3.321 \(\int \frac {e^{\tanh ^{-1}(a x)} (c-a c x)^4}{x^2} \, dx\)

Optimal. Leaf size=106 \[ \frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+3 a c^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {1}{2} a c^4 \sin ^{-1}(a x) \]

[Out]

1/3*a*c^4*(-a^2*x^2+1)^(3/2)-c^4*(-a^2*x^2+1)^(3/2)/x+1/2*a*c^4*arcsin(a*x)+3*a*c^4*arctanh((-a^2*x^2+1)^(1/2)
)-1/2*a*c^4*(-a*x+6)*(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {6128, 1807, 1809, 815, 844, 216, 266, 63, 208} \[ \frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+3 a c^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {1}{2} a c^4 \sin ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x)^4)/x^2,x]

[Out]

-(a*c^4*(6 - a*x)*Sqrt[1 - a^2*x^2])/2 + (a*c^4*(1 - a^2*x^2)^(3/2))/3 - (c^4*(1 - a^2*x^2)^(3/2))/x + (a*c^4*
ArcSin[a*x])/2 + 3*a*c^4*ArcTanh[Sqrt[1 - a^2*x^2]]

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 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 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 815

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[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 1809

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x,
Expon[Pq, x]]}, Simp[(f*(c*x)^(m + q - 1)*(a + b*x^2)^(p + 1))/(b*c^(q - 1)*(m + q + 2*p + 1)), x] + Dist[1/(
b*(m + q + 2*p + 1)), Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*x^q
- a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x]
&& PolyQ[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -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)} (c-a c x)^4}{x^2} \, dx &=c \int \frac {(c-a c x)^3 \sqrt {1-a^2 x^2}}{x^2} \, dx\\ &=-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-c \int \frac {\sqrt {1-a^2 x^2} \left (3 a c^3-a^2 c^3 x+a^3 c^3 x^2\right )}{x} \, dx\\ &=\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {c \int \frac {\left (-9 a^3 c^3+3 a^4 c^3 x\right ) \sqrt {1-a^2 x^2}}{x} \, dx}{3 a^2}\\ &=-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\frac {c \int \frac {18 a^5 c^3-3 a^6 c^3 x}{x \sqrt {1-a^2 x^2}} \, dx}{6 a^4}\\ &=-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}-\left (3 a c^4\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx+\frac {1}{2} \left (a^2 c^4\right ) \int \frac {1}{\sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {1}{2} a c^4 \sin ^{-1}(a x)-\frac {1}{2} \left (3 a c^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {1}{2} a c^4 \sin ^{-1}(a x)+\frac {\left (3 c^4\right ) \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}\\ &=-\frac {1}{2} a c^4 (6-a x) \sqrt {1-a^2 x^2}+\frac {1}{3} a c^4 \left (1-a^2 x^2\right )^{3/2}-\frac {c^4 \left (1-a^2 x^2\right )^{3/2}}{x}+\frac {1}{2} a c^4 \sin ^{-1}(a x)+3 a c^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 152, normalized size = 1.43 \[ -\frac {c^4 \left (-2 a^5 x^5+9 a^4 x^4-14 a^3 x^3-15 a^2 x^2+9 a x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)+24 a x \sqrt {1-a^2 x^2} \sin ^{-1}\left (\frac {\sqrt {1-a x}}{\sqrt {2}}\right )-18 a x \sqrt {1-a^2 x^2} \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+16 a x+6\right )}{6 x \sqrt {1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x)^4)/x^2,x]

[Out]

-1/6*(c^4*(6 + 16*a*x - 15*a^2*x^2 - 14*a^3*x^3 + 9*a^4*x^4 - 2*a^5*x^5 + 9*a*x*Sqrt[1 - a^2*x^2]*ArcSin[a*x]
+ 24*a*x*Sqrt[1 - a^2*x^2]*ArcSin[Sqrt[1 - a*x]/Sqrt[2]] - 18*a*x*Sqrt[1 - a^2*x^2]*ArcTanh[Sqrt[1 - a^2*x^2]]
))/(x*Sqrt[1 - a^2*x^2])

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fricas [A]  time = 0.47, size = 116, normalized size = 1.09 \[ -\frac {6 \, a c^{4} x \arctan \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{a x}\right ) + 18 \, a c^{4} x \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + 16 \, a c^{4} x + {\left (2 \, a^{3} c^{4} x^{3} - 9 \, a^{2} c^{4} x^{2} + 16 \, a c^{4} x + 6 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a*c^4*x*arctan((sqrt(-a^2*x^2 + 1) - 1)/(a*x)) + 18*a*c^4*x*log((sqrt(-a^2*x^2 + 1) - 1)/x) + 16*a*c^4
*x + (2*a^3*c^4*x^3 - 9*a^2*c^4*x^2 + 16*a*c^4*x + 6*c^4)*sqrt(-a^2*x^2 + 1))/x

