\(\int \frac {x^2 (a+b \text {arccosh}(c x))}{(d-c^2 d x^2)^{5/2}} \, dx\) [127]

Optimal result

Integrand size = 27, antiderivative size = 133 \[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {b \sqrt {-1+c x} \sqrt {1+c x}}{6 c^3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b \sqrt {-1+c x} \sqrt {1+c x} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt {d-c^2 d x^2}} \]

[Out]

1/3*x^3*(a+b*arccosh(c*x))/d/(-c^2*d*x^2+d)^(3/2)+1/6*b*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(3/2)
+1/6*b*ln(-c^2*x^2+1)*(c*x-1)^(1/2)*(c*x+1)^(1/2)/c^3/d^2/(-c^2*d*x^2+d)^(1/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {5917, 74, 272, 45} \[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {c x-1} \sqrt {c x+1} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt {d-c^2 d x^2}} \]

[In]

Int[(x^2*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(b*Sqrt[-1 + c*x]*Sqrt[1 + c*x])/(6*c^3*d^2*(1 - c^2*x^2)*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcCosh[c*x]))/(3
*d*(d - c^2*d*x^2)^(3/2)) + (b*Sqrt[-1 + c*x]*Sqrt[1 + c*x]*Log[1 - c^2*x^2])/(6*c^3*d^2*Sqrt[d - c^2*d*x^2])

Rule 45

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])

Rule 74

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

Rule 272

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 5917

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

Rubi steps \begin{align*} \text {integral}& = \frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3}{(-1+c x)^2 (1+c x)^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}} \\ & = \frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \int \frac {x^3}{\left (-1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt {d-c^2 d x^2}} \\ & = \frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {x}{\left (-1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}} \\ & = \frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {\left (b c \sqrt {-1+c x} \sqrt {1+c x}\right ) \text {Subst}\left (\int \left (\frac {1}{c^2 \left (-1+c^2 x\right )^2}+\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt {d-c^2 d x^2}} \\ & = \frac {b \sqrt {-1+c x} \sqrt {1+c x}}{6 c^3 d^2 \left (1-c^2 x^2\right ) \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \text {arccosh}(c x))}{3 d \left (d-c^2 d x^2\right )^{3/2}}+\frac {b \sqrt {-1+c x} \sqrt {1+c x} \log \left (1-c^2 x^2\right )}{6 c^3 d^2 \sqrt {d-c^2 d x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {-1+c x} \sqrt {1+c x} \left (-\frac {2 x^3 (a+b \text {arccosh}(c x))}{(-1+c x)^{3/2} (1+c x)^{3/2}}+\frac {b \left (\frac {1}{1-c^2 x^2}+\log \left (1-c^2 x^2\right )\right )}{c^3}\right )}{6 d^2 \sqrt {d-c^2 d x^2}} \]

[In]

Integrate[(x^2*(a + b*ArcCosh[c*x]))/(d - c^2*d*x^2)^(5/2),x]

[Out]

(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*((-2*x^3*(a + b*ArcCosh[c*x]))/((-1 + c*x)^(3/2)*(1 + c*x)^(3/2)) + (b*((1 - c^2
*x^2)^(-1) + Log[1 - c^2*x^2]))/c^3))/(6*d^2*Sqrt[d - c^2*d*x^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(457\) vs. \(2(113)=226\).

Time = 1.19 (sec) , antiderivative size = 458, normalized size of antiderivative = 3.44

