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

Optimal. Leaf size=97 \[ \frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d}-\frac{b e^2 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2}{9 d}-\frac{2 b e^2 \sqrt{c+d x-1} \sqrt{c+d x+1}}{9 d} \]

[Out]

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

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Rubi [A]  time = 0.0630765, antiderivative size = 97, 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 = {5866, 12, 5662, 100, 74} \[ \frac{e^2 (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )}{3 d}-\frac{b e^2 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2}{9 d}-\frac{2 b e^2 \sqrt{c+d x-1} \sqrt{c+d x+1}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5866

Int[((a_.) + ArcCosh[(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*ArcCosh[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 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rubi steps

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

Mathematica [A]  time = 0.0715758, size = 71, normalized size = 0.73 \[ \frac{e^2 \left (\frac{1}{3} (c+d x)^3 \left (a+b \cosh ^{-1}(c+d x)\right )-\frac{1}{9} b \sqrt{c+d x-1} \sqrt{c+d x+1} \left (c^2+2 c d x+d^2 x^2+2\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 67, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{3}{e}^{2}a}{3}}+{e}^{2}b \left ({\frac{{\rm arccosh} \left (dx+c\right ) \left ( dx+c \right ) ^{3}}{3}}-{\frac{ \left ( dx+c \right ) ^{2}+2}{9}\sqrt{dx+c-1}\sqrt{dx+c+1}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*(d*x+c)^3*e^2*a+e^2*b*(1/3*arccosh(d*x+c)*(d*x+c)^3-1/9*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*((d*x+c)^2+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*arccosh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42013, size = 365, normalized size = 3.76 \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*arccosh(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 = 1.10139, size = 258, normalized size = 2.66 \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{acosh}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname{acosh}{\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{acosh}{\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{acosh}{\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{acosh}{\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*acosh(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*acosh(c + d*x)/(3*d) + b*c**2*e*
*2*x*acosh(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*acosh(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*acosh(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*acosh(c)), True))

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Giac [B]  time = 2.13946, size = 548, normalized size = 5.65 \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*arccosh(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