3.93 \(\int (c e+d e x)^4 (a+b \cosh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=135 \[ \frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4}{25 d}-\frac{4 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2}{75 d}-\frac{8 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1}}{75 d} \]

[Out]

(-8*b*e^4*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(75*d) - (4*b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c +
d*x])/(75*d) - (b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x])/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcCo
sh[c + d*x]))/(5*d)

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Rubi [A]  time = 0.0855618, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4}{25 d}-\frac{4 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^2}{75 d}-\frac{8 b e^4 \sqrt{c+d x-1} \sqrt{c+d x+1}}{75 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcCosh[c + d*x]),x]

[Out]

(-8*b*e^4*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x])/(75*d) - (4*b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^2*Sqrt[1 + c +
d*x])/(75*d) - (b*e^4*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x])/(25*d) + (e^4*(c + d*x)^5*(a + b*ArcCo
sh[c + d*x]))/(5*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)^4 \left (a+b \cosh ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int e^4 x^4 \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int x^4 \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (b e^4\right ) \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{5 d}\\ &=-\frac{b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (b e^4\right ) \operatorname{Subst}\left (\int \frac{4 x^3}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{25 d}\\ &=-\frac{b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (4 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{25 d}\\ &=-\frac{4 b e^4 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x}}{75 d}-\frac{b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (4 b e^4\right ) \operatorname{Subst}\left (\int \frac{2 x}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{75 d}\\ &=-\frac{4 b e^4 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x}}{75 d}-\frac{b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}-\frac{\left (8 b e^4\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{75 d}\\ &=-\frac{8 b e^4 \sqrt{-1+c+d x} \sqrt{1+c+d x}}{75 d}-\frac{4 b e^4 \sqrt{-1+c+d x} (c+d x)^2 \sqrt{1+c+d x}}{75 d}-\frac{b e^4 \sqrt{-1+c+d x} (c+d x)^4 \sqrt{1+c+d x}}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.11841, size = 103, normalized size = 0.76 \[ \frac{e^4 \left ((c+d x)^5 \left (a+b \cosh ^{-1}(c+d x)\right )-\frac{4}{15} b \sqrt{c+d x-1} \sqrt{c+d x+1} \left (c^2+2 c d x+d^2 x^2+2\right )-\frac{1}{5} b \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^4\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcCosh[c + d*x]),x]

[Out]

(e^4*(-(b*Sqrt[-1 + c + d*x]*(c + d*x)^4*Sqrt[1 + c + d*x])/5 - (4*b*Sqrt[-1 + c + d*x]*Sqrt[1 + c + d*x]*(2 +
c^2 + 2*c*d*x + d^2*x^2))/15 + (c + d*x)^5*(a + b*ArcCosh[c + d*x])))/(5*d)

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Maple [A]  time = 0.013, size = 78, normalized size = 0.6 \begin{align*}{\frac{1}{d} \left ({\frac{ \left ( dx+c \right ) ^{5}{e}^{4}a}{5}}+{e}^{4}b \left ({\frac{ \left ( dx+c \right ) ^{5}{\rm arccosh} \left (dx+c\right )}{5}}-{\frac{3\, \left ( dx+c \right ) ^{4}+4\, \left ( dx+c \right ) ^{2}+8}{75}\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)^4*(a+b*arccosh(d*x+c)),x)

[Out]

1/d*(1/5*(d*x+c)^5*e^4*a+e^4*b*(1/5*(d*x+c)^5*arccosh(d*x+c)-1/75*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*(3*(d*x+c)^4
+4*(d*x+c)^2+8)))

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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)^4*(a+b*arccosh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.46632, size = 608, normalized size = 4.5 \begin{align*} \frac{15 \, a d^{5} e^{4} x^{5} + 75 \, a c d^{4} e^{4} x^{4} + 150 \, a c^{2} d^{3} e^{4} x^{3} + 150 \, a c^{3} d^{2} e^{4} x^{2} + 75 \, a c^{4} d e^{4} x + 15 \,{\left (b d^{5} e^{4} x^{5} + 5 \, b c d^{4} e^{4} x^{4} + 10 \, b c^{2} d^{3} e^{4} x^{3} + 10 \, b c^{3} d^{2} e^{4} x^{2} + 5 \, b c^{4} d e^{4} x + b c^{5} e^{4}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) -{\left (3 \, b d^{4} e^{4} x^{4} + 12 \, b c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, b c^{2} + 2 \, b\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, b c^{3} + 2 \, b c\right )} d e^{4} x +{\left (3 \, b c^{4} + 4 \, b c^{2} + 8 \, b\right )} e^{4}\right )} \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}}{75 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arccosh(d*x+c)),x, algorithm="fricas")

[Out]

