3.94 \(\int (c e+d e x)^3 (a+b \cosh ^{-1}(c+d x)) \, dx\)

Optimal. Leaf size=119 \[ \frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}-\frac{b e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{16 d}-\frac{3 b e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)}{32 d}-\frac{3 b e^3 \cosh ^{-1}(c+d x)}{32 d} \]

[Out]

(-3*b*e^3*Sqrt[-1 + c + d*x]*(c + d*x)*Sqrt[1 + c + d*x])/(32*d) - (b*e^3*Sqrt[-1 + c + d*x]*(c + d*x)^3*Sqrt[
1 + c + d*x])/(16*d) - (3*b*e^3*ArcCosh[c + d*x])/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcCosh[c + d*x]))/(4*d)

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Rubi [A]  time = 0.0705448, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5866, 12, 5662, 100, 90, 52} \[ \frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}-\frac{b e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3}{16 d}-\frac{3 b e^3 \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)}{32 d}-\frac{3 b e^3 \cosh ^{-1}(c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3*(a + b*ArcCosh[c + d*x]),x]

[Out]

(-3*b*e^3*Sqrt[-1 + c + d*x]*(c + d*x)*Sqrt[1 + c + d*x])/(32*d) - (b*e^3*Sqrt[-1 + c + d*x]*(c + d*x)^3*Sqrt[
1 + c + d*x])/(16*d) - (3*b*e^3*ArcCosh[c + d*x])/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcCosh[c + d*x]))/(4*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 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rubi steps

\begin{align*} \int (c e+d e x)^3 \left (a+b \cosh ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int e^3 x^3 \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int x^3 \left (a+b \cosh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac{b e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{16 d}+\frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \frac{3 x^2}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{16 d}\\ &=-\frac{b e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{16 d}+\frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}-\frac{\left (3 b e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{16 d}\\ &=-\frac{3 b e^3 \sqrt{-1+c+d x} (c+d x) \sqrt{1+c+d x}}{32 d}-\frac{b e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{16 d}+\frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}-\frac{\left (3 b e^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x} \sqrt{1+x}} \, dx,x,c+d x\right )}{32 d}\\ &=-\frac{3 b e^3 \sqrt{-1+c+d x} (c+d x) \sqrt{1+c+d x}}{32 d}-\frac{b e^3 \sqrt{-1+c+d x} (c+d x)^3 \sqrt{1+c+d x}}{16 d}-\frac{3 b e^3 \cosh ^{-1}(c+d x)}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.124771, size = 115, normalized size = 0.97 \[ \frac{e^3 \left ((c+d x)^4 \left (a+b \cosh ^{-1}(c+d x)\right )-\frac{1}{4} b \sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)^3-\frac{3}{8} b \left (\sqrt{c+d x-1} \sqrt{c+d x+1} (c+d x)+2 \tanh ^{-1}\left (\sqrt{\frac{c+d x-1}{c+d x+1}}\right )\right )\right )}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^3*(a + b*ArcCosh[c + d*x]),x]

[Out]

(e^3*(-(b*Sqrt[-1 + c + d*x]*(c + d*x)^3*Sqrt[1 + c + d*x])/4 + (c + d*x)^4*(a + b*ArcCosh[c + d*x]) - (3*b*(S
qrt[-1 + c + d*x]*(c + d*x)*Sqrt[1 + c + d*x] + 2*ArcTanh[Sqrt[(-1 + c + d*x)/(1 + c + d*x)]]))/8))/(4*d)

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Maple [B]  time = 0.004, size = 359, normalized size = 3. \begin{align*}{\frac{{d}^{3}{x}^{4}a{e}^{3}}{4}}+{d}^{2}{x}^{3}ac{e}^{3}+{\frac{3\,d{x}^{2}a{c}^{2}{e}^{3}}{2}}+xa{c}^{3}{e}^{3}+{\frac{a{c}^{4}{e}^{3}}{4\,d}}+{\frac{{d}^{3}{\rm arccosh} \left (dx+c\right ){x}^{4}b{e}^{3}}{4}}+{d}^{2}{\rm arccosh} \left (dx+c\right ){x}^{3}bc{e}^{3}+{\frac{3\,d{\rm arccosh} \left (dx+c\right ){x}^{2}b{c}^{2}{e}^{3}}{2}}+{\rm arccosh} \left (dx+c\right )xb{c}^{3}{e}^{3}+{\frac{b{\rm arccosh} \left (dx+c\right ){c}^{4}{e}^{3}}{4\,d}}-{\frac{{d}^{2}{x}^{3}b{e}^{3}}{16}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{3\,d{x}^{2}bc{e}^{3}}{16}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{3\,bx{c}^{2}{e}^{3}}{16}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{b{c}^{3}{e}^{3}}{16\,d}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{3\,bx{e}^{3}}{32}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{3\,bc{e}^{3}}{32\,d}\sqrt{dx+c-1}\sqrt{dx+c+1}}-{\frac{3\,{e}^{3}b}{32\,d}\sqrt{dx+c-1}\sqrt{dx+c+1}\ln \left ( dx+c+\sqrt{ \left ( dx+c \right ) ^{2}-1} \right ){\frac{1}{\sqrt{ \left ( dx+c \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3*(a+b*arccosh(d*x+c)),x)

[Out]

