\(\int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx\) [8]

Optimal result

Integrand size = 12, antiderivative size = 132 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b}{2 x^2}+\frac {a b (1+a (a+b x)) \sqrt {1+(a+b x)^2}}{2 \left (1+a^2\right )^2 x^2}-\frac {\left (1+(a+b x)^2\right )^{3/2}}{3 \left (1+a^2\right ) x^3}+\frac {a b^3 \text {arctanh}\left (\frac {1+a (a+b x)}{\sqrt {1+a^2} \sqrt {1+(a+b x)^2}}\right )}{2 \left (1+a^2\right )^{5/2}} \] Output:

-1/3*a/x^3-1/2*b/x^2+1/2*a*b*(1+a*(b*x+a))*(1+(b*x+a)^2)^(1/2)/(a^2+1)^2/x 
^2-1/3*(1+(b*x+a)^2)^(3/2)/(a^2+1)/x^3+1/2*a*b^3*arctanh((1+a*(b*x+a))/(a^ 
2+1)^(1/2)/(1+(b*x+a)^2)^(1/2))/(a^2+1)^(5/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx=\frac {1}{6} \left (-\frac {2 a}{x^3}-\frac {3 b}{x^2}-\frac {\sqrt {1+a^2+2 a b x+b^2 x^2} \left (2+2 a^4+a b x+a^3 b x+2 b^2 x^2+a^2 \left (4-b^2 x^2\right )\right )}{\left (1+a^2\right )^2 x^3}-\frac {3 a b^3 \log (x)}{\left (1+a^2\right )^{5/2}}+\frac {3 a b^3 \log \left (1+a^2+a b x+\sqrt {1+a^2} \sqrt {1+a^2+2 a b x+b^2 x^2}\right )}{\left (1+a^2\right )^{5/2}}\right ) \] Input:

Integrate[E^ArcSinh[a + b*x]/x^4,x]
 

Output:

((-2*a)/x^3 - (3*b)/x^2 - (Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(2 + 2*a^4 + 
a*b*x + a^3*b*x + 2*b^2*x^2 + a^2*(4 - b^2*x^2)))/((1 + a^2)^2*x^3) - (3*a 
*b^3*Log[x])/(1 + a^2)^(5/2) + (3*a*b^3*Log[1 + a^2 + a*b*x + Sqrt[1 + a^2 
]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]])/(1 + a^2)^(5/2))/6
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6293, 2010, 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.

\(\displaystyle \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx\)

\(\Big \downarrow \) 6293

\(\displaystyle \int \frac {\sqrt {(a+b x)^2+1}+a+b x}{x^4}dx\)

\(\Big \downarrow \) 2010

\(\displaystyle \int \left (\frac {\sqrt {a^2+2 a b x+b^2 x^2+1}}{x^4}+\frac {a}{x^4}+\frac {b}{x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a b^3 \text {arctanh}\left (\frac {a^2+a b x+1}{\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{5/2}}+\frac {a b \left (a^2+a b x+1\right ) \sqrt {a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right )^2 x^2}-\frac {\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{3 \left (a^2+1\right ) x^3}-\frac {a}{3 x^3}-\frac {b}{2 x^2}\)

Input:

Int[E^ArcSinh[a + b*x]/x^4,x]
 

Output:

-1/3*a/x^3 - b/(2*x^2) + (a*b*(1 + a^2 + a*b*x)*Sqrt[1 + a^2 + 2*a*b*x + b 
^2*x^2])/(2*(1 + a^2)^2*x^2) - (1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(3*(1 + 
 a^2)*x^3) + (a*b^3*ArcTanh[(1 + a^2 + a*b*x)/(Sqrt[1 + a^2]*Sqrt[1 + a^2 
+ 2*a*b*x + b^2*x^2])])/(2*(1 + a^2)^(5/2))
 

Defintions of rubi rules used

rule 2009

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

rule 2010

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] 
, x] /; FreeQ[{c, m}, x] && SumQ[u] &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) 
+ (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
 

rule 6293

Int[E^(ArcSinh[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(u + Sqrt[1 + u^ 
2])^n, x] /; RationalQ[m] && IntegerQ[n] && PolyQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(501\) vs. \(2(112)=224\).

