3.3.71 \(\int \frac {(1+a^2+2 a b x+b^2 x^2)^{3/2}}{\sinh ^{-1}(a+b x)^3} \, dx\) [271]
Optimal. Leaf size=84 \[ -\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left (1+(a+b x)^2\right )^{3/2}}{b \sinh ^{-1}(a+b x)}+\frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{b} \]
[Out]
-1/2*(1+(b*x+a)^2)^2/b/arcsinh(b*x+a)^2-2*(b*x+a)*(1+(b*x+a)^2)^(3/2)/b/arcsinh(b*x+a)+Chi(2*arcsinh(b*x+a))/b
+Chi(4*arcsinh(b*x+a))/b
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Rubi [A]
time = 0.20, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5860, 5790,
5814, 5791, 3393, 3382, 5819, 5556} \begin {gather*} \frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{b}-\frac {\left ((a+b x)^2+1\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left ((a+b x)^2+1\right )^{3/2}}{b \sinh ^{-1}(a+b x)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/ArcSinh[a + b*x]^3,x]
[Out]
-1/2*(1 + (a + b*x)^2)^2/(b*ArcSinh[a + b*x]^2) - (2*(a + b*x)*(1 + (a + b*x)^2)^(3/2))/(b*ArcSinh[a + b*x]) +
CoshIntegral[2*ArcSinh[a + b*x]]/b + CoshIntegral[4*ArcSinh[a + b*x]]/b
Rule 3382
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]
Rule 3393
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)
Rule 5556
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]
Rule 5790
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[Simp[Sqrt[1 + c^2*
x^2]*(d + e*x^2)^p]*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] - Dist[c*((2*p + 1)/(b*(n + 1)))*Simp[(d
+ e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(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] && LtQ[n, -1]
Rule 5791
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*c))*Simp[(d
+ e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Cosh[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSinh[c*x]], x] /; Free
Q[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p, 0]
Rule 5814
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[(f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[f*(m/(b*c*(
n + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(
n + 1), x], x] - Dist[c*((m + 2*p + 1)/(b*f*(n + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m + 1)*(
1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d]
&& LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1, 0] && IGtQ[m, -3]
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 5860
Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> D
ist[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]
Rubi steps
\begin {align*} \int \frac {\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{\sinh ^{-1}(a+b x)^3} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^{3/2}}{\sinh ^{-1}(x)^3} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}+\frac {2 \text {Subst}\left (\int \frac {x \left (1+x^2\right )}{\sinh ^{-1}(x)^2} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left (1+(a+b x)^2\right )^{3/2}}{b \sinh ^{-1}(a+b x)}+\frac {2 \text {Subst}\left (\int \frac {\sqrt {1+x^2}}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}+\frac {8 \text {Subst}\left (\int \frac {x^2 \sqrt {1+x^2}}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left (1+(a+b x)^2\right )^{3/2}}{b \sinh ^{-1}(a+b x)}+\frac {2 \text {Subst}\left (\int \frac {\cosh ^2(x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}+\frac {8 \text {Subst}\left (\int \frac {\cosh ^2(x) \sinh ^2(x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left (1+(a+b x)^2\right )^{3/2}}{b \sinh ^{-1}(a+b x)}+\frac {2 \text {Subst}\left (\int \left (\frac {1}{2 x}+\frac {\cosh (2 x)}{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}+\frac {8 \text {Subst}\left (\int \left (-\frac {1}{8 x}+\frac {\cosh (4 x)}{8 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left (1+(a+b x)^2\right )^{3/2}}{b \sinh ^{-1}(a+b x)}+\frac {\text {Subst}\left (\int \frac {\cosh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Subst}\left (\int \frac {\cosh (4 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {\left (1+(a+b x)^2\right )^2}{2 b \sinh ^{-1}(a+b x)^2}-\frac {2 (a+b x) \left (1+(a+b x)^2\right )^{3/2}}{b \sinh ^{-1}(a+b x)}+\frac {\text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )}{b}+\frac {\text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{b}\\ \end {align*}
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Mathematica [A]
time = 0.24, size = 108, normalized size = 1.29 \begin {gather*} \frac {-\frac {\left (1+a^2+2 a b x+b^2 x^2\right ) \left (1+a^2+2 a b x+b^2 x^2+4 (a+b x) \sqrt {1+a^2+2 a b x+b^2 x^2} \sinh ^{-1}(a+b x)\right )}{\sinh ^{-1}(a+b x)^2}+2 \text {Chi}\left (2 \sinh ^{-1}(a+b x)\right )+2 \text {Chi}\left (4 \sinh ^{-1}(a+b x)\right )}{2 b} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/ArcSinh[a + b*x]^3,x]
[Out]
(-(((1 + a^2 + 2*a*b*x + b^2*x^2)*(1 + a^2 + 2*a*b*x + b^2*x^2 + 4*(a + b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]
*ArcSinh[a + b*x]))/ArcSinh[a + b*x]^2) + 2*CoshIntegral[2*ArcSinh[a + b*x]] + 2*CoshIntegral[4*ArcSinh[a + b*
x]])/(2*b)
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Maple [A]
time = 3.88, size = 110, normalized size = 1.31
| | |
method |
result |
size |
| | |
default |
\(\frac {16 \hyperbolicCosineIntegral \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )^{2}+16 \hyperbolicCosineIntegral \left (4 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )^{2}-8 \sinh \left (2 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-4 \sinh \left (4 \arcsinh \left (b x +a \right )\right ) \arcsinh \left (b x +a \right )-4 \cosh \left (2 \arcsinh \left (b x +a \right )\right )-\cosh \left (4 \arcsinh \left (b x +a \right )\right )-3}{16 b \arcsinh \left (b x +a \right )^{2}}\) |
\(110\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^3,x,method=_RETURNVERBOSE)
[Out]
1/16/b*(16*Chi(2*arcsinh(b*x+a))*arcsinh(b*x+a)^2+16*Chi(4*arcsinh(b*x+a))*arcsinh(b*x+a)^2-8*sinh(2*arcsinh(b
*x+a))*arcsinh(b*x+a)-4*sinh(4*arcsinh(b*x+a))*arcsinh(b*x+a)-4*cosh(2*arcsinh(b*x+a))-cosh(4*arcsinh(b*x+a))-
3)/arcsinh(b*x+a)^2
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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((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^3,x, algorithm="maxima")
[Out]
-1/2*((b^6*x^6 + 6*a*b^5*x^5 + a^6 + (15*a^2*b^4 + 2*b^4)*x^4 + 2*a^4 + 4*(5*a^3*b^3 + 2*a*b^3)*x^3 + (15*a^4*