1/75*(15*a*d^5*e^4*x^5 + 75*a*c*d^4*e^4*x^4 + 150*a*c^2*d^3*e^4*x^3 + 150*a*c^3*d^2*e^4*x^2 + 75*a*c^4*d*e^4*x
+ 15*(b*d^5*e^4*x^5 + 5*b*c*d^4*e^4*x^4 + 10*b*c^2*d^3*e^4*x^3 + 10*b*c^3*d^2*e^4*x^2 + 5*b*c^4*d*e^4*x + b*c
^5*e^4)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1)) - (3*b*d^4*e^4*x^4 + 12*b*c*d^3*e^4*x^3 + 2*(9*b*c^2
+ 2*b)*d^2*e^4*x^2 + 4*(3*b*c^3 + 2*b*c)*d*e^4*x + (3*b*c^4 + 4*b*c^2 + 8*b)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2
- 1))/d

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Sympy [A]  time = 4.6963, size = 527, normalized size = 3.9 \begin{align*} \begin{cases} a c^{4} e^{4} x + 2 a c^{3} d e^{4} x^{2} + 2 a c^{2} d^{2} e^{4} x^{3} + a c d^{3} e^{4} x^{4} + \frac{a d^{4} e^{4} x^{5}}{5} + \frac{b c^{5} e^{4} \operatorname{acosh}{\left (c + d x \right )}}{5 d} + b c^{4} e^{4} x \operatorname{acosh}{\left (c + d x \right )} - \frac{b c^{4} e^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{25 d} + 2 b c^{3} d e^{4} x^{2} \operatorname{acosh}{\left (c + d x \right )} - \frac{4 b c^{3} e^{4} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{25} + 2 b c^{2} d^{2} e^{4} x^{3} \operatorname{acosh}{\left (c + d x \right )} - \frac{6 b c^{2} d e^{4} x^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{25} - \frac{4 b c^{2} e^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{75 d} + b c d^{3} e^{4} x^{4} \operatorname{acosh}{\left (c + d x \right )} - \frac{4 b c d^{2} e^{4} x^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{25} - \frac{8 b c e^{4} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{75} + \frac{b d^{4} e^{4} x^{5} \operatorname{acosh}{\left (c + d x \right )}}{5} - \frac{b d^{3} e^{4} x^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{25} - \frac{4 b d e^{4} x^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{75} - \frac{8 b e^{4} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{75 d} & \text{for}\: d \neq 0 \\c^{4} e^{4} 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)**4*(a+b*acosh(d*x+c)),x)

[Out]

Piecewise((a*c**4*e**4*x + 2*a*c**3*d*e**4*x**2 + 2*a*c**2*d**2*e**4*x**3 + a*c*d**3*e**4*x**4 + a*d**4*e**4*x
**5/5 + b*c**5*e**4*acosh(c + d*x)/(5*d) + b*c**4*e**4*x*acosh(c + d*x) - b*c**4*e**4*sqrt(c**2 + 2*c*d*x + d*
*2*x**2 - 1)/(25*d) + 2*b*c**3*d*e**4*x**2*acosh(c + d*x) - 4*b*c**3*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 -
1)/25 + 2*b*c**2*d**2*e**4*x**3*acosh(c + d*x) - 6*b*c**2*d*e**4*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/25
- 4*b*c**2*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(75*d) + b*c*d**3*e**4*x**4*acosh(c + d*x) - 4*b*c*d**2*e
**4*x**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/25 - 8*b*c*e**4*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/75 + b*d*
*4*e**4*x**5*acosh(c + d*x)/5 - b*d**3*e**4*x**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/25 - 4*b*d*e**4*x**2*sqr
t(c**2 + 2*c*d*x + d**2*x**2 - 1)/75 - 8*b*e**4*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(75*d), Ne(d, 0)), (c**4*
e**4*x*(a + b*acosh(c)), True))

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Giac [B]  time = 2.8434, size = 1110, normalized size = 8.22 \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)^4*(a+b*arccosh(d*x+c)),x, algorithm="giac")

[Out]

1/600*(120*a*d^4*x^5 + 600*a*c*d^3*x^4 + 1200*a*c^2*d^2*x^3 + 1200*a*c^3*d*x^2 - 600*(d*(c*log(abs(-c*d - (x*a
bs(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*lo
g(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1)))*b*c^4 + 600*(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^3*d + 200*(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*c^2*d^2 + 25*(2
4*x^4*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)*((2*x*(3*x/d^2 - 7
*c/d^3) + (26*c^2*d^3 + 9*d^3)/d^7)*x - 5*(10*c^3*d^2 + 11*c*d^2)/d^7) - 3*(8*c^4 + 24*c^2 + 3)*log(abs(-c*d -
(x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1))*abs(d)))/(d^4*abs(d)))*d)*b*c*d^3 + (120*x^5*log(d*x + c + sqr
t(d^2*x^2 + 2*c*d*x + c^2 - 1)) - (sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1)*((2*(3*x*(4*x/d^2 - 9*c/d^3) + (47*c^2*d^
5 + 16*d^5)/d^9)*x - 7*(22*c^3*d^4 + 23*c*d^4)/d^9)*x + (274*c^4*d^3 + 607*c^2*d^3 + 64*d^3)/d^9) + 15*(8*c^5
+ 40*c^3 + 15*c)*log(abs(-c*d - (x*abs(d) - sqrt(d^2*x^2 + 2*c*d*x + c^2 - 1))*abs(d)))/(d^5*abs(d)))*d)*b*d^4
+ 600*a*c^4*x)*e^4