\(\int \frac {e^{3 \text {arctanh}(a x)} (c-a^2 c x^2)}{x^4} \, dx\) [1166]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=-\frac {c \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a c \sqrt {1-a^2 x^2}}{2 x^2}-\frac {11 a^2 c \sqrt {1-a^2 x^2}}{3 x}-\frac {5}{2} a^3 c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.64 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=-\frac {c \sqrt {1-a^2 x^2} \left (2+9 a x+22 a^2 x^2\right )}{6 x^3}-\frac {5}{2} a^3 c \text {arctanh}\left (\sqrt {1-a^2 x^2}\right ) \] Input:

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

Output:

-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
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {6698, 540, 25, 2338, 25, 27, 534, 243, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx\)

\(\Big \downarrow \) 6698

\(\displaystyle c \int \frac {(a x+1)^3}{x^4 \sqrt {1-a^2 x^2}}dx\)

\(\Big \downarrow \) 540

\(\displaystyle c \left (-\frac {1}{3} \int -\frac {3 x^2 a^3+11 x a^2+9 a}{x^3 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {1}{3} \int \frac {3 x^2 a^3+11 x a^2+9 a}{x^3 \sqrt {1-a^2 x^2}}dx-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 2338

\(\displaystyle c \left (\frac {1}{3} \left (-\frac {1}{2} \int -\frac {a^2 (15 a x+22)}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {a^2 (15 a x+22)}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (\frac {1}{3} \left (\frac {1}{2} a^2 \int \frac {15 a x+22}{x^2 \sqrt {1-a^2 x^2}}dx-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 534

\(\displaystyle c \left (\frac {1}{3} \left (\frac {1}{2} a^2 \left (15 a \int \frac {1}{x \sqrt {1-a^2 x^2}}dx-\frac {22 \sqrt {1-a^2 x^2}}{x}\right )-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle c \left (\frac {1}{3} \left (\frac {1}{2} a^2 \left (\frac {15}{2} a \int \frac {1}{x^2 \sqrt {1-a^2 x^2}}dx^2-\frac {22 \sqrt {1-a^2 x^2}}{x}\right )-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c \left (\frac {1}{3} \left (\frac {1}{2} a^2 \left (-\frac {15 \int \frac {1}{\frac {1}{a^2}-\frac {x^4}{a^2}}d\sqrt {1-a^2 x^2}}{a}-\frac {22 \sqrt {1-a^2 x^2}}{x}\right )-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c \left (\frac {1}{3} \left (\frac {1}{2} a^2 \left (-15 a \text {arctanh}\left (\sqrt {1-a^2 x^2}\right )-\frac {22 \sqrt {1-a^2 x^2}}{x}\right )-\frac {9 a \sqrt {1-a^2 x^2}}{2 x^2}\right )-\frac {\sqrt {1-a^2 x^2}}{3 x^3}\right )\)

Input:

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

Output:

c*(-1/3*Sqrt[1 - a^2*x^2]/x^3 + ((-9*a*Sqrt[1 - a^2*x^2])/(2*x^2) + (a^2*( 
(-22*Sqrt[1 - a^2*x^2])/x - 15*a*ArcTanh[Sqrt[1 - a^2*x^2]]))/2)/3)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 2338

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]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[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] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

rule 6698

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x 
_Symbol] :> Simp[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] || GtQ[c, 
 0]) && IGtQ[(n + 1)/2, 0] &&  !IntegerQ[p - n/2]
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {\left (22 a^{4} x^{4}+9 a^{3} x^{3}-20 a^{2} x^{2}-9 a x -2\right ) c}{6 x^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {5 a^{3} \operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right ) c}{2}\)	\(69\)
default	\(-c \left (\frac {a^{3}}{\sqrt {-a^{2} x^{2}+1}}+\frac {3 a^{4} x}{\sqrt {-a^{2} x^{2}+1}}+\frac {1}{3 x^{3} \sqrt {-a^{2} x^{2}+1}}-\frac {10 a^{2} \left (-\frac {1}{x \sqrt {-a^{2} x^{2}+1}}+\frac {2 a^{2} x}{\sqrt {-a^{2} x^{2}+1}}\right )}{3}-3 a \left (-\frac {1}{2 x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 a^{2} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}\right )+2 a^{3} \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\right )\)	\(184\)
meijerg	\(-\frac {2 a^{2} c \left (-2 a^{2} x^{2}+1\right )}{x \sqrt {-a^{2} x^{2}+1}}-\frac {c \left (-8 a^{4} x^{4}+4 a^{2} x^{2}+1\right )}{3 x^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {a^{3} c \left (\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {\pi }}-\frac {2 a^{3} c \left (\frac {\left (2-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{2}-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {-a^{2} x^{2}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )\right )}{\sqrt {\pi }}-\frac {3 a^{4} c x}{\sqrt {-a^{2} x^{2}+1}}-\frac {3 a^{3} c \left (\frac {\sqrt {\pi }}{2 x^{2} a^{2}}-\frac {3 \left (\frac {5}{3}-2 \ln \left (2\right )+2 \ln \left (x \right )+\ln \left (-a^{2}\right )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }\, \left (-20 a^{2} x^{2}+8\right )}{16 a^{2} x^{2}}+\frac {\sqrt {\pi }\, \left (-24 a^{2} x^{2}+8\right )}{16 a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}+\frac {3 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-a^{2} x^{2}+1}}{2}\right )}{2}\right )}{\sqrt {\pi }}\)	\(309\)