b^2 + 12*a^2*b^2 + b^2)*x^2 + a^2 + 2*(3*a^5*b + 4*a^3*b + a*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + (3*b^7*x^
7 + 21*a*b^6*x^6 + 3*a^7 + (63*a^2*b^5 + 8*b^5)*x^5 + 8*a^5 + 5*(21*a^3*b^4 + 8*a*b^4)*x^4 + (105*a^4*b^3 + 80
*a^2*b^3 + 7*b^3)*x^3 + 7*a^3 + (63*a^5*b^2 + 80*a^3*b^2 + 21*a*b^2)*x^2 + (21*a^6*b + 40*a^4*b + 21*a^2*b + 2
*b)*x + 2*a)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + (3*b^8*x^8 + 24*a*b^7*x^7 + 3*a^8 + 2*(42*a^2*b^6 + 5*b^6)*
x^6 + 10*a^6 + 12*(14*a^3*b^5 + 5*a*b^5)*x^5 + 6*(35*a^4*b^4 + 25*a^2*b^4 + 2*b^4)*x^4 + 12*a^4 + 8*(21*a^5*b^
3 + 25*a^3*b^3 + 6*a*b^3)*x^3 + 6*(14*a^6*b^2 + 25*a^4*b^2 + 12*a^2*b^2 + b^2)*x^2 + 6*a^2 + 12*(2*a^7*b + 5*a
^5*b + 4*a^3*b + a*b)*x + 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + ((4*b^6*x^6 + 24*a*b^5*x^5 + 4*a^6 + (60*a^2*b^4
+ 7*b^4)*x^4 + 7*a^4 + 4*(20*a^3*b^3 + 7*a*b^3)*x^3 + 2*(30*a^4*b^2 + 21*a^2*b^2 + b^2)*x^2 + 2*a^2 + 4*(6*a^5
*b + 7*a^3*b + a*b)*x - 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 3*(4*b^7*x^7 + 28*a*b^6*x^6 + 4*a^7 + 3*(28*a^2*b
^5 + 3*b^5)*x^5 + 9*a^5 + 5*(28*a^3*b^4 + 9*a*b^4)*x^4 + 2*(70*a^4*b^3 + 45*a^2*b^3 + 3*b^3)*x^3 + 6*a^3 + 6*(
14*a^5*b^2 + 15*a^3*b^2 + 3*a*b^2)*x^2 + (28*a^6*b + 45*a^4*b + 18*a^2*b + b)*x + a)*(b^2*x^2 + 2*a*b*x + a^2
+ 1)^(3/2) + (12*b^8*x^8 + 96*a*b^7*x^7 + 12*a^8 + 3*(112*a^2*b^6 + 11*b^6)*x^6 + 33*a^6 + 6*(112*a^3*b^5 + 33
*a*b^5)*x^5 + (840*a^4*b^4 + 495*a^2*b^4 + 31*b^4)*x^4 + 31*a^4 + 4*(168*a^5*b^3 + 165*a^3*b^3 + 31*a*b^3)*x^3
+ (336*a^6*b^2 + 495*a^4*b^2 + 186*a^2*b^2 + 11*b^2)*x^2 + 11*a^2 + 2*(48*a^7*b + 99*a^5*b + 62*a^3*b + 11*a*
b)*x + 1)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (4*b^9*x^9 + 36*a*b^8*x^8 + 4*a^9 + (144*a^2*b^7 + 13*b^7)*x^7 + 13*
a^7 + 7*(48*a^3*b^6 + 13*a*b^6)*x^6 + 3*(168*a^4*b^5 + 91*a^2*b^5 + 5*b^5)*x^5 + 15*a^5 + (504*a^5*b^4 + 455*a
^3*b^4 + 75*a*b^4)*x^4 + (336*a^6*b^3 + 455*a^4*b^3 + 150*a^2*b^3 + 7*b^3)*x^3 + 7*a^3 + 3*(48*a^7*b^2 + 91*a^
5*b^2 + 50*a^3*b^2 + 7*a*b^2)*x^2 + (36*a^8*b + 91*a^6*b + 75*a^4*b + 21*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a*
b*x + a^2 + 1))*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + (b^9*x^9 + 9*a*b^8*x^8 + a^9 + 4*(9*a^2*b^7
+ b^7)*x^7 + 4*a^7 + 28*(3*a^3*b^6 + a*b^6)*x^6 + 6*(21*a^4*b^5 + 14*a^2*b^5 + b^5)*x^5 + 6*a^5 + 2*(63*a^5*b
^4 + 70*a^3*b^4 + 15*a*b^4)*x^4 + 4*(21*a^6*b^3 + 35*a^4*b^3 + 15*a^2*b^3 + b^3)*x^3 + 4*a^3 + 12*(3*a^7*b^2 +
7*a^5*b^2 + 5*a^3*b^2 + a*b^2)*x^2 + (9*a^8*b + 28*a^6*b + 30*a^4*b + 12*a^2*b + b)*x + a)*sqrt(b^2*x^2 + 2*a
*b*x + a^2 + 1))/((b^7*x^6 + 6*a*b^6*x^5 + a^6*b + 3*a^4*b + 3*(5*a^2*b^5 + b^5)*x^4 + 4*(5*a^3*b^4 + 3*a*b^4)
*x^3 + 3*a^2*b + 3*(5*a^4*b^3 + 6*a^2*b^3 + b^3)*x^2 + (b^4*x^3 + 3*a*b^3*x^2 + 3*a^2*b^2*x + a^3*b)*(b^2*x^2
+ 2*a*b*x + a^2 + 1)^(3/2) + 3*(b^5*x^4 + 4*a*b^4*x^3 + a^4*b + a^2*b + (6*a^2*b^3 + b^3)*x^2 + 2*(2*a^3*b^2 +
a*b^2)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + 6*(a^5*b^2 + 2*a^3*b^2 + a*b^2)*x + 3*(b^6*x^5 + 5*a*b^5*x^4 + a^5*
b + 2*a^3*b + 2*(5*a^2*b^4 + b^4)*x^3 + 2*(5*a^3*b^3 + 3*a*b^3)*x^2 + a*b + (5*a^4*b^2 + 6*a^2*b^2 + b^2)*x)*s
qrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + b)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2) + integrate(1/2*((16
*b^6*x^6 + 96*a*b^5*x^5 + 16*a^6 + 10*(24*a^2*b^4 + b^4)*x^4 + 10*a^4 + 40*(8*a^3*b^3 + a*b^3)*x^3 + 3*(80*a^4
*b^2 + 20*a^2*b^2 - b^2)*x^2 - 3*a^2 + 2*(48*a^5*b + 20*a^3*b - 3*a*b)*x + 3)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(5
/2) + 4*(16*b^7*x^7 + 112*a*b^6*x^6 + 16*a^7 + (336*a^2*b^5 + 23*b^5)*x^5 + 23*a^5 + 5*(112*a^3*b^4 + 23*a*b^4
)*x^4 + (560*a^4*b^3 + 230*a^2*b^3 + 7*b^3)*x^3 + 7*a^3 + (336*a^5*b^2 + 230*a^3*b^2 + 21*a*b^2)*x^2 + (112*a^
6*b + 115*a^4*b + 21*a^2*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^2 + 12*(8*b^8*x^8 + 64*a*b^7*x^7 + 8*a^8 + 2*(112
*a^2*b^6 + 9*b^6)*x^6 + 18*a^6 + 4*(112*a^3*b^5 + 27*a*b^5)*x^5 + (560*a^4*b^4 + 270*a^2*b^4 + 13*b^4)*x^4 + 1
3*a^4 + 4*(112*a^5*b^3 + 90*a^3*b^3 + 13*a*b^3)*x^3 + (224*a^6*b^2 + 270*a^4*b^2 + 78*a^2*b^2 + 3*b^2)*x^2 + 3
*a^2 + 2*(32*a^7*b + 54*a^5*b + 26*a^3*b + 3*a*b)*x)*(b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 4*(16*b^9*x^9 + 144
*a*b^8*x^8 + 16*a^9 + (576*a^2*b^7 + 49*b^7)*x^7 + 49*a^7 + 7*(192*a^3*b^6 + 49*a*b^6)*x^6 + 3*(672*a^4*b^5 +
343*a^2*b^5 + 18*b^5)*x^5 + 54*a^5 + (2016*a^5*b^4 + 1715*a^3*b^4 + 270*a*b^4)*x^4 + (1344*a^6*b^3 + 1715*a^4*
b^3 + 540*a^2*b^3 + 25*b^3)*x^3 + 25*a^3 + 3*(192*a^7*b^2 + 343*a^5*b^2 + 180*a^3*b^2 + 25*a*b^2)*x^2 + (144*a
^8*b + 343*a^6*b + 270*a^4*b + 75*a^2*b + 4*b)*x + 4*a)*(b^2*x^2 + 2*a*b*x + a^2 + 1) + (16*b^10*x^10 + 160*a*
b^9*x^9 + 16*a^10 + 2*(360*a^2*b^8 + 31*b^8)*x^8 + 62*a^8 + 16*(120*a^3*b^7 + 31*a*b^7)*x^7 + 7*(480*a^4*b^6 +
248*a^2*b^6 + 13*b^6)*x^6 + 91*a^6 + 14*(288*a^5*b^5 + 248*a^3*b^5 + 39*a*b^5)*x^5 + (3360*a^6*b^4 + 4340*a^4
*b^4 + 1365*a^2*b^4 + 61*b^4)*x^4 + 61*a^4 + 4*(480*a^7*b^3 + 868*a^5*b^3 + 455*a^3*b^3 + 61*a*b^3)*x^3 + (720
*a^8*b^2 + 1736*a^6*b^2 + 1365*a^4*b^2 + 366*a^2*b^2 + 17*b^2)*x^2 + 17*a^2 + 2*(80*a^9*b + 248*a^7*b + 273*a^
5*b + 122*a^3*b + 17*a*b)*x + 1)*sqrt(b^2*x^2 +...
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Fricas [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((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^3,x, algorithm="fricas")
[Out]
integral((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a)^3, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1\right )^{\frac {3}{2}}}{\operatorname {asinh}^{3}{\left (a + b x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b**2*x**2+2*a*b*x+a**2+1)**(3/2)/asinh(b*x+a)**3,x)
[Out]
Integral((a**2 + 2*a*b*x + b**2*x**2 + 1)**(3/2)/asinh(a + b*x)**3, x)
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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((b^2*x^2+2*a*b*x+a^2+1)^(3/2)/arcsinh(b*x+a)^3,x, algorithm="giac")
[Out]
integrate((b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2)/arcsinh(b*x + a)^3, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}^{3/2}}{{\mathrm {asinh}\left (a+b\,x\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2)/asinh(a + b*x)^3,x)
[Out]
int((a^2 + b^2*x^2 + 2*a*b*x + 1)^(3/2)/asinh(a + b*x)^3, x)
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