3.3.26 \(\int \frac {1}{(a+b \sinh ^{-1}(c+d x))^{7/2}} \, dx\) [226]

Optimal. Leaf size=195 \[ -\frac {2 \sqrt {1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac {4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {8 \sqrt {1+(c+d x)^2}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {4 e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.30, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5858, 5773, 5818, 5819, 3389, 2211, 2236, 2235} \begin {gather*} -\frac {4 \sqrt {\pi } e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}+\frac {4 \sqrt {\pi } e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c+d x)}}{\sqrt {b}}\right )}{15 b^{7/2} d}-\frac {8 \sqrt {(c+d x)^2+1}}{15 b^3 d \sqrt {a+b \sinh ^{-1}(c+d x)}}-\frac {4 (c+d x)}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac {2 \sqrt {(c+d x)^2+1}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.15, size = 238, normalized size = 1.22 \begin {gather*} \frac {-6 b^2 e^{\sinh ^{-1}(c+d x)}-2 e^{-\sinh ^{-1}(c+d x)} \left (4 a^2+2 a b \left (-1+4 \sinh ^{-1}(c+d x)\right )+b^2 \left (3-2 \sinh ^{-1}(c+d x)+4 \sinh ^{-1}(c+d x)^2\right )\right )+8 e^{a/b} \sqrt {\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right )^2 \Gamma \left (\frac {1}{2},\frac {a}{b}+\sinh ^{-1}(c+d x)\right )-4 e^{-\frac {a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right ) \left (e^{\frac {a}{b}+\sinh ^{-1}(c+d x)} \left (2 a+b+2 b \sinh ^{-1}(c+d x)\right )+2 b \left (-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {a+b \sinh ^{-1}(c+d x)}{b}\right )\right )}{30 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \arcsinh \left (d x +c \right )\right )^{\frac {7}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \operatorname {asinh}{\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*asinh(c + d*x))^(7/2),x)

[Out]

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

________________________________________________________________________________________