3.139 \(\int (c e+d e x)^2 (a+b \sinh ^{-1}(c+d x))^3 \, dx\)

Optimal. Leaf size=227 \[ \frac{2 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}-\frac{4}{3} a b^2 e^2 x+\frac{2 b e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}-\frac{2 b^3 e^2 \left ((c+d x)^2+1\right )^{3/2}}{27 d}+\frac{14 b^3 e^2 \sqrt{(c+d x)^2+1}}{9 d}-\frac{4 b^3 e^2 (c+d x) \sinh ^{-1}(c+d x)}{3 d} \]

[Out]

(-4*a*b^2*e^2*x)/3 + (14*b^3*e^2*Sqrt[1 + (c + d*x)^2])/(9*d) - (2*b^3*e^2*(1 + (c + d*x)^2)^(3/2))/(27*d) - (
4*b^3*e^2*(c + d*x)*ArcSinh[c + d*x])/(3*d) + (2*b^2*e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x]))/(9*d) + (2*b*e^
2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(3*d) - (b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*Ar
cSinh[c + d*x])^2)/(3*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^3)/(3*d)

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Rubi [A]  time = 0.299909, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ \frac{2 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}-\frac{4}{3} a b^2 e^2 x+\frac{2 b e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}-\frac{2 b^3 e^2 \left ((c+d x)^2+1\right )^{3/2}}{27 d}+\frac{14 b^3 e^2 \sqrt{(c+d x)^2+1}}{9 d}-\frac{4 b^3 e^2 (c+d x) \sinh ^{-1}(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2*(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(-4*a*b^2*e^2*x)/3 + (14*b^3*e^2*Sqrt[1 + (c + d*x)^2])/(9*d) - (2*b^3*e^2*(1 + (c + d*x)^2)^(3/2))/(27*d) - (
4*b^3*e^2*(c + d*x)*ArcSinh[c + d*x])/(3*d) + (2*b^2*e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x]))/(9*d) + (2*b*e^
2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^2)/(3*d) - (b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*(a + b*Ar
cSinh[c + d*x])^2)/(3*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^3)/(3*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)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (b e^2\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 )}{d}\\ &=-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}+\frac{\left (2 b e^2\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 )}{3 d}+\frac{\left (2 b^2 e^2\right ) \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{3 d}\\ &=\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac{2 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (4 b^2 e^2\right ) \operatorname{Subst}\left (\int \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{3 d}-\frac{\left (2 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{9 d}\\ &=-\frac{4}{3} a b^2 e^2 x+\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac{2 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x}} \, dx,x,(c+d x)^2\right )}{9 d}-\frac{\left (4 b^3 e^2\right ) \operatorname{Subst}\left (\int \sinh ^{-1}(x) \, dx,x,c+d x\right )}{3 d}\\ &=-\frac{4}{3} a b^2 e^2 x-\frac{4 b^3 e^2 (c+d x) \sinh ^{-1}(c+d x)}{3 d}+\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac{2 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}-\frac{\left (b^3 e^2\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{\sqrt{1+x}}+\sqrt{1+x}\right ) \, dx,x,(c+d x)^2\right )}{9 d}+\frac{\left (4 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=-\frac{4}{3} a b^2 e^2 x+\frac{14 b^3 e^2 \sqrt{1+(c+d x)^2}}{9 d}-\frac{2 b^3 e^2 \left (1+(c+d x)^2\right )^{3/2}}{27 d}-\frac{4 b^3 e^2 (c+d x) \sinh ^{-1}(c+d x)}{3 d}+\frac{2 b^2 e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac{2 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.311478, size = 258, normalized size = 1.14 \[ \frac{e^2 \left (a \left (3 a^2+2 b^2\right ) (c+d x)^3+\frac{1}{3} b \sqrt{(c+d x)^2+1} \left (-\left (9 a^2+2 b^2\right ) (c+d x)^2+18 a^2+40 b^2\right )-b \sinh ^{-1}(c+d x) \left (-9 a^2 (c+d x)^3+6 a b \sqrt{(c+d x)^2+1} (c+d x)^2-12 a b \sqrt{(c+d x)^2+1}-2 b^2 (c+d x)^3+12 b^2 (c+d x)\right )-12 a b^2 (c+d x)-3 b^2 \sinh ^{-1}(c+d x)^2 \left (-3 a (c+d x)^3+b \sqrt{(c+d x)^2+1} (c+d x)^2-2 b \sqrt{(c+d x)^2+1}\right )+3 b^3 (c+d x)^3 \sinh ^{-1}(c+d x)^3\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^2*(a + b*ArcSinh[c + d*x])^3,x]

