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*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)

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Rubi [A]  time = 0.204961, antiderivative size = 136, normalized size of antiderivative = 1., 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

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*}

Mathematica [A]  time = 0.166324, 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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Maple [A]  time = 0.033, size = 192, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{3}{e}^{2}{a}^{2}}{3}}+{e}^{2}{b}^{2} \left ({\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{3}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) }{3}}-{\frac{2\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}}{9}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{4\,{\it Arcsinh} \left ( dx+c \right ) }{9}\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 ) +2\,{e}^{2}ab \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))^2,x)

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a^2+e^2*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)+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(-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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.89639, size = 767, normalized size = 5.64 \begin{align*} \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} \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))^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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Sympy [A]  time = 3.18346, size = 610, normalized size = 4.49 \begin{align*} \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}{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \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))**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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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 )}^{2}\,{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))^2,x, algorithm="giac")

[Out]

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