac {e^{\tanh ^{-1}(a x)} x^7}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac {\left (1-a^2 x^2\right )^{7/2} \int \frac {x^7}{(1-a x)^4 (1+a x)^3} \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac {\left (1-a^2 x^2\right )^{7/2} \int \left (\frac {1}{a^7}+\frac {1}{8 a^7 (-1+a x)^4}+\frac {11}{16 a^7 (-1+a x)^3}+\frac {3}{2 a^7 (-1+a x)^2}+\frac {51}{32 a^7 (-1+a x)}-\frac {1}{16 a^7 (1+a x)^3}+\frac {5}{16 a^7 (1+a x)^2}-\frac {19}{32 a^7 (1+a x)}\right ) \, dx}{\left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}\\ &=\frac {\left (1-a^2 x^2\right )^{7/2}}{a^7 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^6}+\frac {\left (1-a^2 x^2\right )^{7/2}}{24 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)^3}-\frac {11 \left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)^2}+\frac {3 \left (1-a^2 x^2\right )^{7/2}}{2 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1-a x)}+\frac {\left (1-a^2 x^2\right )^{7/2}}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)^2}-\frac {5 \left (1-a^2 x^2\right )^{7/2}}{16 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7 (1+a x)}+\frac {51 \left (1-a^2 x^2\right )^{7/2} \log (1-a x)}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}-\frac {19 \left (1-a^2 x^2\right )^{7/2} \log (1+a x)}{32 a^8 \left (c-\frac {c}{a^2 x^2}\right )^{7/2} x^7}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 147, normalized size = 0.41 \[ \frac {\sqrt {1-a^2 x^2} \left (-96 a^6 x^6+96 a^5 x^5+366 a^4 x^4-222 a^3 x^3-338 a^2 x^2+122 a x-153 (a x-1)^3 (a x+1)^2 \log (1-a x)+57 (a x-1)^3 (a x+1)^2 \log (a x+1)+88\right )}{96 a^2 c^3 x (a x-1)^3 (a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - a^2*x^2]*(88 + 122*a*x - 338*a^2*x^2 - 222*a^3*x^3 + 366*a^4*x^4 + 96*a^5*x^5 - 96*a^6*x^6 - 153*(-1
+ a*x)^3*(1 + a*x)^2*Log[1 - a*x] + 57*(-1 + a*x)^3*(1 + a*x)^2*Log[1 + a*x]))/(96*a^2*c^3*Sqrt[c - c/(a^2*x^
2)]*x*(-1 + a*x)^3*(1 + a*x)^2)

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fricas [F]  time = 0.79, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} a^{8} x^{8} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{9} c^{4} x^{9} - a^{8} c^{4} x^{8} - 4 \, a^{7} c^{4} x^{7} + 4 \, a^{6} c^{4} x^{6} + 6 \, a^{5} c^{4} x^{5} - 6 \, a^{4} c^{4} x^{4} - 4 \, a^{3} c^{4} x^{3} + 4 \, a^{2} c^{4} x^{2} + a c^{4} x - c^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*a^8*x^8*sqrt((a^2*c*x^2 - c)/(a^2*x^2))/(a^9*c^4*x^9 - a^8*c^4*x^8 - 4*a^7*c^4*x^
7 + 4*a^6*c^4*x^6 + 6*a^5*c^4*x^5 - 6*a^4*c^4*x^4 - 4*a^3*c^4*x^3 + 4*a^2*c^4*x^2 + a*c^4*x - c^4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 239, normalized size = 0.66 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \left (a x +1\right ) \left (96 x^{6} a^{6}+153 \ln \left (a x -1\right ) x^{5} a^{5}-57 \ln \left (a x +1\right ) x^{5} a^{5}-96 x^{5} a^{5}-153 \ln \left (a x -1\right ) x^{4} a^{4}+57 \ln \left (a x +1\right ) x^{4} a^{4}-366 x^{4} a^{4}-306 \ln \left (a x -1\right ) x^{3} a^{3}+114 a^{3} x^{3} \ln \left (a x +1\right )+222 x^{3} a^{3}+306 \ln \left (a x -1\right ) x^{2} a^{2}-114 \ln \left (a x +1\right ) x^{2} a^{2}+338 a^{2} x^{2}+153 \ln \left (a x -1\right ) x a -57 a x \ln \left (a x +1\right )-122 a x -153 \ln \left (a x -1\right )+57 \ln \left (a x +1\right )-88\right )}{96 a^{8} x^{7} \left (\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(-a^2*x^2+1)^(1/2)*(a*x+1)*(96*x^6*a^6+153*ln(a*x-1)*x^5*a^5-57*ln(a*x+1)*x^5*a^5-96*x^5*a^5-153*ln(a*x-
1)*x^4*a^4+57*ln(a*x+1)*x^4*a^4-366*x^4*a^4-306*ln(a*x-1)*x^3*a^3+114*a^3*x^3*ln(a*x+1)+222*x^3*a^3+306*ln(a*x
-1)*x^2*a^2-114*ln(a*x+1)*x^2*a^2+338*a^2*x^2+153*ln(a*x-1)*x*a-57*a*x*ln(a*x+1)-122*a*x-153*ln(a*x-1)+57*ln(a
*x+1)-88)/a^8/x^7/(c*(a^2*x^2-1)/a^2/x^2)^(7/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*x + 1)/((c - c/(a^2*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 \[ \int \frac {a x + 1}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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