3.22 \(\int x (d-c^2 d x^2)^3 (a+b \cosh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=166 \[ -\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}+\frac {35 b d^3 \cosh ^{-1}(c x)}{1024 c^2}+\frac {b d^3 x (c x-1)^{7/2} (c x+1)^{7/2}}{64 c}-\frac {7 b d^3 x (c x-1)^{5/2} (c x+1)^{5/2}}{384 c}+\frac {35 b d^3 x (c x-1)^{3/2} (c x+1)^{3/2}}{1536 c}-\frac {35 b d^3 x \sqrt {c x-1} \sqrt {c x+1}}{1024 c} \]

[Out]

35/1536*b*d^3*x*(c*x-1)^(3/2)*(c*x+1)^(3/2)/c-7/384*b*d^3*x*(c*x-1)^(5/2)*(c*x+1)^(5/2)/c+1/64*b*d^3*x*(c*x-1)
^(7/2)*(c*x+1)^(7/2)/c+35/1024*b*d^3*arccosh(c*x)/c^2-1/8*d^3*(-c^2*x^2+1)^4*(a+b*arccosh(c*x))/c^2-35/1024*b*
d^3*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c

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Rubi [A]  time = 0.08, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5716, 38, 52} \[ -\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}+\frac {35 b d^3 \cosh ^{-1}(c x)}{1024 c^2}+\frac {b d^3 x (c x-1)^{7/2} (c x+1)^{7/2}}{64 c}-\frac {7 b d^3 x (c x-1)^{5/2} (c x+1)^{5/2}}{384 c}+\frac {35 b d^3 x (c x-1)^{3/2} (c x+1)^{3/2}}{1536 c}-\frac {35 b d^3 x \sqrt {c x-1} \sqrt {c x+1}}{1024 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(d - c^2*d*x^2)^3*(a + b*ArcCosh[c*x]),x]

[Out]

(-35*b*d^3*x*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(1024*c) + (35*b*d^3*x*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2))/(1536*c) -
(7*b*d^3*x*(-1 + c*x)^(5/2)*(1 + c*x)^(5/2))/(384*c) + (b*d^3*x*(-1 + c*x)^(7/2)*(1 + c*x)^(7/2))/(64*c) + (3
5*b*d^3*ArcCosh[c*x])/(1024*c^2) - (d^3*(1 - c^2*x^2)^4*(a + b*ArcCosh[c*x]))/(8*c^2)

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(x*(a + b*x)^m*(c + d*x)^m)/(2*m + 1)
, x] + Dist[(2*a*c*m)/(2*m + 1), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 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]

Rule 5716

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

Rubi steps

\begin {align*} \int x \left (d-c^2 d x^2\right )^3 \left (a+b \cosh ^{-1}(c x)\right ) \, dx &=-\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}+\frac {\left (b d^3\right ) \int (-1+c x)^{7/2} (1+c x)^{7/2} \, dx}{8 c}\\ &=\frac {b d^3 x (-1+c x)^{7/2} (1+c x)^{7/2}}{64 c}-\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}-\frac {\left (7 b d^3\right ) \int (-1+c x)^{5/2} (1+c x)^{5/2} \, dx}{64 c}\\ &=-\frac {7 b d^3 x (-1+c x)^{5/2} (1+c x)^{5/2}}{384 c}+\frac {b d^3 x (-1+c x)^{7/2} (1+c x)^{7/2}}{64 c}-\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}+\frac {\left (35 b d^3\right ) \int (-1+c x)^{3/2} (1+c x)^{3/2} \, dx}{384 c}\\ &=\frac {35 b d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}{1536 c}-\frac {7 b d^3 x (-1+c x)^{5/2} (1+c x)^{5/2}}{384 c}+\frac {b d^3 x (-1+c x)^{7/2} (1+c x)^{7/2}}{64 c}-\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}-\frac {\left (35 b d^3\right ) \int \sqrt {-1+c x} \sqrt {1+c x} \, dx}{512 c}\\ &=-\frac {35 b d^3 x \sqrt {-1+c x} \sqrt {1+c x}}{1024 c}+\frac {35 b d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}{1536 c}-\frac {7 b d^3 x (-1+c x)^{5/2} (1+c x)^{5/2}}{384 c}+\frac {b d^3 x (-1+c x)^{7/2} (1+c x)^{7/2}}{64 c}-\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}+\frac {\left (35 b d^3\right ) \int \frac {1}{\sqrt {-1+c x} \sqrt {1+c x}} \, dx}{1024 c}\\ &=-\frac {35 b d^3 x \sqrt {-1+c x} \sqrt {1+c x}}{1024 c}+\frac {35 b d^3 x (-1+c x)^{3/2} (1+c x)^{3/2}}{1536 c}-\frac {7 b d^3 x (-1+c x)^{5/2} (1+c x)^{5/2}}{384 c}+\frac {b d^3 x (-1+c x)^{7/2} (1+c x)^{7/2}}{64 c}+\frac {35 b d^3 \cosh ^{-1}(c x)}{1024 c^2}-\frac {d^3 \left (1-c^2 x^2\right )^4 \left (a+b \cosh ^{-1}(c x)\right )}{8 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 150, normalized size = 0.90 \[ -\frac {d^3 \left (c x \left (384 a c x \left (c^6 x^6-4 c^4 x^4+6 c^2 x^2-4\right )+b \sqrt {c x-1} \sqrt {c x+1} \left (-48 c^6 x^6+200 c^4 x^4-326 c^2 x^2+279\right )\right )+384 b c^2 x^2 \left (c^6 x^6-4 c^4 x^4+6 c^2 x^2-4\right ) \cosh ^{-1}(c x)+558 b \tanh ^{-1}\left (\sqrt {\frac {c x-1}{c x+1}}\right )\right )}{3072 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(d - c^2*d*x^2)^3*(a + b*ArcCosh[c*x]),x]