Input:

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

Output:

1/6*(22*a^4*x^4+9*a^3*x^3-20*a^2*x^2-9*a*x-2)/x^3/(-a^2*x^2+1)^(1/2)*c-5/2 
*a^3*arctanh(1/(-a^2*x^2+1)^(1/2))*c
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=\frac {15 \, a^{3} c x^{3} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) - {\left (22 \, a^{2} c x^{2} + 9 \, a c x + 2 \, c\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, x^{3}} \] Input:

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

Output:

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
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.82 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.82 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=a^{3} c \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{a x} \right )} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{a x} \right )} & \text {otherwise} \end {cases}\right ) + 3 a^{2} c \left (\begin {cases} - \frac {i \sqrt {a^{2} x^{2} - 1}}{x} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {\sqrt {- a^{2} x^{2} + 1}}{x} & \text {otherwise} \end {cases}\right ) + 3 a c \left (\begin {cases} - \frac {a^{2} \operatorname {acosh}{\left (\frac {1}{a x} \right )}}{2} + \frac {a}{2 x \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{2 a x^{3} \sqrt {-1 + \frac {1}{a^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{a^{2} x^{2}}\right |} > 1 \\\frac {i a^{2} \operatorname {asin}{\left (\frac {1}{a x} \right )}}{2} - \frac {i a \sqrt {1 - \frac {1}{a^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right ) + c \left (\begin {cases} - \frac {2 i a^{2} \sqrt {a^{2} x^{2} - 1}}{3 x} - \frac {i \sqrt {a^{2} x^{2} - 1}}{3 x^{3}} & \text {for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac {2 a^{2} \sqrt {- a^{2} x^{2} + 1}}{3 x} - \frac {\sqrt {- a^{2} x^{2} + 1}}{3 x^{3}} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

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*sqrt(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/(2*x*sqrt(-1 + 1/(a**2*x**2))) - 1/(2*a*x**3*sqrt(-1 + 1/(a**2*x* 
*2))), 1/Abs(a**2*x**2) > 1), (I*a**2*asin(1/(a*x))/2 - I*a*sqrt(1 - 1/(a* 
*2*x**2))/(2*x), 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))
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.36 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=\frac {11 \, a^{4} c x}{3 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {5}{2} \, a^{3} c \log \left (\frac {2 \, \sqrt {-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {3 \, a^{3} c}{2 \, \sqrt {-a^{2} x^{2} + 1}} - \frac {10 \, a^{2} c}{3 \, \sqrt {-a^{2} x^{2} + 1} x} - \frac {3 \, a c}{2 \, \sqrt {-a^{2} x^{2} + 1} x^{2}} - \frac {c}{3 \, \sqrt {-a^{2} x^{2} + 1} x^{3}} \] Input:

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

Output:

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)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (78) = 156\).

Time = 0.14 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.32 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=\frac {{\left (a^{4} c + \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{2} c}{x} + \frac {45 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} c}{x^{2}}\right )} a^{6} x^{3}}{24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} {\left | a \right |}} - \frac {5 \, a^{4} c \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{2 \, {\left | a \right |}} - \frac {\frac {45 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{4} c}{x} + \frac {9 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{2} c}{x^{2}} + \frac {{\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} c}{x^{3}}}{24 \, a^{2} {\left | a \right |}} \] Input:

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

Output:

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*abs( 
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))
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=-\frac {c\,\sqrt {1-a^2\,x^2}}{3\,x^3}-\frac {11\,a^2\,c\,\sqrt {1-a^2\,x^2}}{3\,x}-\frac {3\,a\,c\,\sqrt {1-a^2\,x^2}}{2\,x^2}+\frac {a^3\,c\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{2} \] Input:

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

Output:

(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)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {e^{3 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )}{x^4} \, dx=\frac {c \left (-22 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}-9 \sqrt {-a^{2} x^{2}+1}\, a x -2 \sqrt {-a^{2} x^{2}+1}+15 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (a x \right )}{2}\right )\right ) a^{3} x^{3}\right )}{6 x^{3}} \] Input:

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

Output:

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