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

Optimal. Leaf size=115 \[ d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b x \left (6 c^2 d-e\right ) \tan ^{-1}\left (\frac {c x}{\sqrt {-c^2 x^2-1}}\right )}{6 c^2 \sqrt {-c^2 x^2}}+\frac {b e x^2 \sqrt {-c^2 x^2-1}}{6 c \sqrt {-c^2 x^2}} \]

[Out]

d*x*(a+b*arccsch(c*x))+1/3*e*x^3*(a+b*arccsch(c*x))-1/6*b*(6*c^2*d-e)*x*arctan(c*x/(-c^2*x^2-1)^(1/2))/c^2/(-c
^2*x^2)^(1/2)+1/6*b*e*x^2*(-c^2*x^2-1)^(1/2)/c/(-c^2*x^2)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6292, 12, 388, 217, 203} \[ d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b x \left (6 c^2 d-e\right ) \tan ^{-1}\left (\frac {c x}{\sqrt {-c^2 x^2-1}}\right )}{6 c^2 \sqrt {-c^2 x^2}}+\frac {b e x^2 \sqrt {-c^2 x^2-1}}{6 c \sqrt {-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e*x^2*Sqrt[-1 - c^2*x^2])/(6*c*Sqrt[-(c^2*x^2)]) + d*x*(a + b*ArcCsch[c*x]) + (e*x^3*(a + b*ArcCsch[c*x]))/
3 - (b*(6*c^2*d - e)*x*ArcTan[(c*x)/Sqrt[-1 - c^2*x^2]])/(6*c^2*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 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 388

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

Rule 6292

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

Rubi steps

\begin {align*} \int \left (d+e x^2\right ) \left (a+b \text {csch}^{-1}(c x)\right ) \, dx &=d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \int \frac {3 d+e x^2}{3 \sqrt {-1-c^2 x^2}} \, dx}{\sqrt {-c^2 x^2}}\\ &=d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {(b c x) \int \frac {3 d+e x^2}{\sqrt {-1-c^2 x^2}} \, dx}{3 \sqrt {-c^2 x^2}}\\ &=\frac {b e x^2 \sqrt {-1-c^2 x^2}}{6 c \sqrt {-c^2 x^2}}+d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {\left (b c \left (6 d-\frac {e}{c^2}\right ) x\right ) \int \frac {1}{\sqrt {-1-c^2 x^2}} \, dx}{6 \sqrt {-c^2 x^2}}\\ &=\frac {b e x^2 \sqrt {-1-c^2 x^2}}{6 c \sqrt {-c^2 x^2}}+d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {\left (b c \left (6 d-\frac {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 )}{6 \sqrt {-c^2 x^2}}\\ &=\frac {b e x^2 \sqrt {-1-c^2 x^2}}{6 c \sqrt {-c^2 x^2}}+d x \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{3} e x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b \left (6 d-\frac {e}{c^2}\right ) x \tan ^{-1}\left (\frac {c x}{\sqrt {-1-c^2 x^2}}\right )}{6 \sqrt {-c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 135, normalized size = 1.17 \[ a d x+\frac {1}{3} a e x^3+\frac {b d x \sqrt {\frac {1}{c^2 x^2}+1} \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+\frac {b e x^2 \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}}{6 c}-\frac {b e \log \left (x \left (\sqrt {\frac {c^2 x^2+1}{c^2 x^2}}+1\right )\right )}{6 c^3}+b d x \text {csch}^{-1}(c x)+\frac {1}{3} b e x^3 \text {csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (a*e*x^3)/3 + (b*e*x^2*Sqrt[(1 + c^2*x^2)/(c^2*x^2)])/(6*c) + b*d*x*ArcCsch[c*x] + (b*e*x^3*ArcCsch[c*
x])/3 + (b*d*Sqrt[1 + 1/(c^2*x^2)]*x*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - (b*e*Log[x*(1 + Sqrt[(1 + c^2*x^2)/(c^2
*x^2)])])/(6*c^3)

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fricas [B]  time = 1.01, size = 245, normalized size = 2.13 \[ \frac {2 \, a c^{3} e x^{3} + b c^{2} e x^{2} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 6 \, a c^{3} d x + 2 \, {\left (3 \, b c^{3} d + b c^{3} e\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) - {\left (6 \, b c^{2} d - b e\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right ) - 2 \, {\left (3 \, b c^{3} d + b c^{3} e\right )} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) + 2 \, {\left (b c^{3} e x^{3} + 3 \, b c^{3} d x - 3 \, b c^{3} d - b c^{3} e\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )}{6 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*e*x^3 + b*c^2*e*x^2*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 6*a*c^3*d*x + 2*(3*b*c^3*d + b*c^3*e)*log(c*x
*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x + 1) - (6*b*c^2*d - b*e)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x) - 2
*(3*b*c^3*d + b*c^3*e)*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x - 1) + 2*(b*c^3*e*x^3 + 3*b*c^3*d*x - 3*b*c
^3*d - b*c^3*e)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)))/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 126, normalized size = 1.10 \[ \frac {\frac {a \left (\frac {1}{3} e \,c^{3} x^{3}+x \,c^{3} d \right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccsch}\left (c x \right ) e \,c^{3} x^{3}}{3}+\mathrm {arccsch}\left (c x \right ) c^{3} d x +\frac {\sqrt {c^{2} x^{2}+1}\, \left (6 c^{2} d \arcsinh \left (c x \right )+e c x \sqrt {c^{2} x^{2}+1}-e \arcsinh \left (c x \right )\right )}{6 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 148, normalized size = 1.29 \[ \frac {1}{3} \, a e x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \operatorname {arcsch}\left (c x\right ) + \frac {\frac {2 \, \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} - c^{2}} - \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} + \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b e + a d x + \frac {{\left (2 \, c x \operatorname {arcsch}\left (c x\right ) + \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right ) - \log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right )\right )} b d}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e*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*e + a*d*x + 1/2*(2*c*x*arccsch(c*x) + log(sq
rt(1/(c^2*x^2) + 1) + 1) - log(sqrt(1/(c^2*x^2) + 1) - 1))*b*d/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((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 \[ \int \left (a + b \operatorname {acsch}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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