\(\int \frac {(d+c^2 d x^2)^{3/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx\) [142]

Optimal result

Integrand size = 26, antiderivative size = 177 \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=-\frac {b c^3 d x^2 \sqrt {d+c^2 d x^2}}{4 \sqrt {1+c^2 x^2}}+\frac {3}{2} c^2 d x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))-\frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x}+\frac {3 c d \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{4 b \sqrt {1+c^2 x^2}}+\frac {b c d \sqrt {d+c^2 d x^2} \log (x)}{\sqrt {1+c^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.13 \[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {1}{8} \left (\frac {4 a d \left (-2+c^2 x^2\right ) \sqrt {d+c^2 d x^2}}{x}+\frac {4 b d \sqrt {d+c^2 d x^2} \left (-2 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+c x \text {arcsinh}(c x)^2+2 c x \log (c x)\right )}{x \sqrt {1+c^2 x^2}}+12 a c d^{3/2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+\frac {b c d \sqrt {d+c^2 d x^2} (-\cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) (\text {arcsinh}(c x)+\sinh (2 \text {arcsinh}(c x))))}{\sqrt {1+c^2 x^2}}\right ) \] Input:

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

Output:

((4*a*d*(-2 + c^2*x^2)*Sqrt[d + c^2*d*x^2])/x + (4*b*d*Sqrt[d + c^2*d*x^2] 
*(-2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + c*x*ArcSinh[c*x]^2 + 2*c*x*Log[c*x]) 
)/(x*Sqrt[1 + c^2*x^2]) + 12*a*c*d^(3/2)*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2* 
d*x^2]] + (b*c*d*Sqrt[d + c^2*d*x^2]*(-Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c* 
x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])))/Sqrt[1 + c^2*x^2])/8
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6222, 244, 2009, 6200, 15, 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.

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

\(\Big \downarrow \) 6222

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6200

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 6198

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

Input:

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

Output:

-(((d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x) + 3*c^2*d*(-1/4*(b*c*x^2 
*Sqrt[d + c^2*d*x^2])/Sqrt[1 + c^2*x^2] + (x*Sqrt[d + c^2*d*x^2]*(a + b*Ar 
cSinh[c*x]))/2 + (Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b*c*Sqrt[ 
1 + c^2*x^2])) + (b*c*d*Sqrt[d + c^2*d*x^2]*((c^2*x^2)/2 + Log[x]))/Sqrt[1 
 + c^2*x^2]
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

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]
 

rule 6222

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^p*((a + b*Arc 
Sinh[c*x])^n/(f*(m + 1))), x] + (-Simp[2*e*(p/(f^2*(m + 1)))   Int[(f*x)^(m 
 + 2)*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Simp[b*c*(n/(f*( 
m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 + c^2*x 
^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e 
, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.30

method	result	size
default	\(-\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{d x}+a \,c^{2} x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}+\frac {3 \sqrt {c^{2} d \,x^{2}+d}\, a \,c^{2} d x}{2}+\frac {3 a \,c^{2} d^{2} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) x^{2} c^{2}-2 x^{3} c^{3}+6 \operatorname {arcsinh}\left (x c \right )^{2} x c -8 x c \,\operatorname {arcsinh}\left (x c \right )+8 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x c -8 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}-x c \right ) d}{8 \sqrt {c^{2} x^{2}+1}\, x}\)	\(230\)
parts	\(-\frac {a \left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}{d x}+a \,c^{2} x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}+\frac {3 \sqrt {c^{2} d \,x^{2}+d}\, a \,c^{2} d x}{2}+\frac {3 a \,c^{2} d^{2} \ln \left (\frac {x \,c^{2} d}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 \sqrt {c^{2} d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) x^{2} c^{2}-2 x^{3} c^{3}+6 \operatorname {arcsinh}\left (x c \right )^{2} x c -8 x c \,\operatorname {arcsinh}\left (x c \right )+8 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x c -8 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}-x c \right ) d}{8 \sqrt {c^{2} x^{2}+1}\, x}\)	\(230\)

Input:

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

Output:

-a/d/x*(c^2*d*x^2+d)^(5/2)+a*c^2*x*(c^2*d*x^2+d)^(3/2)+3/2*(c^2*d*x^2+d)^( 
1/2)*a*c^2*d*x+3/2*a*c^2*d^2*ln(x*c^2*d/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2)) 
/(c^2*d)^(1/2)+1/8*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/x*(4*(c^2*x^2 
+1)^(1/2)*arcsinh(x*c)*x^2*c^2-2*x^3*c^3+6*arcsinh(x*c)^2*x*c-8*x*c*arcsin 
h(x*c)+8*ln((x*c+(c^2*x^2+1)^(1/2))^2-1)*x*c-8*arcsinh(x*c)*(c^2*x^2+1)^(1 
/2)-x*c)*d
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{x^2} \, dx=\frac {\sqrt {d}\, d \left (4 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-8 \sqrt {c^{2} x^{2}+1}\, a +8 \left (\int \frac {\sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right )}{x^{2}}d x \right ) b x +8 \left (\int \sqrt {c^{2} x^{2}+1}\, \mathit {asinh} \left (c x \right )d x \right ) b \,c^{2} x +12 \,\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right ) a c x -9 a c x \right )}{8 x} \] Input:

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

Output:

(sqrt(d)*d*(4*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - 8*sqrt(c**2*x**2 + 1)*a + 
8*int((sqrt(c**2*x**2 + 1)*asinh(c*x))/x**2,x)*b*x + 8*int(sqrt(c**2*x**2 
+ 1)*asinh(c*x),x)*b*c**2*x + 12*log(sqrt(c**2*x**2 + 1) + c*x)*a*c*x - 9* 
a*c*x))/(8*x)