\(\int \frac {a+b \text {arcsinh}(c x)}{x (\pi +c^2 \pi x^2)^{3/2}} \, dx\) [106]

Optimal result

Integrand size = 26, antiderivative size = 94 \[ \int \frac {a+b \text {arcsinh}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \arctan (c x)}{\pi ^{3/2}}-\frac {2 (a+b \text {arcsinh}(c x)) \text {arctanh}\left (e^{\text {arcsinh}(c x)}\right )}{\pi ^{3/2}}-\frac {b \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(c x)}\right )}{\pi ^{3/2}}+\frac {b \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(c x)}\right )}{\pi ^{3/2}} \] Output:

(a+b*arcsinh(c*x))/Pi/(Pi*c^2*x^2+Pi)^(1/2)-b*arctan(c*x)/Pi^(3/2)-2*(a+b* 
arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))/Pi^(3/2)-b*polylog(2,-c*x-(c^ 
2*x^2+1)^(1/2))/Pi^(3/2)+b*polylog(2,c*x+(c^2*x^2+1)^(1/2))/Pi^(3/2)
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.52 \[ \int \frac {a+b \text {arcsinh}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\frac {\frac {a}{\sqrt {1+c^2 x^2}}+\frac {b \text {arcsinh}(c x)}{\sqrt {1+c^2 x^2}}-2 b \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+b \text {arcsinh}(c x) \log \left (1-e^{-\text {arcsinh}(c x)}\right )-b \text {arcsinh}(c x) \log \left (1+e^{-\text {arcsinh}(c x)}\right )+a \log (x)-a \log \left (\pi \left (1+\sqrt {1+c^2 x^2}\right )\right )+b \operatorname {PolyLog}\left (2,-e^{-\text {arcsinh}(c x)}\right )-b \operatorname {PolyLog}\left (2,e^{-\text {arcsinh}(c x)}\right )}{\pi ^{3/2}} \] Input:

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

Output:

(a/Sqrt[1 + c^2*x^2] + (b*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - 2*b*ArcTan[Tan 
h[ArcSinh[c*x]/2]] + b*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - b*ArcSinh 
[c*x]*Log[1 + E^(-ArcSinh[c*x])] + a*Log[x] - a*Log[Pi*(1 + Sqrt[1 + c^2*x 
^2])] + b*PolyLog[2, -E^(-ArcSinh[c*x])] - b*PolyLog[2, E^(-ArcSinh[c*x])] 
)/Pi^(3/2)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6226, 216, 6231, 3042, 26, 4670, 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.

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

\(\Big \downarrow \) 6226

\(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c x)}{x \sqrt {c^2 \pi x^2+\pi }}dx}{\pi }-\frac {b c \int \frac {1}{c^2 x^2+1}dx}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c x)}{x \sqrt {c^2 \pi x^2+\pi }}dx}{\pi }+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

\(\Big \downarrow \) 6231

\(\displaystyle \frac {\int \frac {a+b \text {arcsinh}(c x)}{c x}d\text {arcsinh}(c x)}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int i (a+b \text {arcsinh}(c x)) \csc (i \text {arcsinh}(c x))d\text {arcsinh}(c x)}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {i \int (a+b \text {arcsinh}(c x)) \csc (i \text {arcsinh}(c x))d\text {arcsinh}(c x)}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

\(\Big \downarrow \) 4670

\(\displaystyle \frac {i \left (i b \int \log \left (1-e^{\text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)-i b \int \log \left (1+e^{\text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)+2 i \text {arctanh}\left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))\right )}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {i \left (i b \int e^{-\text {arcsinh}(c x)} \log \left (1-e^{\text {arcsinh}(c x)}\right )de^{\text {arcsinh}(c x)}-i b \int e^{-\text {arcsinh}(c x)} \log \left (1+e^{\text {arcsinh}(c x)}\right )de^{\text {arcsinh}(c x)}+2 i \text {arctanh}\left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))\right )}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {i \left (2 i \text {arctanh}\left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))+i b \operatorname {PolyLog}\left (2,-e^{\text {arcsinh}(c x)}\right )-i b \operatorname {PolyLog}\left (2,e^{\text {arcsinh}(c x)}\right )\right )}{\pi ^{3/2}}+\frac {a+b \text {arcsinh}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {b \arctan (c x)}{\pi ^{3/2}}\)

Input:

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

Output:

(a + b*ArcSinh[c*x])/(Pi*Sqrt[Pi + c^2*Pi*x^2]) - (b*ArcTan[c*x])/Pi^(3/2) 
 + (I*((2*I)*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c*x]] + I*b*PolyLog[2, 
 -E^ArcSinh[c*x]] - I*b*PolyLog[2, E^ArcSinh[c*x]]))/Pi^(3/2)
 

Defintions of rubi rules used

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 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 4670

