3.77 \(\int x^2 (d+e x^2) (a+b \text {csch}^{-1}(c x)) \, dx\)

Optimal. Leaf size=167 \[ \frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b e x^4 \sqrt {-c^2 x^2-1}}{20 c \sqrt {-c^2 x^2}}+\frac {b x \left (20 c^2 d-9 e\right ) \tan ^{-1}\left (\frac {c x}{\sqrt {-c^2 x^2-1}}\right )}{120 c^4 \sqrt {-c^2 x^2}}+\frac {b x^2 \sqrt {-c^2 x^2-1} \left (20 c^2 d-9 e\right )}{120 c^3 \sqrt {-c^2 x^2}} \]

[Out]

1/3*d*x^3*(a+b*arccsch(c*x))+1/5*e*x^5*(a+b*arccsch(c*x))+1/120*b*(20*c^2*d-9*e)*x*arctan(c*x/(-c^2*x^2-1)^(1/
2))/c^4/(-c^2*x^2)^(1/2)+1/120*b*(20*c^2*d-9*e)*x^2*(-c^2*x^2-1)^(1/2)/c^3/(-c^2*x^2)^(1/2)+1/20*b*e*x^4*(-c^2
*x^2-1)^(1/2)/c/(-c^2*x^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {14, 6302, 12, 459, 321, 217, 203} \[ \frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b x^2 \sqrt {-c^2 x^2-1} \left (20 c^2 d-9 e\right )}{120 c^3 \sqrt {-c^2 x^2}}+\frac {b x \left (20 c^2 d-9 e\right ) \tan ^{-1}\left (\frac {c x}{\sqrt {-c^2 x^2-1}}\right )}{120 c^4 \sqrt {-c^2 x^2}}+\frac {b e x^4 \sqrt {-c^2 x^2-1}}{20 c \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x^2)*(a + b*ArcCsch[c*x]),x]

[Out]

(b*(20*c^2*d - 9*e)*x^2*Sqrt[-1 - c^2*x^2])/(120*c^3*Sqrt[-(c^2*x^2)]) + (b*e*x^4*Sqrt[-1 - c^2*x^2])/(20*c*Sq
rt[-(c^2*x^2)]) + (d*x^3*(a + b*ArcCsch[c*x]))/3 + (e*x^5*(a + b*ArcCsch[c*x]))/5 + (b*(20*c^2*d - 9*e)*x*ArcT
an[(c*x)/Sqrt[-1 - c^2*x^2]])/(120*c^4*Sqrt[-(c^2*x^2)])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 6302

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsch[c*x], u, x] - Dist[(b*c*x)/Sqrt[-(c^2*x^2)], Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&
!(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0]))
|| (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int x^2 \left (d+e x^2\right ) \left (a+b \text {csch}^{-1}(c x)\right ) \, dx &=\frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \int \frac {x^2 \left (5 d+3 e x^2\right )}{15 \sqrt {-1-c^2 x^2}} \, dx}{\sqrt {-c^2 x^2}}\\ &=\frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \int \frac {x^2 \left (5 d+3 e x^2\right )}{\sqrt {-1-c^2 x^2}} \, dx}{15 \sqrt {-c^2 x^2}}\\ &=\frac {b e x^4 \sqrt {-1-c^2 x^2}}{20 c \sqrt {-c^2 x^2}}+\frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {\left (b c \left (20 d-\frac {9 e}{c^2}\right ) x\right ) \int \frac {x^2}{\sqrt {-1-c^2 x^2}} \, dx}{60 \sqrt {-c^2 x^2}}\\ &=\frac {b \left (20 c^2 d-9 e\right ) x^2 \sqrt {-1-c^2 x^2}}{120 c^3 \sqrt {-c^2 x^2}}+\frac {b e x^4 \sqrt {-1-c^2 x^2}}{20 c \sqrt {-c^2 x^2}}+\frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {\left (b \left (20 d-\frac {9 e}{c^2}\right ) x\right ) \int \frac {1}{\sqrt {-1-c^2 x^2}} \, dx}{120 c \sqrt {-c^2 x^2}}\\ &=\frac {b \left (20 c^2 d-9 e\right ) x^2 \sqrt {-1-c^2 x^2}}{120 c^3 \sqrt {-c^2 x^2}}+\frac {b e x^4 \sqrt {-1-c^2 x^2}}{20 c \sqrt {-c^2 x^2}}+\frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {\left (b \left (20 d-\frac {9 e}{c^2}\right ) x\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x^2} \, dx,x,\frac {x}{\sqrt {-1-c^2 x^2}}\right )}{120 c \sqrt {-c^2 x^2}}\\ &=\frac {b \left (20 c^2 d-9 e\right ) x^2 \sqrt {-1-c^2 x^2}}{120 c^3 \sqrt {-c^2 x^2}}+\frac {b e x^4 \sqrt {-1-c^2 x^2}}{20 c \sqrt {-c^2 x^2}}+\frac {1}{3} d x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{5} e x^5 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b \left (20 c^2 d-9 e\right ) x \tan ^{-1}\left (\frac {c x}{\sqrt {-1-c^2 x^2}}\right )}{120 c^4 \sqrt {-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 119, normalized size = 0.71 \[ \frac {c^2 x^2 \left (8 a c^3 x \left (5 d+3 e x^2\right )+b \sqrt {\frac {1}{c^2 x^2}+1} \left (c^2 \left (20 d+6 e x^2\right )-9 e\right )\right )+8 b c^5 x^3 \text {csch}^{-1}(c x) \left (5 d+3 e x^2\right )+b \left (9 e-20 c^2 d\right ) \log \left (x \left (\sqrt {\frac {1}{c^2 x^2}+1}+1\right )\right )}{120 c^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d + e*x^2)*(a + b*ArcCsch[c*x]),x]

