\(\int x (a+b \text {arcsinh}(c x))^{5/2} \, dx\) [158]

Optimal result

Integrand size = 14, antiderivative size = 223 \[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\frac {15 b^2 \sqrt {a+b \text {arcsinh}(c x)}}{64 c^2}+\frac {15}{32} b^2 x^2 \sqrt {a+b \text {arcsinh}(c x)}-\frac {5 b x \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^{3/2}}{8 c}+\frac {(a+b \text {arcsinh}(c x))^{5/2}}{4 c^2}+\frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {15 b^{5/2} e^{\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{256 c^2}-\frac {15 b^{5/2} e^{-\frac {2 a}{b}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{256 c^2} \] Output:

15/64*b^2*(a+b*arcsinh(c*x))^(1/2)/c^2+15/32*b^2*x^2*(a+b*arcsinh(c*x))^(1 
/2)-5/8*b*x*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^(3/2)/c+1/4*(a+b*arcsinh( 
c*x))^(5/2)/c^2+1/2*x^2*(a+b*arcsinh(c*x))^(5/2)-15/512*b^(5/2)*exp(2*a/b) 
*2^(1/2)*Pi^(1/2)*erf(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))/c^2-15/512 
*b^(5/2)*2^(1/2)*Pi^(1/2)*erfi(2^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))/c 
^2/exp(2*a/b)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.52 \[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\frac {e^{-\frac {2 a}{b}} \left (-b^3 \sqrt {-\frac {a+b \text {arcsinh}(c x)}{b}} \Gamma \left (\frac {7}{2},-\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )+b^3 e^{\frac {4 a}{b}} \sqrt {\frac {a}{b}+\text {arcsinh}(c x)} \Gamma \left (\frac {7}{2},\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )\right )}{32 \sqrt {2} c^2 \sqrt {a+b \text {arcsinh}(c x)}} \] Input:

Integrate[x*(a + b*ArcSinh[c*x])^(5/2),x]
 

Output:

(-(b^3*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[7/2, (-2*(a + b*ArcSinh[c*x]) 
)/b]) + b^3*E^((4*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[7/2, (2*(a + b*ArcS 
inh[c*x]))/b])/(32*Sqrt[2]*c^2*E^((2*a)/b)*Sqrt[a + b*ArcSinh[c*x]])
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {6192, 6227, 6192, 6198, 6234, 3042, 25, 3793, 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.

\(\displaystyle \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx\)

\(\Big \downarrow \) 6192

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \int \frac {x^2 (a+b \text {arcsinh}(c x))^{3/2}}{\sqrt {c^2 x^2+1}}dx\)

\(\Big \downarrow \) 6227

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {\int \frac {(a+b \text {arcsinh}(c x))^{3/2}}{\sqrt {c^2 x^2+1}}dx}{2 c^2}-\frac {3 b \int x \sqrt {a+b \text {arcsinh}(c x)}dx}{4 c}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 6192

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}-\frac {1}{4} b c \int \frac {x^2}{\sqrt {c^2 x^2+1} \sqrt {a+b \text {arcsinh}(c x)}}dx\right )}{4 c}-\frac {\int \frac {(a+b \text {arcsinh}(c x))^{3/2}}{\sqrt {c^2 x^2+1}}dx}{2 c^2}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 6198

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}-\frac {1}{4} b c \int \frac {x^2}{\sqrt {c^2 x^2+1} \sqrt {a+b \text {arcsinh}(c x)}}dx\right )}{4 c}-\frac {(a+b \text {arcsinh}(c x))^{5/2}}{5 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 6234

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}-\frac {\int \frac {\sinh ^2\left (\frac {a}{b}-\frac {a+b \text {arcsinh}(c x)}{b}\right )}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{4 c^2}\right )}{4 c}-\frac {(a+b \text {arcsinh}(c x))^{5/2}}{5 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}-\frac {\int -\frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )^2}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{4 c^2}\right )}{4 c}-\frac {(a+b \text {arcsinh}(c x))^{5/2}}{5 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}+\frac {\int \frac {\sin \left (\frac {i a}{b}-\frac {i (a+b \text {arcsinh}(c x))}{b}\right )^2}{\sqrt {a+b \text {arcsinh}(c x)}}d(a+b \text {arcsinh}(c x))}{4 c^2}\right )}{4 c}-\frac {(a+b \text {arcsinh}(c x))^{5/2}}{5 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 3793