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giac [A]  time = 0.25, size = 164, normalized size = 1.55 \[ \frac {a^{4} c^{4} x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left | a \right |}} + \frac {a^{2} c^{4} \arcsin \left (a x\right ) \mathrm {sgn}\relax (a)}{2 \, {\left | a \right |}} + \frac {3 \, a^{2} c^{4} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} - \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} c^{4}}{2 \, x {\left | a \right |}} - \frac {1}{6} \, {\left (16 \, a c^{4} + {\left (2 \, a^{3} c^{4} x - 9 \, a^{2} c^{4}\right )} x\right )} \sqrt {-a^{2} x^{2} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^4*c^4*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*abs(a)) + 1/2*a^2*c^4*arcsin(a*x)*sgn(a)/abs(a) + 3*a^2*c^4*log
(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*c^4/(x
*abs(a)) - 1/6*(16*a*c^4 + (2*a^3*c^4*x - 9*a^2*c^4)*x)*sqrt(-a^2*x^2 + 1)

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maple [A]  time = 0.04, size = 136, normalized size = 1.28 \[ -\frac {c^{4} a^{3} x^{2} \sqrt {-a^{2} x^{2}+1}}{3}-\frac {8 c^{4} a \sqrt {-a^{2} x^{2}+1}}{3}+\frac {3 c^{4} a^{2} x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {c^{4} a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}+3 c^{4} a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {c^{4} \sqrt {-a^{2} x^{2}+1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.46, size = 125, normalized size = 1.18 \[ -\frac {1}{3} \, \sqrt {-a^{2} x^{2} + 1} a^{3} c^{4} x^{2} + \frac {3}{2} \, \sqrt {-a^{2} x^{2} + 1} a^{2} c^{4} x + \frac {1}{2} \, a c^{4} \arcsin \left (a x\right ) + 3 \, a c^{4} \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {8}{3} \, \sqrt {-a^{2} x^{2} + 1} a c^{4} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{4}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(-a^2*x^2 + 1)*a^3*c^4*x^2 + 3/2*sqrt(-a^2*x^2 + 1)*a^2*c^4*x + 1/2*a*c^4*arcsin(a*x) + 3*a*c^4*log(2
*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) - 8/3*sqrt(-a^2*x^2 + 1)*a*c^4 - sqrt(-a^2*x^2 + 1)*c^4/x

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mupad [B]  time = 0.04, size = 131, normalized size = 1.24 \[ \frac {3\,a^2\,c^4\,x\,\sqrt {1-a^2\,x^2}}{2}-\frac {c^4\,\sqrt {1-a^2\,x^2}}{x}-\frac {8\,a\,c^4\,\sqrt {1-a^2\,x^2}}{3}+\frac {a^2\,c^4\,\mathrm {asinh}\left (x\,\sqrt {-a^2}\right )}{2\,\sqrt {-a^2}}-\frac {a^3\,c^4\,x^2\,\sqrt {1-a^2\,x^2}}{3}-a\,c^4\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 6.68, size = 306, normalized size = 2.89 \[ a^{5} c^{4} \left (\begin {cases} - \frac {x^{2} \sqrt {- a^{2} x^{2} + 1}}{3 a^{2}} - \frac {2 \sqrt {- a^{2} x^{2} + 1}}{3 a^{4}} & \text {for}\: a \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) - 3 a^{4} c^{4} \left (\begin {cases} - \frac {i x \sqrt {a^{2} x^{2} - 1}}{2 a^{2}} - \frac {i \operatorname {acosh}{\left (a x \right )}}{2 a^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\\frac {x^{3}}{2 \sqrt {- a^{2} x^{2} + 1}} - \frac {x}{2 a^{2} \sqrt {- a^{2} x^{2} + 1}} + \frac {\operatorname {asin}{\left (a x \right )}}{2 a^{3}} & \text {otherwise} \end {cases}\right ) + 2 a^{3} c^{4} \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a^{2} = 0 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{a^{2}} & \text {otherwise} \end {cases}\right ) + 2 a^{2} c^{4} \left (\begin {cases} \sqrt {\frac {1}{a^{2}}} \operatorname {asin}{\left (x \sqrt {a^{2}} \right )} & \text {for}\: a^{2} > 0 \\\sqrt {- \frac {1}{a^{2}}} \operatorname {asinh}{\left (x \sqrt {- a^{2}} \right )} & \text {for}\: a^{2} < 0 \end {cases}\right ) - 3 a c^{4} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + c^{4} \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**5*c**4*Piecewise((-x**2*sqrt(-a**2*x**2 + 1)/(3*a**2) - 2*sqrt(-a**2*x**2 + 1)/(3*a**4), Ne(a, 0)), (x**4/4
, True)) - 3*a**4*c**4*Piecewise((-I*x*sqrt(a**2*x**2 - 1)/(2*a**2) - I*acosh(a*x)/(2*a**3), Abs(a**2*x**2) >
1), (x**3/(2*sqrt(-a**2*x**2 + 1)) - x/(2*a**2*sqrt(-a**2*x**2 + 1)) + asin(a*x)/(2*a**3), True)) + 2*a**3*c**
4*Piecewise((x**2/2, Eq(a**2, 0)), (-sqrt(-a**2*x**2 + 1)/a**2, True)) + 2*a**2*c**4*Piecewise((sqrt(a**(-2))*
asin(x*sqrt(a**2)), a**2 > 0), (sqrt(-1/a**2)*asinh(x*sqrt(-a**2)), a**2 < 0)) - 3*a*c**4*Piecewise((-acosh(1/
(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + c**4*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x
**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True))

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