3.137 \(\int (c e+d e x)^4 (a+b \sinh ^{-1}(c+d x))^3 \, dx\)

Optimal. Leaf size=326 \[ \frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}-\frac{8 b^2 e^4 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{16}{25} a b^2 e^4 x+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}-\frac{3 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{4 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{8 b e^4 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{6 b^3 e^4 \left ((c+d x)^2+1\right )^{5/2}}{625 d}+\frac{76 b^3 e^4 \left ((c+d x)^2+1\right )^{3/2}}{1125 d}-\frac{298 b^3 e^4 \sqrt{(c+d x)^2+1}}{375 d}+\frac{16 b^3 e^4 (c+d x) \sinh ^{-1}(c+d x)}{25 d} \]

[Out]

(16*a*b^2*e^4*x)/25 - (298*b^3*e^4*Sqrt[1 + (c + d*x)^2])/(375*d) + (76*b^3*e^4*(1 + (c + d*x)^2)^(3/2))/(1125
*d) - (6*b^3*e^4*(1 + (c + d*x)^2)^(5/2))/(625*d) + (16*b^3*e^4*(c + d*x)*ArcSinh[c + d*x])/(25*d) - (8*b^2*e^
4*(c + d*x)^3*(a + b*ArcSinh[c + d*x]))/(75*d) + (6*b^2*e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x]))/(125*d) - (8
*b*e^4*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(25*d) + (4*b*e^4*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(
a + b*ArcSinh[c + d*x])^2)/(25*d) - (3*b*e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(25
*d) + (e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x])^3)/(5*d)

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Rubi [A]  time = 0.473741, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {5865, 12, 5661, 5758, 5717, 5653, 261, 266, 43} \[ \frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}-\frac{8 b^2 e^4 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{16}{25} a b^2 e^4 x+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}-\frac{3 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{4 b e^4 \sqrt{(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{8 b e^4 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{6 b^3 e^4 \left ((c+d x)^2+1\right )^{5/2}}{625 d}+\frac{76 b^3 e^4 \left ((c+d x)^2+1\right )^{3/2}}{1125 d}-\frac{298 b^3 e^4 \sqrt{(c+d x)^2+1}}{375 d}+\frac{16 b^3 e^4 (c+d x) \sinh ^{-1}(c+d x)}{25 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(16*a*b^2*e^4*x)/25 - (298*b^3*e^4*Sqrt[1 + (c + d*x)^2])/(375*d) + (76*b^3*e^4*(1 + (c + d*x)^2)^(3/2))/(1125
*d) - (6*b^3*e^4*(1 + (c + d*x)^2)^(5/2))/(625*d) + (16*b^3*e^4*(c + d*x)*ArcSinh[c + d*x])/(25*d) - (8*b^2*e^
4*(c + d*x)^3*(a + b*ArcSinh[c + d*x]))/(75*d) + (6*b^2*e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x]))/(125*d) - (8
*b*e^4*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(25*d) + (4*b*e^4*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(
a + b*ArcSinh[c + d*x])^2)/(25*d) - (3*b*e^4*(c + d*x)^4*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(25
*d) + (e^4*(c + d*x)^5*(a + b*ArcSinh[c + d*x])^3)/(5*d)

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[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
]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(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 5758

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

Rule 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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] && NeQ[b*c - a*d, 0] && 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])

