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

Optimal. Leaf size=136 \[ \frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac {2 b e^2 \sqrt {(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {4 b e^2 \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {2 b^2 e^2 (c+d x)^3}{27 d}-\frac {4}{9} b^2 e^2 x \]

[Out]

-4/9*b^2*e^2*x+2/27*b^2*e^2*(d*x+c)^3/d+1/3*e^2*(d*x+c)^3*(a+b*arcsinh(d*x+c))^2/d+4/9*b*e^2*(a+b*arcsinh(d*x+
c))*(1+(d*x+c)^2)^(1/2)/d-2/9*b*e^2*(d*x+c)^2*(a+b*arcsinh(d*x+c))*(1+(d*x+c)^2)^(1/2)/d

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Rubi [A]  time = 0.20, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5865, 12, 5661, 5758, 5717, 8, 30} \[ \frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac {2 b e^2 \sqrt {(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {4 b e^2 \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {2 b^2 e^2 (c+d x)^3}{27 d}-\frac {4}{9} b^2 e^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 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)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 \operatorname {Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac {\left (2 b e^2\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=-\frac {2 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}+\frac {\left (4 b e^2\right ) \operatorname {Subst}\left (\int \frac {x \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{9 d}+\frac {\left (2 b^2 e^2\right ) \operatorname {Subst}\left (\int x^2 \, dx,x,c+d x\right )}{9 d}\\ &=\frac {2 b^2 e^2 (c+d x)^3}{27 d}+\frac {4 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}-\frac {2 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}-\frac {\left (4 b^2 e^2\right ) \operatorname {Subst}(\int 1 \, dx,x,c+d x)}{9 d}\\ &=-\frac {4}{9} b^2 e^2 x+\frac {2 b^2 e^2 (c+d x)^3}{27 d}+\frac {4 b e^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}-\frac {2 b e^2 (c+d x)^2 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{9 d}+\frac {e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 147, normalized size = 1.08 \[ \frac {e^2 \left (\left (9 a^2+2 b^2\right ) (c+d x)^3+6 a b \left (2-(c+d x)^2\right ) \sqrt {(c+d x)^2+1}+6 b \sinh ^{-1}(c+d x) \left (3 a (c+d x)^3-b \sqrt {(c+d x)^2+1} (c+d x)^2+2 b \sqrt {(c+d x)^2+1}\right )-12 b^2 (c+d x)+9 b^2 (c+d x)^3 \sinh ^{-1}(c+d x)^2\right )}{27 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(-12*b^2*(c + d*x) + (9*a^2 + 2*b^2)*(c + d*x)^3 + 6*a*b*(2 - (c + d*x)^2)*Sqrt[1 + (c + d*x)^2] + 6*b*(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] + 9*b^2*(c
+ d*x)^3*ArcSinh[c + d*x]^2))/(27*d)