[Out]

(e^2*(-12*a*b^2*(c + d*x) + a*(3*a^2 + 2*b^2)*(c + d*x)^3 + (b*Sqrt[1 + (c + d*x)^2]*(18*a^2 + 40*b^2 - (9*a^2
+ 2*b^2)*(c + d*x)^2))/3 - b*(12*b^2*(c + d*x) - 9*a^2*(c + d*x)^3 - 2*b^2*(c + d*x)^3 - 12*a*b*Sqrt[1 + (c +
d*x)^2] + 6*a*b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x] - 3*b^2*(-3*a*(c + d*x)^3 - 2*b*Sqrt[1 +
(c + d*x)^2] + b*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])*ArcSinh[c + d*x]^2 + 3*b^3*(c + d*x)^3*ArcSinh[c + d*x]^3)
)/(9*d)

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Maple [A]  time = 0.037, size = 360, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{3}{e}^{2}{a}^{3}}{3}}+{e}^{2}{b}^{3} \left ({\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3} \left ( dx+c \right ) \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{3}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{3} \left ( dx+c \right ) }{3}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) ^{2}}{3}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{2\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}}{3}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{ \left ( 2\,dx+2\,c \right ) \left ( 1+ \left ( dx+c \right ) ^{2} \right ){\it Arcsinh} \left ( dx+c \right ) }{9}}-{\frac{ \left ( 14\,dx+14\,c \right ){\it Arcsinh} \left ( dx+c \right ) }{9}}-{\frac{2\, \left ( dx+c \right ) ^{2}}{27}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{40}{27}\sqrt{1+ \left ( dx+c \right ) ^{2}}} \right ) +3\,{e}^{2}a{b}^{2} \left ( 1/3\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) \left ( 1+ \left ( dx+c \right ) ^{2} \right ) -1/3\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) -2/9\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}\sqrt{1+ \left ( dx+c \right ) ^{2}}+4/9\,{\it Arcsinh} \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+{\frac{ \left ( 2+2\, \left ( dx+c \right ) ^{2} \right ) \left ( dx+c \right ) }{27}}-{\frac{14\,dx}{27}}-{\frac{14\,c}{27}} \right ) +3\,{e}^{2}{a}^{2}b \left ( 1/3\, \left ( dx+c \right ) ^{3}{\it Arcsinh} \left ( dx+c \right ) -1/9\, \left ( dx+c \right ) ^{2}\sqrt{1+ \left ( dx+c \right ) ^{2}}+2/9\,\sqrt{1+ \left ( dx+c \right ) ^{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2*(a+b*arcsinh(d*x+c))^3,x)

