\(\int \frac {x^4 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^2} \, dx\) [37]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 171 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\frac {b}{2 c^5 d^2 \sqrt {1+c^2 x^2}}-\frac {b \sqrt {1+c^2 x^2}}{c^5 d^2}+\frac {3 x (a+b \text {arcsinh}(c x))}{2 c^4 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (1+c^2 x^2\right )}-\frac {3 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c^5 d^2}+\frac {3 i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{2 c^5 d^2}-\frac {3 i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{2 c^5 d^2} \] Output:

1/2*b/c^5/d^2/(c^2*x^2+1)^(1/2)-b*(c^2*x^2+1)^(1/2)/c^5/d^2+3/2*x*(a+b*arc 
sinh(c*x))/c^4/d^2-1/2*x^3*(a+b*arcsinh(c*x))/c^2/d^2/(c^2*x^2+1)-3*(a+b*a 
rcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))/c^5/d^2+3/2*I*b*polylog(2,-I*(c 
*x+(c^2*x^2+1)^(1/2)))/c^5/d^2-3/2*I*b*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)) 
)/c^5/d^2
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.57 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\frac {3 a c x+2 a c^3 x^3-b \sqrt {1+c^2 x^2}-2 b c^2 x^2 \sqrt {1+c^2 x^2}+3 b c x \text {arcsinh}(c x)+2 b c^3 x^3 \text {arcsinh}(c x)-3 a \arctan (c x)-3 a c^2 x^2 \arctan (c x)-3 i b \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )-3 i b c^2 x^2 \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )+3 i b \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )+3 i b c^2 x^2 \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )+3 i b \left (1+c^2 x^2\right ) \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )-3 i b \left (1+c^2 x^2\right ) \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{2 c^5 d^2 \left (1+c^2 x^2\right )} \] Input:

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^2,x]
 

Output:

(3*a*c*x + 2*a*c^3*x^3 - b*Sqrt[1 + c^2*x^2] - 2*b*c^2*x^2*Sqrt[1 + c^2*x^ 
2] + 3*b*c*x*ArcSinh[c*x] + 2*b*c^3*x^3*ArcSinh[c*x] - 3*a*ArcTan[c*x] - 3 
*a*c^2*x^2*ArcTan[c*x] - (3*I)*b*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - 
(3*I)*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + (3*I)*b*ArcSinh[c 
*x]*Log[1 + I*E^ArcSinh[c*x]] + (3*I)*b*c^2*x^2*ArcSinh[c*x]*Log[1 + I*E^A 
rcSinh[c*x]] + (3*I)*b*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSinh[c*x]] - (3* 
I)*b*(1 + c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]])/(2*c^5*d^2*(1 + c^2*x^2))
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6225, 27, 243, 53, 2009, 6227, 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.

\(\displaystyle \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (c^2 d x^2+d\right )^2} \, dx\)

\(\Big \downarrow \) 6225

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 243

\(\displaystyle \frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{c^2 x^2+1}dx}{2 c^2 d^2}+\frac {b \int \frac {x^2}{\left (c^2 x^2+1\right )^{3/2}}dx^2}{4 c d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}\)

\(\Big \downarrow \) 53

\(\displaystyle \frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{c^2 x^2+1}dx}{2 c^2 d^2}+\frac {b \int \left (\frac {1}{c^2 \sqrt {c^2 x^2+1}}-\frac {1}{c^2 \left (c^2 x^2+1\right )^{3/2}}\right )dx^2}{4 c d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{c^2 x^2+1}dx}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 6227

\(\displaystyle \frac {3 \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{c^2 x^2+1}dx}{c^2}-\frac {b \int \frac {x}{\sqrt {c^2 x^2+1}}dx}{c}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 241

\(\displaystyle \frac {3 \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{c^2 x^2+1}dx}{c^2}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}-\frac {b \sqrt {c^2 x^2+1}}{c^3}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 6204