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {3 b \left (\frac {\int \left (\frac {1}{2 \sqrt {a+b \text {arcsinh}(c x)}}-\frac {\cosh \left (\frac {2 a}{b}-\frac {2 (a+b \text {arcsinh}(c x))}{b}\right )}{2 \sqrt {a+b \text {arcsinh}(c x)}}\right )d(a+b \text {arcsinh}(c x))}{4 c^2}+\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}\right )}{4 c}-\frac {(a+b \text {arcsinh}(c x))^{5/2}}{5 b c^3}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} x^2 (a+b \text {arcsinh}(c x))^{5/2}-\frac {5}{4} b c \left (-\frac {(a+b \text {arcsinh}(c x))^{5/2}}{5 b c^3}-\frac {3 b \left (\frac {1}{2} x^2 \sqrt {a+b \text {arcsinh}(c x)}-\frac {\frac {1}{4} \sqrt {\frac {\pi }{2}} \sqrt {b} e^{\frac {2 a}{b}} \text {erf}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )+\frac {1}{4} \sqrt {\frac {\pi }{2}} \sqrt {b} e^{-\frac {2 a}{b}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )-\sqrt {a+b \text {arcsinh}(c x)}}{4 c^2}\right )}{4 c}+\frac {x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^{3/2}}{2 c^2}\right )\)

Input:

Int[x*(a + b*ArcSinh[c*x])^(5/2),x]
 

Output:

(x^2*(a + b*ArcSinh[c*x])^(5/2))/2 - (5*b*c*((x*Sqrt[1 + c^2*x^2]*(a + b*A 
rcSinh[c*x])^(3/2))/(2*c^2) - (a + b*ArcSinh[c*x])^(5/2)/(5*b*c^3) - (3*b* 
((x^2*Sqrt[a + b*ArcSinh[c*x]])/2 - (-Sqrt[a + b*ArcSinh[c*x]] + (Sqrt[b]* 
E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/4 
+ (Sqrt[b]*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(4 
*E^((2*a)/b)))/(4*c^2)))/(4*c)))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 6192

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[ 
x^(m + 1)*((a + b*ArcSinh[c*x])^n/(m + 1)), x] - Simp[b*c*(n/(m + 1))   Int 
[x^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; Free 
Q[{a, b, c}, x] && IGtQ[m, 0] && GtQ[n, 0]
 

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 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]
 

rule 6234

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

Maple [F]

\[\int x \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )^{\frac {5}{2}}d x\]

Input:

int(x*(a+b*arcsinh(x*c))^(5/2),x)
 

Output:

int(x*(a+b*arcsinh(x*c))^(5/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x*(a+b*arcsinh(c*x))^(5/2),x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\int x \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {5}{2}}\, dx \] Input:

integrate(x*(a+b*asinh(c*x))**(5/2),x)
 

Output:

Integral(x*(a + b*asinh(c*x))**(5/2), x)
 

Maxima [F]

\[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\int { {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {5}{2}} x \,d x } \] Input:

integrate(x*(a+b*arcsinh(c*x))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*arcsinh(c*x) + a)^(5/2)*x, x)
                                                                                    
                                                                                    
 

Giac [F(-2)]

Exception generated. \[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(x*(a+b*arcsinh(c*x))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=\int x\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{5/2} \,d x \] Input:

int(x*(a + b*asinh(c*x))^(5/2),x)
 

Output:

int(x*(a + b*asinh(c*x))^(5/2), x)
 

Reduce [F]

\[ \int x (a+b \text {arcsinh}(c x))^{5/2} \, dx=2 \left (\int \sqrt {\mathit {asinh} \left (c x \right ) b +a}\, \mathit {asinh} \left (c x \right ) x d x \right ) a b +\left (\int \sqrt {\mathit {asinh} \left (c x \right ) b +a}\, \mathit {asinh} \left (c x \right )^{2} x d x \right ) b^{2}+\left (\int \sqrt {\mathit {asinh} \left (c x \right ) b +a}\, x d x \right ) a^{2} \] Input:

int(x*(a+b*asinh(c*x))^(5/2),x)
 

Output:

2*int(sqrt(asinh(c*x)*b + a)*asinh(c*x)*x,x)*a*b + int(sqrt(asinh(c*x)*b + 
 a)*asinh(c*x)**2*x,x)*b**2 + int(sqrt(asinh(c*x)*b + a)*x,x)*a**2