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a^3+e^2*b^3*(1/3*arcsinh(d*x+c)^3*(d*x+c)*(1+(d*x+c)^2)-1/3*arcsinh(d*x+c)^3*(d*x+c)-1/
3*arcsinh(d*x+c)^2*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+2/3*arcsinh(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+2/9*(d*x+c)*(1+(d*x+
c)^2)*arcsinh(d*x+c)-14/9*(d*x+c)*arcsinh(d*x+c)-2/27*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+40/27*(1+(d*x+c)^2)^(1/2))
+3*e^2*a*b^2*(1/3*arcsinh(d*x+c)^2*(d*x+c)*(1+(d*x+c)^2)-1/3*arcsinh(d*x+c)^2*(d*x+c)-2/9*arcsinh(d*x+c)*(d*x+
c)^2*(1+(d*x+c)^2)^(1/2)+4/9*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)+2/27*(1+(d*x+c)^2)*(d*x+c)-14/27*d*x-14/27*c)+
3*e^2*a^2*b*(1/3*(d*x+c)^3*arcsinh(d*x+c)-1/9*(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+2/9*(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)^2*(a+b*arcsinh(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.87802, size = 1285, normalized size = 5.66 \begin{align*} \frac{3 \,{\left (3 \, a^{3} + 2 \, a b^{2}\right )} d^{3} e^{2} x^{3} + 9 \,{\left (3 \, a^{3} + 2 \, a b^{2}\right )} c d^{2} e^{2} x^{2} - 9 \,{\left (4 \, a b^{2} -{\left (3 \, a^{3} + 2 \, a b^{2}\right )} c^{2}\right )} d e^{2} x + 9 \,{\left (b^{3} d^{3} e^{2} x^{3} + 3 \, b^{3} c d^{2} e^{2} x^{2} + 3 \, b^{3} c^{2} d e^{2} x + b^{3} c^{3} e^{2}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + 9 \,{\left (3 \, a b^{2} d^{3} e^{2} x^{3} + 9 \, a b^{2} c d^{2} e^{2} x^{2} + 9 \, a b^{2} c^{2} d e^{2} x + 3 \, a b^{2} c^{3} e^{2} -{\left (b^{3} d^{2} e^{2} x^{2} + 2 \, b^{3} c d e^{2} x +{\left (b^{3} c^{2} - 2 \, b^{3}\right )} e^{2}\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} + 3 \,{\left ({\left (9 \, a^{2} b + 2 \, b^{3}\right )} d^{3} e^{2} x^{3} + 3 \,{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c d^{2} e^{2} x^{2} - 3 \,{\left (4 \, b^{3} -{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c^{2}\right )} d e^{2} x -{\left (12 \, b^{3} c -{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c^{3}\right )} e^{2} - 6 \,{\left (a b^{2} d^{2} e^{2} x^{2} + 2 \, a b^{2} c d e^{2} x +{\left (a b^{2} c^{2} - 2 \, a b^{2}\right )} e^{2}\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 ) -{\left ({\left (9 \, a^{2} b + 2 \, b^{3}\right )} d^{2} e^{2} x^{2} + 2 \,{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c d e^{2} x -{\left (18 \, a^{2} b + 40 \, b^{3} -{\left (9 \, a^{2} b + 2 \, b^{3}\right )} c^{2}\right )} e^{2}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{27 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2*(a+b*arcsinh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/27*(3*(3*a^3 + 2*a*b^2)*d^3*e^2*x^3 + 9*(3*a^3 + 2*a*b^2)*c*d^2*e^2*x^2 - 9*(4*a*b^2 - (3*a^3 + 2*a*b^2)*c^2
)*d*e^2*x + 9*(b^3*d^3*e^2*x^3 + 3*b^3*c*d^2*e^2*x^2 + 3*b^3*c^2*d*e^2*x + b^3*c^3*e^2)*log(d*x + c + sqrt(d^2
*x^2 + 2*c*d*x + c^2 + 1))^3 + 9*(3*a*b^2*d^3*e^2*x^3 + 9*a*b^2*c*d^2*e^2*x^2 + 9*a*b^2*c^2*d*e^2*x + 3*a*b^2*
c^3*e^2 - (b^3*d^2*e^2*x^2 + 2*b^3*c*d*e^2*x + (b^3*c^2 - 2*b^3)*e^2)*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 + 3*((9*a^2*b + 2*b^3)*d^3*e^2*x^3 + 3*(9*a^2*b + 2*b^3)*c*d^2*e
^2*x^2 - 3*(4*b^3 - (9*a^2*b + 2*b^3)*c^2)*d*e^2*x - (12*b^3*c - (9*a^2*b + 2*b^3)*c^3)*e^2 - 6*(a*b^2*d^2*e^2
*x^2 + 2*a*b^2*c*d*e^2*x + (a*b^2*c^2 - 2*a*b^2)*e^2)*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)) - ((9*a^2*b + 2*b^3)*d^2*e^2*x^2 + 2*(9*a^2*b + 2*b^3)*c*d*e^2*x - (18*a^2*b + 40*
b^3 - (9*a^2*b + 2*b^3)*c^2)*e^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d

