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

Optimal result

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

[Out]

(d*x+c)*(a+b*arcsinh(d*x+c))^(5/2)/d+15/16*b^(5/2)*exp(a/b)*erf((a+b*arcsinh(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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5858, 5772, 5798, 5819, 3389, 2211, 2236, 2235} \[ \int (a+b \text {arcsinh}(c+d x))^{5/2} \, dx=\frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{16 d}-\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c+d x)}}{\sqrt {b}}\right )}{16 d}+\frac {15 b^2 (c+d x) \sqrt {a+b \text {arcsinh}(c+d x)}}{4 d}-\frac {5 b \sqrt {(c+d x)^2+1} (a+b \text {arcsinh}(c+d x))^{3/2}}{2 d}+\frac {(c+d x) (a+b \text {arcsinh}(c+d x))^{5/2}}{d} \]

[In]

Int[(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(15*b^2*(c + d*x)*Sqrt[a + b*ArcSinh[c + d*x]])/(4*d) - (5*b*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3
/2))/(2*d) + ((c + d*x)*(a + b*ArcSinh[c + d*x])^(5/2))/d + (15*b^(5/2)*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSin
h[c + d*x]]/Sqrt[b]])/(16*d) - (15*b^(5/2)*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(16*d*E^(a/b))

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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 2235

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 2236

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

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

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[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 5798

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] - Dist[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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(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 5858

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+b \text {arcsinh}(x))^{5/2} \, dx,x,c+d x\right )}{d} \\ & = \frac {(c+d x) (a+b \text {arcsinh}(c+d x))^{5/2}}{d}-\frac {(5 b) \text {Subst}\left (\int \frac {x (a+b \text {arcsinh}(x))^{3/2}}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{2 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 {\left (15 b^2\right ) \text {Subst}\left (\int \sqrt {a+b \text {arcsinh}(x)} \, dx,x,c+d x\right )}{4 d} \\ & = \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 {\left (15 b^3\right ) \text {Subst}\left (\int \frac {x}{\sqrt {1+x^2} \sqrt {a+b \text {arcsinh}(x)}} \, dx,x,c+d x\right )}{8 d} \\ & = \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 {\left (15 b^2\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c+d x)\right )}{8 d} \\ & = \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 {\left (15 b^2\right ) \text {Subst}\left (\int \frac {e^{-i \left (\frac {i a}{b}-\frac {i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c+d x)\right )}{16 d}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int \frac {e^{i \left (\frac {i a}{b}-\frac {i x}{b}\right )}}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c+d x)\right )}{16 d} \\ & = \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 {\left (15 b^2\right ) \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c+d x)}\right )}{8 d}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c+d x)}\right )}{8 d} \\ & = \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} \\ \end{align*}

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

[In]

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

[Out]

((8*a^2*Sqrt[a + b*ArcSinh[c + d*x]]*(-((E^((2*a)/b)*Gamma[3/2, a/b + ArcSinh[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[S
qrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]]*(Cosh[a/b] + Sinh[a/b])) + Sqrt[b]*(4*Sqrt[b]*Sqrt[a + b*ArcSinh[c + d*x]
]*(2*Sqrt[1 + (c + d*x)^2]*(a - 5*b*ArcSinh[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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + b*asinh(c + d*x))^(5/2),x)

[Out]

int((a + b*asinh(c + d*x))^(5/2), x)