3.430 \(\int \frac {(1+c^2 x^2)^{5/2}}{(a+b \sinh ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=216 \[ -\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c}-\frac {3 \sinh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}+\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c}+\frac {3 \cosh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c}-\frac {\left (c^2 x^2+1\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-(c^2*x^2+1)^3/b/c/(a+b*arcsinh(c*x))+15/16*cosh(2*a/b)*Shi(2*(a+b*arcsinh(c*x))/b)/b^2/c+3/4*cosh(4*a/b)*Shi(
4*(a+b*arcsinh(c*x))/b)/b^2/c+3/16*cosh(6*a/b)*Shi(6*(a+b*arcsinh(c*x))/b)/b^2/c-15/16*Chi(2*(a+b*arcsinh(c*x)
)/b)*sinh(2*a/b)/b^2/c-3/4*Chi(4*(a+b*arcsinh(c*x))/b)*sinh(4*a/b)/b^2/c-3/16*Chi(6*(a+b*arcsinh(c*x))/b)*sinh
(6*a/b)/b^2/c

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Rubi [A]  time = 0.45, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5696, 5779, 5448, 3303, 3298, 3301} \[ -\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{16 b^2 c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{4 b^2 c}-\frac {3 \sinh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{16 b^2 c}+\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{16 b^2 c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{4 b^2 c}+\frac {3 \cosh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{16 b^2 c}-\frac {\left (c^2 x^2+1\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(1 + c^2*x^2)^(5/2)/(a + b*ArcSinh[c*x])^2,x]

[Out]

-((1 + c^2*x^2)^3/(b*c*(a + b*ArcSinh[c*x]))) - (15*CoshIntegral[(2*a)/b + 2*ArcSinh[c*x]]*Sinh[(2*a)/b])/(16*
b^2*c) - (3*CoshIntegral[(4*a)/b + 4*ArcSinh[c*x]]*Sinh[(4*a)/b])/(4*b^2*c) - (3*CoshIntegral[(6*a)/b + 6*ArcS
inh[c*x]]*Sinh[(6*a)/b])/(16*b^2*c) + (15*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(16*b^2*c) + (
3*Cosh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcSinh[c*x]])/(4*b^2*c) + (3*Cosh[(6*a)/b]*SinhIntegral[(6*a)/b + 6*
ArcSinh[c*x]])/(16*b^2*c)

Rule 3298

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 3301

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 3303

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 5448

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 5696

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> 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)*d^IntPart[p]*(d + e*x^2)^Fr
acPart[p])/(b*(n + 1)*(1 + c^2*x^2)^FracPart[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 5779

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])