1/4*d^3*x^4*a*e^3+d^2*x^3*a*c*e^3+3/2*d*x^2*a*c^2*e^3+x*a*c^3*e^3+1/4/d*a*c^4*e^3+1/4*d^3*arccosh(d*x+c)*x^4*b
*e^3+d^2*arccosh(d*x+c)*x^3*b*c*e^3+3/2*d*arccosh(d*x+c)*x^2*b*c^2*e^3+arccosh(d*x+c)*x*b*c^3*e^3+1/4/d*arccos
h(d*x+c)*b*c^4*e^3-1/16*d^2*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*x^3*b*e^3-3/16*d*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*x
^2*b*c*e^3-3/16*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*x*b*c^2*e^3-1/16/d*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*b*c^3*e^3-3
/32*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*x*b*e^3-3/32/d*(d*x+c-1)^(1/2)*(d*x+c+1)^(1/2)*b*c*e^3-3/32/d*e^3*b*(d*x+c
-1)^(1/2)*(d*x+c+1)^(1/2)/((d*x+c)^2-1)^(1/2)*ln(d*x+c+((d*x+c)^2-1)^(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)^3*(a+b*arccosh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42512, size = 495, normalized size = 4.16 \begin{align*} \frac{8 \, a d^{4} e^{3} x^{4} + 32 \, a c d^{3} e^{3} x^{3} + 48 \, a c^{2} d^{2} e^{3} x^{2} + 32 \, a c^{3} d e^{3} x +{\left (8 \, b d^{4} e^{3} x^{4} + 32 \, b c d^{3} e^{3} x^{3} + 48 \, b c^{2} d^{2} e^{3} x^{2} + 32 \, b c^{3} d e^{3} x +{\left (8 \, b c^{4} - 3 \, b\right )} e^{3}\right )} \log \left (d x + c + \sqrt{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\right ) -{\left (2 \, b d^{3} e^{3} x^{3} + 6 \, b c d^{2} e^{3} x^{2} + 3 \,{\left (2 \, b c^{2} + b\right )} d e^{3} x +{\left (2 \, b c^{3} + 3 \, 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*arccosh(d*x+c)),x, algorithm="fricas")

[Out]

1/32*(8*a*d^4*e^3*x^4 + 32*a*c*d^3*e^3*x^3 + 48*a*c^2*d^2*e^3*x^2 + 32*a*c^3*d*e^3*x + (8*b*d^4*e^3*x^4 + 32*b
*c*d^3*e^3*x^3 + 48*b*c^2*d^2*e^3*x^2 + 32*b*c^3*d*e^3*x + (8*b*c^4 - 3*b)*e^3)*log(d*x + c + sqrt(d^2*x^2 + 2
*c*d*x + c^2 - 1)) - (2*b*d^3*e^3*x^3 + 6*b*c*d^2*e^3*x^2 + 3*(2*b*c^2 + b)*d*e^3*x + (2*b*c^3 + 3*b*c)*e^3)*s
qrt(d^2*x^2 + 2*c*d*x + c^2 - 1))/d

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Sympy [A]  time = 2.35689, size = 394, normalized size = 3.31 \begin{align*} \begin{cases} a c^{3} e^{3} x + \frac{3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac{a d^{3} e^{3} x^{4}}{4} + \frac{b c^{4} e^{3} \operatorname{acosh}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname{acosh}{\left (c + d x \right )} - \frac{b c^{3} e^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{16 d} + \frac{3 b c^{2} d e^{3} x^{2} \operatorname{acosh}{\left (c + d x \right )}}{2} - \frac{3 b c^{2} e^{3} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{16} + b c d^{2} e^{3} x^{3} \operatorname{acosh}{\left (c + d x \right )} - \frac{3 b c d e^{3} x^{2} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{16} - \frac{3 b c e^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{32 d} + \frac{b d^{3} e^{3} x^{4} \operatorname{acosh}{\left (c + d x \right )}}{4} - \frac{b d^{2} e^{3} x^{3} \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{16} - \frac{3 b e^{3} x \sqrt{c^{2} + 2 c d x + d^{2} x^{2} - 1}}{32} - \frac{3 b e^{3} \operatorname{acosh}{\left (c + d x \right )}}{32 d} & \text{for}\: d \neq 0 \\c^{3} e^{3} 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)**3*(a+b*acosh(d*x+c)),x)

[Out]

Piecewise((a*c**3*e**3*x + 3*a*c**2*d*e**3*x**2/2 + a*c*d**2*e**3*x**3 + a*d**3*e**3*x**4/4 + b*c**4*e**3*acos
h(c + d*x)/(4*d) + b*c**3*e**3*x*acosh(c + d*x) - b*c**3*e**3*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/(16*d) + 3*
b*c**2*d*e**3*x**2*acosh(c + d*x)/2 - 3*b*c**2*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/16 + b*c*d**2*e**3*
x**3*acosh(c + d*x) - 3*b*c*d*e**3*x**2*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/16 - 3*b*c*e**3*sqrt(c**2 + 2*c*d
*x + d**2*x**2 - 1)/(32*d) + b*d**3*e**3*x**4*acosh(c + d*x)/4 - b*d**2*e**3*x**3*sqrt(c**2 + 2*c*d*x + d**2*x
**2 - 1)/16 - 3*b*e**3*x*sqrt(c**2 + 2*c*d*x + d**2*x**2 - 1)/32 - 3*b*e**3*acosh(c + d*x)/(32*d), Ne(d, 0)),
(c**3*e**3*x*(a + b*acosh(c)), True))

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Giac [B]  time = 2.4579, size = 807, normalized size = 6.78 \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*arccosh(d*x+c)),x, algorithm="giac")

[Out]

1/96*(24*a*d^3*x^4 + 96*a*c*d^2*x^3 + 144*a*c^2*d*x^2 - 96*(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^3 + 72*(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^2*d + 16*(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*d^2 + (24*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*d^3 + 96*a*c^3*x)*e^3