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Sympy [A]  time = 7.19988, size = 1173, normalized size = 5.17 \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)**2*(a+b*asinh(d*x+c))**3,x)

[Out]

Piecewise((a**3*c**2*e**2*x + a**3*c*d*e**2*x**2 + a**3*d**2*e**2*x**3/3 + a**2*b*c**3*e**2*asinh(c + d*x)/d +
3*a**2*b*c**2*e**2*x*asinh(c + d*x) - a**2*b*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(3*d) + 3*a**2*b*
c*d*e**2*x**2*asinh(c + d*x) - 2*a**2*b*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/3 + a**2*b*d**2*e**2*x**
3*asinh(c + d*x) - a**2*b*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/3 + 2*a**2*b*e**2*sqrt(c**2 + 2*c*d
*x + d**2*x**2 + 1)/(3*d) + a*b**2*c**3*e**2*asinh(c + d*x)**2/d + 3*a*b**2*c**2*e**2*x*asinh(c + d*x)**2 + 2*
a*b**2*c**2*e**2*x/3 - 2*a*b**2*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(3*d) + 3*a*b**2
*c*d*e**2*x**2*asinh(c + d*x)**2 + 2*a*b**2*c*d*e**2*x**2/3 - 4*a*b**2*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**
2 + 1)*asinh(c + d*x)/3 + a*b**2*d**2*e**2*x**3*asinh(c + d*x)**2 + 2*a*b**2*d**2*e**2*x**3/9 - 2*a*b**2*d*e**
2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/3 - 4*a*b**2*e**2*x/3 + 4*a*b**2*e**2*sqrt(c**2 + 2
*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(3*d) + b**3*c**3*e**2*asinh(c + d*x)**3/(3*d) + 2*b**3*c**3*e**2*asinh
(c + d*x)/(9*d) + b**3*c**2*e**2*x*asinh(c + d*x)**3 + 2*b**3*c**2*e**2*x*asinh(c + d*x)/3 - b**3*c**2*e**2*sq
rt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/(3*d) - 2*b**3*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2
+ 1)/(27*d) + b**3*c*d*e**2*x**2*asinh(c + d*x)**3 + 2*b**3*c*d*e**2*x**2*asinh(c + d*x)/3 - 2*b**3*c*e**2*x*s
qrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/3 - 4*b**3*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)
/27 - 4*b**3*c*e**2*asinh(c + d*x)/(3*d) + b**3*d**2*e**2*x**3*asinh(c + d*x)**3/3 + 2*b**3*d**2*e**2*x**3*asi
nh(c + d*x)/9 - b**3*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)**2/3 - 2*b**3*d*e**2*x**2
*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/27 - 4*b**3*e**2*x*asinh(c + d*x)/3 + 2*b**3*e**2*sqrt(c**2 + 2*c*d*x +
d**2*x**2 + 1)*asinh(c + d*x)**2/(3*d) + 40*b**3*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(27*d), Ne(d, 0)),
(c**2*e**2*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 )}^{2}{\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)^2*(a+b*arcsinh(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^2*(b*arcsinh(d*x + c) + a)^3, x)