[Out]

(c^2*x^2*(8*a*c^3*x*(5*d + 3*e*x^2) + b*Sqrt[1 + 1/(c^2*x^2)]*(-9*e + c^2*(20*d + 6*e*x^2))) + 8*b*c^5*x^3*(5*
d + 3*e*x^2)*ArcCsch[c*x] + b*(-20*c^2*d + 9*e)*Log[(1 + Sqrt[1 + 1/(c^2*x^2)])*x])/(120*c^5)

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fricas [A]  time = 3.02, size = 273, normalized size = 1.63 \[ \frac {24 \, a c^{5} e x^{5} + 40 \, a c^{5} d x^{3} + 8 \, {\left (5 \, b c^{5} d + 3 \, b c^{5} e\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) + {\left (20 \, b c^{2} d - 9 \, b e\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right ) - 8 \, {\left (5 \, b c^{5} d + 3 \, b c^{5} e\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) + 8 \, {\left (3 \, b c^{5} e x^{5} + 5 \, b c^{5} d x^{3} - 5 \, b c^{5} d - 3 \, b c^{5} e\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (6 \, b c^{4} e x^{4} + {\left (20 \, b c^{4} d - 9 \, b c^{2} e\right )} x^{2}\right )} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}}}{120 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)*(a+b*arccsch(c*x)),x, algorithm="fricas")

[Out]

1/120*(24*a*c^5*e*x^5 + 40*a*c^5*d*x^3 + 8*(5*b*c^5*d + 3*b*c^5*e)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x
+ 1) + (20*b*c^2*d - 9*b*e)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x) - 8*(5*b*c^5*d + 3*b*c^5*e)*log(c*x*
sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x - 1) + 8*(3*b*c^5*e*x^5 + 5*b*c^5*d*x^3 - 5*b*c^5*d - 3*b*c^5*e)*log((c*x*
sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + (6*b*c^4*e*x^4 + (20*b*c^4*d - 9*b*c^2*e)*x^2)*sqrt((c^2*x^2 + 1)/
(c^2*x^2)))/c^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)*(a+b*arccsch(c*x)),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arccsch(c*x) + a)*x^2, x)

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maple [A]  time = 0.05, size = 171, normalized size = 1.02 \[ \frac {\frac {a \left (\frac {1}{5} c^{5} x^{5} e +\frac {1}{3} c^{5} x^{3} d \right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccsch}\left (c x \right ) c^{5} x^{5} e}{5}+\frac {\mathrm {arccsch}\left (c x \right ) c^{5} x^{3} d}{3}-\frac {\sqrt {c^{2} x^{2}+1}\, \left (-6 e \,c^{3} x^{3} \sqrt {c^{2} x^{2}+1}-20 c^{3} d x \sqrt {c^{2} x^{2}+1}+20 c^{2} d \arcsinh \left (c x \right )+9 e c x \sqrt {c^{2} x^{2}+1}-9 e \arcsinh \left (c x \right )\right )}{120 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x^2+d)*(a+b*arccsch(c*x)),x)

[Out]

1/c^3*(a/c^2*(1/5*c^5*x^5*e+1/3*c^5*x^3*d)+b/c^2*(1/5*arccsch(c*x)*c^5*x^5*e+1/3*arccsch(c*x)*c^5*x^3*d-1/120*
(c^2*x^2+1)^(1/2)*(-6*e*c^3*x^3*(c^2*x^2+1)^(1/2)-20*c^3*d*x*(c^2*x^2+1)^(1/2)+20*c^2*d*arcsinh(c*x)+9*e*c*x*(
c^2*x^2+1)^(1/2)-9*e*arcsinh(c*x))/((c^2*x^2+1)/c^2/x^2)^(1/2)/c/x))

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maxima [A]  time = 0.38, size = 227, normalized size = 1.36 \[ \frac {1}{5} \, a e x^{5} + \frac {1}{3} \, a d x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \operatorname {arcsch}\left (c x\right ) + \frac {\frac {2 \, \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} - c^{2}} - \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} + \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b d + \frac {1}{80} \, {\left (16 \, x^{5} \operatorname {arcsch}\left (c x\right ) - \frac {\frac {2 \, {\left (3 \, {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {\frac {1}{c^{2} x^{2}} + 1}\right )}}{c^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )}^{2} - 2 \, c^{4} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} + c^{4}} - \frac {3 \, \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{4}} + \frac {3 \, \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{4}}}{c}\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x^2+d)*(a+b*arccsch(c*x)),x, algorithm="maxima")

[Out]

1/5*a*e*x^5 + 1/3*a*d*x^3 + 1/12*(4*x^3*arccsch(c*x) + (2*sqrt(1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) + 1) - c^2)
- log(sqrt(1/(c^2*x^2) + 1) + 1)/c^2 + log(sqrt(1/(c^2*x^2) + 1) - 1)/c^2)/c)*b*d + 1/80*(16*x^5*arccsch(c*x)
- (2*(3*(1/(c^2*x^2) + 1)^(3/2) - 5*sqrt(1/(c^2*x^2) + 1))/(c^4*(1/(c^2*x^2) + 1)^2 - 2*c^4*(1/(c^2*x^2) + 1)
+ c^4) - 3*log(sqrt(1/(c^2*x^2) + 1) + 1)/c^4 + 3*log(sqrt(1/(c^2*x^2) + 1) - 1)/c^4)/c)*b*e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(e*x**2+d)*(a+b*acsch(c*x)),x)

[Out]

Integral(x**2*(a + b*acsch(c*x))*(d + e*x**2), x)

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