3.2.50 \(\int \frac {(f x)^m (a+b \text {arccosh}(c x))}{(d-c^2 d x^2)^3} \, dx\) [150]

3.2.50.1 Optimal result

Integrand size = 27, antiderivative size = 27 \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\frac {(f x)^{1+m} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}+\frac {(3-m) (f x)^{1+m} (a+b \text {arccosh}(c x))}{8 d^3 f \left (1-c^2 x^2\right )}-\frac {b c (3-m) (f x)^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{8 d^3 f^2 (2+m) \sqrt {-1+c x} \sqrt {1+c x}}-\frac {b c (f x)^{2+m} \sqrt {1-c^2 x^2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 f^2 (2+m) \sqrt {-1+c x} \sqrt {1+c x}}+\frac {(1-m) (3-m) \text {Int}\left (\frac {(f x)^m (a+b \text {arccosh}(c x))}{d-c^2 d x^2},x\right )}{8 d^2} \]

output

1/4*(f*x)^(1+m)*(a+b*arccosh(c*x))/d^3/f/(-c^2*x^2+1)^2+1/8*(3-m)*(f*x)^(1 
+m)*(a+b*arccosh(c*x))/d^3/f/(-c^2*x^2+1)-1/8*b*c*(3-m)*(f*x)^(2+m)*hyperg 
eom([3/2, 1+1/2*m],[2+1/2*m],c^2*x^2)*(-c^2*x^2+1)^(1/2)/d^3/f^2/(2+m)/(c* 
x-1)^(1/2)/(c*x+1)^(1/2)-1/4*b*c*(f*x)^(2+m)*hypergeom([5/2, 1+1/2*m],[2+1 
/2*m],c^2*x^2)*(-c^2*x^2+1)^(1/2)/d^3/f^2/(2+m)/(c*x-1)^(1/2)/(c*x+1)^(1/2 
)+1/8*(1-m)*(3-m)*Unintegrable((f*x)^m*(a+b*arccosh(c*x))/(-c^2*d*x^2+d),x 
)/d^2
 

3.2.50.2 Mathematica [N/A]

Not integrable

Time = 12.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx \]

input

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

output

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

3.2.50.3 Rubi [N/A]

Not integrable

Time = 0.84 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {6351, 27, 136, 279, 278, 6351, 136, 279, 278, 6375}

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.

\(\displaystyle \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx\)

\(\Big \downarrow \) 6351

\(\displaystyle \frac {(3-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{d^2 \left (1-c^2 x^2\right )^2}dx}{4 d}-\frac {b c \int \frac {(f x)^{m+1}}{(c x-1)^{5/2} (c x+1)^{5/2}}dx}{4 d^3 f}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(3-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (1-c^2 x^2\right )^2}dx}{4 d^3}-\frac {b c \int \frac {(f x)^{m+1}}{(c x-1)^{5/2} (c x+1)^{5/2}}dx}{4 d^3 f}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 136

\(\displaystyle \frac {(3-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (1-c^2 x^2\right )^2}dx}{4 d^3}-\frac {b c \sqrt {c^2 x^2-1} \int \frac {(f x)^{m+1}}{\left (c^2 x^2-1\right )^{5/2}}dx}{4 d^3 f \sqrt {c x-1} \sqrt {c x+1}}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {(3-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (1-c^2 x^2\right )^2}dx}{4 d^3}-\frac {b c \sqrt {1-c^2 x^2} \int \frac {(f x)^{m+1}}{\left (1-c^2 x^2\right )^{5/2}}dx}{4 d^3 f \sqrt {c x-1} \sqrt {c x+1}}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {(3-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (1-c^2 x^2\right )^2}dx}{4 d^3}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{4 d^3 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\)

\(\Big \downarrow \) 6351

\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{1-c^2 x^2}dx+\frac {b c \int \frac {(f x)^{m+1}}{(c x-1)^{3/2} (c x+1)^{3/2}}dx}{2 f}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{2 f \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{4 d^3 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\)