[Out]

-1/3072*(d^3*(c*x*(b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(279 - 326*c^2*x^2 + 200*c^4*x^4 - 48*c^6*x^6) + 384*a*c*x*(
-4 + 6*c^2*x^2 - 4*c^4*x^4 + c^6*x^6)) + 384*b*c^2*x^2*(-4 + 6*c^2*x^2 - 4*c^4*x^4 + c^6*x^6)*ArcCosh[c*x] + 5
58*b*ArcTanh[Sqrt[(-1 + c*x)/(1 + c*x)]]))/c^2

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fricas [A]  time = 0.47, size = 185, normalized size = 1.11 \[ -\frac {384 \, a c^{8} d^{3} x^{8} - 1536 \, a c^{6} d^{3} x^{6} + 2304 \, a c^{4} d^{3} x^{4} - 1536 \, a c^{2} d^{3} x^{2} + 3 \, {\left (128 \, b c^{8} d^{3} x^{8} - 512 \, b c^{6} d^{3} x^{6} + 768 \, b c^{4} d^{3} x^{4} - 512 \, b c^{2} d^{3} x^{2} + 93 \, b d^{3}\right )} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (48 \, b c^{7} d^{3} x^{7} - 200 \, b c^{5} d^{3} x^{5} + 326 \, b c^{3} d^{3} x^{3} - 279 \, b c d^{3} x\right )} \sqrt {c^{2} x^{2} - 1}}{3072 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^3*(a+b*arccosh(c*x)),x, algorithm="fricas")

[Out]

-1/3072*(384*a*c^8*d^3*x^8 - 1536*a*c^6*d^3*x^6 + 2304*a*c^4*d^3*x^4 - 1536*a*c^2*d^3*x^2 + 3*(128*b*c^8*d^3*x
^8 - 512*b*c^6*d^3*x^6 + 768*b*c^4*d^3*x^4 - 512*b*c^2*d^3*x^2 + 93*b*d^3)*log(c*x + sqrt(c^2*x^2 - 1)) - (48*
b*c^7*d^3*x^7 - 200*b*c^5*d^3*x^5 + 326*b*c^3*d^3*x^3 - 279*b*c*d^3*x)*sqrt(c^2*x^2 - 1))/c^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^3*(a+b*arccosh(c*x)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.02, size = 258, normalized size = 1.55 \[ -\frac {c^{6} d^{3} a \,x^{8}}{8}+\frac {c^{4} d^{3} a \,x^{6}}{2}-\frac {3 c^{2} d^{3} a \,x^{4}}{4}+\frac {d^{3} a \,x^{2}}{2}-\frac {c^{6} d^{3} b \,\mathrm {arccosh}\left (c x \right ) x^{8}}{8}+\frac {c^{4} d^{3} b \,\mathrm {arccosh}\left (c x \right ) x^{6}}{2}-\frac {3 c^{2} d^{3} b \,\mathrm {arccosh}\left (c x \right ) x^{4}}{4}+\frac {d^{3} b \,\mathrm {arccosh}\left (c x \right ) x^{2}}{2}+\frac {c^{5} d^{3} b \sqrt {c x -1}\, \sqrt {c x +1}\, x^{7}}{64}-\frac {25 c^{3} d^{3} b \sqrt {c x -1}\, \sqrt {c x +1}\, x^{5}}{384}+\frac {163 c \,d^{3} b \sqrt {c x -1}\, \sqrt {c x +1}\, x^{3}}{1536}-\frac {93 b \,d^{3} x \sqrt {c x -1}\, \sqrt {c x +1}}{1024 c}-\frac {93 d^{3} b \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (c x +\sqrt {c^{2} x^{2}-1}\right )}{1024 c^{2} \sqrt {c^{2} x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(-c^2*d*x^2+d)^3*(a+b*arccosh(c*x)),x)

