3.1.8 \(\int \frac {(d+c^2 d x^2) (a+b \text {arcsinh}(c x))}{x^3} \, dx\) [8]

3.1.8.1 Optimal result

Integrand size = 22, antiderivative size = 128 \[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=-\frac {b c d \sqrt {1+c^2 x^2}}{2 x}+\frac {1}{2} b c^2 d \text {arcsinh}(c x)-\frac {d \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {c^2 d (a+b \text {arcsinh}(c x))^2}{2 b}+c^2 d (a+b \text {arcsinh}(c x)) \log \left (1-e^{-2 \text {arcsinh}(c x)}\right )-\frac {1}{2} b c^2 d \operatorname {PolyLog}\left (2,e^{-2 \text {arcsinh}(c x)}\right ) \]

output

1/2*b*c^2*d*arcsinh(c*x)-1/2*d*(c^2*x^2+1)*(a+b*arcsinh(c*x))/x^2+1/2*c^2* 
d*(a+b*arcsinh(c*x))^2/b+c^2*d*(a+b*arcsinh(c*x))*ln(1-1/(c*x+(c^2*x^2+1)^ 
(1/2))^2)-1/2*b*c^2*d*polylog(2,1/(c*x+(c^2*x^2+1)^(1/2))^2)-1/2*b*c*d*(c^ 
2*x^2+1)^(1/2)/x
 

3.1.8.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.87 \[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=-\frac {a d}{2 x^2}-\frac {b c d \sqrt {1+c^2 x^2}}{2 x}-\frac {b d \text {arcsinh}(c x)}{2 x^2}-\frac {1}{2} b c^2 d \text {arcsinh}(c x)^2+b c^2 d \text {arcsinh}(c x) \log \left (1-e^{2 \text {arcsinh}(c x)}\right )+a c^2 d \log (x)+\frac {1}{2} b c^2 d \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}(c x)}\right ) \]

input

Integrate[((d + c^2*d*x^2)*(a + b*ArcSinh[c*x]))/x^3,x]
 

output

-1/2*(a*d)/x^2 - (b*c*d*Sqrt[1 + c^2*x^2])/(2*x) - (b*d*ArcSinh[c*x])/(2*x 
^2) - (b*c^2*d*ArcSinh[c*x]^2)/2 + b*c^2*d*ArcSinh[c*x]*Log[1 - E^(2*ArcSi 
nh[c*x])] + a*c^2*d*Log[x] + (b*c^2*d*PolyLog[2, E^(2*ArcSinh[c*x])])/2
 

3.1.8.3 Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.69 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.20, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6217, 247, 222, 6190, 25, 3042, 26, 4201, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {\left (c^2 d x^2+d\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx\)

\(\Big \downarrow \) 6217

\(\displaystyle c^2 d \int \frac {a+b \text {arcsinh}(c x)}{x}dx+\frac {1}{2} b c d \int \frac {\sqrt {c^2 x^2+1}}{x^2}dx-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}\)

\(\Big \downarrow \) 247

\(\displaystyle c^2 d \int \frac {a+b \text {arcsinh}(c x)}{x}dx+\frac {1}{2} b c d \left (c^2 \int \frac {1}{\sqrt {c^2 x^2+1}}dx-\frac {\sqrt {c^2 x^2+1}}{x}\right )-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}\)

\(\Big \downarrow \) 222

\(\displaystyle c^2 d \int \frac {a+b \text {arcsinh}(c x)}{x}dx-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 6190

\(\displaystyle \frac {c^2 d \int -\left ((a+b \text {arcsinh}(c x)) \coth \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )\right )d(a+b \text {arcsinh}(c x))}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {c^2 d \int (a+b \text {arcsinh}(c x)) \coth \left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )d(a+b \text {arcsinh}(c x))}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {c^2 d \int -i (a+b \text {arcsinh}(c x)) \tan \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}+\frac {\pi }{2}\right )d(a+b \text {arcsinh}(c x))}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {i c^2 d \int (a+b \text {arcsinh}(c x)) \tan \left (\frac {1}{2} \left (\frac {2 i a}{b}+\pi \right )-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )d(a+b \text {arcsinh}(c x))}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 4201

\(\displaystyle \frac {i c^2 d \left (2 i \int \frac {e^{\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}-i \pi } (a+b \text {arcsinh}(c x))}{1+e^{\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}-i \pi }}d(a+b \text {arcsinh}(c x))-\frac {1}{2} i (a+b \text {arcsinh}(c x))^2\right )}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {i c^2 d \left (2 i \left (\frac {1}{2} b \int \log \left (1+e^{\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}-i \pi }\right )d(a+b \text {arcsinh}(c x))-\frac {1}{2} b (a+b \text {arcsinh}(c x)) \log \left (1+e^{-\frac {2 (a+b \text {arcsinh}(c x))}{b}+\frac {2 a}{b}-i \pi }\right )\right )-\frac {1}{2} i (a+b \text {arcsinh}(c x))^2\right )}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {i c^2 d \left (2 i \left (-\frac {1}{4} b^2 \int e^{-\frac {2 a}{b}+\frac {2 (a+b \text {arcsinh}(c x))}{b}+i \pi } \log \left (1+e^{\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}-i \pi }\right )de^{\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}-i \pi }-\frac {1}{2} b (a+b \text {arcsinh}(c x)) \log \left (1+e^{-\frac {2 (a+b \text {arcsinh}(c x))}{b}+\frac {2 a}{b}-i \pi }\right )\right )-\frac {1}{2} i (a+b \text {arcsinh}(c x))^2\right )}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {i c^2 d \left (2 i \left (\frac {1}{4} b^2 \operatorname {PolyLog}(2,-a-b \text {arcsinh}(c x))-\frac {1}{2} b (a+b \text {arcsinh}(c x)) \log \left (1+e^{-\frac {2 (a+b \text {arcsinh}(c x))}{b}+\frac {2 a}{b}-i \pi }\right )\right )-\frac {1}{2} i (a+b \text {arcsinh}(c x))^2\right )}{b}-\frac {d \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{2 x^2}+\frac {1}{2} b c d \left (c \text {arcsinh}(c x)-\frac {\sqrt {c^2 x^2+1}}{x}\right )\)