Time = 0.19 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.80

method	result	size
default	\(-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{3 \left (a^{2}+1\right ) x^{3}}-\frac {a b \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{2 \left (a^{2}+1\right ) x^{2}}-\frac {a b \left (-\frac {\left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{\left (a^{2}+1\right ) x}+\frac {a b \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{a^{2}+1}+\frac {2 b^{2} \left (\frac {\left (2 b^{2} x +2 a b \right ) \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{4 b^{2}}+\frac {\left (4 b^{2} \left (a^{2}+1\right )-4 a^{2} b^{2}\right ) \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{2} \sqrt {b^{2}}}\right )}{a^{2}+1}\right )}{2 \left (a^{2}+1\right )}+\frac {b^{2} \left (\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+\frac {a b \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{\sqrt {b^{2}}}-\sqrt {a^{2}+1}\, \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{x}\right )\right )}{2 a^{2}+2}\right )}{a^{2}+1}-\frac {b}{2 x^{2}}-\frac {a}{3 x^{3}}\)	\(502\)

Input:

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

Output:

-1/3/(a^2+1)/x^3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-a*b/(a^2+1)*(-1/2/(a^2+1)/x 
^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-1/2*a*b/(a^2+1)*(-1/(a^2+1)/x*(b^2*x^2+2* 
a*b*x+a^2+1)^(3/2)+a*b/(a^2+1)*((b^2*x^2+2*a*b*x+a^2+1)^(1/2)+a*b*ln((b^2* 
x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-(a^2+1)^(1/2 
)*ln((2*a^2+2+2*a*b*x+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/x))+2 
*b^2/(a^2+1)*(1/4*(2*b^2*x+2*a*b)/b^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/8*(4 
*b^2*(a^2+1)-4*a^2*b^2)/b^2*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^ 
2+1)^(1/2))/(b^2)^(1/2)))+1/2*b^2/(a^2+1)*((b^2*x^2+2*a*b*x+a^2+1)^(1/2)+a 
*b*ln((b^2*x+a*b)/(b^2)^(1/2)+(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/(b^2)^(1/2)-( 
a^2+1)^(1/2)*ln((2*a^2+2+2*a*b*x+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^( 
1/2))/x)))-1/2*b/x^2-1/3*a/x^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (112) = 224\).

Time = 0.08 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.74 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx=\frac {3 \, \sqrt {a^{2} + 1} a b^{3} x^{3} \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + \sqrt {a^{2} + 1} a + 1\right )} + {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) - 2 \, a^{7} + {\left (a^{4} - a^{2} - 2\right )} b^{3} x^{3} - 6 \, a^{5} - 6 \, a^{3} - 3 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} b x - {\left (2 \, a^{6} - {\left (a^{4} - a^{2} - 2\right )} b^{2} x^{2} + 6 \, a^{4} + {\left (a^{5} + 2 \, a^{3} + a\right )} b x + 6 \, a^{2} + 2\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, a}{6 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \] Input:

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

Output:

1/6*(3*sqrt(a^2 + 1)*a*b^3*x^3*log(-(a^2*b*x + a^3 + sqrt(b^2*x^2 + 2*a*b* 
x + a^2 + 1)*(a^2 + sqrt(a^2 + 1)*a + 1) + (a*b*x + a^2 + 1)*sqrt(a^2 + 1) 
 + a)/x) - 2*a^7 + (a^4 - a^2 - 2)*b^3*x^3 - 6*a^5 - 6*a^3 - 3*(a^6 + 3*a^ 
4 + 3*a^2 + 1)*b*x - (2*a^6 - (a^4 - a^2 - 2)*b^2*x^2 + 6*a^4 + (a^5 + 2*a 
^3 + a)*b*x + 6*a^2 + 2)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*a)/((a^6 + 
3*a^4 + 3*a^2 + 1)*x^3)
 

Sympy [F]

\[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx=\int \frac {a + b x + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}{x^{4}}\, dx \] Input:

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (112) = 224\).