Rubi steps

\begin{align*} \int (c e+d e x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^4 x^4 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int x^4 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}-\frac{\left (3 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^5 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{5 d}\\ &=-\frac{3 b e^4 (c+d x)^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}+\frac{\left (12 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{25 d}+\frac{\left (6 b^2 e^4\right ) \operatorname{Subst}\left (\int x^4 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{25 d}\\ &=\frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}+\frac{4 b e^4 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{3 b e^4 (c+d x)^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}-\frac{\left (8 b e^4\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{25 d}-\frac{\left (8 b^2 e^4\right ) \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{25 d}-\frac{\left (6 b^3 e^4\right ) \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{125 d}\\ &=-\frac{8 b^2 e^4 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}-\frac{8 b e^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{4 b e^4 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{3 b e^4 (c+d x)^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}+\frac{\left (16 b^2 e^4\right ) \operatorname{Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{25 d}-\frac{\left (3 b^3 e^4\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x}} \, dx,x,(c+d x)^2\right )}{125 d}+\frac{\left (8 b^3 e^4\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{75 d}\\ &=\frac{16}{25} a b^2 e^4 x-\frac{8 b^2 e^4 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}-\frac{8 b e^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{4 b e^4 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{3 b e^4 (c+d x)^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}-\frac{\left (3 b^3 e^4\right ) \operatorname{Subst}\left (\int \left (\frac{1}{\sqrt{1+x}}-2 \sqrt{1+x}+(1+x)^{3/2}\right ) \, dx,x,(c+d x)^2\right )}{125 d}+\frac{\left (4 b^3 e^4\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x}} \, dx,x,(c+d x)^2\right )}{75 d}+\frac{\left (16 b^3 e^4\right ) \operatorname{Subst}\left (\int \sinh ^{-1}(x) \, dx,x,c+d x\right )}{25 d}\\ &=\frac{16}{25} a b^2 e^4 x-\frac{6 b^3 e^4 \sqrt{1+(c+d x)^2}}{125 d}+\frac{4 b^3 e^4 \left (1+(c+d x)^2\right )^{3/2}}{125 d}-\frac{6 b^3 e^4 \left (1+(c+d x)^2\right )^{5/2}}{625 d}+\frac{16 b^3 e^4 (c+d x) \sinh ^{-1}(c+d x)}{25 d}-\frac{8 b^2 e^4 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}-\frac{8 b e^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{4 b e^4 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{3 b e^4 (c+d x)^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}+\frac{\left (4 b^3 e^4\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{\sqrt{1+x}}+\sqrt{1+x}\right ) \, dx,x,(c+d x)^2\right )}{75 d}-\frac{\left (16 b^3 e^4\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{25 d}\\ &=\frac{16}{25} a b^2 e^4 x-\frac{298 b^3 e^4 \sqrt{1+(c+d x)^2}}{375 d}+\frac{76 b^3 e^4 \left (1+(c+d x)^2\right )^{3/2}}{1125 d}-\frac{6 b^3 e^4 \left (1+(c+d x)^2\right )^{5/2}}{625 d}+\frac{16 b^3 e^4 (c+d x) \sinh ^{-1}(c+d x)}{25 d}-\frac{8 b^2 e^4 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{75 d}+\frac{6 b^2 e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )}{125 d}-\frac{8 b e^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{4 b e^4 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}-\frac{3 b e^4 (c+d x)^4 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.465942, size = 355, normalized size = 1.09 \[ \frac{e^4 \left (3 a \left (25 a^2+6 b^2\right ) (c+d x)^5+\frac{1}{15} b \sqrt{(c+d x)^2+1} \left (-27 \left (25 a^2+2 b^2\right ) (c+d x)^4+4 \left (225 a^2+68 b^2\right ) (c+d x)^2-8 \left (225 a^2+518 b^2\right )\right )-b \sinh ^{-1}(c+d x) \left (-225 a^2 (c+d x)^5+90 a b \sqrt{(c+d x)^2+1} (c+d x)^4-120 a b \sqrt{(c+d x)^2+1} (c+d x)^2+240 a b \sqrt{(c+d x)^2+1}-18 b^2 (c+d x)^5+40 b^2 (c+d x)^3-240 b^2 (c+d x)\right )-40 a b^2 (c+d x)^3+240 a b^2 (c+d x)-15 b^2 \sinh ^{-1}(c+d x)^2 \left (-15 a (c+d x)^5+3 b \sqrt{(c+d x)^2+1} (c+d x)^4-4 b \sqrt{(c+d x)^2+1} (c+d x)^2+8 b \sqrt{(c+d x)^2+1}\right )+75 b^3 (c+d x)^5 \sinh ^{-1}(c+d x)^3\right )}{375 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(e^4*(240*a*b^2*(c + d*x) - 40*a*b^2*(c + d*x)^3 + 3*a*(25*a^2 + 6*b^2)*(c + d*x)^5 + (b*Sqrt[1 + (c + d*x)^2]
*(-8*(225*a^2 + 518*b^2) + 4*(225*a^2 + 68*b^2)*(c + d*x)^2 - 27*(25*a^2 + 2*b^2)*(c + d*x)^4))/15 - b*(-240*b
^2*(c + d*x) + 40*b^2*(c + d*x)^3 - 225*a^2*(c + d*x)^5 - 18*b^2*(c + d*x)^5 + 240*a*b*Sqrt[1 + (c + d*x)^2] -
120*a*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2] + 90*a*b*(c + d*x)^4*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] - 15*b
^2*(-15*a*(c + d*x)^5 + 8*b*Sqrt[1 + (c + d*x)^2] - 4*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2] + 3*b*(c + d*x)^4*Sq
rt[1 + (c + d*x)^2])*ArcSinh[c + d*x]^2 + 75*b^3*(c + d*x)^5*ArcSinh[c + d*x]^3))/(375*d)