Rubi steps

\begin {align*} \int \frac {\left (1+c^2 x^2\right )^{5/2}}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac {\left (1+c^2 x^2\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {(6 c) \int \frac {x \left (1+c^2 x^2\right )^2}{a+b \sinh ^{-1}(c x)} \, dx}{b}\\ &=-\frac {\left (1+c^2 x^2\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {6 \operatorname {Subst}\left (\int \frac {\cosh ^5(x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1+c^2 x^2\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {6 \operatorname {Subst}\left (\int \left (\frac {5 \sinh (2 x)}{32 (a+b x)}+\frac {\sinh (4 x)}{8 (a+b x)}+\frac {\sinh (6 x)}{32 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c}\\ &=-\frac {\left (1+c^2 x^2\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (6 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}+\frac {3 \operatorname {Subst}\left (\int \frac {\sinh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c}+\frac {15 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}\\ &=-\frac {\left (1+c^2 x^2\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\left (15 \cosh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}+\frac {\left (3 \cosh \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c}+\frac {\left (3 \cosh \left (\frac {6 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}-\frac {\left (15 \sinh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}-\frac {\left (3 \sinh \left (\frac {4 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c}-\frac {\left (3 \sinh \left (\frac {6 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c}\\ &=-\frac {\left (1+c^2 x^2\right )^3}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {15 \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {2 a}{b}\right )}{16 b^2 c}-\frac {3 \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {4 a}{b}\right )}{4 b^2 c}-\frac {3 \text {Chi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {6 a}{b}\right )}{16 b^2 c}+\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{16 b^2 c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{4 b^2 c}+\frac {3 \cosh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{16 b^2 c}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 311, normalized size = 1.44 \[ -\frac {15 \sinh \left (\frac {2 a}{b}\right ) \left (a+b \sinh ^{-1}(c x)\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+12 \sinh \left (\frac {4 a}{b}\right ) \left (a+b \sinh ^{-1}(c x)\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+3 a \sinh \left (\frac {6 a}{b}\right ) \text {Chi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+3 b \sinh \left (\frac {6 a}{b}\right ) \sinh ^{-1}(c x) \text {Chi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-15 a \cosh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-15 b \cosh \left (\frac {2 a}{b}\right ) \sinh ^{-1}(c x) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-12 a \cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-12 b \cosh \left (\frac {4 a}{b}\right ) \sinh ^{-1}(c x) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-3 a \cosh \left (\frac {6 a}{b}\right ) \text {Shi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-3 b \cosh \left (\frac {6 a}{b}\right ) \sinh ^{-1}(c x) \text {Shi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+16 b c^6 x^6+48 b c^4 x^4+48 b c^2 x^2+16 b}{16 b^2 c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + c^2*x^2)^(5/2)/(a + b*ArcSinh[c*x])^2,x]

[Out]

-1/16*(16*b + 48*b*c^2*x^2 + 48*b*c^4*x^4 + 16*b*c^6*x^6 + 15*(a + b*ArcSinh[c*x])*CoshIntegral[2*(a/b + ArcSi
nh[c*x])]*Sinh[(2*a)/b] + 12*(a + b*ArcSinh[c*x])*CoshIntegral[4*(a/b + ArcSinh[c*x])]*Sinh[(4*a)/b] + 3*a*Cos
hIntegral[6*(a/b + ArcSinh[c*x])]*Sinh[(6*a)/b] + 3*b*ArcSinh[c*x]*CoshIntegral[6*(a/b + ArcSinh[c*x])]*Sinh[(
6*a)/b] - 15*a*Cosh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*x])] - 15*b*ArcSinh[c*x]*Cosh[(2*a)/b]*SinhIntegr
al[2*(a/b + ArcSinh[c*x])] - 12*a*Cosh[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])] - 12*b*ArcSinh[c*x]*Cosh[
(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])] - 3*a*Cosh[(6*a)/b]*SinhIntegral[6*(a/b + ArcSinh[c*x])] - 3*b*A
rcSinh[c*x]*Cosh[(6*a)/b]*SinhIntegral[6*(a/b + ArcSinh[c*x])])/(b^2*c*(a + b*ArcSinh[c*x]))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c^{4} x^{4} + 2 \, c^{2} x^{2} + 1\right )} \sqrt {c^{2} x^{2} + 1}}{b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (c^{2} x^{2} + 1\right )}^{\frac {5}{2}}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

integrate((c^2*x^2 + 1)^(5/2)/(b*arcsinh(c*x) + a)^2, x)

