\(\int \frac {x^5 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [163]

Optimal result

Integrand size = 26, antiderivative size = 212 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {5 b x \sqrt {1+c^2 x^2}}{3 c^5 d \sqrt {d+c^2 d x^2}}-\frac {b x^3 \sqrt {1+c^2 x^2}}{9 c^3 d \sqrt {d+c^2 d x^2}}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}-\frac {2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{c^6 d^2}+\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}+\frac {b \sqrt {1+c^2 x^2} \arctan (c x)}{c^6 d \sqrt {d+c^2 d x^2}} \] Output:

5/3*b*x*(c^2*x^2+1)^(1/2)/c^5/d/(c^2*d*x^2+d)^(1/2)-1/9*b*x^3*(c^2*x^2+1)^ 
(1/2)/c^3/d/(c^2*d*x^2+d)^(1/2)-(a+b*arcsinh(c*x))/c^6/d/(c^2*d*x^2+d)^(1/ 
2)-2*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(c*x))/c^6/d^2+1/3*(c^2*d*x^2+d)^(3/2 
)*(a+b*arcsinh(c*x))/c^6/d^3+b*(c^2*x^2+1)^(1/2)*arctan(c*x)/c^6/d/(c^2*d* 
x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (15 b c x+14 b c^3 x^3-b c^5 x^5-24 a \sqrt {1+c^2 x^2}-12 a c^2 x^2 \sqrt {1+c^2 x^2}+3 a c^4 x^4 \sqrt {1+c^2 x^2}+3 b \sqrt {1+c^2 x^2} \left (-8-4 c^2 x^2+c^4 x^4\right ) \text {arcsinh}(c x)+9 \left (b+b c^2 x^2\right ) \arctan (c x)\right )}{9 c^6 d^2 \left (1+c^2 x^2\right )^{3/2}} \] Input:

Integrate[(x^5*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]
 

Output:

(Sqrt[d + c^2*d*x^2]*(15*b*c*x + 14*b*c^3*x^3 - b*c^5*x^5 - 24*a*Sqrt[1 + 
c^2*x^2] - 12*a*c^2*x^2*Sqrt[1 + c^2*x^2] + 3*a*c^4*x^4*Sqrt[1 + c^2*x^2] 
+ 3*b*Sqrt[1 + c^2*x^2]*(-8 - 4*c^2*x^2 + c^4*x^4)*ArcSinh[c*x] + 9*(b + b 
*c^2*x^2)*ArcTan[c*x]))/(9*c^6*d^2*(1 + c^2*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.73, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6219, 27, 1467, 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.

\(\displaystyle \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (c^2 d x^2+d\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6219

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1467

\(\displaystyle \frac {b \sqrt {c^2 d x^2+d} \int \left (-c^2 x^2+\frac {3}{c^2 x^2+1}+5\right )dx}{3 c^5 d^2 \sqrt {c^2 x^2+1}}+\frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}-\frac {2 \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^2}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {c^2 d x^2+d}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c^6 d^3}-\frac {2 \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{c^6 d^2}-\frac {a+b \text {arcsinh}(c x)}{c^6 d \sqrt {c^2 d x^2+d}}+\frac {b \left (\frac {3 \arctan (c x)}{c}-\frac {1}{3} c^2 x^3+5 x\right ) \sqrt {c^2 d x^2+d}}{3 c^5 d^2 \sqrt {c^2 x^2+1}}\)

Input:

Int[(x^5*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]
 

Output:

-((a + b*ArcSinh[c*x])/(c^6*d*Sqrt[d + c^2*d*x^2])) - (2*Sqrt[d + c^2*d*x^ 
2]*(a + b*ArcSinh[c*x]))/(c^6*d^2) + ((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh 
[c*x]))/(3*c^6*d^3) + (b*Sqrt[d + c^2*d*x^2]*(5*x - (c^2*x^3)/3 + (3*ArcTa 
n[c*x])/c))/(3*c^5*d^2*Sqrt[1 + c^2*x^2])
 

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 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

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

rule 6219

Int[((a_.) + ArcSinh[(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*ArcSi 
nh[c*x])   u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   Int[S 
implifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x 
] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1) 
/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.04 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.12

method	result	size
default	\(a \left (\frac {x^{4}}{3 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {4 \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )}{3 c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (3 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-\sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}-12 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}+9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )-9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )+15 \sqrt {c^{2} x^{2}+1}\, x c -24 \,\operatorname {arcsinh}\left (x c \right )\right )}{9 d^{2} c^{6} \left (c^{2} x^{2}+1\right )}\)	\(238\)
parts	\(a \left (\frac {x^{4}}{3 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {4 \left (\frac {x^{2}}{c^{2} d \sqrt {c^{2} d \,x^{2}+d}}+\frac {2}{d \,c^{4} \sqrt {c^{2} d \,x^{2}+d}}\right )}{3 c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (3 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-\sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}-12 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}+9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )-9 i \sqrt {c^{2} x^{2}+1}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )+15 \sqrt {c^{2} x^{2}+1}\, x c -24 \,\operatorname {arcsinh}\left (x c \right )\right )}{9 d^{2} c^{6} \left (c^{2} x^{2}+1\right )}\)	\(238\)