Time = 0.05 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.67 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx=-\frac {a^{3} b^{3} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{2 \, {\left (a^{2} + 1\right )}^{\frac {5}{2}}} + \frac {a b^{3} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{2 \, {\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a b^{3}}{2 \, {\left (a^{2} + 1\right )}^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} b^{2}}{2 \, {\left (a^{2} + 1\right )}^{2} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a b}{2 \, {\left (a^{2} + 1\right )}^{2} x^{2}} - \frac {b}{2 \, x^{2}} - \frac {a}{3 \, x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}}{3 \, {\left (a^{2} + 1\right )} x^{3}} \] Input:

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

Output:

-1/2*a^3*b^3*arcsinh(2*a*b*x/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 
 2*a^2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2/(sqrt(-4*a^2*b^2 + 
4*(a^2 + 1)*b^2)*abs(x)))/(a^2 + 1)^(5/2) + 1/2*a*b^3*arcsinh(2*a*b*x/(sqr 
t(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)) + 2*a^2/(sqrt(-4*a^2*b^2 + 4*(a^2 
+ 1)*b^2)*abs(x)) + 2/(sqrt(-4*a^2*b^2 + 4*(a^2 + 1)*b^2)*abs(x)))/(a^2 + 
1)^(3/2) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*b^3/(a^2 + 1)^2 - 1/2*s 
qrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2*b^2/((a^2 + 1)^2*x) + 1/2*(b^2*x^2 + 
2*a*b*x + a^2 + 1)^(3/2)*a*b/((a^2 + 1)^2*x^2) - 1/2*b/x^2 - 1/3*a/x^3 - 1 
/3*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/((a^2 + 1)*x^3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 715 vs. \(2 (112) = 224\).

Time = 0.19 (sec) , antiderivative size = 715, normalized size of antiderivative = 5.42 \[ \int \frac {e^{\text {arcsinh}(a+b x)}}{x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*a*b^3*log(abs(-2*x*abs(b) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*s 
qrt(a^2 + 1))/abs(-2*x*abs(b) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 2*sq 
rt(a^2 + 1)))/((a^4 + 2*a^2 + 1)*sqrt(a^2 + 1)) - 1/6*(3*b*x + 2*a)/x^3 + 
1/3*(20*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3*a^5*b^3 + 12*(x*a 
bs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a^7*b^3 + 6*(x*abs(b) - sqrt(b^ 
2*x^2 + 2*a*b*x + a^2 + 1))^4*a^4*b^2*abs(b) + 24*(x*abs(b) - sqrt(b^2*x^2 
 + 2*a*b*x + a^2 + 1))^2*a^6*b^2*abs(b) + 2*a^8*b^2*abs(b) + 3*(x*abs(b) - 
 sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^5*a*b^3 + 32*(x*abs(b) - sqrt(b^2*x^2 
+ 2*a*b*x + a^2 + 1))^3*a^3*b^3 + 33*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + 
a^2 + 1))*a^5*b^3 + 12*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^4*a^ 
2*b^2*abs(b) + 48*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2*a^4*b^2 
*abs(b) + 8*a^6*b^2*abs(b) + 12*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 
 1))^3*a*b^3 + 30*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a^3*b^3 + 
 6*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^4*b^2*abs(b) + 24*(x*abs 
(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2*a^2*b^2*abs(b) + 12*a^4*b^2*abs 
(b) + 9*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a*b^3 + 8*a^2*b^2*a 
bs(b) + 2*b^2*abs(b))/((a^4 + 2*a^2 + 1)*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b 
*x + a^2 + 1))^2 - a^2 - 1)^3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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