[Out]

-1/8*c^6*d^3*a*x^8+1/2*c^4*d^3*a*x^6-3/4*c^2*d^3*a*x^4+1/2*d^3*a*x^2-1/8*c^6*d^3*b*arccosh(c*x)*x^8+1/2*c^4*d^
3*b*arccosh(c*x)*x^6-3/4*c^2*d^3*b*arccosh(c*x)*x^4+1/2*d^3*b*arccosh(c*x)*x^2+1/64*c^5*d^3*b*(c*x-1)^(1/2)*(c
*x+1)^(1/2)*x^7-25/384*c^3*d^3*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)*x^5+163/1536*c*d^3*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)*
x^3-93/1024*b*d^3*x*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c-93/1024/c^2*d^3*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)/(c^2*x^2-1)^(1
/2)*ln(c*x+(c^2*x^2-1)^(1/2))

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maxima [B]  time = 0.37, size = 423, normalized size = 2.55 \[ -\frac {1}{8} \, a c^{6} d^{3} x^{8} + \frac {1}{2} \, a c^{4} d^{3} x^{6} - \frac {1}{3072} \, {\left (384 \, x^{8} \operatorname {arcosh}\left (c x\right ) - {\left (\frac {48 \, \sqrt {c^{2} x^{2} - 1} x^{7}}{c^{2}} + \frac {56 \, \sqrt {c^{2} x^{2} - 1} x^{5}}{c^{4}} + \frac {70 \, \sqrt {c^{2} x^{2} - 1} x^{3}}{c^{6}} + \frac {105 \, \sqrt {c^{2} x^{2} - 1} x}{c^{8}} + \frac {105 \, \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{c^{9}}\right )} c\right )} b c^{6} d^{3} - \frac {3}{4} \, a c^{2} d^{3} x^{4} + \frac {1}{96} \, {\left (48 \, x^{6} \operatorname {arcosh}\left (c x\right ) - {\left (\frac {8 \, \sqrt {c^{2} x^{2} - 1} x^{5}}{c^{2}} + \frac {10 \, \sqrt {c^{2} x^{2} - 1} x^{3}}{c^{4}} + \frac {15 \, \sqrt {c^{2} x^{2} - 1} x}{c^{6}} + \frac {15 \, \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{c^{7}}\right )} c\right )} b c^{4} d^{3} - \frac {3}{32} \, {\left (8 \, x^{4} \operatorname {arcosh}\left (c x\right ) - {\left (\frac {2 \, \sqrt {c^{2} x^{2} - 1} x^{3}}{c^{2}} + \frac {3 \, \sqrt {c^{2} x^{2} - 1} x}{c^{4}} + \frac {3 \, \log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{c^{5}}\right )} c\right )} b c^{2} d^{3} + \frac {1}{2} \, a d^{3} x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {arcosh}\left (c x\right ) - c {\left (\frac {\sqrt {c^{2} x^{2} - 1} x}{c^{2}} + \frac {\log \left (2 \, c^{2} x + 2 \, \sqrt {c^{2} x^{2} - 1} c\right )}{c^{3}}\right )}\right )} b d^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c^2*d*x^2+d)^3*(a+b*arccosh(c*x)),x, algorithm="maxima")

