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

Optimal. Leaf size=76 \[ \frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac{b e^2 \left ((c+d x)^2+1\right )^{3/2}}{9 d}+\frac{b e^2 \sqrt{(c+d x)^2+1}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0656067, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {5865, 12, 5661, 266, 43} \[ \frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac{b e^2 \left ((c+d x)^2+1\right )^{3/2}}{9 d}+\frac{b e^2 \sqrt{(c+d x)^2+1}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^2*Sqrt[1 + (c + d*x)^2])/(3*d) - (b*e^2*(1 + (c + d*x)^2)^(3/2))/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[
c + d*x]))/(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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (c e+d e x)^2 \left (a+b \sinh ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x}} \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{\sqrt{1+x}}+\sqrt{1+x}\right ) \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2}}{3 d}-\frac{b e^2 \left (1+(c+d x)^2\right )^{3/2}}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0440766, size = 64, normalized size = 0.84 \[ \frac{e^2 \left (\frac{1}{3} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )-\frac{1}{9} b \left (c^2+2 c d x+d^2 x^2-2\right ) \sqrt{(c+d x)^2+1}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 73, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{3}{e}^{2}a}{3}}+{e}^{2}b \left ({\frac{ \left ( dx+c \right ) ^{3}{\it Arcsinh} \left ( dx+c \right ) }{3}}-{\frac{ \left ( dx+c \right ) ^{2}}{9}\sqrt{1+ \left ( dx+c \right ) ^{2}}}+{\frac{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)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39488, size = 365, normalized size = 4.8 \begin{align*} \frac{3 \, a d^{3} e^{2} x^{3} + 9 \, a c d^{2} e^{2} x^{2} + 9 \, a c^{2} d e^{2} x + 3 \,{\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x + b c^{3} e^{2}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) -{\left (b d^{2} e^{2} x^{2} + 2 \, b c d e^{2} x +{\left (b c^{2} - 2 \, b\right )} e^{2}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{9 \, 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)),x, algorithm="fricas")

[Out]

1/9*(3*a*d^3*e^2*x^3 + 9*a*c*d^2*e^2*x^2 + 9*a*c^2*d*e^2*x + 3*(b*d^3*e^2*x^3 + 3*b*c*d^2*e^2*x^2 + 3*b*c^2*d*
e^2*x + b*c^3*e^2)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (b*d^2*e^2*x^2 + 2*b*c*d*e^2*x + (b*c^2
- 2*b)*e^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/d

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Sympy [A]  time = 1.21131, size = 258, normalized size = 3.39 \begin{align*} \begin{cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac{a d^{2} e^{2} x^{3}}{3} + \frac{b c^{3} e^{2} \operatorname{asinh}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname{asinh}{\left (c + d x \right )} - \frac{b c^{2} e^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9 d} + b c d e^{2} x^{2} \operatorname{asinh}{\left (c + d x \right )} - \frac{2 b c e^{2} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9} + \frac{b d^{2} e^{2} x^{3} \operatorname{asinh}{\left (c + d x \right )}}{3} - \frac{b d e^{2} x^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9} + \frac{2 b e^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} + 1}}{9 d} & \text{for}\: d \neq 0 \\c^{2} e^{2} x \left (a + b \operatorname{asinh}{\left (c \right )}\right ) & \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)),x)

[Out]

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

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Giac [B]  time = 2.2866, size = 548, normalized size = 7.21 \begin{align*} \frac{1}{18} \,{\left (6 \, a d^{2} x^{3} + 18 \, a c d x^{2} - 18 \,{\left (d{\left (\frac{c \log \left ({\left | -c d -{\left (x{\left | d \right |} - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\left | d \right |} \right |}\right )}{d{\left | d \right |}} + \frac{\sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{d^{2}}\right )} - x \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )\right )} b c^{2} + 9 \,{\left (2 \, x^{2} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) -{\left (\sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}{\left (\frac{x}{d^{2}} - \frac{3 \, c}{d^{3}}\right )} - \frac{{\left (2 \, c^{2} - 1\right )} \log \left ({\left | -c d -{\left (x{\left | d \right |} - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\left | d \right |} \right |}\right )}{d^{2}{\left | d \right |}}\right )} d\right )} b c d +{\left (6 \, x^{3} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) -{\left (\sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}{\left (x{\left (\frac{2 \, x}{d^{2}} - \frac{5 \, c}{d^{3}}\right )} + \frac{11 \, c^{2} d - 4 \, d}{d^{5}}\right )} + \frac{3 \,{\left (2 \, c^{3} - 3 \, c\right )} \log \left ({\left | -c d -{\left (x{\left | d \right |} - \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{\left | d \right |} \right |}\right )}{d^{3}{\left | d \right |}}\right )} d\right )} b d^{2} + 18 \, a c^{2} x\right )} e^{2} \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)),x, algorithm="giac")

[Out]

1/18*(6*a*d^2*x^3 + 18*a*c*d*x^2 - 18*(d*(c*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(
d)))/(d*abs(d)) + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)/d^2) - x*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)))
*b*c^2 + 9*(2*x^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x/d^2
- 3*c/d^3) - (2*c^2 - 1)*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*abs(d)))/(d^2*abs(d)))
*d)*b*c*d + (6*x^3*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*(x*(2
*x/d^2 - 5*c/d^3) + (11*c^2*d - 4*d)/d^5) + 3*(2*c^3 - 3*c)*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x
+ c^2 + 1))*abs(d)))/(d^3*abs(d)))*d)*b*d^2 + 18*a*c^2*x)*e^2