\(\displaystyle \frac {3 \left (-\frac {\int \frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}d\text {arcsinh}(c x)}{c^3}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}-\frac {b \sqrt {c^2 x^2+1}}{c^3}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (-\frac {\int (a+b \text {arcsinh}(c x)) \csc \left (i \text {arcsinh}(c x)+\frac {\pi }{2}\right )d\text {arcsinh}(c x)}{c^3}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}-\frac {b \sqrt {c^2 x^2+1}}{c^3}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {3 \left (-\frac {-i b \int \log \left (1-i e^{\text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)+i b \int \log \left (1+i e^{\text {arcsinh}(c x)}\right )d\text {arcsinh}(c x)+2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{c^3}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}-\frac {b \sqrt {c^2 x^2+1}}{c^3}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {3 \left (-\frac {-i b \int e^{-\text {arcsinh}(c x)} \log \left (1-i e^{\text {arcsinh}(c x)}\right )de^{\text {arcsinh}(c x)}+i b \int e^{-\text {arcsinh}(c x)} \log \left (1+i e^{\text {arcsinh}(c x)}\right )de^{\text {arcsinh}(c x)}+2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{c^3}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}-\frac {b \sqrt {c^2 x^2+1}}{c^3}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {3 \left (-\frac {2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))-i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )+i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^3}+\frac {x (a+b \text {arcsinh}(c x))}{c^2}-\frac {b \sqrt {c^2 x^2+1}}{c^3}\right )}{2 c^2 d^2}-\frac {x^3 (a+b \text {arcsinh}(c x))}{2 c^2 d^2 \left (c^2 x^2+1\right )}+\frac {b \left (\frac {2 \sqrt {c^2 x^2+1}}{c^4}+\frac {2}{c^4 \sqrt {c^2 x^2+1}}\right )}{4 c d^2}\)

Input:

Int[(x^4*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^2,x]
 

Output:

(b*(2/(c^4*Sqrt[1 + c^2*x^2]) + (2*Sqrt[1 + c^2*x^2])/c^4))/(4*c*d^2) - (x 
^3*(a + b*ArcSinh[c*x]))/(2*c^2*d^2*(1 + c^2*x^2)) + (3*(-((b*Sqrt[1 + c^2 
*x^2])/c^3) + (x*(a + b*ArcSinh[c*x]))/c^2 - (2*(a + b*ArcSinh[c*x])*ArcTa 
n[E^ArcSinh[c*x]] - I*b*PolyLog[2, (-I)*E^ArcSinh[c*x]] + I*b*PolyLog[2, I 
*E^ArcSinh[c*x]])/c^3))/(2*c^2*d^2)
 

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 53
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

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 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 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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 6225
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/(2*e*(p + 1))), x] + (-Simp[f^2*((m - 1)/(2*e*(p + 1))) 
   Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - S 
imp[b*f*(n/(2*c*(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]) /; Fre 
eQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IG 
tQ[m, 1]
 

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]
 
Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.31

method result size
derivativedivides \(\frac {\frac {a \left (x c +\frac {x c}{2 c^{2} x^{2}+2}-\frac {3 \arctan \left (x c \right )}{2}\right )}{d^{2}}+\frac {b \left (x c \,\operatorname {arcsinh}\left (x c \right )+\frac {x c \,\operatorname {arcsinh}\left (x c \right )}{2 c^{2} x^{2}+2}-\frac {3 \,\operatorname {arcsinh}\left (x c \right ) \arctan \left (x c \right )}{2}-\frac {c^{2} x^{2}}{\sqrt {c^{2} x^{2}+1}}-\frac {1}{2 \sqrt {c^{2} x^{2}+1}}-\frac {3 \arctan \left (x c \right ) \ln \left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}+\frac {3 \arctan \left (x c \right ) \ln \left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}+\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}-\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}\right )}{d^{2}}}{c^{5}}\) \(224\)
default \(\frac {\frac {a \left (x c +\frac {x c}{2 c^{2} x^{2}+2}-\frac {3 \arctan \left (x c \right )}{2}\right )}{d^{2}}+\frac {b \left (x c \,\operatorname {arcsinh}\left (x c \right )+\frac {x c \,\operatorname {arcsinh}\left (x c \right )}{2 c^{2} x^{2}+2}-\frac {3 \,\operatorname {arcsinh}\left (x c \right ) \arctan \left (x c \right )}{2}-\frac {c^{2} x^{2}}{\sqrt {c^{2} x^{2}+1}}-\frac {1}{2 \sqrt {c^{2} x^{2}+1}}-\frac {3 \arctan \left (x c \right ) \ln \left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}+\frac {3 \arctan \left (x c \right ) \ln \left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}+\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}-\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}\right )}{d^{2}}}{c^{5}}\) \(224\)
parts \(\frac {a \left (\frac {x}{c^{4}}-\frac {-\frac {x}{2 \left (c^{2} x^{2}+1\right )}+\frac {3 \arctan \left (x c \right )}{2 c}}{c^{4}}\right )}{d^{2}}+\frac {b \left (x c \,\operatorname {arcsinh}\left (x c \right )+\frac {x c \,\operatorname {arcsinh}\left (x c \right )}{2 c^{2} x^{2}+2}-\frac {3 \,\operatorname {arcsinh}\left (x c \right ) \arctan \left (x c \right )}{2}-\frac {c^{2} x^{2}}{\sqrt {c^{2} x^{2}+1}}-\frac {1}{2 \sqrt {c^{2} x^{2}+1}}-\frac {3 \arctan \left (x c \right ) \ln \left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}+\frac {3 \arctan \left (x c \right ) \ln \left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}+\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}-\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (i x c +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2}\right )}{d^{2} c^{5}}\) \(233\)

