3.5.11 \(\int \frac {x^3 \sqrt {1+c^2 x^2}}{(a+b \sinh ^{-1}(c x))^2} \, dx\) [411]

Optimal. Leaf size=213 \[ -\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^4}-\frac {3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}+\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4} \]

[Out]

-x^3*(c^2*x^2+1)/b/c/(a+b*arcsinh(c*x))-1/8*Chi((a+b*arcsinh(c*x))/b)*cosh(a/b)/b^2/c^4-3/16*Chi(3*(a+b*arcsin
h(c*x))/b)*cosh(3*a/b)/b^2/c^4+5/16*Chi(5*(a+b*arcsinh(c*x))/b)*cosh(5*a/b)/b^2/c^4+1/8*Shi((a+b*arcsinh(c*x))
/b)*sinh(a/b)/b^2/c^4+3/16*Shi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b^2/c^4-5/16*Shi(5*(a+b*arcsinh(c*x))/b)*si
nh(5*a/b)/b^2/c^4

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Rubi [A]
time = 0.44, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5814, 5780, 5556, 3384, 3379, 3382} \begin {gather*} -\frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^4}-\frac {3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}+\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^4}+\frac {3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^4}-\frac {x^3 \left (c^2 x^2+1\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((x^3*(1 + c^2*x^2))/(b*c*(a + b*ArcSinh[c*x]))) - (Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2*c^
4) - (3*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^4) + (5*Cosh[(5*a)/b]*CoshIntegral[(
5*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^4) + (Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2*c^4) + (3*S
inh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^4) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*
ArcSinh[c*x]))/b])/(16*b^2*c^4)

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

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 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

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 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 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]

Rubi steps

\begin {align*} \int \frac {x^3 \sqrt {1+c^2 x^2}}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {3 \int \frac {x^2}{a+b \sinh ^{-1}(c x)} \, dx}{b c}+\frac {(5 c) \int \frac {x^4}{a+b \sinh ^{-1}(c x)} \, dx}{b}\\ &=-\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {3 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}+\frac {5 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^4(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}\\ &=-\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {3 \text {Subst}\left (\int \left (-\frac {\cosh (x)}{4 (a+b x)}+\frac {\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}+\frac {5 \text {Subst}\left (\int \left (\frac {\cosh (x)}{8 (a+b x)}-\frac {3 \cosh (3 x)}{16 (a+b x)}+\frac {\cosh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^4}\\ &=-\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {5 \text {Subst}\left (\int \frac {\cosh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}+\frac {5 \text {Subst}\left (\int \frac {\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^4}-\frac {3 \text {Subst}\left (\int \frac {\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac {3 \text {Subst}\left (\int \frac {\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}-\frac {15 \text {Subst}\left (\int \frac {\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}\\ &=-\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\left (5 \cosh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^4}-\frac {\left (3 \cosh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac {\left (3 \cosh \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}-\frac {\left (15 \cosh \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}+\frac {\left (5 \cosh \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}-\frac {\left (5 \sinh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^4}+\frac {\left (3 \sinh \left (\frac {a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}-\frac {\left (3 \sinh \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^4}+\frac {\left (15 \sinh \left (\frac {3 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}-\frac {\left (5 \sinh \left (\frac {5 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^4}\\ &=-\frac {x^3 \left (1+c^2 x^2\right )}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {\cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^4}-\frac {3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^4}+\frac {5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^4}+\frac {\sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^4}+\frac {3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^4}-\frac {5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.48, size = 175, normalized size = 0.82 \begin {gather*} -\frac {\frac {16 b c^3 x^3}{a+b \sinh ^{-1}(c x)}+\frac {16 b c^5 x^5}{a+b \sinh ^{-1}(c x)}+2 \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+3 \cosh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-5 \cosh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-2 \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-3 \sinh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+5 \sinh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{16 b^2 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((16*b*c^3*x^3)/(a + b*ArcSinh[c*x]) + (16*b*c^5*x^5)/(a + b*ArcSinh[c*x]) + 2*Cosh[a/b]*CoshIntegral[a/
b + ArcSinh[c*x]] + 3*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSinh[c*x])] - 5*Cosh[(5*a)/b]*CoshIntegral[5*(a/b
+ ArcSinh[c*x])] - 2*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] - 3*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSi
nh[c*x])] + 5*Sinh[(5*a)/b]*SinhIntegral[5*(a/b + ArcSinh[c*x])])/(b^2*c^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(632\) vs. \(2(201)=402\).
time = 6.76, size = 633, normalized size = 2.97

