\(\int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx\) [61]

Optimal result

Integrand size = 27, antiderivative size = 279 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=-\frac {b c \sqrt {d-c^2 d x^2}}{42 x^6 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {b c^3 \sqrt {d-c^2 d x^2}}{140 x^4 \sqrt {-1+c x} \sqrt {1+c x}}+\frac {2 b c^5 \sqrt {d-c^2 d x^2}}{105 x^2 \sqrt {-1+c x} \sqrt {1+c x}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{105 d x^3}-\frac {8 b c^7 \sqrt {d-c^2 d x^2} \log (x)}{105 \sqrt {-1+c x} \sqrt {1+c x}} \] Output:

-1/42*b*c*(-c^2*d*x^2+d)^(1/2)/x^6/(c*x-1)^(1/2)/(c*x+1)^(1/2)+1/140*b*c^3 
*(-c^2*d*x^2+d)^(1/2)/x^4/(c*x-1)^(1/2)/(c*x+1)^(1/2)+2/105*b*c^5*(-c^2*d* 
x^2+d)^(1/2)/x^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)-1/7*(-c^2*d*x^2+d)^(3/2)*(a+b 
*arccosh(c*x))/d/x^7-4/35*c^2*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x))/d/x^ 
5-8/105*c^4*(-c^2*d*x^2+d)^(3/2)*(a+b*arccosh(c*x))/d/x^3-8/105*b*c^7*(-c^ 
2*d*x^2+d)^(1/2)*ln(x)/(c*x-1)^(1/2)/(c*x+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.52 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=\frac {\sqrt {d-c^2 d x^2} \left (60 (-1+c x)^{3/2} (1+c x)^{3/2} (a+b \text {arccosh}(c x))+16 c^2 x^2 (-1+c x)^{3/2} (1+c x)^{3/2} \left (3+2 c^2 x^2\right ) (a+b \text {arccosh}(c x))-b c x \left (10-3 c^2 x^2-8 c^4 x^4+32 c^6 x^6 \log (x)\right )\right )}{420 x^7 \sqrt {-1+c x} \sqrt {1+c x}} \] Input:

Integrate[(Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/x^8,x]
 

Output:

(Sqrt[d - c^2*d*x^2]*(60*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)*(a + b*ArcCosh[c 
*x]) + 16*c^2*x^2*(-1 + c*x)^(3/2)*(1 + c*x)^(3/2)*(3 + 2*c^2*x^2)*(a + b* 
ArcCosh[c*x]) - b*c*x*(10 - 3*c^2*x^2 - 8*c^4*x^4 + 32*c^6*x^6*Log[x])))/( 
420*x^7*Sqrt[-1 + c*x]*Sqrt[1 + c*x])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.64, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6337, 27, 2010, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx\)

\(\Big \downarrow \) 6337

\(\displaystyle -\frac {b c \sqrt {d-c^2 d x^2} \int -\frac {-8 c^6 x^6-4 c^4 x^4-3 c^2 x^2+15}{105 x^7}dx}{\sqrt {c x-1} \sqrt {c x+1}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{105 d x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \frac {-8 c^6 x^6-4 c^4 x^4-3 c^2 x^2+15}{x^7}dx}{105 \sqrt {c x-1} \sqrt {c x+1}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{105 d x^3}\)

\(\Big \downarrow \) 2010

\(\displaystyle \frac {b c \sqrt {d-c^2 d x^2} \int \left (-\frac {8 c^6}{x}-\frac {4 c^4}{x^3}-\frac {3 c^2}{x^5}+\frac {15}{x^7}\right )dx}{105 \sqrt {c x-1} \sqrt {c x+1}}-\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{105 d x^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{7 d x^7}-\frac {4 c^2 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{35 d x^5}-\frac {8 c^4 \left (d-c^2 d x^2\right )^{3/2} (a+b \text {arccosh}(c x))}{105 d x^3}+\frac {b c \sqrt {d-c^2 d x^2} \left (-8 c^6 \log (x)+\frac {2 c^4}{x^2}+\frac {3 c^2}{4 x^4}-\frac {5}{2 x^6}\right )}{105 \sqrt {c x-1} \sqrt {c x+1}}\)

