3.1.30 \(\int \frac {x^2 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx\) [30]

3.1.30.1 Optimal result
3.1.30.2 Mathematica [A] (verified)
3.1.30.3 Rubi [A] (verified)
3.1.30.4 Maple [A] (verified)
3.1.30.5 Fricas [F]
3.1.30.6 Sympy [F]
3.1.30.7 Maxima [F]
3.1.30.8 Giac [F]
3.1.30.9 Mupad [F(-1)]

3.1.30.1 Optimal result

Integrand size = 24, antiderivative size = 108 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=-\frac {b \sqrt {1+c^2 x^2}}{c^3 d}+\frac {x (a+b \text {arcsinh}(c x))}{c^2 d}-\frac {2 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c^3 d}+\frac {i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c^3 d}-\frac {i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^3 d} \]

output
x*(a+b*arcsinh(c*x))/c^2/d-2*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/ 
2))/c^3/d+I*b*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c^3/d-I*b*polylog(2,I* 
(c*x+(c^2*x^2+1)^(1/2)))/c^3/d-b*(c^2*x^2+1)^(1/2)/c^3/d
 
3.1.30.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.12 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {a c x-b \sqrt {1+c^2 x^2}+b c x \text {arcsinh}(c x)-a \arctan (c x)-i b \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )+i b \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )+i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )-i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^3 d} \]

input
Integrate[(x^2*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2),x]
 
output
(a*c*x - b*Sqrt[1 + c^2*x^2] + b*c*x*ArcSinh[c*x] - a*ArcTan[c*x] - I*b*Ar 
cSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + I*b*ArcSinh[c*x]*Log[1 + I*E^ArcSin 
h[c*x]] + I*b*PolyLog[2, (-I)*E^ArcSinh[c*x]] - I*b*PolyLog[2, I*E^ArcSinh 
[c*x]])/(c^3*d)
 
3.1.30.3 Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6227, 27, 241, 6204, 3042, 4668, 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.

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

\(\Big \downarrow \) 6227

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 241

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

\(\Big \downarrow \) 6204

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4668

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))-i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )+i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^3 d}+\frac {x (a+b \text {arcsinh}(c x))}{c^2 d}-\frac {b \sqrt {c^2 x^2+1}}{c^3 d}\)

input
Int[(x^2*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2),x]
 
output
-((b*Sqrt[1 + c^2*x^2])/(c^3*d)) + (x*(a + b*ArcSinh[c*x]))/(c^2*d) - (2*( 
a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]] - I*b*PolyLog[2, (-I)*E^ArcSinh 
[c*x]] + I*b*PolyLog[2, I*E^ArcSinh[c*x]])/(c^3*d)
 

3.1.30.3.1 Defintions of rubi rules used

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 241
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ 
(2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
 

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 4668
Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 6204
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symb 
ol] :> Simp[1/(c*d)   Subst[Int[(a + b*x)^n*Sech[x], x], x, ArcSinh[c*x]], 
x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]
 

rule 6227
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a 
+ b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Simp[f^2*((m - 1)/(c^2*(m + 
2*p + 1)))   Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] 
 - Simp[b*f*(n/(c*(m + 2*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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[ 
m, 1] && NeQ[m + 2*p + 1, 0]
 
3.1.30.4 Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.57

method result size
derivativedivides \(\frac {\frac {a \left (c x -\arctan \left (c x \right )\right )}{d}+\frac {b \left (-\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )+\operatorname {arcsinh}\left (c x \right ) c x -\sqrt {c^{2} x^{2}+1}-\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}}{c^{3}}\) \(170\)
default \(\frac {\frac {a \left (c x -\arctan \left (c x \right )\right )}{d}+\frac {b \left (-\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )+\operatorname {arcsinh}\left (c x \right ) c x -\sqrt {c^{2} x^{2}+1}-\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}}{c^{3}}\) \(170\)
parts \(\frac {a \left (\frac {x}{c^{2}}-\frac {\arctan \left (c x \right )}{c^{3}}\right )}{d}+\frac {b \left (-\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )+\operatorname {arcsinh}\left (c x \right ) c x -\sqrt {c^{2} x^{2}+1}-\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d \,c^{3}}\) \(174\)

input
int(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x,method=_RETURNVERBOSE)
 
output
1/c^3*(a/d*(c*x-arctan(c*x))+b/d*(-arcsinh(c*x)*arctan(c*x)+arcsinh(c*x)*c 
*x-(c^2*x^2+1)^(1/2)-arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+arcta 
n(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*dilog(1+I*(1+I*c*x)/(c^2*x^2+ 
1)^(1/2))-I*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))))
 
3.1.30.5 Fricas [F]

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

input
integrate(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="fricas")
 
output
integral((b*x^2*arcsinh(c*x) + a*x^2)/(c^2*d*x^2 + d), x)
 
3.1.30.6 Sympy [F]

\[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {\int \frac {a x^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac {b x^{2} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]

input
integrate(x**2*(a+b*asinh(c*x))/(c**2*d*x**2+d),x)
 
output
(Integral(a*x**2/(c**2*x**2 + 1), x) + Integral(b*x**2*asinh(c*x)/(c**2*x* 
*2 + 1), x))/d
 
3.1.30.7 Maxima [F]

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

input
integrate(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="maxima")
 
output
a*(x/(c^2*d) - arctan(c*x)/(c^3*d)) + b*integrate(x^2*log(c*x + sqrt(c^2*x 
^2 + 1))/(c^2*d*x^2 + d), x)
 
3.1.30.8 Giac [F]

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

input
integrate(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d),x, algorithm="giac")
 
output
integrate((b*arcsinh(c*x) + a)*x^2/(c^2*d*x^2 + d), x)
 
3.1.30.9 Mupad [F(-1)]

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

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