\(\int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {d+c^2 d x^2}} \, dx\) [162]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {d+c^2 d x^2}} \, dx=-\frac {b c \sqrt {1+c^2 x^2}}{6 x^2 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{3 d x}-\frac {2 b c^3 \sqrt {1+c^2 x^2} \log (x)}{3 \sqrt {d+c^2 d x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {d+c^2 d x^2}} \, dx=\frac {\sqrt {d+c^2 d x^2} \left (-b c x+6 b c^3 x^3-2 a \sqrt {1+c^2 x^2}+4 a c^2 x^2 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (-1+2 c^2 x^2\right ) \text {arcsinh}(c x)-4 b c^3 x^3 \log (x)\right )}{6 d x^3 \sqrt {1+c^2 x^2}} \] Input:

Integrate[(a + b*ArcSinh[c*x])/(x^4*Sqrt[d + c^2*d*x^2]),x]
 

Output:

(Sqrt[d + c^2*d*x^2]*(-(b*c*x) + 6*b*c^3*x^3 - 2*a*Sqrt[1 + c^2*x^2] + 4*a 
*c^2*x^2*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 + c^2*x^2]*(-1 + 2*c^2*x^2)*ArcSin 
h[c*x] - 4*b*c^3*x^3*Log[x]))/(6*d*x^3*Sqrt[1 + c^2*x^2])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6224, 15, 6215, 14}

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.

\(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {c^2 d x^2+d}} \, dx\)

\(\Big \downarrow \) 6224

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 6215

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

\(\Big \downarrow \) 14

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

Input:

Int[(a + b*ArcSinh[c*x])/(x^4*Sqrt[d + c^2*d*x^2]),x]
 

Output:

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

Defintions of rubi rules used

rule 14

Int[(a_.)/(x_), x_Symbol] :> Simp[a*Log[x], x] /; FreeQ[a, x]
 

rule 15

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

rule 6215

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 + 1)*((a + 
b*ArcSinh[c*x])^n/(d*f*(m + 1))), 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, m, p}, x] && EqQ 
[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 6224

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 + 1)*((a + 
b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] + (-Simp[c^2*((m + 2*p + 3)/(f^2*(m + 
1)))   Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Sim 
p[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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[m, -1]
 

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.18

method	result	size
default	\(a \left (-\frac {\sqrt {c^{2} d \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 c^{2} \sqrt {c^{2} d \,x^{2}+d}}{3 d x}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 \,\operatorname {arcsinh}\left (x c \right ) x^{3} c^{3}-4 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x^{3} c^{3}+4 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) x^{2} c^{2}-2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}-x c \right )}{6 \sqrt {c^{2} x^{2}+1}\, d \,x^{3}}\)	\(166\)
parts	\(a \left (-\frac {\sqrt {c^{2} d \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 c^{2} \sqrt {c^{2} d \,x^{2}+d}}{3 d x}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (4 \,\operatorname {arcsinh}\left (x c \right ) x^{3} c^{3}-4 \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right ) x^{3} c^{3}+4 \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) x^{2} c^{2}-2 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}-x c \right )}{6 \sqrt {c^{2} x^{2}+1}\, d \,x^{3}}\)	\(166\)

Input:

int((a+b*arcsinh(x*c))/x^4/(c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.57 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {d+c^2 d x^2}} \, dx=\frac {2 \, {\left (2 \, b c^{4} x^{4} + b c^{2} x^{2} - b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (b c^{5} x^{5} + b c^{3} x^{3}\right )} \sqrt {d} \log \left (\frac {c^{2} d x^{6} + c^{2} d x^{2} + d x^{4} - \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} {\left (x^{4} - 1\right )} \sqrt {d} + d}{c^{2} x^{4} + x^{2}}\right ) + {\left (4 \, a c^{4} x^{4} + 2 \, a c^{2} x^{2} + {\left (b c x^{3} - b c x\right )} \sqrt {c^{2} x^{2} + 1} - 2 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{6 \, {\left (c^{2} d x^{5} + d x^{3}\right )}} \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

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

integrate((a+b*asinh(c*x))/x**4/(c**2*d*x**2+d)**(1/2),x)
 

Output:

Integral((a + b*asinh(c*x))/(x**4*sqrt(d*(c**2*x**2 + 1))), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {d+c^2 d x^2}} \, dx=-\frac {1}{6} \, {\left (\frac {4 \, c^{2} \log \left (x\right )}{\sqrt {d}} + \frac {1}{\sqrt {d} x^{2}}\right )} b c + \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} c^{2}}{d x} - \frac {\sqrt {c^{2} d x^{2} + d}}{d x^{3}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} c^{2}}{d x} - \frac {\sqrt {c^{2} d x^{2} + d}}{d x^{3}}\right )} \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima" 
)
 

Output:

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

Giac [F]

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

integrate((a+b*arcsinh(c*x))/x^4/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")
 

Output:

integrate((b*arcsinh(c*x) + a)/(sqrt(c^2*d*x^2 + d)*x^4), x)
 

Mupad [F(-1)]

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

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^(1/2)),x)
 

Output:

int((a + b*asinh(c*x))/(x^4*(d + c^2*d*x^2)^(1/2)), x)
 

Reduce [F]

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

int((a+b*asinh(c*x))/x^4/(c^2*d*x^2+d)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(2*sqrt(c**2*x**2 + 1)*a*c**2*x**2 - sqrt(c**2*x**2 + 1)*a + 3*int(asinh(c 
*x)/(sqrt(c**2*x**2 + 1)*x**4),x)*b*x**3 - 2*a*c**3*x**3)/(3*sqrt(d)*x**3)