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maple [B]  time = 0.71, size = 704, normalized size = 3.26 \[ -\frac {5}{16 b c \left (a +b \arcsinh \left (c x \right )\right )}-\frac {32 c^{6} x^{6}-32 c^{5} x^{5} \sqrt {c^{2} x^{2}+1}+48 c^{4} x^{4}-32 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+18 c^{2} x^{2}-6 c x \sqrt {c^{2} x^{2}+1}+1}{64 c \left (a +b \arcsinh \left (c x \right )\right ) b}+\frac {3 \,{\mathrm e}^{\frac {6 a}{b}} \Ei \left (1, 6 \arcsinh \left (c x \right )+\frac {6 a}{b}\right )}{32 c \,b^{2}}-\frac {3 \left (8 c^{4} x^{4}-8 c^{3} x^{3} \sqrt {c^{2} x^{2}+1}+8 c^{2} x^{2}-4 c x \sqrt {c^{2} x^{2}+1}+1\right )}{32 c \left (a +b \arcsinh \left (c x \right )\right ) b}+\frac {3 \,{\mathrm e}^{\frac {4 a}{b}} \Ei \left (1, 4 \arcsinh \left (c x \right )+\frac {4 a}{b}\right )}{8 c \,b^{2}}-\frac {15 \left (2 c^{2} x^{2}-2 c x \sqrt {c^{2} x^{2}+1}+1\right )}{64 c \left (a +b \arcsinh \left (c x \right )\right ) b}+\frac {15 \,{\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{32 c \,b^{2}}-\frac {15 \left (2 x^{2} b \,c^{2}+2 b c \sqrt {c^{2} x^{2}+1}\, x +2 \arcsinh \left (c x \right ) {\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right ) b +2 \,{\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right ) a +b \right )}{64 c \,b^{2} \left (a +b \arcsinh \left (c x \right )\right )}-\frac {3 \left (8 x^{4} b \,c^{4}+8 \sqrt {c^{2} x^{2}+1}\, x^{3} b \,c^{3}+8 x^{2} b \,c^{2}+4 b c \sqrt {c^{2} x^{2}+1}\, x +4 \arcsinh \left (c x \right ) \Ei \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right ) {\mathrm e}^{-\frac {4 a}{b}} b +4 \Ei \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right ) {\mathrm e}^{-\frac {4 a}{b}} a +b \right )}{32 c \,b^{2} \left (a +b \arcsinh \left (c x \right )\right )}-\frac {32 x^{6} b \,c^{6}+32 \sqrt {c^{2} x^{2}+1}\, x^{5} b \,c^{5}+48 x^{4} b \,c^{4}+32 \sqrt {c^{2} x^{2}+1}\, x^{3} b \,c^{3}+18 x^{2} b \,c^{2}+6 b c \sqrt {c^{2} x^{2}+1}\, x +6 \arcsinh \left (c x \right ) \Ei \left (1, -6 \arcsinh \left (c x \right )-\frac {6 a}{b}\right ) {\mathrm e}^{-\frac {6 a}{b}} b +6 \Ei \left (1, -6 \arcsinh \left (c x \right )-\frac {6 a}{b}\right ) {\mathrm e}^{-\frac {6 a}{b}} a +b}{64 c \,b^{2} \left (a +b \arcsinh \left (c x \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x))^2,x)

[Out]