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

rule 6226

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(f*x)^(m + 1))*(d + e*x^2)^(p + 1)*((a 
 + b*ArcSinh[c*x])^n/(2*d*f*(p + 1))), x] + (Simp[(m + 2*p + 3)/(2*d*(p + 1 
))   Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Simp[ 
b*c*(n/(2*f*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p]   Int[(f*x)^(m + 
1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{ 
a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !G 
tQ[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])
 

rule 6231

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.) 
*(x_)^2], x_Symbol] :> Simp[(1/c^(m + 1))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e 
*x^2]]   Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; FreeQ 
[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && IntegerQ[m]
 

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.65

method	result	size
default	\(a \left (\frac {1}{\pi \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{\pi ^{\frac {3}{2}}}\right )+b \left (\frac {\operatorname {arcsinh}\left (x c \right )}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}}-\frac {2 \arctan \left (x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {\operatorname {dilog}\left (1+x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {\operatorname {arcsinh}\left (x c \right ) \ln \left (1+x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {\operatorname {dilog}\left (x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}\right )\)	\(155\)
parts	\(a \left (\frac {1}{\pi \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{\pi ^{\frac {3}{2}}}\right )+b \left (\frac {\operatorname {arcsinh}\left (x c \right )}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}}-\frac {2 \arctan \left (x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {\operatorname {dilog}\left (1+x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {\operatorname {arcsinh}\left (x c \right ) \ln \left (1+x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {\operatorname {dilog}\left (x c +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}\right )\)	\(155\)

Input:

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

Output:

a*(1/Pi/(Pi*c^2*x^2+Pi)^(1/2)-1/Pi^(3/2)*arctanh(Pi^(1/2)/(Pi*c^2*x^2+Pi)^ 
(1/2)))+b*(1/Pi^(3/2)*arcsinh(x*c)/(c^2*x^2+1)^(1/2)-2/Pi^(3/2)*arctan(x*c 
+(c^2*x^2+1)^(1/2))-1/Pi^(3/2)*dilog(1+x*c+(c^2*x^2+1)^(1/2))-1/Pi^(3/2)*a 
rcsinh(x*c)*ln(1+x*c+(c^2*x^2+1)^(1/2))-1/Pi^(3/2)*dilog(x*c+(c^2*x^2+1)^( 
1/2)))
 

Fricas [F]

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

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

Output:

integral(sqrt(pi + pi*c^2*x^2)*(b*arcsinh(c*x) + a)/(pi^2*c^4*x^5 + 2*pi^2 
*c^2*x^3 + pi^2*x), x)
 

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\frac {\int \frac {a}{c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} + x \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} + x \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {3}{2}}} \] Input:

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

Output:

(Integral(a/(c**2*x**3*sqrt(c**2*x**2 + 1) + x*sqrt(c**2*x**2 + 1)), x) + 
Integral(b*asinh(c*x)/(c**2*x**3*sqrt(c**2*x**2 + 1) + x*sqrt(c**2*x**2 + 
1)), x))/pi**(3/2)
 

Maxima [F]

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

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

Output:

-a*(arcsinh(1/(c*abs(x)))/pi^(3/2) - 1/(pi*sqrt(pi + pi*c^2*x^2))) + b*int 
egrate(log(c*x + sqrt(c^2*x^2 + 1))/((pi + pi*c^2*x^2)^(3/2)*x), x)
 

Giac [F]

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

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

Output:

integrate((b*arcsinh(c*x) + a)/((pi + pi*c^2*x^2)^(3/2)*x), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^{2} x^{2}+1}\, a +\left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{2} x^{3}+\sqrt {c^{2} x^{2}+1}\, x}d x \right ) b \,c^{2} x^{2}+\left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{2} x^{3}+\sqrt {c^{2} x^{2}+1}\, x}d x \right ) b +\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x -1\right ) a \,c^{2} x^{2}+\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x -1\right ) a -\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x +1\right ) a \,c^{2} x^{2}-\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x +1\right ) a}{\sqrt {\pi }\, \pi \left (c^{2} x^{2}+1\right )} \] Input:

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

Output:

(sqrt(c**2*x**2 + 1)*a + int(asinh(c*x)/(sqrt(c**2*x**2 + 1)*c**2*x**3 + s 
qrt(c**2*x**2 + 1)*x),x)*b*c**2*x**2 + int(asinh(c*x)/(sqrt(c**2*x**2 + 1) 
*c**2*x**3 + sqrt(c**2*x**2 + 1)*x),x)*b + log(sqrt(c**2*x**2 + 1) + c*x - 
 1)*a*c**2*x**2 + log(sqrt(c**2*x**2 + 1) + c*x - 1)*a - log(sqrt(c**2*x** 
2 + 1) + c*x + 1)*a*c**2*x**2 - log(sqrt(c**2*x**2 + 1) + c*x + 1)*a)/(sqr 
t(pi)*pi*(c**2*x**2 + 1))