\(\int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx\) [58]

Optimal result

Integrand size = 21, antiderivative size = 105 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx=\frac {3 b e^3 (c+d x) \sqrt {1+(c+d x)^2}}{32 d}-\frac {b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{16 d}-\frac {3 b e^3 \text {arcsinh}(c+d x)}{32 d}+\frac {e^3 (c+d x)^4 (a+b \text {arcsinh}(c+d x))}{4 d} \] Output:

3/32*b*e^3*(d*x+c)*(1+(d*x+c)^2)^(1/2)/d-1/16*b*e^3*(d*x+c)^3*(1+(d*x+c)^2 
)^(1/2)/d-3/32*b*e^3*arcsinh(d*x+c)/d+1/4*e^3*(d*x+c)^4*(a+b*arcsinh(d*x+c 
))/d
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.79 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx=\frac {e^3 \left (3 b (c+d x) \sqrt {1+(c+d x)^2}-2 b (c+d x)^3 \sqrt {1+(c+d x)^2}-3 b \text {arcsinh}(c+d x)+8 (c+d x)^4 (a+b \text {arcsinh}(c+d x))\right )}{32 d} \] Input:

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

Output:

(e^3*(3*b*(c + d*x)*Sqrt[1 + (c + d*x)^2] - 2*b*(c + d*x)^3*Sqrt[1 + (c + 
d*x)^2] - 3*b*ArcSinh[c + d*x] + 8*(c + d*x)^4*(a + b*ArcSinh[c + d*x])))/ 
(32*d)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6274, 27, 6191, 262, 262, 222}

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.

\(\displaystyle \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx\)

\(\Big \downarrow \) 6274

\(\displaystyle \frac {\int e^3 (c+d x)^3 (a+b \text {arcsinh}(c+d x))d(c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^3 \int (c+d x)^3 (a+b \text {arcsinh}(c+d x))d(c+d x)}{d}\)

\(\Big \downarrow \) 6191

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

(e^3*(-1/4*(b*(((c + d*x)^3*Sqrt[1 + (c + d*x)^2])/4 - (3*(((c + d*x)*Sqrt 
[1 + (c + d*x)^2])/2 - ArcSinh[c + d*x]/2))/4)) + ((c + d*x)^4*(a + b*ArcS 
inh[c + d*x]))/4))/d
 

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 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 262

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

rule 6191

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

rule 6274

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^( 
m_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b* 
ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {a \,e^{3} \left (d x +c \right )^{4}}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \,\operatorname {arcsinh}\left (d x +c \right )}{32}\right )}{d}\)	\(86\)
default	\(\frac {\frac {a \,e^{3} \left (d x +c \right )^{4}}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \,\operatorname {arcsinh}\left (d x +c \right )}{32}\right )}{d}\)	\(86\)
parts	\(\frac {a \,e^{3} \left (d x +c \right )^{4}}{4 d}+\frac {e^{3} b \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \,\operatorname {arcsinh}\left (d x +c \right )}{32}\right )}{d}\)	\(88\)
orering	\(\frac {\left (14 d^{4} x^{4}+56 c \,d^{3} x^{3}+84 c^{2} d^{2} x^{2}+56 c^{3} d x +14 c^{4}-3 d^{2} x^{2}-6 c d x -3 c^{2}-12\right ) \left (d e x +c e \right )^{3} \left (a +b \,\operatorname {arcsinh}\left (d x +c \right )\right )}{32 \left (d x +c \right )^{3} d}-\frac {\left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) \left (2 d^{2} x^{2}+4 c d x +2 c^{2}-3\right ) \left (3 \left (d e x +c e \right )^{2} \left (a +b \,\operatorname {arcsinh}\left (d x +c \right )\right ) d e +\frac {\left (d e x +c e \right )^{3} b d}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{32 d^{2} \left (d x +c \right )^{2}}\)	\(192\)

Input:

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

Output:

1/d*(1/4*a*e^3*(d*x+c)^4+e^3*b*(1/4*(d*x+c)^4*arcsinh(d*x+c)-1/16*(d*x+c)^ 
3*(1+(d*x+c)^2)^(1/2)+3/32*(d*x+c)*(1+(d*x+c)^2)^(1/2)-3/32*arcsinh(d*x+c) 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (93) = 186\).

