3.335 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^4 (c-a c x)} \, dx\)

Optimal. Leaf size=125 \[ -\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}+\frac {2 a^3 (a x+1)}{c \sqrt {1-a^2 x^2}}-\frac {3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 125, 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, 1807, 807, 266, 63, 208} \[ \frac {2 a^3 (a x+1)}{c \sqrt {1-a^2 x^2}}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]/(x^4*(c - a*c*x)),x]

[Out]

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

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 807

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

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 1807

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

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^4 (c-a c x)} \, dx &=c \int \frac {\sqrt {1-a^2 x^2}}{x^4 (c-a c x)^2} \, dx\\ &=\frac {\int \frac {(c+a c x)^2}{x^4 \left (1-a^2 x^2\right )^{3/2}} \, dx}{c^3}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\int \frac {-c^2-2 a c^2 x-2 a^2 c^2 x^2-2 a^3 c^2 x^3}{x^4 \sqrt {1-a^2 x^2}} \, dx}{c^3}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}+\frac {\int \frac {6 a c^2+8 a^2 c^2 x+6 a^3 c^2 x^2}{x^3 \sqrt {1-a^2 x^2}} \, dx}{3 c^3}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {\int \frac {-16 a^2 c^2-18 a^3 c^2 x}{x^2 \sqrt {1-a^2 x^2}} \, dx}{6 c^3}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}+\frac {\left (3 a^3\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx}{c}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}+\frac {\left (3 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )}{2 c}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {(3 a) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )}{c}\\ &=\frac {2 a^3 (1+a x)}{c \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{3 c x^3}-\frac {a \sqrt {1-a^2 x^2}}{c x^2}-\frac {8 a^2 \sqrt {1-a^2 x^2}}{3 c x}-\frac {3 a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 91, normalized size = 0.73 \[ -\frac {-14 a^4 x^4-9 a^3 x^3+7 a^2 x^2+9 a^3 x^3 \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+3 a x+1}{3 c x^3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]/(x^4*(c - a*c*x)),x]

[Out]

-1/3*(1 + 3*a*x + 7*a^2*x^2 - 9*a^3*x^3 - 14*a^4*x^4 + 9*a^3*x^3*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]])
/(c*x^3*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.41, size = 107, normalized size = 0.86 \[ \frac {6 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 9 \, {\left (a^{4} x^{4} - a^{3} x^{3}\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (14 \, a^{3} x^{3} - 5 \, a^{2} x^{2} - 2 \, a x - 1\right )} \sqrt {-a^{2} x^{2} + 1}}{3 \, {\left (a c x^{4} - c x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x, algorithm="fricas")

[Out]

1/3*(6*a^4*x^4 - 6*a^3*x^3 + 9*(a^4*x^4 - a^3*x^3)*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (14*a^3*x^3 - 5*a^2*x^2 -
2*a*x - 1)*sqrt(-a^2*x^2 + 1))/(a*c*x^4 - c*x^3)

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giac [B]  time = 0.19, size = 283, normalized size = 2.26 \[ -\frac {{\left (a^{4} + \frac {5 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2}}{x} + \frac {27 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2}}{x^{2}} - \frac {129 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a^{2} x^{3}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} - 1\right )} {\left | a \right |}} - \frac {3 \, a^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{c {\left | a \right |}} - \frac {\frac {33 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4} c^{2}}{x} + \frac {6 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2} c^{2}}{x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c^{2}}{x^{3}}}{24 \, a^{2} c^{3} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x, algorithm="giac")

[Out]

-1/24*(a^4 + 5*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2/x + 27*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/x^2 - 129*(sqrt(-a
^2*x^2 + 1)*abs(a) + a)^3/(a^2*x^3))*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c*((sqrt(-a^2*x^2 + 1)*abs(a)
+ a)/(a^2*x) - 1)*abs(a)) - 3*a^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/(c*abs(a)) - 1
/24*(33*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c^2/x + 6*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c^2/x^2 + (sqrt(-a
^2*x^2 + 1)*abs(a) + a)^3*c^2/x^3)/(a^2*c^3*abs(a))

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maple [A]  time = 0.04, size = 142, normalized size = 1.14 \[ -\frac {2 a^{3} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )+\frac {8 a^{2} \sqrt {-a^{2} x^{2}+1}}{3 x}+\frac {\sqrt {-a^{2} x^{2}+1}}{3 x^{3}}+\frac {2 a^{2} \sqrt {-a^{2} \left (x -\frac {1}{a}\right )^{2}-2 a \left (x -\frac {1}{a}\right )}}{x -\frac {1}{a}}-2 a \left (-\frac {\sqrt {-a^{2} x^{2}+1}}{2 x^{2}}-\frac {a^{2} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)/x^4/(-a*c*x+c),x, algorithm="maxima")

[Out]

-integrate((a*x + 1)/(sqrt(-a^2*x^2 + 1)*(a*c*x - c)*x^4), x)

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mupad [B]  time = 0.06, size = 140, normalized size = 1.12 \[ -\frac {\sqrt {1-a^2\,x^2}}{3\,c\,x^3}-\frac {a\,\sqrt {1-a^2\,x^2}}{c\,x^2}-\frac {8\,a^2\,\sqrt {1-a^2\,x^2}}{3\,c\,x}-\frac {2\,a^4\,\sqrt {1-a^2\,x^2}}{\left (\frac {c\,\sqrt {-a^2}}{a}-c\,x\,\sqrt {-a^2}\right )\,\sqrt {-a^2}}+\frac {a^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + 1)/(x^4*(1 - a^2*x^2)^(1/2)*(c - a*c*x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)/x**4/(-a*c*x+c),x)

[Out]

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

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