Input:

int(x^4*(a+b*arcsinh(x*c))/(c^2*d*x^2+d)^2,x,method=_RETURNVERBOSE)
 

Output:

1/c^5*(a/d^2*(x*c+1/2*x*c/(c^2*x^2+1)-3/2*arctan(x*c))+b/d^2*(x*c*arcsinh( 
x*c)+1/2*arcsinh(x*c)*x*c/(c^2*x^2+1)-3/2*arcsinh(x*c)*arctan(x*c)-c^2*x^2 
/(c^2*x^2+1)^(1/2)-1/2/(c^2*x^2+1)^(1/2)-3/2*arctan(x*c)*ln(1+I*(1+I*x*c)/ 
(c^2*x^2+1)^(1/2))+3/2*arctan(x*c)*ln(1-I*(1+I*x*c)/(c^2*x^2+1)^(1/2))+3/2 
*I*dilog(1+I*(1+I*x*c)/(c^2*x^2+1)^(1/2))-3/2*I*dilog(1-I*(1+I*x*c)/(c^2*x 
^2+1)^(1/2))))
 

Fricas [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{2}} \,d x } \] Input:

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x, algorithm="fricas")
 

Output:

integral((b*x^4*arcsinh(c*x) + a*x^4)/(c^4*d^2*x^4 + 2*c^2*d^2*x^2 + d^2), 
 x)
 

Sympy [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\frac {\int \frac {a x^{4}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \] Input:

integrate(x**4*(a+b*asinh(c*x))/(c**2*d*x**2+d)**2,x)
 

Output:

(Integral(a*x**4/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*x**4*asinh 
(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1), x))/d**2
 

Maxima [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{2}} \,d x } \] Input:

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,x, algorithm="maxima")
 

Output:

1/2*a*(x/(c^6*d^2*x^2 + c^4*d^2) + 2*x/(c^4*d^2) - 3*arctan(c*x)/(c^5*d^2) 
) + b*integrate(x^4*log(c*x + sqrt(c^2*x^2 + 1))/(c^4*d^2*x^4 + 2*c^2*d^2* 
x^2 + d^2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^4*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^2,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
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \] Input:

int((x^4*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^2,x)
 

Output:

int((x^4*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^2} \, dx=\frac {-3 \mathit {atan} \left (c x \right ) a \,c^{2} x^{2}-3 \mathit {atan} \left (c x \right ) a +2 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{4}}{c^{4} x^{4}+2 c^{2} x^{2}+1}d x \right ) b \,c^{7} x^{2}+2 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{4}}{c^{4} x^{4}+2 c^{2} x^{2}+1}d x \right ) b \,c^{5}+2 a \,c^{3} x^{3}+3 a c x}{2 c^{5} d^{2} \left (c^{2} x^{2}+1\right )} \] Input:

int(x^4*(a+b*asinh(c*x))/(c^2*d*x^2+d)^2,x)
                                                                                    
                                                                                    
 

Output:

( - 3*atan(c*x)*a*c**2*x**2 - 3*atan(c*x)*a + 2*int((asinh(c*x)*x**4)/(c** 
4*x**4 + 2*c**2*x**2 + 1),x)*b*c**7*x**2 + 2*int((asinh(c*x)*x**4)/(c**4*x 
**4 + 2*c**2*x**2 + 1),x)*b*c**5 + 2*a*c**3*x**3 + 3*a*c*x)/(2*c**5*d**2*( 
c**2*x**2 + 1))