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fricas [B]  time = 0.57, size = 358, normalized size = 2.63 \[ \frac {{\left (9 \, a^{2} + 2 \, b^{2}\right )} d^{3} e^{2} x^{3} + 3 \, {\left (9 \, a^{2} + 2 \, b^{2}\right )} c d^{2} e^{2} x^{2} + 3 \, {\left ({\left (9 \, a^{2} + 2 \, b^{2}\right )} c^{2} - 4 \, b^{2}\right )} d e^{2} x + 9 \, {\left (b^{2} d^{3} e^{2} x^{3} + 3 \, b^{2} c d^{2} e^{2} x^{2} + 3 \, b^{2} c^{2} d e^{2} x + b^{2} c^{3} e^{2}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 6 \, {\left (3 \, a b d^{3} e^{2} x^{3} + 9 \, a b c d^{2} e^{2} x^{2} + 9 \, a b c^{2} d e^{2} x + 3 \, a b c^{3} e^{2} - {\left (b^{2} d^{2} e^{2} x^{2} + 2 \, b^{2} c d e^{2} x + {\left (b^{2} c^{2} - 2 \, b^{2}\right )} e^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 6 \, {\left (a b d^{2} e^{2} x^{2} + 2 \, a b c d e^{2} x + {\left (a b c^{2} - 2 \, a b\right )} e^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{27 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*((9*a^2 + 2*b^2)*d^3*e^2*x^3 + 3*(9*a^2 + 2*b^2)*c*d^2*e^2*x^2 + 3*((9*a^2 + 2*b^2)*c^2 - 4*b^2)*d*e^2*x
+ 9*(b^2*d^3*e^2*x^3 + 3*b^2*c*d^2*e^2*x^2 + 3*b^2*c^2*d*e^2*x + b^2*c^3*e^2)*log(d*x + c + sqrt(d^2*x^2 + 2*c
*d*x + c^2 + 1))^2 + 6*(3*a*b*d^3*e^2*x^3 + 9*a*b*c*d^2*e^2*x^2 + 9*a*b*c^2*d*e^2*x + 3*a*b*c^3*e^2 - (b^2*d^2
*e^2*x^2 + 2*b^2*c*d*e^2*x + (b^2*c^2 - 2*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)) - 6*(a*b*d^2*e^2*x^2 + 2*a*b*c*d*e^2*x + (a*b*c^2 - 2*a*b)*e^2)*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 \[ \int {\left (d e x + c e\right )}^{2} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 163, normalized size = 1.20 \[ \frac {\frac {\left (d x +c \right )^{3} e^{2} a^{2}}{3}+e^{2} b^{2} \left (\frac {\left (d x +c \right )^{3} \arcsinh \left (d x +c \right )^{2}}{3}+\frac {4 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{9}-\frac {2 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right )^{2}}{9}-\frac {4 d x}{9}-\frac {4 c}{9}+\frac {2 \left (d x +c \right )^{3}}{27}\right )+2 e^{2} a b \left (\frac {\left (d x +c \right )^{3} \arcsinh \left (d x +c \right )}{3}-\frac {\left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{9}+\frac {2 \sqrt {1+\left (d x +c \right )^{2}}}{9}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a^2+e^2*b^2*(1/3*(d*x+c)^3*arcsinh(d*x+c)^2+4/9*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)-2/9*
arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)*(d*x+c)^2-4/9*d*x-4/9*c+2/27*(d*x+c)^3)+2*e^2*a*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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a^{2} d^{2} e^{2} x^{3} + a^{2} c d e^{2} x^{2} + {\left (2 \, x^{2} \operatorname {arsinh}\left (d x + c\right ) - d {\left (\frac {3 \, c^{2} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{3}} + \frac {\sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} x}{d^{2}} - \frac {{\left (c^{2} + 1\right )} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{3}} - \frac {3 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c}{d^{3}}\right )}\right )} a b c d e^{2} + \frac {1}{9} \, {\left (6 \, x^{3} \operatorname {arsinh}\left (d x + c\right ) - d {\left (\frac {2 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} x^{2}}{d^{2}} - \frac {15 \, c^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{4}} - \frac {5 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c x}{d^{3}} + \frac {9 \, {\left (c^{2} + 1\right )} c \operatorname {arsinh}\left (\frac {2 \, {\left (d^{2} x + c d\right )}}{\sqrt {-4 \, c^{2} d^{2} + 4 \, {\left (c^{2} + 1\right )} d^{2}}}\right )}{d^{4}} + \frac {15 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} c^{2}}{d^{4}} - \frac {4 \, \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} {\left (c^{2} + 1\right )}}{d^{4}}\right )}\right )} a b d^{2} e^{2} + a^{2} c^{2} e^{2} x + \frac {2 \, {\left ({\left (d x + c\right )} \operatorname {arsinh}\left (d x + c\right ) - \sqrt {{\left (d x + c\right )}^{2} + 1}\right )} a b c^{2} e^{2}}{d} + \frac {1}{3} \, {\left (b^{2} d^{2} e^{2} x^{3} + 3 \, b^{2} c d e^{2} x^{2} + 3 \, b^{2} c^{2} e^{2} x\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} - \int \frac {2 \, {\left (b^{2} d^{5} e^{2} x^{5} + 5 \, b^{2} c d^{4} e^{2} x^{4} + {\left (10 \, c^{2} d^{3} e^{2} + d^{3} e^{2}\right )} b^{2} x^{3} + 3 \, {\left (3 \, c^{3} d^{2} e^{2} + c d^{2} e^{2}\right )} b^{2} x^{2} + 3 \, {\left (c^{4} d e^{2} + c^{2} d e^{2}\right )} b^{2} x + {\left (b^{2} d^{4} e^{2} x^{4} + 4 \, b^{2} c d^{3} e^{2} x^{3} + 6 \, b^{2} c^{2} d^{2} e^{2} x^{2} + 3 \, b^{2} c^{3} d e^{2} x\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 )}{3 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + c^{3} + {\left (3 \, c^{2} d + d\right )} x + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}^{\frac {3}{2}} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*d^2*e^2*x^3 + a^2*c*d*e^2*x^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)/sqr
t(-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*d*e^2 + 1/9*(6*x^3*ar
csinh(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*d^2*e^2 + a^2*c^2*e^2*x + 2*((d*x + c)*arcsinh(d*x + c) - sqrt((d*x + c)
^2 + 1))*a*b*c^2*e^2/d + 1/3*(b^2*d^2*e^2*x^3 + 3*b^2*c*d*e^2*x^2 + 3*b^2*c^2*e^2*x)*log(d*x + c + sqrt(d^2*x^
2 + 2*c*d*x + c^2 + 1))^2 - integrate(2/3*(b^2*d^5*e^2*x^5 + 5*b^2*c*d^4*e^2*x^4 + (10*c^2*d^3*e^2 + d^3*e^2)*
b^2*x^3 + 3*(3*c^3*d^2*e^2 + c*d^2*e^2)*b^2*x^2 + 3*(c^4*d*e^2 + c^2*d*e^2)*b^2*x + (b^2*d^4*e^2*x^4 + 4*b^2*c
*d^3*e^2*x^3 + 6*b^2*c^2*d^2*e^2*x^2 + 3*b^2*c^3*d*e^2*x)*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))/(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 \[ \int {\left (c\,e+d\,e\,x\right )}^2\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.86, size = 610, normalized size = 4.49 \[ \begin {cases} a^{2} c^{2} e^{2} x + a^{2} c d e^{2} x^{2} + \frac {a^{2} d^{2} e^{2} x^{3}}{3} + \frac {2 a b c^{3} e^{2} \operatorname {asinh}{\left (c + d x \right )}}{3 d} + 2 a b c^{2} e^{2} x \operatorname {asinh}{\left (c + d x \right )} - \frac {2 a b c^{2} e^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9 d} + 2 a b c d e^{2} x^{2} \operatorname {asinh}{\left (c + d x \right )} - \frac {4 a b c e^{2} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9} + \frac {2 a b d^{2} e^{2} x^{3} \operatorname {asinh}{\left (c + d x \right )}}{3} - \frac {2 a b d e^{2} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9} + \frac {4 a b e^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9 d} + \frac {b^{2} c^{3} e^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{3 d} + b^{2} c^{2} e^{2} x \operatorname {asinh}^{2}{\left (c + d x \right )} + \frac {2 b^{2} c^{2} e^{2} x}{9} - \frac {2 b^{2} c^{2} e^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{9 d} + b^{2} c d e^{2} x^{2} \operatorname {asinh}^{2}{\left (c + d x \right )} + \frac {2 b^{2} c d e^{2} x^{2}}{9} - \frac {4 b^{2} c e^{2} x \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{9} + \frac {b^{2} d^{2} e^{2} x^{3} \operatorname {asinh}^{2}{\left (c + d x \right )}}{3} + \frac {2 b^{2} d^{2} e^{2} x^{3}}{27} - \frac {2 b^{2} d e^{2} x^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{9} - \frac {4 b^{2} e^{2} x}{9} + \frac {4 b^{2} e^{2} \sqrt {c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname {asinh}{\left (c + d x \right )}}{9 d} & \text {for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname {asinh}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c**2*e**2*x + a**2*c*d*e**2*x**2 + a**2*d**2*e**2*x**3/3 + 2*a*b*c**3*e**2*asinh(c + d*x)/(3*d
) + 2*a*b*c**2*e**2*x*asinh(c + d*x) - 2*a*b*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/(9*d) + 2*a*b*c*d*
e**2*x**2*asinh(c + d*x) - 4*a*b*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/9 + 2*a*b*d**2*e**2*x**3*asinh(
c + d*x)/3 - 2*a*b*d*e**2*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)/9 + 4*a*b*e**2*sqrt(c**2 + 2*c*d*x + d**2*
x**2 + 1)/(9*d) + b**2*c**3*e**2*asinh(c + d*x)**2/(3*d) + b**2*c**2*e**2*x*asinh(c + d*x)**2 + 2*b**2*c**2*e*
*2*x/9 - 2*b**2*c**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/(9*d) + b**2*c*d*e**2*x**2*asinh
(c + d*x)**2 + 2*b**2*c*d*e**2*x**2/9 - 4*b**2*c*e**2*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c + d*x)/9
+ b**2*d**2*e**2*x**3*asinh(c + d*x)**2/3 + 2*b**2*d**2*e**2*x**3/27 - 2*b**2*d*e**2*x**2*sqrt(c**2 + 2*c*d*x
+ d**2*x**2 + 1)*asinh(c + d*x)/9 - 4*b**2*e**2*x/9 + 4*b**2*e**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 + 1)*asinh(c
+ d*x)/(9*d), Ne(d, 0)), (c**2*e**2*x*(a + b*asinh(c))**2, True))

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