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Maple [A]  time = 0.05, size = 548, normalized size = 1.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c))^3,x)

[Out]

1/d*(1/5*(d*x+c)^5*e^4*a^3+e^4*b^3*(1/5*(d*x+c)^3*arcsinh(d*x+c)^3*(1+(d*x+c)^2)-1/5*arcsinh(d*x+c)^3*(d*x+c)*
(1+(d*x+c)^2)+1/5*arcsinh(d*x+c)^3*(d*x+c)-3/25*arcsinh(d*x+c)^2*(d*x+c)^2*(1+(d*x+c)^2)^(3/2)+7/25*arcsinh(d*
x+c)^2*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-8/25*arcsinh(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+6/125*arcsinh(d*x+c)*(d*x+c)*(1
+(d*x+c)^2)^2+298/375*(d*x+c)*arcsinh(d*x+c)-76/375*(d*x+c)*(1+(d*x+c)^2)*arcsinh(d*x+c)-6/625*(d*x+c)^2*(1+(d
*x+c)^2)^(3/2)+326/5625*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-4144/5625*(1+(d*x+c)^2)^(1/2))+3*e^4*a*b^2*(1/5*(d*x+c)^
3*arcsinh(d*x+c)^2*(1+(d*x+c)^2)-1/5*arcsinh(d*x+c)^2*(d*x+c)*(1+(d*x+c)^2)+1/5*arcsinh(d*x+c)^2*(d*x+c)-2/25*
arcsinh(d*x+c)*(d*x+c)^2*(1+(d*x+c)^2)^(3/2)+14/75*arcsinh(d*x+c)*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)-16/75*arcsinh(
d*x+c)*(1+(d*x+c)^2)^(1/2)+2/125*(1+(d*x+c)^2)^2*(d*x+c)+298/1125*d*x+298/1125*c-76/1125*(1+(d*x+c)^2)*(d*x+c)
)+3*e^4*a^2*b*(1/5*(d*x+c)^5*arcsinh(d*x+c)-1/25*(d*x+c)^4*(1+(d*x+c)^2)^(1/2)+4/75*(d*x+c)^2*(1+(d*x+c)^2)^(1
/2)-8/75*(1+(d*x+c)^2)^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.16982, size = 2325, normalized size = 7.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/5625*(45*(25*a^3 + 6*a*b^2)*d^5*e^4*x^5 + 225*(25*a^3 + 6*a*b^2)*c*d^4*e^4*x^4 - 150*(4*a*b^2 - 3*(25*a^3 +
6*a*b^2)*c^2)*d^3*e^4*x^3 - 450*(4*a*b^2*c - (25*a^3 + 6*a*b^2)*c^3)*d^2*e^4*x^2 - 225*(8*a*b^2*c^2 - (25*a^3
+ 6*a*b^2)*c^4 - 16*a*b^2)*d*e^4*x + 1125*(b^3*d^5*e^4*x^5 + 5*b^3*c*d^4*e^4*x^4 + 10*b^3*c^2*d^3*e^4*x^3 + 10
*b^3*c^3*d^2*e^4*x^2 + 5*b^3*c^4*d*e^4*x + b^3*c^5*e^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^3 + 2
25*(15*a*b^2*d^5*e^4*x^5 + 75*a*b^2*c*d^4*e^4*x^4 + 150*a*b^2*c^2*d^3*e^4*x^3 + 150*a*b^2*c^3*d^2*e^4*x^2 + 75
*a*b^2*c^4*d*e^4*x + 15*a*b^2*c^5*e^4 - (3*b^3*d^4*e^4*x^4 + 12*b^3*c*d^3*e^4*x^3 + 2*(9*b^3*c^2 - 2*b^3)*d^2*
e^4*x^2 + 4*(3*b^3*c^3 - 2*b^3*c)*d*e^4*x + (3*b^3*c^4 - 4*b^3*c^2 + 8*b^3)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2
+ 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2 + 15*(9*(25*a^2*b + 2*b^3)*d^5*e^4*x^5 + 45*(25*a^2*b
+ 2*b^3)*c*d^4*e^4*x^4 - 10*(4*b^3 - 9*(25*a^2*b + 2*b^3)*c^2)*d^3*e^4*x^3 - 30*(4*b^3*c - 3*(25*a^2*b + 2*b^
3)*c^3)*d^2*e^4*x^2 - 15*(8*b^3*c^2 - 3*(25*a^2*b + 2*b^3)*c^4 - 16*b^3)*d*e^4*x - (40*b^3*c^3 - 9*(25*a^2*b +
2*b^3)*c^5 - 240*b^3*c)*e^4 - 30*(3*a*b^2*d^4*e^4*x^4 + 12*a*b^2*c*d^3*e^4*x^3 + 2*(9*a*b^2*c^2 - 2*a*b^2)*d^
2*e^4*x^2 + 4*(3*a*b^2*c^3 - 2*a*b^2*c)*d*e^4*x + (3*a*b^2*c^4 - 4*a*b^2*c^2 + 8*a*b^2)*e^4)*sqrt(d^2*x^2 + 2*
c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (27*(25*a^2*b + 2*b^3)*d^4*e^4*x^4 + 108*
(25*a^2*b + 2*b^3)*c*d^3*e^4*x^3 - 2*(450*a^2*b + 136*b^3 - 81*(25*a^2*b + 2*b^3)*c^2)*d^2*e^4*x^2 + 4*(27*(25
*a^2*b + 2*b^3)*c^3 - 2*(225*a^2*b + 68*b^3)*c)*d*e^4*x + (27*(25*a^2*b + 2*b^3)*c^4 + 1800*a^2*b + 4144*b^3 -
4*(225*a^2*b + 68*b^3)*c^2)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d

