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*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)

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Rubi [A]  time = 0.257649, antiderivative size = 172, normalized size of antiderivative = 1., 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 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 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 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)^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*}

Mathematica [A]  time = 0.189873, 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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Maple [A]  time = 0.043, size = 229, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{4}{e}^{3}{a}^{2}}{4}}+{e}^{3}{b}^{2} \left ({\frac{ \left ( dx+c \right ) ^{2} \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{4}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{4}}-{\frac{{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) }{8} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) }{16}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{5\, \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}}{32}}+{\frac{ \left ( dx+c \right ) ^{2} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{32}}-{\frac{ \left ( dx+c \right ) ^{2}}{8}}-{\frac{1}{8}} \right ) +2\,{e}^{3}ab \left ( 1/4\, \left ( dx+c \right ) ^{4}{\it Arcsinh} \left ( dx+c \right ) -1/16\, \left ( dx+c \right ) ^{3}\sqrt{1+ \left ( dx+c \right ) ^{2}}+{\frac{ \left ( 3\,dx+3\,c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}}{32}}-{\frac{3\,{\it Arcsinh} \left ( dx+c \right ) }{32}} \right ) \right ) } \end{align*}

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)^2*arcsinh(d*x+c)^2*(1+(d*x+c)^2)-1/4*arcsinh(d*x+c)^2*(1+(d*x+
c)^2)-1/8*arcsinh(d*x+c)*(d*x+c)*(1+(d*x+c)^2)^(3/2)+5/16*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)*(d*x+c)+5/32*arcs
inh(d*x+c)^2+1/32*(d*x+c)^2*(1+(d*x+c)^2)-1/8*(d*x+c)^2-1/8)+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(-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)^3*(a+b*arcsinh(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.78894, size = 1034, normalized size = 6.01 \begin{align*} \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} \end{align*}

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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Sympy [A]  time = 6.29332, size = 916, normalized size = 5.33 \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)**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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{3}{\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)^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)