\(\int \frac {x^3 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [176]

Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {b x}{6 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \text {arcsinh}(c x)}{3 c^4 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{c^4 d^2 \sqrt {d+c^2 d x^2}}+\frac {5 b \sqrt {1+c^2 x^2} \arctan (c x)}{6 c^4 d^2 \sqrt {d+c^2 d x^2}} \] Output:

-1/6*b*x/c^3/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)+1/3*(a+b*arcsinh(c* 
x))/c^4/d/(c^2*d*x^2+d)^(3/2)-(a+b*arcsinh(c*x))/c^4/d^2/(c^2*d*x^2+d)^(1/ 
2)+5/6*b*(c^2*x^2+1)^(1/2)*arctan(c*x)/c^4/d^2/(c^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d+c^2 d x^2} \left (b c x+b c^3 x^3+4 a \sqrt {1+c^2 x^2}+6 a c^2 x^2 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (2+3 c^2 x^2\right ) \text {arcsinh}(c x)-5 b \left (1+c^2 x^2\right )^2 \arctan (c x)\right )}{6 c^4 d^3 \left (1+c^2 x^2\right )^{5/2}} \] Input:

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

Output:

-1/6*(Sqrt[d + c^2*d*x^2]*(b*c*x + b*c^3*x^3 + 4*a*Sqrt[1 + c^2*x^2] + 6*a 
*c^2*x^2*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(2 + 3*c^2*x^2)*ArcSinh 
[c*x] - 5*b*(1 + c^2*x^2)^2*ArcTan[c*x]))/(c^4*d^3*(1 + c^2*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6219, 27, 298, 216}

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.

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

\(\Big \downarrow \) 6219

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(a + b*ArcSinh[c*x])/(3*c^4*d*(d + c^2*d*x^2)^(3/2)) - (a + b*ArcSinh[c*x] 
)/(c^4*d^2*Sqrt[d + c^2*d*x^2]) + (b*Sqrt[d + c^2*d*x^2]*(-1/2*x/(1 + c^2* 
x^2) + (5*ArcTan[c*x])/(2*c)))/(3*c^3*d^3*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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

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 = 0.89 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.83

method	result	size
default	\(a \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right ) x^{2}}{\left (c^{2} x^{2}+1\right )^{2} d^{3} c^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c^{3}}-\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{4}}+\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{4}}-\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{4}}\)	\(263\)
parts	\(a \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right ) x^{2}}{\left (c^{2} x^{2}+1\right )^{2} d^{3} c^{2}}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} d^{3} c^{3}}-\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )}{3 \left (c^{2} x^{2}+1\right )^{2} d^{3} c^{4}}+\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{4}}-\frac {5 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (x c +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, d^{3} c^{4}}\)	\(263\)

Input:

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

Output:

a*(-x^2/c^2/d/(c^2*d*x^2+d)^(3/2)-2/3/d/c^4/(c^2*d*x^2+d)^(3/2))-b*(d*(c^2 
*x^2+1))^(1/2)/(c^2*x^2+1)^2/d^3/c^2*arcsinh(x*c)*x^2-1/6*b*(d*(c^2*x^2+1) 
)^(1/2)/(c^2*x^2+1)^(3/2)/d^3/c^3*x-2/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1 
)^2/d^3/c^4*arcsinh(x*c)+5/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d 
^3/c^4*ln(x*c+(c^2*x^2+1)^(1/2)+I)-5/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+ 
1)^(1/2)/d^3/c^4*ln(x*c+(c^2*x^2+1)^(1/2)-I)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.31 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {5 \, {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) + 4 \, {\left (3 \, b c^{2} x^{2} + 2 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (6 \, a c^{2} x^{2} + \sqrt {c^{2} x^{2} + 1} b c x + 4 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{12 \, {\left (c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \] Input:

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

Output:

-1/12*(5*(b*c^4*x^4 + 2*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)) + 4*(3*b*c^2*x^2 + 2*b)*s 
qrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(6*a*c^2*x^2 + sqrt(c^ 
2*x^2 + 1)*b*c*x + 4*a)*sqrt(c^2*d*x^2 + d))/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 
+ c^4*d^3)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.96 \[ \int \frac {x^3 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {x}{c^{6} d^{\frac {5}{2}} x^{2} + c^{4} d^{\frac {5}{2}}} - \frac {5 \, \arctan \left (c x\right )}{c^{5} d^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} + \frac {2}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {3 \, x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d} + \frac {2}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{4} d}\right )} \] Input:

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

Output:

-1/6*b*c*(x/(c^6*d^(5/2)*x^2 + c^4*d^(5/2)) - 5*arctan(c*x)/(c^5*d^(5/2))) 
 - 1/3*b*(3*x^2/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 2/((c^2*d*x^2 + d)^(3/2)*c 
^4*d))*arcsinh(c*x) - 1/3*a*(3*x^2/((c^2*d*x^2 + d)^(3/2)*c^2*d) + 2/((c^2 
*d*x^2 + d)^(3/2)*c^4*d))
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 3*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - 2*sqrt(c**2*x**2 + 1)*a + 3*int((a 
sinh(c*x)*x**3)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c**2*x**2 + 1)*c** 
2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**8*x**4 + 6*int((asinh(c*x)*x**3)/(sq 
rt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2* 
x**2 + 1)),x)*b*c**6*x**2 + 3*int((asinh(c*x)*x**3)/(sqrt(c**2*x**2 + 1)*c 
**4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c** 
4)/(3*sqrt(d)*c**4*d**2*(c**4*x**4 + 2*c**2*x**2 + 1))