\(\Big \downarrow \) 136

\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{1-c^2 x^2}dx+\frac {b c \sqrt {c^2 x^2-1} \int \frac {(f x)^{m+1}}{\left (c^2 x^2-1\right )^{3/2}}dx}{2 f \sqrt {c x-1} \sqrt {c x+1}}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{2 f \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{4 d^3 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{1-c^2 x^2}dx-\frac {b c \sqrt {1-c^2 x^2} \int \frac {(f x)^{m+1}}{\left (1-c^2 x^2\right )^{3/2}}dx}{2 f \sqrt {c x-1} \sqrt {c x+1}}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{2 f \left (1-c^2 x^2\right )}\right )}{4 d^3}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{4 d^3 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{1-c^2 x^2}dx+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{2 f \left (1-c^2 x^2\right )}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{2 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\right )}{4 d^3}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{4 d^3 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\)

\(\Big \downarrow \) 6375

\(\displaystyle \frac {(3-m) \left (\frac {1}{2} (1-m) \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{1-c^2 x^2}dx+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{2 f \left (1-c^2 x^2\right )}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{2 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\right )}{4 d^3}+\frac {(f x)^{m+1} (a+b \text {arccosh}(c x))}{4 d^3 f \left (1-c^2 x^2\right )^2}-\frac {b c \sqrt {1-c^2 x^2} (f x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},c^2 x^2\right )}{4 d^3 f^2 (m+2) \sqrt {c x-1} \sqrt {c x+1}}\)

input

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

output

$Aborted
 

3.2.50.3.1 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 136

Int[((f_.)*(x_))^(p_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), 
x_] :> Simp[(a + b*x)^FracPart[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^Fr 
acPart[m])   Int[(a*c + b*d*x^2)^m*(f*x)^p, x], x] /; FreeQ[{a, b, c, d, f, 
 m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 6351

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/(2*d*f*(p + 1))), x] + (Simp[(m + 2*p + 3)/(2*d*(p + 1 
))   Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcCosh[c*x])^n, x], x] - Simp[ 
b*c*(n/(2*f*(p + 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*ArcCosh[c*x]) 
^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] & 
& GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] || IntegerQ[p] || 
EqQ[n, 1])
 

rule 6375

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_.)*((d_) + (e 
_.)*(x_)^2)^(p_.), x_Symbol] :> Unintegrable[(f*x)^m*(d + e*x^2)^p*(a + b*A 
rcCosh[c*x])^n, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x]
 

3.2.50.4 Maple [N/A] (verified)

Not integrable

Time = 1.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00

\[\int \frac {\left (f x \right )^{m} \left (a +b \,\operatorname {arccosh}\left (c x \right )\right )}{\left (-c^{2} d \,x^{2}+d \right )^{3}}d x\]

input

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

output

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

3.2.50.5 Fricas [N/A]

Not integrable

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.11 \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]

input

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

output

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

3.2.50.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.50.7 Maxima [N/A]

Not integrable

Time = 0.72 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]

input

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

output

-integrate((b*arccosh(c*x) + a)*(f*x)^m/(c^2*d*x^2 - d)^3, x)
 

3.2.50.8 Giac [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int { -\frac {{\left (b \operatorname {arcosh}\left (c x\right ) + a\right )} \left (f x\right )^{m}}{{\left (c^{2} d x^{2} - d\right )}^{3}} \,d x } \]

input

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

output

integrate(-(b*arccosh(c*x) + a)*(f*x)^m/(c^2*d*x^2 - d)^3, x)
 

3.2.50.9 Mupad [N/A]

Not integrable

Time = 3.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {(f x)^m (a+b \text {arccosh}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx=\int \frac {\left (a+b\,\mathrm {acosh}\left (c\,x\right )\right )\,{\left (f\,x\right )}^m}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]

input

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

output

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