Input:

int(x^5*(a+b*arcsinh(x*c))/(c^2*d*x^2+d)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

a*(1/3*x^4/c^2/d/(c^2*d*x^2+d)^(1/2)-4/3/c^2*(x^2/c^2/d/(c^2*d*x^2+d)^(1/2 
)+2/d/c^4/(c^2*d*x^2+d)^(1/2)))+1/9*b*(d*(c^2*x^2+1))^(1/2)*(3*arcsinh(x*c 
)*c^4*x^4-(c^2*x^2+1)^(1/2)*c^3*x^3-12*arcsinh(x*c)*c^2*x^2+9*I*(c^2*x^2+1 
)^(1/2)*ln(x*c+(c^2*x^2+1)^(1/2)+I)-9*I*(c^2*x^2+1)^(1/2)*ln(x*c+(c^2*x^2+ 
1)^(1/2)-I)+15*(c^2*x^2+1)^(1/2)*x*c-24*arcsinh(x*c))/d^2/c^6/(c^2*x^2+1)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.93 \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=-\frac {9 \, {\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) - 6 \, {\left (b c^{4} x^{4} - 4 \, b c^{2} x^{2} - 8 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (3 \, a c^{4} x^{4} - 12 \, a c^{2} x^{2} - {\left (b c^{3} x^{3} - 15 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 24 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{18 \, {\left (c^{8} d^{2} x^{2} + c^{6} d^{2}\right )}} \] Input:

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="fricas" 
)
 

Output:

-1/18*(9*(b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(c^2*d*x^2 + d)*sqrt(c^2*x^2 
 + 1)*c*sqrt(d)*x/(c^4*d*x^4 - d)) - 6*(b*c^4*x^4 - 4*b*c^2*x^2 - 8*b)*sqr 
t(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(3*a*c^4*x^4 - 12*a*c^2* 
x^2 - (b*c^3*x^3 - 15*b*c*x)*sqrt(c^2*x^2 + 1) - 24*a)*sqrt(c^2*d*x^2 + d) 
)/(c^8*d^2*x^2 + c^6*d^2)
 

Sympy [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^{5} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**5*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(3/2),x)
 

Output:

Integral(x**5*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{5}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="maxima" 
)
 

Output:

1/3*a*(x^4/(sqrt(c^2*d*x^2 + d)*c^2*d) - 4*x^2/(sqrt(c^2*d*x^2 + d)*c^4*d) 
 - 8/(sqrt(c^2*d*x^2 + d)*c^6*d)) + 1/3*b*((c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d 
)*x^2 - 8*sqrt(d))*log(c*x + sqrt(c^2*x^2 + 1))/(sqrt(c^2*x^2 + 1)*c^6*d^2 
) - integrate((c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))/(sqrt(c^2* 
x^2 + 1)*x), x)/(c^6*d^2) + 3*integrate(1/3*(c^4*sqrt(d)*x^4 - 4*c^2*sqrt( 
d)*x^2 - 8*sqrt(d))/(c^9*d^2*x^4 + c^7*d^2*x^2 + (c^8*d^2*x^3 + c^6*d^2*x) 
*sqrt(c^2*x^2 + 1)), x))
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),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. \[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\int \frac {x^5\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,d x \] Input:

int((x^5*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2),x)
 

Output:

int((x^5*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^5 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}-4 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-8 \sqrt {c^{2} x^{2}+1}\, a +3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{8} x^{2}+3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{5}}{\sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{6}}{3 \sqrt {d}\, c^{6} d \left (c^{2} x^{2}+1\right )} \] Input:

int(x^5*(a+b*asinh(c*x))/(c^2*d*x^2+d)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(c**2*x**2 + 1)*a*c**4*x**4 - 4*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - 8*s 
qrt(c**2*x**2 + 1)*a + 3*int((asinh(c*x)*x**5)/(sqrt(c**2*x**2 + 1)*c**2*x 
**2 + sqrt(c**2*x**2 + 1)),x)*b*c**8*x**2 + 3*int((asinh(c*x)*x**5)/(sqrt( 
c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**6)/(3*sqrt(d)*c**6 
*d*(c**2*x**2 + 1))