Time = 0.09 (sec) , antiderivative size = 228, normalized size of antiderivative = 2.17 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx=\frac {8 \, a d^{4} e^{3} x^{4} + 32 \, a c d^{3} e^{3} x^{3} + 48 \, a c^{2} d^{2} e^{3} x^{2} + 32 \, a c^{3} d e^{3} x + {\left (8 \, b d^{4} e^{3} x^{4} + 32 \, b c d^{3} e^{3} x^{3} + 48 \, b c^{2} d^{2} e^{3} x^{2} + 32 \, b c^{3} d e^{3} x + {\left (8 \, b c^{4} - 3 \, b\right )} e^{3}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - {\left (2 \, b d^{3} e^{3} x^{3} + 6 \, b c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, b c^{2} - b\right )} d e^{3} x + {\left (2 \, b c^{3} - 3 \, b c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{32 \, d} \] Input:

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

Output:

1/32*(8*a*d^4*e^3*x^4 + 32*a*c*d^3*e^3*x^3 + 48*a*c^2*d^2*e^3*x^2 + 32*a*c 
^3*d*e^3*x + (8*b*d^4*e^3*x^4 + 32*b*c*d^3*e^3*x^3 + 48*b*c^2*d^2*e^3*x^2 
+ 32*b*c^3*d*e^3*x + (8*b*c^4 - 3*b)*e^3)*log(d*x + c + sqrt(d^2*x^2 + 2*c 
*d*x + c^2 + 1)) - (2*b*d^3*e^3*x^3 + 6*b*c*d^2*e^3*x^2 + 3*(2*b*c^2 - b)* 
d*e^3*x + (2*b*c^3 - 3*b*c)*e^3)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (94) = 188\).

Time = 0.29 (sec) , antiderivative size = 394, normalized size of antiderivative = 3.75 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx=\begin {cases} a c^{3} e^{3} x + \frac {3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac {a d^{3} e^{3} x^{4}}{4} + \frac {b c^{4} e^{3} \operatorname {asinh}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname {asinh}{\left (c + d x \right )} - \frac {b c^{3} e^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{16 d} + \frac {3 b c^{2} d e^{3} x^{2} \operatorname {asinh}{\left (c + d x \right )}}{2} - \frac {3 b c^{2} e^{3} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{16} + b c d^{2} e^{3} x^{3} \operatorname {asinh}{\left (c + d x \right )} - \frac {3 b c d e^{3} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{16} + \frac {3 b c e^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{32 d} + \frac {b d^{3} e^{3} x^{4} \operatorname {asinh}{\left (c + d x \right )}}{4} - \frac {b d^{2} e^{3} x^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{16} + \frac {3 b e^{3} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{32} - \frac {3 b e^{3} \operatorname {asinh}{\left (c + d x \right )}}{32 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {asinh}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a*c**3*e**3*x + 3*a*c**2*d*e**3*x**2/2 + a*c*d**2*e**3*x**3 + a 
*d**3*e**3*x**4/4 + b*c**4*e**3*asinh(c + d*x)/(4*d) + b*c**3*e**3*x*asinh 
(c + d*x) - b*c**3*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(16*d) + 3*b* 
c**2*d*e**3*x**2*asinh(c + d*x)/2 - 3*b*c**2*e**3*x*sqrt(c**2 + 2*c*d*x + 
d**2*x**2 + 1)/16 + b*c*d**2*e**3*x**3*asinh(c + d*x) - 3*b*c*d*e**3*x**2* 
sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/16 + 3*b*c*e**3*sqrt(c**2 + 2*c*d*x + 
 d**2*x**2 + 1)/(32*d) + b*d**3*e**3*x**4*asinh(c + d*x)/4 - b*d**2*e**3*x 
**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/16 + 3*b*e**3*x*sqrt(c**2 + 2*c*d 
*x + d**2*x**2 + 1)/32 - 3*b*e**3*asinh(c + d*x)/(32*d), Ne(d, 0)), (c**3* 
e**3*x*(a + b*asinh(c)), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 790 vs. \(2 (93) = 186\).

Time = 0.06 (sec) , antiderivative size = 790, normalized size of antiderivative = 7.52 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*a*d^3*e^3*x^4 + a*c*d^2*e^3*x^3 + 3/2*a*c^2*d*e^3*x^2 + 3/4*(2*x^2*arc 
sinh(d*x + c) - d*(3*c^2*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 
+ 1)*d^2))/d^3 + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x/d^2 - (c^2 + 1)*arcsi 
nh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^3 - 3*sqrt(d^2*x^ 
2 + 2*c*d*x + c^2 + 1)*c/d^3))*b*c^2*d*e^3 + 1/6*(6*x^3*arcsinh(d*x + c) - 
 d*(2*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x^2/d^2 - 15*c^3*arcsinh(2*(d^2*x 
+ c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^4 - 5*sqrt(d^2*x^2 + 2*c*d*x 
+ c^2 + 1)*c*x/d^3 + 9*(c^2 + 1)*c*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 
 + 4*(c^2 + 1)*d^2))/d^4 + 15*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c^2/d^4 - 
4*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)/d^4))*b*c*d^2*e^3 + 1/96*(24 
*x^4*arcsinh(d*x + c) - (6*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*x^3/d^2 - 14* 
sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c*x^2/d^3 + 105*c^4*arcsinh(2*(d^2*x + c 
*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^5 + 35*sqrt(d^2*x^2 + 2*c*d*x + 
c^2 + 1)*c^2*x/d^4 - 90*(c^2 + 1)*c^2*arcsinh(2*(d^2*x + c*d)/sqrt(-4*c^2* 
d^2 + 4*(c^2 + 1)*d^2))/d^5 - 105*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*c^3/d^ 
5 - 9*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)*x/d^4 + 9*(c^2 + 1)^2*ar 
csinh(2*(d^2*x + c*d)/sqrt(-4*c^2*d^2 + 4*(c^2 + 1)*d^2))/d^5 + 55*sqrt(d^ 
2*x^2 + 2*c*d*x + c^2 + 1)*(c^2 + 1)*c/d^5)*d)*b*d^3*e^3 + a*c^3*e^3*x + ( 
(d*x + c)*arcsinh(d*x + c) - sqrt((d*x + c)^2 + 1))*b*c^3*e^3/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 617 vs. \(2 (93) = 186\).

