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

Optimal. Leaf size=172 \[ \frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}-\frac {b e^3 \sqrt {(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{8 d}+\frac {3 b e^3 \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )}{16 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}+\frac {b^2 e^3 (c+d x)^4}{32 d}-\frac {3 b^2 e^3 (c+d x)^2}{32 d} \]

[Out]

-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*ar
csinh(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*arcsi
nh(d*x+c))*(1+(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.26, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5865, 12, 5661, 5758, 5675, 30} \[ \frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}-\frac {b e^3 \sqrt {(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{8 d}+\frac {3 b e^3 \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )}{16 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}+\frac {b^2 e^3 (c+d x)^4}{32 d}-\frac {3 b^2 e^3 (c+d x)^2}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*e^3*(c + d*x)^2)/(32*d) + (b^2*e^3*(c + d*x)^4)/(32*d) + (3*b*e^3*(c + d*x)*Sqrt[1 + (c + d*x)^2]*(a +
b*ArcSinh[c + d*x]))/(16*d) - (b*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(8*d) - (3*e
^3*(a + b*ArcSinh[c + d*x])^2)/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSinh[c + d*x])^2)/(4*d)

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 5675

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

Rubi steps

\begin {align*} \int (c e+d e x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^3 x^3 \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int x^3 \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (b e^3\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=-\frac {b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{8 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac {\left (3 b e^3\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{8 d}+\frac {\left (b^2 e^3\right ) \operatorname {Subst}\left (\int x^3 \, dx,x,c+d x\right )}{8 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} \left (a+b \sinh ^{-1}(c+d x)\right )}{16 d}-\frac {b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{8 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}-\frac {\left (3 b e^3\right ) \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{16 d}-\frac {\left (3 b^2 e^3\right ) \operatorname {Subst}(\int x \, dx,x,c+d x)}{16 d}\\ &=-\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} \left (a+b \sinh ^{-1}(c+d x)\right )}{16 d}-\frac {b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{8 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 170, normalized size = 0.99 \[ \frac {e^3 \left (\left (8 a^2+b^2\right ) (c+d x)^4+2 a b \left (3-2 (c+d x)^2\right ) \sqrt {(c+d x)^2+1} (c+d x)+2 b (c+d x) \sinh ^{-1}(c+d x) \left (8 a (c+d x)^3-2 b \sqrt {(c+d x)^2+1} (c+d x)^2+3 b \sqrt {(c+d x)^2+1}\right )-6 a b \sinh ^{-1}(c+d x)-3 b^2 (c+d x)^2+b^2 \left (8 (c+d x)^4-3\right ) \sinh ^{-1}(c+d x)^2\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 486, normalized size = 2.83 \[ \frac {{\left (8 \, a^{2} + b^{2}\right )} d^{4} e^{3} x^{4} + 4 \, {\left (8 \, a^{2} + b^{2}\right )} c d^{3} e^{3} x^{3} + 3 \, {\left (2 \, {\left (8 \, a^{2} + b^{2}\right )} c^{2} - b^{2}\right )} d^{2} e^{3} x^{2} + 2 \, {\left (2 \, {\left (8 \, a^{2} + b^{2}\right )} c^{3} - 3 \, b^{2} c\right )} d e^{3} x + {\left (8 \, b^{2} d^{4} e^{3} x^{4} + 32 \, b^{2} c d^{3} e^{3} x^{3} + 48 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 32 \, b^{2} c^{3} d e^{3} x + {\left (8 \, b^{2} c^{4} - 3 \, b^{2}\right )} e^{3}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 2 \, {\left (8 \, a b d^{4} e^{3} x^{4} + 32 \, a b c d^{3} e^{3} x^{3} + 48 \, a b c^{2} d^{2} e^{3} x^{2} + 32 \, a b c^{3} d e^{3} x + {\left (8 \, a b c^{4} - 3 \, a b\right )} e^{3} - {\left (2 \, b^{2} d^{3} e^{3} x^{3} + 6 \, b^{2} c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, b^{2} c^{2} - b^{2}\right )} d e^{3} x + {\left (2 \, b^{2} c^{3} - 3 \, b^{2} c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 2 \, {\left (2 \, a b d^{3} e^{3} x^{3} + 6 \, a b c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, a b c^{2} - a b\right )} d e^{3} x + {\left (2 \, a b c^{3} - 3 \, a b c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 + sqrt(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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d e x + c e\right )}^{3} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 194, normalized size = 1.13 \[ \frac {\frac {\left (d x +c \right )^{4} e^{3} a^{2}}{4}+e^{3} b^{2} \left (\frac {\left (d x +c \right )^{4} \arcsinh \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{8}+\frac {3 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right )}{16}-\frac {3 \arcsinh \left (d x +c \right )^{2}}{32}+\frac {\left (d x +c \right )^{4}}{32}-\frac {3 \left (d x +c \right )^{2}}{32}-\frac {3}{32}\right )+2 e^{3} a b \left (\frac {\left (d x +c \right )^{4} \arcsinh \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \arcsinh \left (d x +c \right )}{32}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*arcsi
nh(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*arcsi
nh(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*ar
csinh(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*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*x^2 + 4*b^2*c^3*e^3*x)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c
^2 + 1))^2 - integrate(1/2*(b^2*d^6*e^3*x^6 + 6*b^2*c*d^5*e^3*x^5 + (15*c^2*d^4*e^3 + d^4*e^3)*b^2*x^4 + 4*(5*
c^3*d^3*e^3 + c*d^3*e^3)*b^2*x^3 + 2*(7*c^4*d^2*e^3 + 3*c^2*d^2*e^3)*b^2*x^2 + 4*(c^5*d*e^3 + c^3*d*e^3)*b^2*x
+ (b^2*d^5*e^3*x^5 + 5*b^2*c*d^4*e^3*x^4 + 10*b^2*c^2*d^3*e^3*x^3 + 10*b^2*c^3*d^2*e^3*x^2 + 4*b^2*c^4*d*e^3*
x)*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))/(d^3*x^3 + 3*c*d^2*x^2
+ c^3 + (3*c^2*d + d)*x + (d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,e+d\,e\,x\right )}^3\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.59, size = 916, normalized size = 5.33 \[ \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}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*sqrt(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*asinh(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)*a
sinh(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*sqr
t(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 + d*x)/16 - 3*b**2*e**3*asinh(c + d*x)**2/(32*d), Ne(d, 0)), (c**
3*e**3*x*(a + b*asinh(c))**2, True))

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