method	result	size
default	\(-\frac {16 c^{5} x^{5}-16 \sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+20 c^{3} x^{3}-12 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+5 c x -\sqrt {c^{2} x^{2}+1}}{32 c^{4} b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {5 \,{\mathrm e}^{\frac {5 a}{b}} \expIntegral \left (1, 5 \arcsinh \left (c x \right )+\frac {5 a}{b}\right )}{32 c^{4} b^{2}}+\frac {-4 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+4 c^{3} x^{3}-\sqrt {c^{2} x^{2}+1}+3 c x}{32 c^{4} b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {3 \,{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{32 c^{4} b^{2}}+\frac {-\sqrt {c^{2} x^{2}+1}+c x}{16 c^{4} b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{16 c^{4} b^{2}}+\frac {\arcsinh \left (c x \right ) {\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) b +{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right ) a +b c x +\sqrt {c^{2} x^{2}+1}\, b}{16 c^{4} b^{2} \left (a +b \arcsinh \left (c x \right )\right )}+\frac {4 b \,c^{3} x^{3}+4 \sqrt {c^{2} x^{2}+1}\, b \,c^{2} x^{2}+3 \arcsinh \left (c x \right ) {\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right ) b +3 \,{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right ) a +3 b c x +\sqrt {c^{2} x^{2}+1}\, b}{32 c^{4} b^{2} \left (a +b \arcsinh \left (c x \right )\right )}-\frac {16 b \,c^{5} x^{5}+16 \sqrt {c^{2} x^{2}+1}\, b \,c^{4} x^{4}+20 b \,c^{3} x^{3}+12 \sqrt {c^{2} x^{2}+1}\, b \,c^{2} x^{2}+5 \arcsinh \left (c x \right ) \expIntegral \left (1, -5 \arcsinh \left (c x \right )-\frac {5 a}{b}\right ) {\mathrm e}^{-\frac {5 a}{b}} b +5 \expIntegral \left (1, -5 \arcsinh \left (c x \right )-\frac {5 a}{b}\right ) {\mathrm e}^{-\frac {5 a}{b}} a +5 b c x +\sqrt {c^{2} x^{2}+1}\, b}{32 c^{4} b^{2} \left (a +b \arcsinh \left (c x \right )\right )}\)	\(633\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(16*c^5*x^5-16*(c^2*x^2+1)^(1/2)*c^4*x^4+20*c^3*x^3-12*c^2*x^2*(c^2*x^2+1)^(1/2)+5*c*x-(c^2*x^2+1)^(1/2)
)/c^4/b/(a+b*arcsinh(c*x))-5/32/c^4/b^2*exp(5*a/b)*Ei(1,5*arcsinh(c*x)+5*a/b)+1/32*(-4*c^2*x^2*(c^2*x^2+1)^(1/
2)+4*c^3*x^3-(c^2*x^2+1)^(1/2)+3*c*x)/c^4/b/(a+b*arcsinh(c*x))+3/32/c^4/b^2*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a
/b)+1/16*(-(c^2*x^2+1)^(1/2)+c*x)/c^4/b/(a+b*arcsinh(c*x))+1/16/c^4/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)+1/16/c
^4/b^2*(arcsinh(c*x)*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*b+exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)*a+b*c*x+(c^2*x^2+1)
^(1/2)*b)/(a+b*arcsinh(c*x))+1/32/c^4/b^2*(4*b*c^3*x^3+4*(c^2*x^2+1)^(1/2)*b*c^2*x^2+3*arcsinh(c*x)*exp(-3*a/b
)*Ei(1,-3*arcsinh(c*x)-3*a/b)*b+3*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)*a+3*b*c*x+(c^2*x^2+1)^(1/2)*b)/(a+b*
arcsinh(c*x))-1/32/c^4/b^2*(16*b*c^5*x^5+16*(c^2*x^2+1)^(1/2)*b*c^4*x^4+20*b*c^3*x^3+12*(c^2*x^2+1)^(1/2)*b*c^
2*x^2+5*arcsinh(c*x)*Ei(1,-5*arcsinh(c*x)-5*a/b)*exp(-5*a/b)*b+5*Ei(1,-5*arcsinh(c*x)-5*a/b)*exp(-5*a/b)*a+5*b
*c*x+(c^2*x^2+1)^(1/2)*b)/(a+b*arcsinh(c*x))

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

[Out]

-((c^2*x^5 + x^3)*(c^2*x^2 + 1) + (c^3*x^6 + c*x^4)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2 + sqrt(c^2*x^2 + 1)*a*b*c^
2*x + a*b*c + (b^2*c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2*x^2 + 1))) + integrate(((
5*c^3*x^5 + 2*c*x^3)*(c^2*x^2 + 1)^(3/2) + (10*c^4*x^6 + 11*c^2*x^4 + 3*x^2)*(c^2*x^2 + 1) + (5*c^5*x^7 + 9*c^
3*x^5 + 4*c*x^3)*sqrt(c^2*x^2 + 1))/(a*b*c^5*x^4 + (c^2*x^2 + 1)*a*b*c^3*x^2 + 2*a*b*c^3*x^2 + a*b*c + (b^2*c^
5*x^4 + (c^2*x^2 + 1)*b^2*c^3*x^2 + 2*b^2*c^3*x^2 + b^2*c + 2*(b^2*c^4*x^3 + b^2*c^2*x)*sqrt(c^2*x^2 + 1))*log
(c*x + sqrt(c^2*x^2 + 1)) + 2*(a*b*c^4*x^3 + a*b*c^2*x)*sqrt(c^2*x^2 + 1)), x)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \sqrt {c^{2} x^{2} + 1}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [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(x^3*(c^2*x^2+1)^(1/2)/(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\sqrt {c^2\,x^2+1}}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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