3.1.96 \(\int \frac {a+b \sinh ^{-1}(c x)}{x (\pi +c^2 \pi x^2)^{3/2}} \, dx\) [96]

Optimal. Leaf size=94 \[ \frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \text {ArcTan}(c x)}{\pi ^{3/2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}} \]

[Out]

-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)+(a+b*arcsinh(c*x))/Pi/(Pi*c^2*x^2+Pi)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5811, 5816, 4267, 2317, 2438, 209} \begin {gather*} \frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi c^2 x^2+\pi }}-\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{3/2}}-\frac {b \text {ArcTan}(c x)}{\pi ^{3/2}}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2438

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 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[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 5811

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] + (Dist[(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] + Dist[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] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 1])

Rule 5816

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

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {\pi +c^2 \pi x^2}} \, dx}{\pi }-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{\pi \sqrt {\pi +c^2 \pi x^2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}+\frac {\text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{3/2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{3/2}}+\frac {b \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{3/2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}\\ &=\frac {a+b \sinh ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{\pi \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.23, size = 143, normalized size = 1.52 \begin {gather*} \frac {\frac {a}{\sqrt {1+c^2 x^2}}+\frac {b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}-2 b \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+b \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-b \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )+a \log (x)-a \log \left (\pi \left (1+\sqrt {1+c^2 x^2}\right )\right )+b \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-b \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )}{\pi ^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a/Sqrt[1 + c^2*x^2] + (b*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - 2*b*ArcTan[Tanh[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)

________________________________________________________________________________________

Maple [A]
time = 3.54, size = 157, normalized size = 1.67

method	result	size
default	\(a \left (\frac {1}{\pi \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {\arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{\pi ^{\frac {3}{2}}}\right )+\frac {b \arcsinh \left (c x \right )}{\pi ^{\frac {3}{2}} \sqrt {c^{2} x^{2}+1}}-\frac {2 b \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {b \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {b \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}-\frac {b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {3}{2}}}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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/Pi^(3/2)/(c^2*x^2+1)^(1/2)
*arcsinh(c*x)-2*b/Pi^(3/2)*arctan(c*x+(c^2*x^2+1)^(1/2))-b/Pi^(3/2)*dilog(c*x+(c^2*x^2+1)^(1/2))-b/Pi^(3/2)*di
log(1+c*x+(c^2*x^2+1)^(1/2))-b/Pi^(3/2)*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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*sqr
t(c**2*x**2 + 1) + x*sqrt(c**2*x**2 + 1)), x))/pi**(3/2)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________