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

Optimal. Leaf size=97 \[ \frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 a x+5}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {8 (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^3} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6128, 852, 1805, 823, 12, 266, 63, 208} \[ \frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {8 a x+5}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {8 (a x+1)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(1 + a*x))/(5*c^3*(1 - a^2*x^2)^(5/2)) + (4*a*x)/(5*c^3*(1 - a^2*x^2)^(3/2)) + (5 + 8*a*x)/(5*c^3*Sqrt[1 -
a^2*x^2]) - ArcTanh[Sqrt[1 - a^2*x^2]]/c^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 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 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[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]

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{x (c-a c x)^3} \, dx &=c \int \frac {\sqrt {1-a^2 x^2}}{x (c-a c x)^4} \, dx\\ &=\frac {\int \frac {(c+a c x)^4}{x \left (1-a^2 x^2\right )^{7/2}} \, dx}{c^7}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 c^4-12 a c^4 x+5 a^2 c^4 x^2}{x \left (1-a^2 x^2\right )^{5/2}} \, dx}{5 c^7}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {\int \frac {15 c^4+24 a c^4 x}{x \left (1-a^2 x^2\right )^{3/2}} \, dx}{15 c^7}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {5+8 a x}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {\int \frac {15 a^2 c^4}{x \sqrt {1-a^2 x^2}} \, dx}{15 a^2 c^7}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {5+8 a x}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {\int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {5+8 a x}{5 c^3 \sqrt {1-a^2 x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{2 c^3}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {5+8 a x}{5 c^3 \sqrt {1-a^2 x^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )}{a^2 c^3}\\ &=\frac {8 (1+a x)}{5 c^3 \left (1-a^2 x^2\right )^{5/2}}+\frac {4 a x}{5 c^3 \left (1-a^2 x^2\right )^{3/2}}+\frac {5+8 a x}{5 c^3 \sqrt {1-a^2 x^2}}-\frac {\tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c^3}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 71, normalized size = 0.73 \[ \frac {24 a^5 x^5-60 a^3 x^3+3 \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};1-a^2 x^2\right )+5 a^2 x^2+60 a x+16}{15 c^3 \left (1-a^2 x^2\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(16 + 60*a*x + 5*a^2*x^2 - 60*a^3*x^3 + 24*a^5*x^5 + 3*Hypergeometric2F1[-5/2, 1, -3/2, 1 - a^2*x^2])/(15*c^3*
(1 - a^2*x^2)^(5/2))

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fricas [A]  time = 0.59, size = 130, normalized size = 1.34 \[ \frac {13 \, a^{3} x^{3} - 39 \, a^{2} x^{2} + 39 \, a x + 5 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (8 \, a^{2} x^{2} - 19 \, a x + 13\right )} \sqrt {-a^{2} x^{2} + 1} - 13}{5 \, {\left (a^{3} c^{3} x^{3} - 3 \, a^{2} c^{3} x^{2} + 3 \, a c^{3} x - c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(13*a^3*x^3 - 39*a^2*x^2 + 39*a*x + 5*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log((sqrt(-a^2*x^2 + 1) - 1)/x) -
(8*a^2*x^2 - 19*a*x + 13)*sqrt(-a^2*x^2 + 1) - 13)/(a^3*c^3*x^3 - 3*a^2*c^3*x^2 + 3*a*c^3*x - c^3)

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giac [B]  time = 0.22, size = 189, normalized size = 1.95 \[ -\frac {a \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c^{3} {\left | a \right |}} + \frac {2 \, {\left (13 \, a - \frac {45 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{a x} + \frac {75 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{a^{3} x^{2}} - \frac {55 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{5} x^{3}} + \frac {20 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4}}{a^{7} x^{4}}\right )}}{5 \, c^{3} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )}^{5} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c^3*abs(a)) + 2/5*(13*a - 45*(sqrt(-a^2*x^2
+ 1)*abs(a) + a)/(a*x) + 75*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/(a^3*x^2) - 55*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3
/(a^5*x^3) + 20*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4/(a^7*x^4))/(c^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) - 1
)^5*abs(a))

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maple [B]  time = 0.04, size = 275, normalized size = 2.84 \[ -\frac {\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}}{a}+\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{a \left (x -\frac {1}{a}\right )}+\frac {\frac {2 \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{5 a \left (x -\frac {1}{a}\right )^{3}}-\frac {4 a \left (\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 a \left (x -\frac {1}{a}\right )^{2}}-\frac {\sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{3 \left (x -\frac {1}{a}\right )}\right )}{5}}{a^{2}}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 209, normalized size = 2.15 \[ \frac {3\,a^2\,\sqrt {1-a^2\,x^2}}{5\,\left (a^4\,c^3\,x^2-2\,a^3\,c^3\,x+a^2\,c^3\right )}+\frac {8\,a\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}\right )}+\frac {2\,a\,\sqrt {1-a^2\,x^2}}{5\,\sqrt {-a^2}\,\left (3\,c^3\,x\,\sqrt {-a^2}-\frac {c^3\,\sqrt {-a^2}}{a}+a^2\,c^3\,x^3\,\sqrt {-a^2}-3\,a\,c^3\,x^2\,\sqrt {-a^2}\right )}+\frac {\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((1 - a^2*x^2)^(1/2)*1i)*1i)/c^3 + (3*a^2*(1 - a^2*x^2)^(1/2))/(5*(a^2*c^3 - 2*a^3*c^3*x + a^4*c^3*x^2))
+ (8*a*(1 - a^2*x^2)^(1/2))/(5*(-a^2)^(1/2)*(c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a)) + (2*a*(1 - a^2*x^2)^
(1/2))/(5*(-a^2)^(1/2)*(3*c^3*x*(-a^2)^(1/2) - (c^3*(-a^2)^(1/2))/a + a^2*c^3*x^3*(-a^2)^(1/2) - 3*a*c^3*x^2*(
-a^2)^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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