3.1144 \(\int \frac {e^{3 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^4} \, dx\)

Optimal. Leaf size=94 \[ -\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {5}{2} a^3 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

-5/2*a^3*c*arctanh((-a^2*x^2+1)^(1/2))-1/3*c*(-a^2*x^2+1)^(1/2)/x^3-3/2*a*c*(-a^2*x^2+1)^(1/2)/x^2-11/3*a^2*c*
(-a^2*x^2+1)^(1/2)/x

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Rubi [A]  time = 0.20, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {6148, 1807, 807, 266, 63, 208} \[ -\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {5}{2} a^3 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2))/x^4,x]

[Out]

-(c*Sqrt[1 - a^2*x^2])/(3*x^3) - (3*a*c*Sqrt[1 - a^2*x^2])/(2*x^2) - (11*a^2*c*Sqrt[1 - a^2*x^2])/(3*x) - (5*a
^3*c*ArcTanh[Sqrt[1 - a^2*x^2]])/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 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 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 6148

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

Rubi steps

\begin {align*} \int \frac {e^{3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx &=c \int \frac {(1+a x)^3}{x^4 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {1}{3} c \int \frac {-9 a-11 a^2 x-3 a^3 x^2}{x^3 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}+\frac {1}{6} c \int \frac {22 a^2+15 a^3 x}{x^2 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{2} \left (5 a^3 c\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{4} \left (5 a^3 c\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}-\frac {1}{2} (5 a c) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}-\frac {5}{2} a^3 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 60, normalized size = 0.64 \[ -\frac {c \sqrt {1-a^2 x^2} \left (22 a^2 x^2+9 a x+2\right )}{6 x^3}-\frac {5}{2} a^3 c \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(3*ArcTanh[a*x])*(c - a^2*c*x^2))/x^4,x]

[Out]

-1/6*(c*Sqrt[1 - a^2*x^2]*(2 + 9*a*x + 22*a^2*x^2))/x^3 - (5*a^3*c*ArcTanh[Sqrt[1 - a^2*x^2]])/2

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fricas [A]  time = 0.65, size = 66, normalized size = 0.70 \[ \frac {15 \, a^{3} c x^{3} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (22 \, a^{2} c x^{2} + 9 \, a c x + 2 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^4,x, algorithm="fricas")

[Out]

1/6*(15*a^3*c*x^3*log((sqrt(-a^2*x^2 + 1) - 1)/x) - (22*a^2*c*x^2 + 9*a*c*x + 2*c)*sqrt(-a^2*x^2 + 1))/x^3

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giac [B]  time = 0.23, size = 218, normalized size = 2.32 \[ \frac {{\left (a^{4} c + \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2} c}{x} + \frac {45 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c}{x^{2}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left | a \right |}} - \frac {5 \, a^{4} c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} - \frac {\frac {45 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4} c}{x} + \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2} c}{x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c}{x^{3}}}{24 \, a^{2} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^4,x, algorithm="giac")

[Out]

1/24*(a^4*c + 9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2*c/x + 45*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c/x^2)*a^6*x^3/
((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) - 5/2*a^4*c*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*ab
s(x)))/abs(a) - 1/24*(45*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c/x + 9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2*c/x
^2 + (sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c/x^3)/(a^2*abs(a))

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maple [B]  time = 0.04, size = 184, normalized size = 1.96 \[ -c \left (\frac {a^{3}}{\sqrt {-a^{2} x^{2}+1}}+\frac {3 x \,a^{4}}{\sqrt {-a^{2} x^{2}+1}}+2 a^{3} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )-\frac {10 a^{2} \left (-\frac {1}{x \sqrt {-a^{2} x^{2}+1}}+\frac {2 a^{2} x}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}+\frac {1}{3 x^{3} \sqrt {-a^{2} x^{2}+1}}-3 a \left (-\frac {1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 a^{2} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^4,x)

[Out]

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

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maxima [A]  time = 0.33, size = 128, normalized size = 1.36 \[ \frac {11 \, a^{4} c x}{3 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {5}{2} \, a^{3} c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {3 \, a^{3} c}{2 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {10 \, a^{2} c}{3 \, \sqrt {-a^{2} x^{2} + 1} x} - \frac {3 \, a c}{2 \, \sqrt {-a^{2} x^{2} + 1} x^{2}} - \frac {c}{3 \, \sqrt {-a^{2} x^{2} + 1} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a^2*c*x^2+c)/x^4,x, algorithm="maxima")

[Out]

11/3*a^4*c*x/sqrt(-a^2*x^2 + 1) - 5/2*a^3*c*log(2*sqrt(-a^2*x^2 + 1)/abs(x) + 2/abs(x)) + 3/2*a^3*c/sqrt(-a^2*
x^2 + 1) - 10/3*a^2*c/(sqrt(-a^2*x^2 + 1)*x) - 3/2*a*c/(sqrt(-a^2*x^2 + 1)*x^2) - 1/3*c/(sqrt(-a^2*x^2 + 1)*x^
3)

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mupad [B]  time = 0.89, size = 82, normalized size = 0.87 \[ -\frac {c\,\sqrt {1-a^2\,x^2}}{3\,x^3}-\frac {11\,a^2\,c\,\sqrt {1-a^2\,x^2}}{3\,x}-\frac {3\,a\,c\,\sqrt {1-a^2\,x^2}}{2\,x^2}+\frac {a^3\,c\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c - a^2*c*x^2)*(a*x + 1)^3)/(x^4*(1 - a^2*x^2)^(3/2)),x)

[Out]

(a^3*c*atan((1 - a^2*x^2)^(1/2)*1i)*5i)/2 - (c*(1 - a^2*x^2)^(1/2))/(3*x^3) - (11*a^2*c*(1 - a^2*x^2)^(1/2))/(
3*x) - (3*a*c*(1 - a^2*x^2)^(1/2))/(2*x^2)

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sympy [C]  time = 28.08, size = 267, normalized size = 2.84 \[ a^{3} c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + 3 a^{2} c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) + 3 a c \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} - \frac {a \sqrt {-1 + \frac {1}{a^{2} x^{2}}}}{2 x} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a}{2 x \sqrt {1 - \frac {1}{a^{2} x^{2}}}} + \frac {i}{2 a x^{3} \sqrt {1 - \frac {1}{a^{2} x^{2}}}} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c*Piecewise((-acosh(1/(a*x)), 1/Abs(a**2*x**2) > 1), (I*asin(1/(a*x)), True)) + 3*a**2*c*Piecewise((-I*sq
rt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) + 3*a*c*Piecewise((-a**2*acosh(1/(a
*x))/2 - a*sqrt(-1 + 1/(a**2*x**2))/(2*x), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*x))/2 - I*a/(2*x*sqrt(1 -
1/(a**2*x**2))) + I/(2*a*x**3*sqrt(1 - 1/(a**2*x**2))), True)) + c*Piecewise((-2*I*a**2*sqrt(a**2*x**2 - 1)/(3
*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a**2*sqrt(-a**2*x**2 + 1)/(3*x) - sqrt(-a**2*x*
*2 + 1)/(3*x**3), True))

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