\(\int \frac {x^4 (a+b \text {arcsinh}(c x))}{(\pi +c^2 \pi x^2)^{3/2}} \, dx\) [101]

Optimal result

Integrand size = 26, antiderivative size = 131 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=-\frac {b x^2}{4 c^3 \pi ^{3/2}}-\frac {x^3 (a+b \text {arcsinh}(c x))}{c^2 \pi \sqrt {\pi +c^2 \pi x^2}}+\frac {3 x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{2 c^4 \pi ^2}-\frac {3 (a+b \text {arcsinh}(c x))^2}{4 b c^5 \pi ^{3/2}}-\frac {b \log \left (1+c^2 x^2\right )}{2 c^5 \pi ^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.12 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\frac {12 a c x+4 a c^3 x^3-6 b \sqrt {1+c^2 x^2} \text {arcsinh}(c x)^2-b \sqrt {1+c^2 x^2} \cosh (2 \text {arcsinh}(c x))-4 b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )+\text {arcsinh}(c x) \left (9 b c x-12 a \sqrt {1+c^2 x^2}+b \sinh (3 \text {arcsinh}(c x))\right )}{8 c^5 \pi ^{3/2} \sqrt {1+c^2 x^2}} \] Input:

Integrate[(x^4*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(3/2),x]
 

Output:

(12*a*c*x + 4*a*c^3*x^3 - 6*b*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 - b*Sqrt[1 
+ c^2*x^2]*Cosh[2*ArcSinh[c*x]] - 4*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + 
 ArcSinh[c*x]*(9*b*c*x - 12*a*Sqrt[1 + c^2*x^2] + b*Sinh[3*ArcSinh[c*x]])) 
/(8*c^5*Pi^(3/2)*Sqrt[1 + c^2*x^2])
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {6225, 243, 49, 2009, 6227, 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.

\(\displaystyle \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi c^2 x^2+\pi \right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6225

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6227

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 6198

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

Input:

Int[(x^4*(a + b*ArcSinh[c*x]))/(Pi + c^2*Pi*x^2)^(3/2),x]
 

Output:

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

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 49

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 6225

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

rule 6227

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a 
+ b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Simp[f^2*((m - 1)/(c^2*(m + 
2*p + 1)))   Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] 
 - Simp[b*f*(n/(c*(m + 2*p + 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] && IGtQ[ 
m, 1] && NeQ[m + 2*p + 1, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(260\) vs. \(2(113)=226\).

Time = 1.25 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.99

method	result	size
default	\(\frac {a \,x^{3}}{2 \pi \,c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {3 a x}{2 c^{4} \pi \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {3 a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 c^{4} \pi \sqrt {\pi \,c^{2}}}-\frac {b \left (-4 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+2 c^{4} x^{4}+6 \operatorname {arcsinh}\left (x c \right )^{2} x^{2} c^{2}+8 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-8 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-12 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c +3 c^{2} x^{2}+6 \operatorname {arcsinh}\left (x c \right )^{2}+8 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )-8 \,\operatorname {arcsinh}\left (x c \right )+1\right )}{8 c^{5} \pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right )}\)	\(261\)
parts	\(\frac {a \,x^{3}}{2 \pi \,c^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}+\frac {3 a x}{2 c^{4} \pi \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {3 a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{2 c^{4} \pi \sqrt {\pi \,c^{2}}}-\frac {b \left (-4 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+2 c^{4} x^{4}+6 \operatorname {arcsinh}\left (x c \right )^{2} x^{2} c^{2}+8 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-8 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-12 \,\operatorname {arcsinh}\left (x c \right ) \sqrt {c^{2} x^{2}+1}\, x c +3 c^{2} x^{2}+6 \operatorname {arcsinh}\left (x c \right )^{2}+8 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )-8 \,\operatorname {arcsinh}\left (x c \right )+1\right )}{8 c^{5} \pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right )}\)	\(261\)

Input:

int(x^4*(a+b*arcsinh(x*c))/(Pi*c^2*x^2+Pi)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*a*x^3/Pi/c^2/(Pi*c^2*x^2+Pi)^(1/2)+3/2*a/c^4*x/Pi/(Pi*c^2*x^2+Pi)^(1/2 
)-3/2*a/c^4/Pi*ln(Pi*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^ 
(1/2)-1/8*b*(-4*arcsinh(x*c)*(c^2*x^2+1)^(1/2)*x^3*c^3+2*c^4*x^4+6*arcsinh 
(x*c)^2*x^2*c^2+8*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)*x^2*c^2-8*arcsinh(x*c)*c 
^2*x^2-12*arcsinh(x*c)*(c^2*x^2+1)^(1/2)*x*c+3*c^2*x^2+6*arcsinh(x*c)^2+8* 
ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)-8*arcsinh(x*c)+1)/c^5/Pi^(3/2)/(c^2*x^2+1)
 

Fricas [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

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

integrate(x**4*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(3/2),x)
 

Output:

(Integral(a*x**4/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) 
 + Integral(b*x**4*asinh(c*x)/(c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x 
**2 + 1)), x))/pi**(3/2)
 

Maxima [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),x, algorithm="maxim 
a")
 

Output:

1/2*a*(x^3/(pi*sqrt(pi + pi*c^2*x^2)*c^2) + 3*x/(pi*sqrt(pi + pi*c^2*x^2)* 
c^4) - 3*arcsinh(c*x)/(pi^(3/2)*c^5)) + b*integrate(x^4*log(c*x + sqrt(c^2 
*x^2 + 1))/(pi + pi*c^2*x^2)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(3/2),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. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx=\frac {4 \sqrt {c^{2} x^{2}+1}\, a \,c^{3} x^{3}+12 \sqrt {c^{2} x^{2}+1}\, a c x +8 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{4}}{\sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{7} x^{2}+8 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{4}}{\sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{5}-12 \,\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right ) a \,c^{2} x^{2}-12 \,\mathrm {log}\left (\sqrt {c^{2} x^{2}+1}+c x \right ) a +9 a \,c^{2} x^{2}+9 a}{8 \sqrt {\pi }\, c^{5} \pi \left (c^{2} x^{2}+1\right )} \] Input:

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

Output:

(4*sqrt(c**2*x**2 + 1)*a*c**3*x**3 + 12*sqrt(c**2*x**2 + 1)*a*c*x + 8*int( 
(asinh(c*x)*x**4)/(sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x) 
*b*c**7*x**2 + 8*int((asinh(c*x)*x**4)/(sqrt(c**2*x**2 + 1)*c**2*x**2 + sq 
rt(c**2*x**2 + 1)),x)*b*c**5 - 12*log(sqrt(c**2*x**2 + 1) + c*x)*a*c**2*x* 
*2 - 12*log(sqrt(c**2*x**2 + 1) + c*x)*a + 9*a*c**2*x**2 + 9*a)/(8*sqrt(pi 
)*c**5*pi*(c**2*x**2 + 1))