Integrand size = 23, antiderivative size = 172 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=-\frac {3 b^2 e^3 (c+d x)^2}{32 d}+\frac {b^2 e^3 (c+d x)^4}{32 d}+\frac {3 b e^3 (c+d x) \sqrt {1+(c+d x)^2} (a+b \text {arcsinh}(c+d x))}{16 d}-\frac {b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} (a+b \text {arcsinh}(c+d x))}{8 d}-\frac {3 e^3 (a+b \text {arcsinh}(c+d x))^2}{32 d}+\frac {e^3 (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2}{4 d} \]
-3/32*b^2*e^3*(d*x+c)^2/d+1/32*b^2*e^3*(d*x+c)^4/d-3/32*e^3*(a+b*arcsinh(d *x+c))^2/d+1/4*e^3*(d*x+c)^4*(a+b*arcsinh(d*x+c))^2/d+3/16*b*e^3*(d*x+c)*( a+b*arcsinh(d*x+c))*(1+(d*x+c)^2)^(1/2)/d-1/8*b*e^3*(d*x+c)^3*(a+b*arcsinh (d*x+c))*(1+(d*x+c)^2)^(1/2)/d
Time = 0.23 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.99 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=\frac {e^3 \left (-3 b^2 (c+d x)^2+\left (8 a^2+b^2\right ) (c+d x)^4+2 a b (c+d x) \left (3-2 (c+d x)^2\right ) \sqrt {1+(c+d x)^2}-6 a b \text {arcsinh}(c+d x)+2 b (c+d x) \left (8 a (c+d x)^3+3 b \sqrt {1+(c+d x)^2}-2 b (c+d x)^2 \sqrt {1+(c+d x)^2}\right ) \text {arcsinh}(c+d x)+b^2 \left (-3+8 (c+d x)^4\right ) \text {arcsinh}(c+d x)^2\right )}{32 d} \]
(e^3*(-3*b^2*(c + d*x)^2 + (8*a^2 + b^2)*(c + d*x)^4 + 2*a*b*(c + d*x)*(3 - 2*(c + d*x)^2)*Sqrt[1 + (c + d*x)^2] - 6*a*b*ArcSinh[c + d*x] + 2*b*(c + d*x)*(8*a*(c + d*x)^3 + 3*b*Sqrt[1 + (c + d*x)^2] - 2*b*(c + d*x)^2*Sqrt[ 1 + (c + d*x)^2])*ArcSinh[c + d*x] + b^2*(-3 + 8*(c + d*x)^4)*ArcSinh[c + d*x]^2))/(32*d)
Time = 0.65 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.88, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6274, 27, 6191, 6227, 15, 6227, 15, 6198}
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.
| \(\displaystyle \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 6274 |
| \(\displaystyle \frac {\int e^3 (c+d x)^3 (a+b \text {arcsinh}(c+d x))^2d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {e^3 \int (c+d x)^3 (a+b \text {arcsinh}(c+d x))^2d(c+d x)}{d}\) |
| \(\Big \downarrow \) 6191 |
| \(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2-\frac {1}{2} b \int \frac {(c+d x)^4 (a+b \text {arcsinh}(c+d x))}{\sqrt {(c+d x)^2+1}}d(c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 6227 |
| \(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2-\frac {1}{2} b \left (-\frac {3}{4} \int \frac {(c+d x)^2 (a+b \text {arcsinh}(c+d x))}{\sqrt {(c+d x)^2+1}}d(c+d x)-\frac {1}{4} b \int (c+d x)^3d(c+d x)+\frac {1}{4} \sqrt {(c+d x)^2+1} (c+d x)^3 (a+b \text {arcsinh}(c+d x))\right )\right )}{d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2-\frac {1}{2} b \left (-\frac {3}{4} \int \frac {(c+d x)^2 (a+b \text {arcsinh}(c+d x))}{\sqrt {(c+d x)^2+1}}d(c+d x)+\frac {1}{4} \sqrt {(c+d x)^2+1} (c+d x)^3 (a+b \text {arcsinh}(c+d x))-\frac {1}{16} b (c+d x)^4\right )\right )}{d}\) |
| \(\Big \downarrow \) 6227 |
| \(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2-\frac {1}{2} b \left (-\frac {3}{4} \left (-\frac {1}{2} \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {(c+d x)^2+1}}d(c+d x)-\frac {1}{2} b \int (c+d x)d(c+d x)+\frac {1}{2} (c+d x) \sqrt {(c+d x)^2+1} (a+b \text {arcsinh}(c+d x))\right )+\frac {1}{4} \sqrt {(c+d x)^2+1} (c+d x)^3 (a+b \text {arcsinh}(c+d x))-\frac {1}{16} b (c+d x)^4\right )\right )}{d}\) |
| \(\Big \downarrow \) 15 |
| \(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2-\frac {1}{2} b \left (-\frac {3}{4} \left (-\frac {1}{2} \int \frac {a+b \text {arcsinh}(c+d x)}{\sqrt {(c+d x)^2+1}}d(c+d x)+\frac {1}{2} \sqrt {(c+d x)^2+1} (c+d x) (a+b \text {arcsinh}(c+d x))-\frac {1}{4} b (c+d x)^2\right )+\frac {1}{4} \sqrt {(c+d x)^2+1} (c+d x)^3 (a+b \text {arcsinh}(c+d x))-\frac {1}{16} b (c+d x)^4\right )\right )}{d}\) |
