\(\int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {\pi +c^2 \pi x^2}} \, dx\) [99]

Optimal result

Integrand size = 26, antiderivative size = 97 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {\pi +c^2 \pi x^2}} \, dx=-\frac {b c}{6 \sqrt {\pi } x^2}-\frac {\sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{3 \pi x^3}+\frac {2 c^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{3 \pi x}-\frac {2 b c^3 \log (x)}{3 \sqrt {\pi }} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {-b c x+6 b c^3 x^3-2 a \sqrt {1+c^2 x^2}+4 a c^2 x^2 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (-1+2 c^2 x^2\right ) \text {arcsinh}(c x)-4 b c^3 x^3 \log (x)}{6 \sqrt {\pi } x^3} \] Input:

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

Output:

(-(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)*ArcSinh[c*x] - 4*b*c^3*x^3* 
Log[x])/(6*Sqrt[Pi]*x^3)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, 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.

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

\(\Big \downarrow \) 6224

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 6215

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

\(\Big \downarrow \) 14

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

Input:

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

Output:

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

Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(81)=162\).

Time = 1.20 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.85

method	result	size
default	\(a \left (-\frac {\sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,x^{3}}+\frac {2 c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi x}\right )+\frac {4 b \,c^{3} \operatorname {arcsinh}\left (x c \right )}{3 \sqrt {\pi }}-\frac {2 b \,x^{4} c^{7}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b \,x^{2} \left (c^{2} x^{2}+1\right ) c^{5}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {2 b \,x^{2} \operatorname {arcsinh}\left (x c \right ) c^{5}}{\sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b x \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) c^{4}}{\sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {2 b \left (c^{2} x^{2}+1\right ) c^{3}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b \,\operatorname {arcsinh}\left (x c \right ) c^{3}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {5 b \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) c^{2}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x}+\frac {b \left (c^{2} x^{2}+1\right ) c}{6 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x^{2}}+\frac {b \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right )}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x^{3}}-\frac {2 b \,c^{3} \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )}{3 \sqrt {\pi }}\)	\(373\)
parts	\(a \left (-\frac {\sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi \,x^{3}}+\frac {2 c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}{3 \pi x}\right )+\frac {4 b \,c^{3} \operatorname {arcsinh}\left (x c \right )}{3 \sqrt {\pi }}-\frac {2 b \,x^{4} c^{7}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b \,x^{2} \left (c^{2} x^{2}+1\right ) c^{5}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {2 b \,x^{2} \operatorname {arcsinh}\left (x c \right ) c^{5}}{\sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b x \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) c^{4}}{\sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {2 b \left (c^{2} x^{2}+1\right ) c^{3}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}+\frac {2 b \,\operatorname {arcsinh}\left (x c \right ) c^{3}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right )}-\frac {5 b \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right ) c^{2}}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x}+\frac {b \left (c^{2} x^{2}+1\right ) c}{6 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x^{2}}+\frac {b \sqrt {c^{2} x^{2}+1}\, \operatorname {arcsinh}\left (x c \right )}{3 \sqrt {\pi }\, \left (3 c^{2} x^{2}-1\right ) x^{3}}-\frac {2 b \,c^{3} \ln \left (\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )}{3 \sqrt {\pi }}\)	\(373\)

Input:

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

Output:

a*(-1/3/Pi/x^3*(Pi*c^2*x^2+Pi)^(1/2)+2/3/Pi*c^2/x*(Pi*c^2*x^2+Pi)^(1/2))+4 
/3*b/Pi^(1/2)*c^3*arcsinh(x*c)-2/3*b/Pi^(1/2)/(3*c^2*x^2-1)*x^4*c^7+2/3*b/ 
Pi^(1/2)/(3*c^2*x^2-1)*x^2*(c^2*x^2+1)*c^5-2*b/Pi^(1/2)/(3*c^2*x^2-1)*x^2* 
arcsinh(x*c)*c^5+2*b/Pi^(1/2)/(3*c^2*x^2-1)*x*(c^2*x^2+1)^(1/2)*arcsinh(x* 
c)*c^4-2/3*b/Pi^(1/2)/(3*c^2*x^2-1)*(c^2*x^2+1)*c^3+2/3*b/Pi^(1/2)/(3*c^2* 
x^2-1)*arcsinh(x*c)*c^3-5/3*b/Pi^(1/2)/(3*c^2*x^2-1)/x*(c^2*x^2+1)^(1/2)*a 
rcsinh(x*c)*c^2+1/6*b/Pi^(1/2)/(3*c^2*x^2-1)/x^2*(c^2*x^2+1)*c+1/3*b/Pi^(1 
/2)/(3*c^2*x^2-1)/x^3*(c^2*x^2+1)^(1/2)*arcsinh(x*c)-2/3*b/Pi^(1/2)*c^3*ln 
((x*c+(c^2*x^2+1)^(1/2))^2-1)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (81) = 162\).

Time = 0.12 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.29 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (2 \, b c^{4} x^{4} + b c^{2} x^{2} - b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi } {\left (b c^{5} x^{5} + b c^{3} x^{3}\right )} \log \left (\frac {\pi + \pi c^{2} x^{6} + \pi c^{2} x^{2} + \pi x^{4} - \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} {\left (x^{4} - 1\right )}}{c^{2} x^{4} + x^{2}}\right ) + \sqrt {\pi + \pi c^{2} x^{2}} {\left (4 \, a c^{4} x^{4} + 2 \, a c^{2} x^{2} + {\left (b c x^{3} - b c x\right )} \sqrt {c^{2} x^{2} + 1} - 2 \, a\right )}}{6 \, {\left (\pi c^{2} x^{5} + \pi x^{3}\right )}} \] Input:

integrate((a+b*arcsinh(c*x))/x^4/(pi*c^2*x^2+pi)^(1/2),x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.25 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^4 \sqrt {\pi +c^2 \pi x^2}} \, dx=-\frac {1}{6} \, {\left (\frac {4 \, c^{2} \log \left (x\right )}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi } x^{2}}\right )} b c + \frac {1}{3} \, b {\left (\frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}} c^{2}}{\pi x} - \frac {\sqrt {\pi + \pi c^{2} x^{2}}}{\pi x^{3}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {\pi + \pi c^{2} x^{2}} c^{2}}{\pi x} - \frac {\sqrt {\pi + \pi c^{2} x^{2}}}{\pi x^{3}}\right )} \] Input:

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

Output:

-1/6*(4*c^2*log(x)/sqrt(pi) + 1/(sqrt(pi)*x^2))*b*c + 1/3*b*(2*sqrt(pi + p 
i*c^2*x^2)*c^2/(pi*x) - sqrt(pi + pi*c^2*x^2)/(pi*x^3))*arcsinh(c*x) + 1/3 
*a*(2*sqrt(pi + pi*c^2*x^2)*c^2/(pi*x) - sqrt(pi + pi*c^2*x^2)/(pi*x^3))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((a+b*asinh(c*x))/x^4/(Pi*c^2*x^2+Pi)^(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(pi)*x**3 
)