input

Int[((d + c^2*d*x^2)*(a + b*ArcSinh[c*x]))/x^3,x]
 

output

-1/2*(d*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/x^2 + (b*c*d*(-(Sqrt[1 + c^2*x 
^2]/x) + c*ArcSinh[c*x]))/2 + (I*c^2*d*((-1/2*I)*(a + b*ArcSinh[c*x])^2 + 
(2*I)*(-1/2*(b*(a + b*ArcSinh[c*x])*Log[1 + E^((2*a)/b - I*Pi - (2*(a + b* 
ArcSinh[c*x]))/b)]) + (b^2*PolyLog[2, -a - b*ArcSinh[c*x]])/4)))/b
 

3.1.8.3.1 Defintions of rubi rules used

rule 25

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

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 247

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

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 6190

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Simp[1/b 
 Subst[Int[x^n*Coth[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a 
, b, c}, x] && IGtQ[n, 0]
 

rule 6217

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

3.1.8.4 Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.18

method	result	size
parts	\(d a \left (-\frac {1}{2 x^{2}}+c^{2} \ln \left (x \right )\right )+d b \,c^{2} \left (-\frac {\operatorname {arcsinh}\left (c x \right )^{2}}{2}-\frac {c x \sqrt {c^{2} x^{2}+1}-c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )}{2 c^{2} x^{2}}+\operatorname {arcsinh}\left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+\operatorname {arcsinh}\left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(151\)
derivativedivides	\(c^{2} \left (d a \left (\ln \left (c x \right )-\frac {1}{2 c^{2} x^{2}}\right )+d b \left (-\frac {\operatorname {arcsinh}\left (c x \right )^{2}}{2}-\frac {c x \sqrt {c^{2} x^{2}+1}-c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )}{2 c^{2} x^{2}}+\operatorname {arcsinh}\left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+\operatorname {arcsinh}\left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\right )\)	\(153\)
default	\(c^{2} \left (d a \left (\ln \left (c x \right )-\frac {1}{2 c^{2} x^{2}}\right )+d b \left (-\frac {\operatorname {arcsinh}\left (c x \right )^{2}}{2}-\frac {c x \sqrt {c^{2} x^{2}+1}-c^{2} x^{2}+\operatorname {arcsinh}\left (c x \right )}{2 c^{2} x^{2}}+\operatorname {arcsinh}\left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+\operatorname {arcsinh}\left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+\operatorname {polylog}\left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\right )\)	\(153\)

input

int((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^3,x,method=_RETURNVERBOSE)
 

output

d*a*(-1/2/x^2+c^2*ln(x))+d*b*c^2*(-1/2*arcsinh(c*x)^2-1/2*(c*x*(c^2*x^2+1) 
^(1/2)-c^2*x^2+arcsinh(c*x))/c^2/x^2+arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/ 
2))+polylog(2,-c*x-(c^2*x^2+1)^(1/2))+arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1 
/2))+polylog(2,c*x+(c^2*x^2+1)^(1/2)))
 

3.1.8.5 Fricas [F]

\[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=\int { \frac {{\left (c^{2} d x^{2} + d\right )} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input

integrate((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^3,x, algorithm="fricas")
 

output

integral((a*c^2*d*x^2 + a*d + (b*c^2*d*x^2 + b*d)*arcsinh(c*x))/x^3, x)
 

3.1.8.6 Sympy [F]

\[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=d \left (\int \frac {a}{x^{3}}\, dx + \int \frac {a c^{2}}{x}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {b c^{2} \operatorname {asinh}{\left (c x \right )}}{x}\, dx\right ) \]

input

integrate((c**2*d*x**2+d)*(a+b*asinh(c*x))/x**3,x)
 

output

d*(Integral(a/x**3, x) + Integral(a*c**2/x, x) + Integral(b*asinh(c*x)/x** 
3, x) + Integral(b*c**2*asinh(c*x)/x, x))
 

3.1.8.7 Maxima [F]

\[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=\int { \frac {{\left (c^{2} d x^{2} + d\right )} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{x^{3}} \,d x } \]

input

integrate((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^3,x, algorithm="maxima")
 

output

b*c^2*d*integrate(log(c*x + sqrt(c^2*x^2 + 1))/x, x) + a*c^2*d*log(x) - 1/ 
2*b*d*(sqrt(c^2*x^2 + 1)*c/x + arcsinh(c*x)/x^2) - 1/2*a*d/x^2
 

3.1.8.8 Giac [F(-2)]

Exception generated. \[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=\text {Exception raised: TypeError} \]

input

integrate((c^2*d*x^2+d)*(a+b*arcsinh(c*x))/x^3,x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

3.1.8.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+c^2 d x^2\right ) (a+b \text {arcsinh}(c x))}{x^3} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\left (d\,c^2\,x^2+d\right )}{x^3} \,d x \]

input

int(((a + b*asinh(c*x))*(d + c^2*d*x^2))/x^3,x)
 

output

int(((a + b*asinh(c*x))*(d + c^2*d*x^2))/x^3, x)