| \(\Big \downarrow \) 6198 |
| \(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \text {arcsinh}(c+d x))^2-\frac {1}{2} b \left (\frac {1}{4} \sqrt {(c+d x)^2+1} (c+d x)^3 (a+b \text {arcsinh}(c+d x))-\frac {3}{4} \left (\frac {1}{2} \sqrt {(c+d x)^2+1} (c+d x) (a+b \text {arcsinh}(c+d x))-\frac {(a+b \text {arcsinh}(c+d x))^2}{4 b}-\frac {1}{4} b (c+d x)^2\right )-\frac {1}{16} b (c+d x)^4\right )\right )}{d}\) |
(e^3*(((c + d*x)^4*(a + b*ArcSinh[c + d*x])^2)/4 - (b*(-1/16*(b*(c + d*x)^ 4) + ((c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/4 - (3*( -1/4*(b*(c + d*x)^2) + ((c + d*x)*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/2 - (a + b*ArcSinh[c + d*x])^2/(4*b)))/4))/2))/d
3.2.28.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ Symbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*( a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c ^2*d] && NeQ[n, -1]
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]
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]
Time = 0.41 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {\frac {e^{3} a^{2} \left (d x +c \right )^{4}}{4}+e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \operatorname {arcsinh}\left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{8}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right ) \operatorname {arcsinh}\left (d x +c \right )}{16}-\frac {3 \operatorname {arcsinh}\left (d x +c \right )^{2}}{32}+\frac {\left (d x +c \right )^{4}}{32}-\frac {3 \left (d x +c \right )^{2}}{32}-\frac {3}{32}\right )+2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \,\operatorname {arcsinh}\left (d x +c \right )}{32}\right )}{d}\) | \(194\) |
| default | \(\frac {\frac {e^{3} a^{2} \left (d x +c \right )^{4}}{4}+e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \operatorname {arcsinh}\left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{8}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right ) \operatorname {arcsinh}\left (d x +c \right )}{16}-\frac {3 \operatorname {arcsinh}\left (d x +c \right )^{2}}{32}+\frac {\left (d x +c \right )^{4}}{32}-\frac {3 \left (d x +c \right )^{2}}{32}-\frac {3}{32}\right )+2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \,\operatorname {arcsinh}\left (d x +c \right )}{32}\right )}{d}\) | \(194\) |
| parts | \(\frac {e^{3} a^{2} \left (d x +c \right )^{4}}{4 d}+\frac {e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \operatorname {arcsinh}\left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{8}+\frac {3 \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right ) \operatorname {arcsinh}\left (d x +c \right )}{16}-\frac {3 \operatorname {arcsinh}\left (d x +c \right )^{2}}{32}+\frac {\left (d x +c \right )^{4}}{32}-\frac {3 \left (d x +c \right )^{2}}{32}-\frac {3}{32}\right )}{d}+\frac {2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \operatorname {arcsinh}\left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \,\operatorname {arcsinh}\left (d x +c \right )}{32}\right )}{d}\) | \(199\) |
1/d*(1/4*e^3*a^2*(d*x+c)^4+e^3*b^2*(1/4*(d*x+c)^4*arcsinh(d*x+c)^2-1/8*(d* x+c)^3*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+3/16*(1+(d*x+c)^2)^(1/2)*(d*x+c) *arcsinh(d*x+c)-3/32*arcsinh(d*x+c)^2+1/32*(d*x+c)^4-3/32*(d*x+c)^2-3/32)+ 2*e^3*a*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)))
Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (156) = 312\).
