3.471 \(\int \frac {e^{3 \tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\)

Optimal. Leaf size=128 \[ \frac {(a x+1)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {2 (a x+1)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {10 (a x+1)^2}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {5 \sqrt {1-a^2 x^2}}{a c^2}-\frac {5 \sin ^{-1}(a x)}{a c^2} \]

[Out]

1/5*(a*x+1)^5/a/c^2/(-a^2*x^2+1)^(5/2)-2/3*(a*x+1)^4/a/c^2/(-a^2*x^2+1)^(3/2)-5*arcsin(a*x)/a/c^2+10/3*(a*x+1)
^2/a/c^2/(-a^2*x^2+1)^(1/2)+5*(-a^2*x^2+1)^(1/2)/a/c^2

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Rubi [A]  time = 0.26, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6131, 6128, 852, 1635, 21, 669, 641, 216} \[ \frac {(a x+1)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {2 (a x+1)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {10 (a x+1)^2}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {5 \sqrt {1-a^2 x^2}}{a c^2}-\frac {5 \sin ^{-1}(a x)}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + a*x)^5/(5*a*c^2*(1 - a^2*x^2)^(5/2)) - (2*(1 + a*x)^4)/(3*a*c^2*(1 - a^2*x^2)^(3/2)) + (10*(1 + a*x)^2)/(
3*a*c^2*Sqrt[1 - a^2*x^2]) + (5*Sqrt[1 - a^2*x^2])/(a*c^2) - (5*ArcSin[a*x])/(a*c^2)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 216

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 669

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

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 6128

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

Rule 6131

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

Rubi steps

\begin {align*} \int \frac {e^{3 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx &=\frac {a^2 \int \frac {e^{3 \tanh ^{-1}(a x)} x^2}{(1-a x)^2} \, dx}{c^2}\\ &=\frac {a^2 \int \frac {x^2 \left (1-a^2 x^2\right )^{3/2}}{(1-a x)^5} \, dx}{c^2}\\ &=\frac {a^2 \int \frac {x^2 (1+a x)^5}{\left (1-a^2 x^2\right )^{7/2}} \, dx}{c^2}\\ &=\frac {(1+a x)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {a^2 \int \frac {\left (\frac {5}{a^2}+\frac {5 x}{a}\right ) (1+a x)^4}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^2}\\ &=\frac {(1+a x)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {\int \frac {(1+a x)^5}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac {(1+a x)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {2 (1+a x)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {5 \int \frac {(1+a x)^3}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=\frac {(1+a x)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {2 (1+a x)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {10 (1+a x)^2}{3 a c^2 \sqrt {1-a^2 x^2}}-\frac {5 \int \frac {1+a x}{\sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=\frac {(1+a x)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {2 (1+a x)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {10 (1+a x)^2}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {5 \sqrt {1-a^2 x^2}}{a c^2}-\frac {5 \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx}{c^2}\\ &=\frac {(1+a x)^5}{5 a c^2 \left (1-a^2 x^2\right )^{5/2}}-\frac {2 (1+a x)^4}{3 a c^2 \left (1-a^2 x^2\right )^{3/2}}+\frac {10 (1+a x)^2}{3 a c^2 \sqrt {1-a^2 x^2}}+\frac {5 \sqrt {1-a^2 x^2}}{a c^2}-\frac {5 \sin ^{-1}(a x)}{a c^2}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 61, normalized size = 0.48 \[ \frac {\frac {\sqrt {1-a^2 x^2} \left (15 a^3 x^3-188 a^2 x^2+279 a x-118\right )}{(a x-1)^3}-75 \sin ^{-1}(a x)}{15 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[1 - a^2*x^2]*(-118 + 279*a*x - 188*a^2*x^2 + 15*a^3*x^3))/(-1 + a*x)^3 - 75*ArcSin[a*x])/(15*a*c^2)

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fricas [A]  time = 0.49, size = 143, normalized size = 1.12 \[ \frac {118 \, a^{3} x^{3} - 354 \, a^{2} x^{2} + 354 \, a x + 150 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + {\left (15 \, a^{3} x^{3} - 188 \, a^{2} x^{2} + 279 \, a x - 118\right )} \sqrt {-a^{2} x^{2} + 1} - 118}{15 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(118*a^3*x^3 - 354*a^2*x^2 + 354*a*x + 150*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*arctan((sqrt(-a^2*x^2 + 1) -
1)/(a*x)) + (15*a^3*x^3 - 188*a^2*x^2 + 279*a*x - 118)*sqrt(-a^2*x^2 + 1) - 118)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2
+ 3*a^2*c^2*x - a*c^2)