-5/16/b/c/(a+b*arcsinh(c*x))-1/64*(32*c^6*x^6-32*c^5*x^5*(c^2*x^2+1)^(1/2)+48*c^4*x^4-32*c^3*x^3*(c^2*x^2+1)^(
1/2)+18*c^2*x^2-6*c*x*(c^2*x^2+1)^(1/2)+1)/c/(a+b*arcsinh(c*x))/b+3/32/c/b^2*exp(6*a/b)*Ei(1,6*arcsinh(c*x)+6*
a/b)-3/32*(8*c^4*x^4-8*c^3*x^3*(c^2*x^2+1)^(1/2)+8*c^2*x^2-4*c*x*(c^2*x^2+1)^(1/2)+1)/c/(a+b*arcsinh(c*x))/b+3
/8/c/b^2*exp(4*a/b)*Ei(1,4*arcsinh(c*x)+4*a/b)-15/64*(2*c^2*x^2-2*c*x*(c^2*x^2+1)^(1/2)+1)/c/(a+b*arcsinh(c*x)
)/b+15/32/c/b^2*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)-15/64/c/b^2*(2*x^2*b*c^2+2*b*c*(c^2*x^2+1)^(1/2)*x+2*arc
sinh(c*x)*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*b+2*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*a+b)/(a+b*arcsin
h(c*x))-3/32/c/b^2*(8*x^4*b*c^4+8*(c^2*x^2+1)^(1/2)*x^3*b*c^3+8*x^2*b*c^2+4*b*c*(c^2*x^2+1)^(1/2)*x+4*arcsinh(
c*x)*Ei(1,-4*arcsinh(c*x)-4*a/b)*exp(-4*a/b)*b+4*Ei(1,-4*arcsinh(c*x)-4*a/b)*exp(-4*a/b)*a+b)/(a+b*arcsinh(c*x
))-1/64/c/b^2*(32*x^6*b*c^6+32*(c^2*x^2+1)^(1/2)*x^5*b*c^5+48*x^4*b*c^4+32*(c^2*x^2+1)^(1/2)*x^3*b*c^3+18*x^2*
b*c^2+6*b*c*(c^2*x^2+1)^(1/2)*x+6*arcsinh(c*x)*Ei(1,-6*arcsinh(c*x)-6*a/b)*exp(-6*a/b)*b+6*Ei(1,-6*arcsinh(c*x
)-6*a/b)*exp(-6*a/b)*a+b)/(a+b*arcsinh(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (c^{6} x^{6} + 3 \, c^{4} x^{4} + 3 \, c^{2} x^{2} + 1\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (c^{7} x^{7} + 3 \, c^{5} x^{5} + 3 \, c^{3} x^{3} + c x\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} a b c^{2} x + a b c + {\left (b^{2} c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )} + \int \frac {{\left (6 \, c^{6} x^{6} + 11 \, c^{4} x^{4} + 4 \, c^{2} x^{2} - 1\right )} {\left (c^{2} x^{2} + 1\right )}^{\frac {3}{2}} + 6 \, {\left (2 \, c^{7} x^{7} + 5 \, c^{5} x^{5} + 4 \, c^{3} x^{3} + c x\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (6 \, c^{8} x^{8} + 19 \, c^{6} x^{6} + 21 \, c^{4} x^{4} + 9 \, c^{2} x^{2} + 1\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{4} x^{4} + {\left (c^{2} x^{2} + 1\right )} a b c^{2} x^{2} + 2 \, a b c^{2} x^{2} + a b + {\left (b^{2} c^{4} x^{4} + {\left (c^{2} x^{2} + 1\right )} b^{2} c^{2} x^{2} + 2 \, b^{2} c^{2} x^{2} + b^{2} + 2 \, {\left (b^{2} c^{3} x^{3} + b^{2} c x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{3} x^{3} + a b c x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c^2*x^2+1)^(5/2)/(a+b*arcsinh(c*x))^2,x, algorithm="maxima")

[Out]

-((c^6*x^6 + 3*c^4*x^4 + 3*c^2*x^2 + 1)*(c^2*x^2 + 1) + (c^7*x^7 + 3*c^5*x^5 + 3*c^3*x^3 + c*x)*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(((6*c^6*x^6 + 11*c^4*x^4 + 4*c^2*x^2 - 1)*(c^2*x^2 + 1)^(3/2) + 6*(2
*c^7*x^7 + 5*c^5*x^5 + 4*c^3*x^3 + c*x)*(c^2*x^2 + 1) + (6*c^8*x^8 + 19*c^6*x^6 + 21*c^4*x^4 + 9*c^2*x^2 + 1)*
sqrt(c^2*x^2 + 1))/(a*b*c^4*x^4 + (c^2*x^2 + 1)*a*b*c^2*x^2 + 2*a*b*c^2*x^2 + a*b + (b^2*c^4*x^4 + (c^2*x^2 +
1)*b^2*c^2*x^2 + 2*b^2*c^2*x^2 + b^2 + 2*(b^2*c^3*x^3 + b^2*c*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1
)) + 2*(a*b*c^3*x^3 + a*b*c*x)*sqrt(c^2*x^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c^2\,x^2+1\right )}^{5/2}}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*x^2 + 1)^(5/2)/(a + b*asinh(c*x))^2,x)

[Out]

int((c^2*x^2 + 1)^(5/2)/(a + b*asinh(c*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c**2*x**2+1)**(5/2)/(a+b*asinh(c*x))**2,x)

[Out]

Integral((c**2*x**2 + 1)**(5/2)/(a + b*asinh(c*x))**2, x)

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