Time = 0.26 (sec) , antiderivative size = 486, normalized size of antiderivative = 2.83 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=\frac {{\left (8 \, a^{2} + b^{2}\right )} d^{4} e^{3} x^{4} + 4 \, {\left (8 \, a^{2} + b^{2}\right )} c d^{3} e^{3} x^{3} + 3 \, {\left (2 \, {\left (8 \, a^{2} + b^{2}\right )} c^{2} - b^{2}\right )} d^{2} e^{3} x^{2} + 2 \, {\left (2 \, {\left (8 \, a^{2} + b^{2}\right )} c^{3} - 3 \, b^{2} c\right )} d e^{3} x + {\left (8 \, b^{2} d^{4} e^{3} x^{4} + 32 \, b^{2} c d^{3} e^{3} x^{3} + 48 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 32 \, b^{2} c^{3} d e^{3} x + {\left (8 \, b^{2} c^{4} - 3 \, b^{2}\right )} e^{3}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 2 \, {\left (8 \, a b d^{4} e^{3} x^{4} + 32 \, a b c d^{3} e^{3} x^{3} + 48 \, a b c^{2} d^{2} e^{3} x^{2} + 32 \, a b c^{3} d e^{3} x + {\left (8 \, a b c^{4} - 3 \, a b\right )} e^{3} - {\left (2 \, b^{2} d^{3} e^{3} x^{3} + 6 \, b^{2} c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, b^{2} c^{2} - b^{2}\right )} d e^{3} x + {\left (2 \, b^{2} c^{3} - 3 \, b^{2} c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 2 \, {\left (2 \, a b d^{3} e^{3} x^{3} + 6 \, a b c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, a b c^{2} - a b\right )} d e^{3} x + {\left (2 \, a b c^{3} - 3 \, a b c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{32 \, d} \]
1/32*((8*a^2 + b^2)*d^4*e^3*x^4 + 4*(8*a^2 + b^2)*c*d^3*e^3*x^3 + 3*(2*(8* a^2 + b^2)*c^2 - b^2)*d^2*e^3*x^2 + 2*(2*(8*a^2 + b^2)*c^3 - 3*b^2*c)*d*e^ 3*x + (8*b^2*d^4*e^3*x^4 + 32*b^2*c*d^3*e^3*x^3 + 48*b^2*c^2*d^2*e^3*x^2 + 32*b^2*c^3*d*e^3*x + (8*b^2*c^4 - 3*b^2)*e^3)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2 + 2*(8*a*b*d^4*e^3*x^4 + 32*a*b*c*d^3*e^3*x^3 + 48 *a*b*c^2*d^2*e^3*x^2 + 32*a*b*c^3*d*e^3*x + (8*a*b*c^4 - 3*a*b)*e^3 - (2*b ^2*d^3*e^3*x^3 + 6*b^2*c*d^2*e^3*x^2 + 3*(2*b^2*c^2 - b^2)*d*e^3*x + (2*b^ 2*c^3 - 3*b^2*c)*e^3)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqr t(d^2*x^2 + 2*c*d*x + c^2 + 1)) - 2*(2*a*b*d^3*e^3*x^3 + 6*a*b*c*d^2*e^3*x ^2 + 3*(2*a*b*c^2 - a*b)*d*e^3*x + (2*a*b*c^3 - 3*a*b*c)*e^3)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 916 vs. \(2 (155) = 310\).
Time = 0.47 (sec) , antiderivative size = 916, normalized size of antiderivative = 5.33 \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=\begin {cases} a^{2} c^{3} e^{3} x + \frac {3 a^{2} c^{2} d e^{3} x^{2}}{2} + a^{2} c d^{2} e^{3} x^{3} + \frac {a^{2} d^{3} e^{3} x^{4}}{4} + \frac {a b c^{4} e^{3} \operatorname {asinh}{\left (c + d x \right )}}{2 d} + 2 a b c^{3} e^{3} x \operatorname {asinh}{\left (c + d x \right )} - \frac {a b c^{3} e^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{8 d} + 3 a b c^{2} d e^{3} x^{2} \operatorname {asinh}{\left (c + d x \right )} - \frac {3 a b c^{2} e^{3} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{8} + 2 a b c d^{2} e^{3} x^{3} \operatorname {asinh}{\left (c + d x \right )} - \frac {3 a b c d e^{3} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{8} + \frac {3 a b c e^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{16 d} + \frac {a b d^{3} e^{3} x^{4} \operatorname {asinh}{\left (c + d x \right )}}{2} - \frac {a b d^{2} e^{3} x^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{8} + \frac {3 a b e^{3} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{16} - \frac {3 a b e^{3} \operatorname {asinh}{\left (c + d x \right )}}{16 d} + \frac {b^{2} c^{4} e^{3} \operatorname {asinh}^{2}{\left (c + d x \right )}}{4 d} + b^{2} c^{3} e^{3} x \operatorname {asinh}^{2}{\left (c + d x \right )} + \frac {b^{2} c^{3} e^{3} x}{8} - \frac {b^{2} c^{3} e^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{8 