3.2.4 \(\int (a+b \text {arcsinh}(c+d x))^{5/2} \, dx\) [104]

3.2.4.1 Optimal result

Integrand size = 14, antiderivative size = 179 \[ \int (a+b \text {arcsinh}(c+d x))^{5/2} \, dx=\frac {15 b^2 (c+d x) \sqrt {a+b \text {arcsinh}(c+d x)}}{4 d}-\frac {5 b \sqrt {1+(c+d x)^2} (a+b \text {arcsinh}(c+d x))^{3/2}}{2 d}+\frac {(c+d x) (a+b \text {arcsinh}(c+d x))^{5/2}}{d}+\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{16 d}-\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{16 d} \]

output

(d*x+c)*(a+b*arcsinh(d*x+c))^(5/2)/d+15/16*b^(5/2)*exp(a/b)*erf((a+b*arcsi 
nh(d*x+c))^(1/2)/b^(1/2))*Pi^(1/2)/d-15/16*b^(5/2)*erfi((a+b*arcsinh(d*x+c 
))^(1/2)/b^(1/2))*Pi^(1/2)/d/exp(a/b)-5/2*b*(a+b*arcsinh(d*x+c))^(3/2)*(1+ 
(d*x+c)^2)^(1/2)/d+15/4*b^2*(d*x+c)*(a+b*arcsinh(d*x+c))^(1/2)/d
 

3.2.4.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(458\) vs. \(2(179)=358\).

Time = 1.45 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.56 \[ \int (a+b \text {arcsinh}(c+d x))^{5/2} \, dx=\frac {8 a^2 e^{-\frac {a}{b}} \sqrt {a+b \text {arcsinh}(c+d x)} \left (-\frac {e^{\frac {2 a}{b}} \Gamma \left (\frac {3}{2},\frac {a}{b}+\text {arcsinh}(c+d x)\right )}{\sqrt {\frac {a}{b}+\text {arcsinh}(c+d x)}}+\frac {\Gamma \left (\frac {3}{2},-\frac {a+b \text {arcsinh}(c+d x)}{b}\right )}{\sqrt {-\frac {a+b \text {arcsinh}(c+d x)}{b}}}\right )+4 a \sqrt {b} \left (4 \sqrt {b} \sqrt {a+b \text {arcsinh}(c+d x)} \left (-3 \sqrt {1+(c+d x)^2}+2 (c+d x) \text {arcsinh}(c+d x)\right )+(2 a+3 b) \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {a}{b}\right )-\sinh \left (\frac {a}{b}\right )\right )+(-2 a+3 b) \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {a}{b}\right )+\sinh \left (\frac {a}{b}\right )\right )\right )+\sqrt {b} \left (4 \sqrt {b} \sqrt {a+b \text {arcsinh}(c+d x)} \left (2 \sqrt {1+(c+d x)^2} (a-5 b \text {arcsinh}(c+d x))+b (c+d x) \left (15+4 \text {arcsinh}(c+d x)^2\right )\right )+\left (4 a^2+12 a b+15 b^2\right ) \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right ) \left (-\cosh \left (\frac {a}{b}\right )+\sinh \left (\frac {a}{b}\right )\right )+\left (4 a^2-12 a b+15 b^2\right ) \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right ) \left (\cosh \left (\frac {a}{b}\right )+\sinh \left (\frac {a}{b}\right )\right )\right )}{16 d} \]

input

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

output

((8*a^2*Sqrt[a + b*ArcSinh[c + d*x]]*(-((E^((2*a)/b)*Gamma[3/2, a/b + ArcS 
inh[c + d*x]])/Sqrt[a/b + ArcSinh[c + d*x]]) + Gamma[3/2, -((a + b*ArcSinh 
[c + d*x])/b)]/Sqrt[-((a + b*ArcSinh[c + d*x])/b)]))/E^(a/b) + 4*a*Sqrt[b] 
*(4*Sqrt[b]*Sqrt[a + b*ArcSinh[c + d*x]]*(-3*Sqrt[1 + (c + d*x)^2] + 2*(c 
+ d*x)*ArcSinh[c + d*x]) + (2*a + 3*b)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c 
+ d*x]]/Sqrt[b]]*(Cosh[a/b] - Sinh[a/b]) + (-2*a + 3*b)*Sqrt[Pi]*Erf[Sqrt[ 
a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(Cosh[a/b] + Sinh[a/b])) + Sqrt[b]*(4*Sqr 
t[b]*Sqrt[a + b*ArcSinh[c + d*x]]*(2*Sqrt[1 + (c + d*x)^2]*(a - 5*b*ArcSin 
h[c + d*x]) + b*(c + d*x)*(15 + 4*ArcSinh[c + d*x]^2)) + (4*a^2 + 12*a*b + 
 15*b^2)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(-Cosh[a/b] + 
 Sinh[a/b]) + (4*a^2 - 12*a*b + 15*b^2)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c 
+ d*x]]/Sqrt[b]]*(Cosh[a/b] + Sinh[a/b])))/(16*d)
 

3.2.4.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.91 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {6273, 6187, 6213, 6187, 6234, 25, 3042, 26, 3789, 2611, 2633, 2634}

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.

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

\(\Big \downarrow \) 6273

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

\(\Big \downarrow \) 6187

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

\(\Big \downarrow \) 6213

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

\(\Big \downarrow \) 6187

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

\(\Big \downarrow \) 6234

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3789

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

\(\Big \downarrow \) 2611

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

\(\Big \downarrow \) 2633

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

\(\Big \downarrow \) 2634

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

input

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

output

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

3.2.4.3.1 Defintions of rubi rules used

rule 25

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

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2611

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : 
> Simp[2/d   Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d 
*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2633

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ 
F, a, b, c, d}, x] && PosQ[b]
 

rule 2634

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 3042

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

rule 3789

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
/2   Int[(c + d*x)^m/E^(I*(e + f*x)), x], x] - Simp[I/2   Int[(c + d*x)^m*E 
^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]
 

rule 6187

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

rule 6213

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

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]
 

rule 6273

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[1/d 
   Subst[Int[(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d 
, n}, x]
 

3.2.4.4 Maple [F]

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

input

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

output

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

3.2.4.5 Fricas [F(-2)]

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

input

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

output

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

3.2.4.6 Sympy [F]

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

input

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

output

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

3.2.4.7 Maxima [F]

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

input

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

output

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

3.2.4.8 Giac [F]

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

input

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

output

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

3.2.4.9 Mupad [F(-1)]

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

input

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

output

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