\(\int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx\) [156]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 178 \[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=-\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}-\frac {4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}+\frac {4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c} \]

[Out]

-4/15*x/b^2/(a+b*arcsinh(c*x))^(3/2)-4/15*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/c+4/
15*erfi((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(7/2)/c/exp(a/b)-2/5*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsinh(c
*x))^(5/2)-8/15*(c^2*x^2+1)^(1/2)/b^3/c/(a+b*arcsinh(c*x))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {5773, 5818, 5819, 3389, 2211, 2236, 2235} \[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=-\frac {4 \sqrt {\pi } e^{a/b} \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}+\frac {4 \sqrt {\pi } e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}-\frac {8 \sqrt {c^2 x^2+1}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {2 \sqrt {c^2 x^2+1}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}} \]

[In]

Int[(a + b*ArcSinh[c*x])^(-7/2),x]

[Out]

(-2*Sqrt[1 + c^2*x^2])/(5*b*c*(a + b*ArcSinh[c*x])^(5/2)) - (4*x)/(15*b^2*(a + b*ArcSinh[c*x])^(3/2)) - (8*Sqr
t[1 + c^2*x^2])/(15*b^3*c*Sqrt[a + b*ArcSinh[c*x]]) - (4*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]
])/(15*b^(7/2)*c) + (4*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(15*b^(7/2)*c*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 5773

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Sqrt[1 + c^2*x^2]*((a + b*ArcSinh[c*x])^(n + 1
)/(b*c*(n + 1))), x] - Dist[c/(b*(n + 1)), Int[x*((a + b*ArcSinh[c*x])^(n + 1)/Sqrt[1 + c^2*x^2]), x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[(f*x)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x]
 /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}+\frac {(2 c) \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^{5/2}} \, dx}{5 b} \\ & = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}+\frac {4 \int \frac {1}{(a+b \text {arcsinh}(c x))^{3/2}} \, dx}{15 b^2} \\ & = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}+\frac {(8 c) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \text {arcsinh}(c x)}} \, dx}{15 b^3} \\ & = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}-\frac {x}{b}\right )}{\sqrt {x}} \, dx,x,a+b \text {arcsinh}(c x)\right )}{15 b^4 c} \\ & = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}-\frac {4 \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 x)\right )}{15 b^4 c}+\frac {4 \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 x)\right )}{15 b^4 c} \\ & = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}-\frac {8 \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c x)}\right )}{15 b^4 c}+\frac {8 \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \text {arcsinh}(c x)}\right )}{15 b^4 c} \\ & = -\frac {2 \sqrt {1+c^2 x^2}}{5 b c (a+b \text {arcsinh}(c x))^{5/2}}-\frac {4 x}{15 b^2 (a+b \text {arcsinh}(c x))^{3/2}}-\frac {8 \sqrt {1+c^2 x^2}}{15 b^3 c \sqrt {a+b \text {arcsinh}(c x)}}-\frac {4 e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c}+\frac {4 e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \text {arcsinh}(c x)}}{\sqrt {b}}\right )}{15 b^{7/2} c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=\frac {-6 b^2 e^{\text {arcsinh}(c x)}-2 e^{-\text {arcsinh}(c x)} \left (4 a^2+2 a b (-1+4 \text {arcsinh}(c x))+b^2 \left (3-2 \text {arcsinh}(c x)+4 \text {arcsinh}(c x)^2\right )\right )+8 e^{a/b} \sqrt {\frac {a}{b}+\text {arcsinh}(c x)} (a+b \text {arcsinh}(c x))^2 \Gamma \left (\frac {1}{2},\frac {a}{b}+\text {arcsinh}(c x)\right )-4 e^{-\frac {a}{b}} (a+b \text {arcsinh}(c x)) \left (e^{\frac {a}{b}+\text {arcsinh}(c x)} (2 a+b+2 b \text {arcsinh}(c x))+2 b \left (-\frac {a+b \text {arcsinh}(c x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \text {arcsinh}(c x)}{b}\right )\right )}{30 b^3 c (a+b \text {arcsinh}(c x))^{5/2}} \]

[In]

Integrate[(a + b*ArcSinh[c*x])^(-7/2),x]

[Out]

(-6*b^2*E^ArcSinh[c*x] - (2*(4*a^2 + 2*a*b*(-1 + 4*ArcSinh[c*x]) + b^2*(3 - 2*ArcSinh[c*x] + 4*ArcSinh[c*x]^2)
))/E^ArcSinh[c*x] + 8*E^(a/b)*Sqrt[a/b + ArcSinh[c*x]]*(a + b*ArcSinh[c*x])^2*Gamma[1/2, a/b + ArcSinh[c*x]] -
 (4*(a + b*ArcSinh[c*x])*(E^(a/b + ArcSinh[c*x])*(2*a + b + 2*b*ArcSinh[c*x]) + 2*b*(-((a + b*ArcSinh[c*x])/b)
)^(3/2)*Gamma[1/2, -((a + b*ArcSinh[c*x])/b)]))/E^(a/b))/(30*b^3*c*(a + b*ArcSinh[c*x])^(5/2))

Maple [F]

\[\int \frac {1}{\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{\frac {7}{2}}}d x\]

[In]

int(1/(a+b*arcsinh(c*x))^(7/2),x)

[Out]

int(1/(a+b*arcsinh(c*x))^(7/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a+b*arcsinh(c*x))^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=\int \frac {1}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(1/(a+b*asinh(c*x))**(7/2),x)

[Out]

Integral((a + b*asinh(c*x))**(-7/2), x)

Maxima [F]

\[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arcsinh(c*x))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*arcsinh(c*x) + a)^(-7/2), x)

Giac [F]

\[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=\int { \frac {1}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(a+b*arcsinh(c*x))^(7/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^(-7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \text {arcsinh}(c x))^{7/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(1/(a + b*asinh(c*x))^(7/2),x)

[Out]

int(1/(a + b*asinh(c*x))^(7/2), x)