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

Optimal. Leaf size=103 \[ \frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d}-\frac{b e \sqrt{(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{b^2 e (c+d x)^2}{4 d} \]

[Out]

(b^2*e*(c + d*x)^2)/(4*d) - (b*e*(c + d*x)*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x]))/(2*d) + (e*(a + b*A
rcSinh[c + d*x])^2)/(4*d) + (e*(c + d*x)^2*(a + b*ArcSinh[c + d*x])^2)/(2*d)

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Rubi [A]  time = 0.144862, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5865, 12, 5661, 5758, 5675, 30} \[ \frac{e (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{2 d}-\frac{b e \sqrt{(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )}{2 d}+\frac{e \left (a+b \sinh ^{-1}(c+d x)\right )^2}{4 d}+\frac{b^2 e (c+d x)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.177152, size = 120, normalized size = 1.17 \[ \frac{e \left (\left (2 a^2+b^2\right ) (c+d x)^2-2 a b \sqrt{(c+d x)^2+1} (c+d x)+2 b (c+d x) \sinh ^{-1}(c+d x) \left (2 a (c+d x)-b \sqrt{(c+d x)^2+1}\right )+2 a b \sinh ^{-1}(c+d x)+b^2 \left (2 (c+d x)^2+1\right ) \sinh ^{-1}(c+d x)^2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.027, size = 135, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{2}e{a}^{2}}{2}}+e{b}^{2} \left ({\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( 1+ \left ( dx+c \right ) ^{2} \right ) }{2}}-{\frac{{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) }{2}\sqrt{1+ \left ( dx+c \right ) ^{2}}}-{\frac{ \left ({\it Arcsinh} \left ( dx+c \right ) \right ) ^{2}}{4}}+{\frac{ \left ( dx+c \right ) ^{2}}{4}}+{\frac{1}{4}} \right ) +2\,eab \left ( 1/2\,{\it Arcsinh} \left ( dx+c \right ) \left ( dx+c \right ) ^{2}-1/4\, \left ( dx+c \right ) \sqrt{1+ \left ( dx+c \right ) ^{2}}+1/4\,{\it Arcsinh} \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.96909, size = 533, normalized size = 5.17 \begin{align*} \frac{{\left (2 \, a^{2} + b^{2}\right )} d^{2} e x^{2} + 2 \,{\left (2 \, a^{2} + b^{2}\right )} c d e x +{\left (2 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} c d e x +{\left (2 \, b^{2} c^{2} + b^{2}\right )} e\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 2 \,{\left (2 \, a b d^{2} e x^{2} + 4 \, a b c d e x +{\left (2 \, a b c^{2} + a b\right )} e -{\left (b^{2} d e x + b^{2} c e\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 (a b d e x + a b c e\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*a^2 + b^2)*d^2*e*x^2 + 2*(2*a^2 + b^2)*c*d*e*x + (2*b^2*d^2*e*x^2 + 4*b^2*c*d*e*x + (2*b^2*c^2 + b^2)*
e)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2 + 2*(2*a*b*d^2*e*x^2 + 4*a*b*c*d*e*x + (2*a*b*c^2 + a*b)
*e - (b^2*d*e*x + b^2*c*e)*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*(a*b*d*e*x + a*b*c*e)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d

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Sympy [A]  time = 1.09358, size = 335, normalized size = 3.25 \begin{align*} \begin{cases} a^{2} c e x + \frac{a^{2} d e x^{2}}{2} + \frac{a b c^{2} e \operatorname{asinh}{\left (c + d x \right )}}{d} + 2 a b c e x \operatorname{asinh}{\left (c + d x \right )} - \frac{a b c e \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{2 d} + a b d e x^{2} \operatorname{asinh}{\left (c + d x \right )} - \frac{a b e x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{2} + \frac{a b e \operatorname{asinh}{\left (c + d x \right )}}{2 d} + \frac{b^{2} c^{2} e \operatorname{asinh}^{2}{\left (c + d x \right )}}{2 d} + b^{2} c e x \operatorname{asinh}^{2}{\left (c + d x \right )} + \frac{b^{2} c e x}{2} - \frac{b^{2} c e \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}{\left (c + d x \right )}}{2 d} + \frac{b^{2} d e x^{2} \operatorname{asinh}^{2}{\left (c + d x \right )}}{2} + \frac{b^{2} d e x^{2}}{4} - \frac{b^{2} e x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1} \operatorname{asinh}{\left (c + d x \right )}}{2} + \frac{b^{2} e \operatorname{asinh}^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\c e 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)*(a+b*asinh(d*x+c))**2,x)

[Out]

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

[Out]

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