method	result	size
default	\(a \left (\frac {x}{2 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {\frac {x}{3 d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {-c^{2} d \,x^{2}+d}}}{2 c^{2}}\right )+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{3} x^{3}+\sqrt {c x -1}\, \sqrt {c x +1}\, c^{2} x^{2}-\sqrt {c x -1}\, \sqrt {c x +1}\right ) \left (-2 \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{5} c^{5}+2 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{6} c^{6}+6 c^{4} x^{4} \operatorname {arccosh}\left (c x \right )+2 \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{3} c^{3}-6 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{4} c^{4}+\sqrt {c x -1}\, \sqrt {c x +1}\, c^{3} x^{3}-c^{4} x^{4}-6 c^{2} x^{2} \operatorname {arccosh}\left (c x \right )+6 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{2} c^{2}+2 c^{2} x^{2}+2 \,\operatorname {arccosh}\left (c x \right )-2 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right )-1\right )}{6 \left (3 c^{8} x^{8}-9 c^{6} x^{6}+10 c^{4} x^{4}-5 c^{2} x^{2}+1\right ) d^{3} c^{3}}\)	\(458\)
parts	\(a \left (\frac {x}{2 c^{2} d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {\frac {x}{3 d \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {-c^{2} d \,x^{2}+d}}}{2 c^{2}}\right )+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{3} x^{3}+\sqrt {c x -1}\, \sqrt {c x +1}\, c^{2} x^{2}-\sqrt {c x -1}\, \sqrt {c x +1}\right ) \left (-2 \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{5} c^{5}+2 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{6} c^{6}+6 c^{4} x^{4} \operatorname {arccosh}\left (c x \right )+2 \sqrt {c x -1}\, \sqrt {c x +1}\, \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{3} c^{3}-6 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{4} c^{4}+\sqrt {c x -1}\, \sqrt {c x +1}\, c^{3} x^{3}-c^{4} x^{4}-6 c^{2} x^{2} \operatorname {arccosh}\left (c x \right )+6 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right ) x^{2} c^{2}+2 c^{2} x^{2}+2 \,\operatorname {arccosh}\left (c x \right )-2 \ln \left (\left (c x +\sqrt {c x -1}\, \sqrt {c x +1}\right )^{2}-1\right )-1\right )}{6 \left (3 c^{8} x^{8}-9 c^{6} x^{6}+10 c^{4} x^{4}-5 c^{2} x^{2}+1\right ) d^{3} c^{3}}\)	\(458\)

[In]

int(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

a*(1/2*x/c^2/d/(-c^2*d*x^2+d)^(3/2)-1/2/c^2*(1/3/d*x/(-c^2*d*x^2+d)^(3/2)+2/3/d^2*x/(-c^2*d*x^2+d)^(1/2)))+1/6
*b*(-d*(c^2*x^2-1))^(1/2)*(c^3*x^3+(c*x-1)^(1/2)*(c*x+1)^(1/2)*c^2*x^2-(c*x-1)^(1/2)*(c*x+1)^(1/2))*(-2*(c*x-1
)^(1/2)*(c*x+1)^(1/2)*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2-1)*x^5*c^5+2*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))
^2-1)*x^6*c^6+6*c^4*x^4*arccosh(c*x)+2*(c*x-1)^(1/2)*(c*x+1)^(1/2)*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2-1)*x
^3*c^3-6*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2-1)*x^4*c^4+(c*x-1)^(1/2)*(c*x+1)^(1/2)*c^3*x^3-c^4*x^4-6*c^2*x
^2*arccosh(c*x)+6*ln((c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2-1)*x^2*c^2+2*c^2*x^2+2*arccosh(c*x)-2*ln((c*x+(c*x-1)
^(1/2)*(c*x+1)^(1/2))^2-1)-1)/(3*c^8*x^8-9*c^6*x^6+10*c^4*x^4-5*c^2*x^2+1)/d^3/c^3

Fricas [F]

\[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(b*x^2*arccosh(c*x) + a*x^2)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3
), x)

Sympy [F]

\[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^{2} \left (a + b \operatorname {acosh}{\left (c x \right )}\right )}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**2*(a+b*acosh(c*x))/(-c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x**2*(a + b*acosh(c*x))/(-d*(c*x - 1)*(c*x + 1))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.27 \[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\frac {1}{6} \, b c {\left (\frac {\sqrt {-d}}{c^{6} d^{3} x^{2} - c^{4} d^{3}} - \frac {\sqrt {-d} \log \left (c x + 1\right )}{c^{4} d^{3}} - \frac {\sqrt {-d} \log \left (c x - 1\right )}{c^{4} d^{3}}\right )} - \frac {1}{3} \, b {\left (\frac {x}{\sqrt {-c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \operatorname {arcosh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {x}{\sqrt {-c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \]

[In]

integrate(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/6*b*c*(sqrt(-d)/(c^6*d^3*x^2 - c^4*d^3) - sqrt(-d)*log(c*x + 1)/(c^4*d^3) - sqrt(-d)*log(c*x - 1)/(c^4*d^3))
- 1/3*b*(x/(sqrt(-c^2*d*x^2 + d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d))*arccosh(c*x) - 1/3*a*(x/(sqrt(-
c^2*d*x^2 + d)*c^2*d^2) - x/((-c^2*d*x^2 + d)^(3/2)*c^2*d))

Giac [F]

\[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} x^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x^2*(a+b*arccosh(c*x))/(-c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccosh(c*x) + a)*x^2/(-c^2*d*x^2 + d)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^{5/2}} \, dx=\int \frac {x^2\,\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^{5/2}} \,d x \]

[In]

int((x^2*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2),x)

[Out]

int((x^2*(a + b*acosh(c*x)))/(d - c^2*d*x^2)^(5/2), x)