\(\int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx\) [267]

Optimal result

Integrand size = 25, antiderivative size = 184 \[ \int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {1}{4} b^2 x \sqrt {d+c^2 d x^2}-\frac {b^2 \sqrt {d+c^2 d x^2} \text {arcsinh}(c x)}{4 c \sqrt {1+c^2 x^2}}-\frac {b c x^2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{2 \sqrt {1+c^2 x^2}}+\frac {1}{2} x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2+\frac {\sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^3}{6 b c \sqrt {1+c^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.09 \[ \int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {1}{24} \left (12 a^2 x \sqrt {d+c^2 d x^2}+\frac {12 a^2 \sqrt {d} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{c}+\frac {b^2 \sqrt {d+c^2 d x^2} \left (4 \text {arcsinh}(c x)^3-6 \text {arcsinh}(c x) \cosh (2 \text {arcsinh}(c x))+\left (3+6 \text {arcsinh}(c x)^2\right ) \sinh (2 \text {arcsinh}(c x))\right )}{c \sqrt {1+c^2 x^2}}+\frac {6 a b \sqrt {d+c^2 d x^2} (-\cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) (\text {arcsinh}(c x)+\sinh (2 \text {arcsinh}(c x))))}{c \sqrt {1+c^2 x^2}}\right ) \] Input:

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

Output:

(12*a^2*x*Sqrt[d + c^2*d*x^2] + (12*a^2*Sqrt[d]*Log[c*d*x + Sqrt[d]*Sqrt[d 
 + c^2*d*x^2]])/c + (b^2*Sqrt[d + c^2*d*x^2]*(4*ArcSinh[c*x]^3 - 6*ArcSinh 
[c*x]*Cosh[2*ArcSinh[c*x]] + (3 + 6*ArcSinh[c*x]^2)*Sinh[2*ArcSinh[c*x]])) 
/(c*Sqrt[1 + c^2*x^2]) + (6*a*b*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] 
 + 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])))/(c*Sqrt[1 + c^2* 
x^2]))/24
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6200, 6191, 262, 222, 6198}

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.

\(\displaystyle \int \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))^2 \, dx\)

\(\Big \downarrow \) 6200

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

\(\Big \downarrow \) 6191

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 222

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

\(\Big \downarrow \) 6198

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 222

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

rule 262

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

rule 6191

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
 :> Simp[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^n/(d*(m + 1))), x] - Simp[b*c* 
(n/(d*(m + 1)))   Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + 
c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
 

rule 6198

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_ 
Symbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*( 
a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c 
^2*d] && NeQ[n, -1]
 

rule 6200

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_ 
Symbol] :> Simp[x*Sqrt[d + e*x^2]*((a + b*ArcSinh[c*x])^n/2), x] + (Simp[(1 
/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   Int[(a + b*ArcSinh[c*x])^n/Sq 
rt[1 + c^2*x^2], x], x] - Simp[b*c*(n/2)*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2* 
x^2]]   Int[x*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e 
}, x] && EqQ[e, c^2*d] && GtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(479\) vs. \(2(158)=316\).

Time = 0.74 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.61

method	result	size
default	\(\frac {x \sqrt {c^{2} d \,x^{2}+d}\, a^{2}}{2}+\frac {a^{2} d \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 \sqrt {c^{2} d}}+b^{2} \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )^{3}}{6 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}+2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c +\sqrt {c^{2} x^{2}+1}\right ) \left (2 \operatorname {arcsinh}\left (x c \right )^{2}-2 \,\operatorname {arcsinh}\left (x c \right )+1\right )}{16 \left (c^{2} x^{2}+1\right ) c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}-2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c -\sqrt {c^{2} x^{2}+1}\right ) \left (2 \operatorname {arcsinh}\left (x c \right )^{2}+2 \,\operatorname {arcsinh}\left (x c \right )+1\right )}{16 \left (c^{2} x^{2}+1\right ) c}\right )+2 a b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )^{2}}{4 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}+2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \,\operatorname {arcsinh}\left (x c \right )\right )}{16 \left (c^{2} x^{2}+1\right ) c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}-2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \,\operatorname {arcsinh}\left (x c \right )\right )}{16 \left (c^{2} x^{2}+1\right ) c}\right )\)	\(480\)
parts	\(\frac {x \sqrt {c^{2} d \,x^{2}+d}\, a^{2}}{2}+\frac {a^{2} d \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 \sqrt {c^{2} d}}+b^{2} \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )^{3}}{6 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}+2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c +\sqrt {c^{2} x^{2}+1}\right ) \left (2 \operatorname {arcsinh}\left (x c \right )^{2}-2 \,\operatorname {arcsinh}\left (x c \right )+1\right )}{16 \left (c^{2} x^{2}+1\right ) c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}-2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c -\sqrt {c^{2} x^{2}+1}\right ) \left (2 \operatorname {arcsinh}\left (x c \right )^{2}+2 \,\operatorname {arcsinh}\left (x c \right )+1\right )}{16 \left (c^{2} x^{2}+1\right ) c}\right )+2 a b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (x c \right )^{2}}{4 \sqrt {c^{2} x^{2}+1}\, c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}+2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c +\sqrt {c^{2} x^{2}+1}\right ) \left (-1+2 \,\operatorname {arcsinh}\left (x c \right )\right )}{16 \left (c^{2} x^{2}+1\right ) c}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 x^{3} c^{3}-2 x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+2 x c -\sqrt {c^{2} x^{2}+1}\right ) \left (1+2 \,\operatorname {arcsinh}\left (x c \right )\right )}{16 \left (c^{2} x^{2}+1\right ) c}\right )\)	\(480\)