d} + \frac {3 b^{2} c^{2} d e^{3} x^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{2} + \frac {3 b^{2} c^{2} d e^{3} x^{2}}{16} - \frac {3 b^{2} c^{2} e^{3} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{8} + b^{2} c d^{2} e^{3} x^{3} \operatorname {asinh}^{2}{\left (c + d x \right )} + \frac {b^{2} c d^{2} e^{3} x^{3}}{8} - \frac {3 b^{2} c d e^{3} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{8} - \frac {3 b^{2} c e^{3} x}{16} + \frac {3 b^{2} c e^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{16 d} + \frac {b^{2} d^{3} e^{3} x^{4} \operatorname {asinh}^{2}{\left (c + d x \right )}}{4} + \frac {b^{2} d^{3} e^{3} x^{4}}{32} - \frac {b^{2} d^{2} e^{3} x^{3} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{8} - \frac {3 b^{2} d e^{3} x^{2}}{32} + \frac {3 b^{2} e^{3} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{16} - \frac {3 b^{2} e^{3} \operatorname {asinh}^{2}{\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 )^{2} & \text {otherwise} \end {cases} \]
Piecewise((a**2*c**3*e**3*x + 3*a**2*c**2*d*e**3*x**2/2 + a**2*c*d**2*e**3 *x**3 + a**2*d**3*e**3*x**4/4 + a*b*c**4*e**3*asinh(c + d*x)/(2*d) + 2*a*b *c**3*e**3*x*asinh(c + d*x) - a*b*c**3*e**3*sqrt(c**2 + 2*c*d*x + d**2*x** 2 + 1)/(8*d) + 3*a*b*c**2*d*e**3*x**2*asinh(c + d*x) - 3*a*b*c**2*e**3*x*s qrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/8 + 2*a*b*c*d**2*e**3*x**3*asinh(c + d *x) - 3*a*b*c*d*e**3*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/8 + 3*a*b*c *e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(16*d) + a*b*d**3*e**3*x**4*asi nh(c + d*x)/2 - a*b*d**2*e**3*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/8 + 3*a*b*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/16 - 3*a*b*e**3*asinh( c + d*x)/(16*d) + b**2*c**4*e**3*asinh(c + d*x)**2/(4*d) + b**2*c**3*e**3* x*asinh(c + d*x)**2 + b**2*c**3*e**3*x/8 - b**2*c**3*e**3*sqrt(c**2 + 2*c* d*x + d**2*x**2 + 1)*asinh(c + d*x)/(8*d) + 3*b**2*c**2*d*e**3*x**2*asinh( c + d*x)**2/2 + 3*b**2*c**2*d*e**3*x**2/16 - 3*b**2*c**2*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/8 + b**2*c*d**2*e**3*x**3*asinh( c + d*x)**2 + b**2*c*d**2*e**3*x**3/8 - 3*b**2*c*d*e**3*x**2*sqrt(c**2 + 2 *c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/8 - 3*b**2*c*e**3*x/16 + 3*b**2*c*e **3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(16*d) + b**2*d**3 *e**3*x**4*asinh(c + d*x)**2/4 + b**2*d**3*e**3*x**4/32 - b**2*d**2*e**3*x **3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/8 - 3*b**2*d*e**3* x**2/32 + 3*b**2*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + ...
\[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{3} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2} \,d x } \]
1/4*a^2*d^3*e^3*x^4 + a^2*c*d^2*e^3*x^3 + 3/2*a^2*c^2*d*e^3*x^2 + 3/2*(2*x ^2*arcsinh(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) *arcsinh(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))*a*b*c^2*d*e^3 + 1/3*(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))*a*b*c*d^2*e^3 + 1/48*(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)/sq rt(-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*arcsinh(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)*a*b*d^3*e^3 + a^2 *c^3*e^3*x + 2*((d*x + c)*arcsinh(d*x + c) - sqrt((d*x + c)^2 + 1))*a*b*c^ 3*e^3/d + 1/4*(b^2*d^3*e^3*x^4 + 4*b^2*c*d^2*e^3*x^3 + 6*b^2*c^2*d*e^3*...
\[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=\int { {\left (d e x + c e\right )}^{3} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (c e+d e x)^3 (a+b \text {arcsinh}(c+d x))^2 \, dx=\int {\left (c\,e+d\,e\,x\right )}^3\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \]