3.3.78 \(\int \frac {\text {arcsinh}(a+b x)^3}{(1+a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [278]

3.3.78.1 Optimal result

Integrand size = 30, antiderivative size = 115 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\text {arcsinh}(a+b x)^3}{b}+\frac {(a+b x) \text {arcsinh}(a+b x)^3}{b \sqrt {1+(a+b x)^2}}-\frac {3 \text {arcsinh}(a+b x)^2 \log \left (1+e^{2 \text {arcsinh}(a+b x)}\right )}{b}-\frac {3 \text {arcsinh}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(a+b x)}\right )}{b}+\frac {3 \operatorname {PolyLog}\left (3,-e^{2 \text {arcsinh}(a+b x)}\right )}{2 b} \]

output

arcsinh(b*x+a)^3/b-3*arcsinh(b*x+a)^2*ln(1+(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/ 
b-3*arcsinh(b*x+a)*polylog(2,-(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b+3/2*polylog 
(3,-(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b+(b*x+a)*arcsinh(b*x+a)^3/b/(1+(b*x+a) 
^2)^(1/2)
 

3.3.78.2 Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11 \[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \text {arcsinh}(a+b x)^2 \left (\frac {\left (a+b x-\sqrt {1+a^2+2 a b x+b^2 x^2}\right ) \text {arcsinh}(a+b x)}{\sqrt {1+a^2+2 a b x+b^2 x^2}}-3 \log \left (1+e^{-2 \text {arcsinh}(a+b x)}\right )\right )+6 \text {arcsinh}(a+b x) \operatorname {PolyLog}\left (2,-e^{-2 \text {arcsinh}(a+b x)}\right )+3 \operatorname {PolyLog}\left (3,-e^{-2 \text {arcsinh}(a+b x)}\right )}{2 b} \]

input

Integrate[ArcSinh[a + b*x]^3/(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

output

(2*ArcSinh[a + b*x]^2*(((a + b*x - Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])*ArcS 
inh[a + b*x])/Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2] - 3*Log[1 + E^(-2*ArcSinh[ 
a + b*x])]) + 6*ArcSinh[a + b*x]*PolyLog[2, -E^(-2*ArcSinh[a + b*x])] + 3* 
PolyLog[3, -E^(-2*ArcSinh[a + b*x])])/(2*b)
 

3.3.78.3 Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6275, 6202, 6212, 3042, 26, 4201, 2620, 3011, 2720, 7143}

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.

\(\displaystyle \int \frac {\text {arcsinh}(a+b x)^3}{\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 6275

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

\(\Big \downarrow \) 6202

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

\(\Big \downarrow \) 6212

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^3}{\sqrt {(a+b x)^2+1}}+3 i \int \text {arcsinh}(a+b x)^2 \tan (i \text {arcsinh}(a+b x))d\text {arcsinh}(a+b x)}{b}\)

\(\Big \downarrow \) 4201

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^3}{\sqrt {(a+b x)^2+1}}+3 i \left (2 i \left (-\frac {1}{2} \int \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(a+b x)}\right )d\text {arcsinh}(a+b x)+\frac {1}{2} \text {arcsinh}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(a+b x)}\right )+\frac {1}{2} \text {arcsinh}(a+b x)^2 \log \left (e^{2 \text {arcsinh}(a+b x)}+1\right )\right )-\frac {1}{3} i \text {arcsinh}(a+b x)^3\right )}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^3}{\sqrt {(a+b x)^2+1}}+3 i \left (2 i \left (-\frac {1}{4} \int e^{-2 \text {arcsinh}(a+b x)} \operatorname {PolyLog}(2,-a-b x)de^{2 \text {arcsinh}(a+b x)}+\frac {1}{2} \text {arcsinh}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(a+b x)}\right )+\frac {1}{2} \text {arcsinh}(a+b x)^2 \log \left (e^{2 \text {arcsinh}(a+b x)}+1\right )\right )-\frac {1}{3} i \text {arcsinh}(a+b x)^3\right )}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\frac {(a+b x) \text {arcsinh}(a+b x)^3}{\sqrt {(a+b x)^2+1}}+3 i \left (2 i \left (\frac {1}{2} \text {arcsinh}(a+b x) \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(a+b x)}\right )+\frac {1}{2} \text {arcsinh}(a+b x)^2 \log \left (e^{2 \text {arcsinh}(a+b x)}+1\right )-\frac {1}{4} \operatorname {PolyLog}(3,-a-b x)\right )-\frac {1}{3} i \text {arcsinh}(a+b x)^3\right )}{b}\)

input

Int[ArcSinh[a + b*x]^3/(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

output

(((a + b*x)*ArcSinh[a + b*x]^3)/Sqrt[1 + (a + b*x)^2] + (3*I)*((-1/3*I)*Ar 
cSinh[a + b*x]^3 + (2*I)*((ArcSinh[a + b*x]^2*Log[1 + E^(2*ArcSinh[a + b*x 
])])/2 + (ArcSinh[a + b*x]*PolyLog[2, -E^(2*ArcSinh[a + b*x])])/2 - PolyLo 
g[3, -a - b*x]/4)))/b
 

3.3.78.3.1 Defintions of rubi rules used

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 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 6202

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), 
x_Symbol] :> Simp[x*((a + b*ArcSinh[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp 
[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]   Int[x*((a + b*ArcSinh[ 
c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, 
 c^2*d] && GtQ[n, 0]
 

rule 6212

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

rule 6275

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + 
(C_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/d   Subst[Int[(C/d^2 + (C/d^2)*x^2 
)^p*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B, C 
, n, p}, x] && EqQ[B*(1 + c^2) - 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

3.3.78.4 Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.77

method	result	size
default	\(-\frac {\left (b^{2} x^{2}-\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, b x +2 a b x -a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}+a^{2}+1\right ) \operatorname {arcsinh}\left (b x +a \right )^{3}}{b \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}+\frac {2 \operatorname {arcsinh}\left (b x +a \right )^{3}}{b}-\frac {3 \operatorname {arcsinh}\left (b x +a \right )^{2} \ln \left (1+\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}-\frac {3 \,\operatorname {arcsinh}\left (b x +a \right ) \operatorname {polylog}\left (2, -\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{b}+\frac {3 \operatorname {polylog}\left (3, -\left (b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )^{2}\right )}{2 b}\)	\(203\)

input

int(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x,method=_RETURNVERBOSE 
)
 

output

-(b^2*x^2-(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*b*x+2*a*b*x-a*(b^2*x^2+2*a*b*x+a^2 
+1)^(1/2)+a^2+1)/b/(b^2*x^2+2*a*b*x+a^2+1)*arcsinh(b*x+a)^3+2*arcsinh(b*x+ 
a)^3/b-3*arcsinh(b*x+a)^2*ln(1+(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b-3*arcsinh( 
b*x+a)*polylog(2,-(b*x+a+(1+(b*x+a)^2)^(1/2))^2)/b+3/2*polylog(3,-(b*x+a+( 
1+(b*x+a)^2)^(1/2))^2)/b
 

3.3.78.5 Fricas [F]

\[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="fri 
cas")
 

output

integral(sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*arcsinh(b*x + a)^3/(b^4*x^4 + 4 
*a*b^3*x^3 + 2*(3*a^2 + 1)*b^2*x^2 + a^4 + 4*(a^3 + a)*b*x + 2*a^2 + 1), x 
)
 

3.3.78.6 Sympy [F]

\[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\operatorname {asinh}^{3}{\left (a + b x \right )}}{\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]

input

integrate(asinh(b*x+a)**3/(b**2*x**2+2*a*b*x+a**2+1)**(3/2),x)
 

output

Integral(asinh(a + b*x)**3/(a**2 + 2*a*b*x + b**2*x**2 + 1)**(3/2), x)
 

3.3.78.7 Maxima [F]

\[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="max 
ima")
 

output

integrate(arcsinh(b*x + a)^3/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2), x)
 

3.3.78.8 Giac [F]

\[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\operatorname {arsinh}\left (b x + a\right )^{3}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate(arcsinh(b*x+a)^3/(b^2*x^2+2*a*b*x+a^2+1)^(3/2),x, algorithm="gia 
c")
 

output

integrate(arcsinh(b*x + a)^3/(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2), x)
 

3.3.78.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arcsinh}(a+b x)^3}{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\mathrm {asinh}\left (a+b\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}} \,d x \]

input

int(asinh(a + b*x)^3/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2),x)
 

output

int(asinh(a + b*x)^3/(a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2), x)