Input:

Int[(Sqrt[d - c^2*d*x^2]*(a + b*ArcCosh[c*x]))/x^8,x]
 

Output:

-1/7*((d - c^2*d*x^2)^(3/2)*(a + b*ArcCosh[c*x]))/(d*x^7) - (4*c^2*(d - c^ 
2*d*x^2)^(3/2)*(a + b*ArcCosh[c*x]))/(35*d*x^5) - (8*c^4*(d - c^2*d*x^2)^( 
3/2)*(a + b*ArcCosh[c*x]))/(105*d*x^3) + (b*c*Sqrt[d - c^2*d*x^2]*(-5/(2*x 
^6) + (3*c^2)/(4*x^4) + (2*c^4)/x^2 - 8*c^6*Log[x]))/(105*Sqrt[-1 + c*x]*S 
qrt[1 + c*x])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

rule 6337

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_ 
), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcCo 
sh[c*x])   u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/(Sqrt[1 + c*x]*Sqrt[-1 + c 
*x])]   Int[SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b 
, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] 
 && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2536\) vs. \(2(235)=470\).

Time = 0.42 (sec) , antiderivative size = 2537, normalized size of antiderivative = 9.09

method	result	size
default	\(\text {Expression too large to display}\)	\(2537\)
parts	\(\text {Expression too large to display}\)	\(2537\)

Input:

int((-c^2*d*x^2+d)^(1/2)*(a+b*arccosh(c*x))/x^8,x,method=_RETURNVERBOSE)
 

Output:

-120/7*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^ 
2*x^2+225)/(c*x-1)^(1/2)/(c*x+1)^(1/2)*arccosh(c*x)*c^7-128/105*b*(-d*(c^2 
*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^13/( 
c*x-1)/(c*x+1)*c^20+16/105*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x 
^6-21*c^4*x^4-315*c^2*x^2+225)*x^11/(c*x-1)/(c*x+1)*c^18+40/21*b*(-d*(c^2* 
x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^9/(c* 
x-1)/(c*x+1)*c^16+214/105*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^ 
6-21*c^4*x^4-315*c^2*x^2+225)*x^7/(c*x-1)/(c*x+1)*c^14-152/105*b*(-d*(c^2* 
x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^5/(c* 
x-1)/(c*x+1)*c^12-30/7*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-2 
1*c^4*x^4-315*c^2*x^2+225)*x^3/(c*x-1)/(c*x+1)*c^10+20/7*b*(-d*(c^2*x^2-1) 
)^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x/(c*x-1)/(c* 
x+1)*c^8+225/7*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^ 
4-315*c^2*x^2+225)/x^7/(c*x-1)/(c*x+1)*arccosh(c*x)+16/3*b*(-d*(c^2*x^2-1) 
)^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^6/(c*x-1)^( 
1/2)/(c*x+1)^(1/2)*c^13-469/60*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c 
^6*x^6-21*c^4*x^4-315*c^2*x^2+225)*x^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)*c^9+71/ 
28*b*(-d*(c^2*x^2-1))^(1/2)/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^ 
2+225)/x^2/(c*x-1)^(1/2)/(c*x+1)^(1/2)*c^5+255/28*b*(-d*(c^2*x^2-1))^(1/2) 
/(280*c^8*x^8-105*c^6*x^6-21*c^4*x^4-315*c^2*x^2+225)/x^4/(c*x-1)^(1/2)...
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 616, normalized size of antiderivative = 2.21 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=\left [\frac {4 \, {\left (8 \, b c^{8} x^{8} - 4 \, b c^{6} x^{6} - b c^{4} x^{4} - 18 \, b c^{2} x^{2} + 15 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) + 16 \, {\left (b c^{9} x^{9} - b c^{7} x^{7}\right )} \sqrt {-d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} - d x^{4} + \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} {\left (x^{4} - 1\right )} \sqrt {-d} - d}{c^{2} x^{4} - x^{2}}\right ) + {\left (8 \, b c^{5} x^{5} - {\left (8 \, b c^{5} + 3 \, b c^{3} - 10 \, b c\right )} x^{7} + 3 \, b c^{3} x^{3} - 10 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} + 4 \, {\left (8 \, a c^{8} x^{8} - 4 \, a c^{6} x^{6} - a c^{4} x^{4} - 18 \, a c^{2} x^{2} + 15 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{420 \, {\left (c^{2} x^{9} - x^{7}\right )}}, -\frac {32 \, {\left (b c^{9} x^{9} - b c^{7} x^{7}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} {\left (x^{2} - 1\right )} \sqrt {d}}{c^{2} d x^{4} + {\left (c^{2} - 1\right )} d x^{2} - d}\right ) - 4 \, {\left (8 \, b c^{8} x^{8} - 4 \, b c^{6} x^{6} - b c^{4} x^{4} - 18 \, b c^{2} x^{2} + 15 \, b\right )} \sqrt {-c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} - 1}\right ) - {\left (8 \, b c^{5} x^{5} - {\left (8 \, b c^{5} + 3 \, b c^{3} - 10 \, b c\right )} x^{7} + 3 \, b c^{3} x^{3} - 10 \, b c x\right )} \sqrt {-c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} - 1} - 4 \, {\left (8 \, a c^{8} x^{8} - 4 \, a c^{6} x^{6} - a c^{4} x^{4} - 18 \, a c^{2} x^{2} + 15 \, a\right )} \sqrt {-c^{2} d x^{2} + d}}{420 \, {\left (c^{2} x^{9} - x^{7}\right )}}\right ] \] Input:

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arccosh(c*x))/x^8,x, algorithm="fricas 
")
 

Output:

[1/420*(4*(8*b*c^8*x^8 - 4*b*c^6*x^6 - b*c^4*x^4 - 18*b*c^2*x^2 + 15*b)*sq 
rt(-c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 - 1)) + 16*(b*c^9*x^9 - b*c^7*x^ 
7)*sqrt(-d)*log((c^2*d*x^6 + c^2*d*x^2 - d*x^4 + sqrt(-c^2*d*x^2 + d)*sqrt 
(c^2*x^2 - 1)*(x^4 - 1)*sqrt(-d) - d)/(c^2*x^4 - x^2)) + (8*b*c^5*x^5 - (8 
*b*c^5 + 3*b*c^3 - 10*b*c)*x^7 + 3*b*c^3*x^3 - 10*b*c*x)*sqrt(-c^2*d*x^2 + 
 d)*sqrt(c^2*x^2 - 1) + 4*(8*a*c^8*x^8 - 4*a*c^6*x^6 - a*c^4*x^4 - 18*a*c^ 
2*x^2 + 15*a)*sqrt(-c^2*d*x^2 + d))/(c^2*x^9 - x^7), -1/420*(32*(b*c^9*x^9 
 - b*c^7*x^7)*sqrt(d)*arctan(sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1)*(x^2 - 
 1)*sqrt(d)/(c^2*d*x^4 + (c^2 - 1)*d*x^2 - d)) - 4*(8*b*c^8*x^8 - 4*b*c^6* 
x^6 - b*c^4*x^4 - 18*b*c^2*x^2 + 15*b)*sqrt(-c^2*d*x^2 + d)*log(c*x + sqrt 
(c^2*x^2 - 1)) - (8*b*c^5*x^5 - (8*b*c^5 + 3*b*c^3 - 10*b*c)*x^7 + 3*b*c^3 
*x^3 - 10*b*c*x)*sqrt(-c^2*d*x^2 + d)*sqrt(c^2*x^2 - 1) - 4*(8*a*c^8*x^8 - 
 4*a*c^6*x^6 - a*c^4*x^4 - 18*a*c^2*x^2 + 15*a)*sqrt(-c^2*d*x^2 + d))/(c^2 
*x^9 - x^7)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=\text {Timed out} \] Input:

integrate((-c**2*d*x**2+d)**(1/2)*(a+b*acosh(c*x))/x**8,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=-\frac {1}{420} \, {\left (32 \, c^{6} \sqrt {-d} \log \left (x\right ) - \frac {8 \, c^{4} \sqrt {-d} x^{4} + 3 \, c^{2} \sqrt {-d} x^{2} - 10 \, \sqrt {-d}}{x^{6}}\right )} b c - \frac {1}{105} \, {\left (\frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4}}{d x^{3}} + \frac {12 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{5}} + \frac {15 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{7}}\right )} b \operatorname {arcosh}\left (c x\right ) - \frac {1}{105} \, {\left (\frac {8 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4}}{d x^{3}} + \frac {12 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2}}{d x^{5}} + \frac {15 \, {\left (-c^{2} d x^{2} + d\right )}^{\frac {3}{2}}}{d x^{7}}\right )} a \] Input:

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arccosh(c*x))/x^8,x, algorithm="maxima 
")
 

Output:

-1/420*(32*c^6*sqrt(-d)*log(x) - (8*c^4*sqrt(-d)*x^4 + 3*c^2*sqrt(-d)*x^2 
- 10*sqrt(-d))/x^6)*b*c - 1/105*(8*(-c^2*d*x^2 + d)^(3/2)*c^4/(d*x^3) + 12 
*(-c^2*d*x^2 + d)^(3/2)*c^2/(d*x^5) + 15*(-c^2*d*x^2 + d)^(3/2)/(d*x^7))*b 
*arccosh(c*x) - 1/105*(8*(-c^2*d*x^2 + d)^(3/2)*c^4/(d*x^3) + 12*(-c^2*d*x 
^2 + d)^(3/2)*c^2/(d*x^5) + 15*(-c^2*d*x^2 + d)^(3/2)/(d*x^7))*a
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arccosh(c*x))/x^8,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=\int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,\sqrt {d-c^2\,d\,x^2}}{x^8} \,d x \] Input:

int(((a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2))/x^8,x)
 

Output:

int(((a + b*acosh(c*x))*(d - c^2*d*x^2)^(1/2))/x^8, x)
 

Reduce [F]

\[ \int \frac {\sqrt {d-c^2 d x^2} (a+b \text {arccosh}(c x))}{x^8} \, dx=\frac {\sqrt {d}\, \left (8 \sqrt {-c^{2} x^{2}+1}\, a \,c^{6} x^{6}+4 \sqrt {-c^{2} x^{2}+1}\, a \,c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, a \,c^{2} x^{2}-15 \sqrt {-c^{2} x^{2}+1}\, a +105 \left (\int \frac {\sqrt {-c^{2} x^{2}+1}\, \mathit {acosh} \left (c x \right )}{x^{8}}d x \right ) b \,x^{7}\right )}{105 x^{7}} \] Input:

int((-c^2*d*x^2+d)^(1/2)*(a+b*acosh(c*x))/x^8,x)
                                                                                    
                                                                                    
 

Output:

(sqrt(d)*(8*sqrt( - c**2*x**2 + 1)*a*c**6*x**6 + 4*sqrt( - c**2*x**2 + 1)* 
a*c**4*x**4 + 3*sqrt( - c**2*x**2 + 1)*a*c**2*x**2 - 15*sqrt( - c**2*x**2 
+ 1)*a + 105*int((sqrt( - c**2*x**2 + 1)*acosh(c*x))/x**8,x)*b*x**7))/(105 
*x**7)