[Out]

-1/8*a*c^6*d^3*x^8 + 1/2*a*c^4*d^3*x^6 - 1/3072*(384*x^8*arccosh(c*x) - (48*sqrt(c^2*x^2 - 1)*x^7/c^2 + 56*sqr
t(c^2*x^2 - 1)*x^5/c^4 + 70*sqrt(c^2*x^2 - 1)*x^3/c^6 + 105*sqrt(c^2*x^2 - 1)*x/c^8 + 105*log(2*c^2*x + 2*sqrt
(c^2*x^2 - 1)*c)/c^9)*c)*b*c^6*d^3 - 3/4*a*c^2*d^3*x^4 + 1/96*(48*x^6*arccosh(c*x) - (8*sqrt(c^2*x^2 - 1)*x^5/
c^2 + 10*sqrt(c^2*x^2 - 1)*x^3/c^4 + 15*sqrt(c^2*x^2 - 1)*x/c^6 + 15*log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^7)
*c)*b*c^4*d^3 - 3/32*(8*x^4*arccosh(c*x) - (2*sqrt(c^2*x^2 - 1)*x^3/c^2 + 3*sqrt(c^2*x^2 - 1)*x/c^4 + 3*log(2*
c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^5)*c)*b*c^2*d^3 + 1/2*a*d^3*x^2 + 1/4*(2*x^2*arccosh(c*x) - c*(sqrt(c^2*x^2 -
1)*x/c^2 + log(2*c^2*x + 2*sqrt(c^2*x^2 - 1)*c)/c^3))*b*d^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*acosh(c*x))*(d - c^2*d*x^2)^3,x)

[Out]

int(x*(a + b*acosh(c*x))*(d - c^2*d*x^2)^3, x)

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sympy [A]  time = 11.61, size = 260, normalized size = 1.57 \[ \begin {cases} - \frac {a c^{6} d^{3} x^{8}}{8} + \frac {a c^{4} d^{3} x^{6}}{2} - \frac {3 a c^{2} d^{3} x^{4}}{4} + \frac {a d^{3} x^{2}}{2} - \frac {b c^{6} d^{3} x^{8} \operatorname {acosh}{\left (c x \right )}}{8} + \frac {b c^{5} d^{3} x^{7} \sqrt {c^{2} x^{2} - 1}}{64} + \frac {b c^{4} d^{3} x^{6} \operatorname {acosh}{\left (c x \right )}}{2} - \frac {25 b c^{3} d^{3} x^{5} \sqrt {c^{2} x^{2} - 1}}{384} - \frac {3 b c^{2} d^{3} x^{4} \operatorname {acosh}{\left (c x \right )}}{4} + \frac {163 b c d^{3} x^{3} \sqrt {c^{2} x^{2} - 1}}{1536} + \frac {b d^{3} x^{2} \operatorname {acosh}{\left (c x \right )}}{2} - \frac {93 b d^{3} x \sqrt {c^{2} x^{2} - 1}}{1024 c} - \frac {93 b d^{3} \operatorname {acosh}{\left (c x \right )}}{1024 c^{2}} & \text {for}\: c \neq 0 \\\frac {d^{3} x^{2} \left (a + \frac {i \pi b}{2}\right )}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(-c**2*d*x**2+d)**3*(a+b*acosh(c*x)),x)

[Out]

Piecewise((-a*c**6*d**3*x**8/8 + a*c**4*d**3*x**6/2 - 3*a*c**2*d**3*x**4/4 + a*d**3*x**2/2 - b*c**6*d**3*x**8*
acosh(c*x)/8 + b*c**5*d**3*x**7*sqrt(c**2*x**2 - 1)/64 + b*c**4*d**3*x**6*acosh(c*x)/2 - 25*b*c**3*d**3*x**5*s
qrt(c**2*x**2 - 1)/384 - 3*b*c**2*d**3*x**4*acosh(c*x)/4 + 163*b*c*d**3*x**3*sqrt(c**2*x**2 - 1)/1536 + b*d**3
*x**2*acosh(c*x)/2 - 93*b*d**3*x*sqrt(c**2*x**2 - 1)/(1024*c) - 93*b*d**3*acosh(c*x)/(1024*c**2), Ne(c, 0)), (
d**3*x**2*(a + I*pi*b/2)/2, True))

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