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Sympy [A]  time = 25.807, size = 2518, normalized size = 7.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4*(a+b*asinh(d*x+c))**3,x)

[Out]

Piecewise((a**3*c**4*e**4*x + 2*a**3*c**3*d*e**4*x**2 + 2*a**3*c**2*d**2*e**4*x**3 + a**3*c*d**3*e**4*x**4 + a
**3*d**4*e**4*x**5/5 + 3*a**2*b*c**5*e**4*asinh(c + d*x)/(5*d) + 3*a**2*b*c**4*e**4*x*asinh(c + d*x) - 3*a**2*
b*c**4*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(25*d) + 6*a**2*b*c**3*d*e**4*x**2*asinh(c + d*x) - 12*a**2*b
*c**3*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 6*a**2*b*c**2*d**2*e**4*x**3*asinh(c + d*x) - 18*a**2*b
*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 4*a**2*b*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2
+ 1)/(25*d) + 3*a**2*b*c*d**3*e**4*x**4*asinh(c + d*x) - 12*a**2*b*c*d**2*e**4*x**3*sqrt(c**2 + 2*c*d*x + d**
2*x**2 + 1)/25 + 8*a**2*b*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 3*a**2*b*d**4*e**4*x**5*asinh(c +
d*x)/5 - 3*a**2*b*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/25 + 4*a**2*b*d*e**4*x**2*sqrt(c**2 + 2
*c*d*x + d**2*x**2 + 1)/25 - 8*a**2*b*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(25*d) + 3*a*b**2*c**5*e**4*as
inh(c + d*x)**2/(5*d) + 3*a*b**2*c**4*e**4*x*asinh(c + d*x)**2 + 6*a*b**2*c**4*e**4*x/25 - 6*a*b**2*c**4*e**4*
sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(25*d) + 6*a*b**2*c**3*d*e**4*x**2*asinh(c + d*x)**2 + 12*
a*b**2*c**3*d*e**4*x**2/25 - 24*a*b**2*c**3*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/25 + 6*
a*b**2*c**2*d**2*e**4*x**3*asinh(c + d*x)**2 + 12*a*b**2*c**2*d**2*e**4*x**3/25 - 36*a*b**2*c**2*d*e**4*x**2*s
qrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/25 - 8*a*b**2*c**2*e**4*x/25 + 8*a*b**2*c**2*e**4*sqrt(c**2
+ 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(25*d) + 3*a*b**2*c*d**3*e**4*x**4*asinh(c + d*x)**2 + 6*a*b**2*c*d
**3*e**4*x**4/25 - 24*a*b**2*c*d**2*e**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/25 - 8*a*b**
2*c*d*e**4*x**2/25 + 16*a*b**2*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/25 + 3*a*b**2*d**4
*e**4*x**5*asinh(c + d*x)**2/5 + 6*a*b**2*d**4*e**4*x**5/125 - 6*a*b**2*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d
**2*x**2 + 1)*asinh(c + d*x)/25 - 8*a*b**2*d**2*e**4*x**3/75 + 8*a*b**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2
*x**2 + 1)*asinh(c + d*x)/25 + 16*a*b**2*e**4*x/25 - 16*a*b**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh
(c + d*x)/(25*d) + b**3*c**5*e**4*asinh(c + d*x)**3/(5*d) + 6*b**3*c**5*e**4*asinh(c + d*x)/(125*d) + b**3*c**
4*e**4*x*asinh(c + d*x)**3 + 6*b**3*c**4*e**4*x*asinh(c + d*x)/25 - 3*b**3*c**4*e**4*sqrt(c**2 + 2*c*d*x + d**
2*x**2 + 1)*asinh(c + d*x)**2/(25*d) - 6*b**3*c**4*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(625*d) + 2*b**3*
c**3*d*e**4*x**2*asinh(c + d*x)**3 + 12*b**3*c**3*d*e**4*x**2*asinh(c + d*x)/25 - 12*b**3*c**3*e**4*x*sqrt(c**
2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/25 - 24*b**3*c**3*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/6
25 - 8*b**3*c**3*e**4*asinh(c + d*x)/(75*d) + 2*b**3*c**2*d**2*e**4*x**3*asinh(c + d*x)**3 + 12*b**3*c**2*d**2
*e**4*x**3*asinh(c + d*x)/25 - 18*b**3*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2
/25 - 36*b**3*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/625 - 8*b**3*c**2*e**4*x*asinh(c + d*x)/25
+ 4*b**3*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/(25*d) + 272*b**3*c**2*e**4*sqrt(c*
*2 + 2*c*d*x + d**2*x**2 + 1)/(5625*d) + b**3*c*d**3*e**4*x**4*asinh(c + d*x)**3 + 6*b**3*c*d**3*e**4*x**4*asi
nh(c + d*x)/25 - 12*b**3*c*d**2*e**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/25 - 24*b**3*
c*d**2*e**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/625 - 8*b**3*c*d*e**4*x**2*asinh(c + d*x)/25 + 8*b**3*c*
e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/25 + 544*b**3*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**
2*x**2 + 1)/5625 + 16*b**3*c*e**4*asinh(c + d*x)/(25*d) + b**3*d**4*e**4*x**5*asinh(c + d*x)**3/5 + 6*b**3*d**
4*e**4*x**5*asinh(c + d*x)/125 - 3*b**3*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/
25 - 6*b**3*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/625 - 8*b**3*d**2*e**4*x**3*asinh(c + d*x)/75
+ 4*b**3*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/25 + 272*b**3*d*e**4*x**2*sqrt(c**
2 + 2*c*d*x + d**2*x**2 + 1)/5625 + 16*b**3*e**4*x*asinh(c + d*x)/25 - 8*b**3*e**4*sqrt(c**2 + 2*c*d*x + d**2*
x**2 + 1)*asinh(c + d*x)**2/(25*d) - 4144*b**3*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(5625*d), Ne(d, 0)),
(c**4*e**4*x*(a + b*asinh(c))**3, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{4}{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsinh(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^4*(b*arcsinh(d*x + c) + a)^3, x)