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giac [A]  time = 0.21, size = 180, normalized size = 1.41 \[ -\frac {5 \, \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{c^{2} {\left | a \right |}} + \frac {\sqrt {-a^{2} x^{2} + 1}}{a c^{2}} - \frac {2 \, {\left (\frac {440 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a^{2} x} - \frac {670 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{4} x^{2}} + \frac {360 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{6} x^{3}} - \frac {75 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{8} x^{4}} - 103\right )}}{15 \, c^{2} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*arcsin(a*x)*sgn(a)/(c^2*abs(a)) + sqrt(-a^2*x^2 + 1)/(a*c^2) - 2/15*(440*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a
^2*x) - 670*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^4*x^2) + 360*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3/(a^6*x^3) - 75
*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^8*x^4) - 103)/(c^2*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1)^5*abs(a
))

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maple [A]  time = 0.05, size = 212, normalized size = 1.66 \[ -\frac {a \,x^{2}}{c^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {14}{a \,c^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {25 x}{c^{2} \sqrt {-a^{2} x^{2}+1}}-\frac {5 \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{c^{2} \sqrt {a^{2}}}+\frac {8}{5 a^{3} c^{2} \left (x -\frac {1}{a}\right )^{2} \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}+\frac {116}{15 a^{2} c^{2} \left (x -\frac {1}{a}\right ) \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}-\frac {232 x}{15 c^{2} \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/c^2*x^2/(-a^2*x^2+1)^(1/2)+14/a/c^2/(-a^2*x^2+1)^(1/2)+25*x/c^2/(-a^2*x^2+1)^(1/2)-5/c^2/(a^2)^(1/2)*arctan
((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+8/5/a^3/c^2/(x-1/a)^2/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)+116/15/a^2/c^2/(x-
1/a)/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)-232/15/c^2/(-a^2*(x-1/a)^2-2*a*(x-1/a))^(1/2)*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a x + 1\right )}^{3}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (c - \frac {c}{a x}\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.82, size = 270, normalized size = 2.11 \[ \frac {8\,a^2\,\sqrt {1-a^2\,x^2}}{15\,\left (a^5\,c^2\,x^2-2\,a^4\,c^2\,x+a^3\,c^2\right )}-\frac {5\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{c^2\,\sqrt {-a^2}}-\frac {4\,a\,\sqrt {1-a^2\,x^2}}{a^4\,c^2\,x^2-2\,a^3\,c^2\,x+a^2\,c^2}+\frac {\sqrt {1-a^2\,x^2}}{a\,c^2}+\frac {143\,\sqrt {1-a^2\,x^2}}{15\,\sqrt {-a^2}\,\left (c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}\right )}+\frac {4\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^2\,x\,\sqrt {-a^2}-\frac {c^2\,\sqrt {-a^2}}{a}+a^2\,c^2\,x^3\,\sqrt {-a^2}-3\,a\,c^2\,x^2\,\sqrt {-a^2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*a^2*(1 - a^2*x^2)^(1/2))/(15*(a^3*c^2 - 2*a^4*c^2*x + a^5*c^2*x^2)) - (5*asinh(x*(-a^2)^(1/2)))/(c^2*(-a^2)
^(1/2)) - (4*a*(1 - a^2*x^2)^(1/2))/(a^2*c^2 - 2*a^3*c^2*x + a^4*c^2*x^2) + (1 - a^2*x^2)^(1/2)/(a*c^2) + (143
*(1 - a^2*x^2)^(1/2))/(15*(-a^2)^(1/2)*(c^2*x*(-a^2)^(1/2) - (c^2*(-a^2)^(1/2))/a)) + (4*(1 - a^2*x^2)^(1/2))/
(5*(-a^2)^(1/2)*(3*c^2*x*(-a^2)^(1/2) - (c^2*(-a^2)^(1/2))/a + a^2*c^2*x^3*(-a^2)^(1/2) - 3*a*c^2*x^2*(-a^2)^(
1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{2} \left (\int \frac {x^{2}}{- a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} + 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {3 a x^{3}}{- a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} + 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {3 a^{2} x^{4}}{- a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} + 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx + \int \frac {a^{3} x^{5}}{- a^{4} x^{4} \sqrt {- a^{2} x^{2} + 1} + 2 a^{3} x^{3} \sqrt {- a^{2} x^{2} + 1} - 2 a x \sqrt {- a^{2} x^{2} + 1} + \sqrt {- a^{2} x^{2} + 1}}\, dx\right )}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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