Time = 0.61 (sec) , antiderivative size = 617, normalized size of antiderivative = 5.88 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*a*d^3*e^3*x^4 + a*c*d^2*e^3*x^3 + 3/2*a*c^2*d*e^3*x^2 - (d*(c*log(abs( 
-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d)))/(d*abs(d)) 
+ sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2) - x*log(d*x + c + sqrt(d^2*x^2 + 
2*c*d*x + c^2 + 1)))*b*c^3*e^3 + 3/4*(2*x^2*log(d*x + c + sqrt(d^2*x^2 + 2 
*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x/d^2 - 3*c/d^3) 
- (2*c^2 - 1)*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) 
)*abs(d)))/(d^2*abs(d)))*d)*b*c^2*d*e^3 + 1/6*(6*x^3*log(d*x + c + sqrt(d^ 
2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x*(2*x/d 
^2 - 5*c/d^3) + (11*c^2*d - 4*d)/d^5) + 3*(2*c^3 - 3*c)*log(abs(-c*d - (x* 
abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d)))/(d^3*abs(d)))*d)*b*c* 
d^2*e^3 + 1/96*(24*x^4*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - 
(sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*((2*x*(3*x/d^2 - 7*c/d^3) + (26*c^2*d^3 
 - 9*d^3)/d^7)*x - 5*(10*c^3*d^2 - 11*c*d^2)/d^7) - 3*(8*c^4 - 24*c^2 + 3) 
*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d)))/(d 
^4*abs(d)))*d)*b*d^3*e^3 + a*c^3*e^3*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.14 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x)) \, dx=\frac {e^{3} \left (32 \mathit {asinh} \left (d x +c \right ) b \,c^{4}+32 \mathit {asinh} \left (d x +c \right ) b \,c^{3} d x +48 \mathit {asinh} \left (d x +c \right ) b \,c^{2} d^{2} x^{2}+32 \mathit {asinh} \left (d x +c \right ) b c \,d^{3} x^{3}+8 \mathit {asinh} \left (d x +c \right ) b \,d^{4} x^{4}-2 \sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}\, b \,c^{3}-6 \sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}\, b \,c^{2} d x -6 \sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}\, b c \,d^{2} x^{2}+3 \sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}\, b c -2 \sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}\, b \,d^{3} x^{3}+3 \sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}\, b d x -24 \,\mathrm {log}\left (\sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}+c +d x \right ) b \,c^{4}-3 \,\mathrm {log}\left (\sqrt {d^{2} x^{2}+2 c d x +c^{2}+1}+c +d x \right ) b +32 a \,c^{3} d x +48 a \,c^{2} d^{2} x^{2}+32 a c \,d^{3} x^{3}+8 a \,d^{4} x^{4}\right )}{32 d} \] Input:

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

Output:

(e**3*(32*asinh(c + d*x)*b*c**4 + 32*asinh(c + d*x)*b*c**3*d*x + 48*asinh( 
c + d*x)*b*c**2*d**2*x**2 + 32*asinh(c + d*x)*b*c*d**3*x**3 + 8*asinh(c + 
d*x)*b*d**4*x**4 - 2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*c**3 - 6*sqrt( 
c**2 + 2*c*d*x + d**2*x**2 + 1)*b*c**2*d*x - 6*sqrt(c**2 + 2*c*d*x + d**2* 
x**2 + 1)*b*c*d**2*x**2 + 3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*c - 2*s 
qrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*b*d**3*x**3 + 3*sqrt(c**2 + 2*c*d*x + 
d**2*x**2 + 1)*b*d*x - 24*log(sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1) + c + d 
*x)*b*c**4 - 3*log(sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1) + c + d*x)*b + 32* 
a*c**3*d*x + 48*a*c**2*d**2*x**2 + 32*a*c*d**3*x**3 + 8*a*d**4*x**4))/(32* 
d)