Input:

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

Output:

1/2*x*(c^2*d*x^2+d)^(1/2)*a^2+1/2*a^2*d*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^ 
2+d)^(1/2))/(c^2*d)^(1/2)+b^2*(1/6*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2) 
/c*arcsinh(x*c)^3+1/16*(d*(c^2*x^2+1))^(1/2)*(2*x^3*c^3+2*x^2*c^2*(c^2*x^2 
+1)^(1/2)+2*x*c+(c^2*x^2+1)^(1/2))*(2*arcsinh(x*c)^2-2*arcsinh(x*c)+1)/(c^ 
2*x^2+1)/c+1/16*(d*(c^2*x^2+1))^(1/2)*(2*x^3*c^3-2*x^2*c^2*(c^2*x^2+1)^(1/ 
2)+2*x*c-(c^2*x^2+1)^(1/2))*(2*arcsinh(x*c)^2+2*arcsinh(x*c)+1)/(c^2*x^2+1 
)/c)+2*a*b*(1/4*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*arcsinh(x*c)^2+1 
/16*(d*(c^2*x^2+1))^(1/2)*(2*x^3*c^3+2*x^2*c^2*(c^2*x^2+1)^(1/2)+2*x*c+(c^ 
2*x^2+1)^(1/2))*(-1+2*arcsinh(x*c))/(c^2*x^2+1)/c+1/16*(d*(c^2*x^2+1))^(1/ 
2)*(2*x^3*c^3-2*x^2*c^2*(c^2*x^2+1)^(1/2)+2*x*c-(c^2*x^2+1)^(1/2))*(1+2*ar 
csinh(x*c))/(c^2*x^2+1)/c)
 

Fricas [F]

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

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

Output:

integral(sqrt(c^2*d*x^2 + d)*(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^ 
2), x)
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate((c**2*d*x**2+d)**(1/2)*(a+b*asinh(c*x))**2,x)
 

Output:

Integral(sqrt(d*(c**2*x**2 + 1))*(a + b*asinh(c*x))**2, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [F(-2)]

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

integrate((c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(c*x))^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. \[ \int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx=\int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {d\,c^2\,x^2+d} \,d x \] Input:

int((a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(1/2),x)
 

Output:

int((a + b*asinh(c*x))^2*(d + c^2*d*x^2)^(1/2), x)
 

Reduce [F]

\[ \int \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2 \, dx=\frac {\sqrt {d}\, \left (\sqrt {c^{2} x^{2}+1}\, a^{2} c x +4 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right )d x \right ) a b c +2 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right )^{2}d x \right ) b^{2} c +\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right ) a^{2}\right )}{2 c} \] Input:

int((c^2*d*x^2+d)^(1/2)*(a+b*asinh(c*x))^2,x)
 

Output:

(sqrt(d)*(sqrt(c**2*x**2 + 1)*a**2*c*x + 4*int(sqrt(c**2*x**2 + 1)*asinh(c 
*x),x)*a*b*c + 2*int(sqrt(c**2*x**2 + 1)*asinh(c*x)**2,x)*b**2*c + log(sqr 
t(c**2*x**2 + 1) + c*x)*a**2))/(2*c)