3.164 \(\int \frac {a+b \text {csch}^{-1}(c x)}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=278 \[ \frac {2 x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b c \sqrt {e} x \sqrt {-c^2 x^2-1} E\left (\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )|1-\frac {c^2 d}{e}\right )}{3 d^{3/2} \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2} \sqrt {\frac {d \left (c^2 x^2+1\right )}{d+e x^2}}}-\frac {b x \left (3 c^2 d-2 e\right ) \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d^3 \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right ) \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}} \]

[Out]

1/3*x*(a+b*arccsch(c*x))/d/(e*x^2+d)^(3/2)+2/3*x*(a+b*arccsch(c*x))/d^2/(e*x^2+d)^(1/2)-1/3*b*c*x*(1/(1+e*x^2/
d))^(1/2)*(1+e*x^2/d)^(1/2)*EllipticE(x*e^(1/2)/d^(1/2)/(1+e*x^2/d)^(1/2),(1-c^2*d/e)^(1/2))*e^(1/2)*(-c^2*x^2
-1)^(1/2)/d^(3/2)/(c^2*d-e)/(-c^2*x^2)^(1/2)/(d*(c^2*x^2+1)/(e*x^2+d))^(1/2)/(e*x^2+d)^(1/2)-1/3*b*(3*c^2*d-2*
e)*x*(1/(c^2*x^2+1))^(1/2)*(c^2*x^2+1)^(1/2)*EllipticF(c*x/(c^2*x^2+1)^(1/2),(1-e/c^2/d)^(1/2))*(e*x^2+d)^(1/2
)/d^3/(c^2*d-e)/(-c^2*x^2)^(1/2)/(-c^2*x^2-1)^(1/2)/((e*x^2+d)/d/(c^2*x^2+1))^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {192, 191, 6292, 12, 525, 418, 411} \[ \frac {2 x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b x \left (3 c^2 d-2 e\right ) \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d^3 \sqrt {-c^2 x^2} \sqrt {-c^2 x^2-1} \left (c^2 d-e\right ) \sqrt {\frac {d+e x^2}{d \left (c^2 x^2+1\right )}}}-\frac {b c \sqrt {e} x \sqrt {-c^2 x^2-1} E\left (\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )|1-\frac {c^2 d}{e}\right )}{3 d^{3/2} \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2} \sqrt {\frac {d \left (c^2 x^2+1\right )}{d+e x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/(d + e*x^2)^(5/2),x]

[Out]

(x*(a + b*ArcCsch[c*x]))/(3*d*(d + e*x^2)^(3/2)) + (2*x*(a + b*ArcCsch[c*x]))/(3*d^2*Sqrt[d + e*x^2]) - (b*c*S
qrt[e]*x*Sqrt[-1 - c^2*x^2]*EllipticE[ArcTan[(Sqrt[e]*x)/Sqrt[d]], 1 - (c^2*d)/e])/(3*d^(3/2)*(c^2*d - e)*Sqrt
[-(c^2*x^2)]*Sqrt[(d*(1 + c^2*x^2))/(d + e*x^2)]*Sqrt[d + e*x^2]) - (b*(3*c^2*d - 2*e)*x*Sqrt[d + e*x^2]*Ellip
ticF[ArcTan[c*x], 1 - e/(c^2*d)])/(3*d^3*(c^2*d - e)*Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2]*Sqrt[(d + e*x^2)/(d*(
1 + 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 191

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

Rule 192

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

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

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 \frac {a+b \text {csch}^{-1}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {3 d+2 e x^2}{3 d^2 \sqrt {-1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{\sqrt {-c^2 x^2}}\\ &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {3 d+2 e x^2}{\sqrt {-1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 d^2 \sqrt {-c^2 x^2}}\\ &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}-\frac {\left (b c \left (3 c^2 d-2 e\right ) x\right ) \int \frac {1}{\sqrt {-1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{3 d^2 \left (c^2 d-e\right ) \sqrt {-c^2 x^2}}-\frac {(b c e x) \int \frac {\sqrt {-1-c^2 x^2}}{\left (d+e x^2\right )^{3/2}} \, dx}{3 d \left (c^2 d-e\right ) \sqrt {-c^2 x^2}}\\ &=\frac {x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \text {csch}^{-1}(c x)\right )}{3 d^2 \sqrt {d+e x^2}}-\frac {b c \sqrt {e} x \sqrt {-1-c^2 x^2} E\left (\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )|1-\frac {c^2 d}{e}\right )}{3 d^{3/2} \left (c^2 d-e\right ) \sqrt {-c^2 x^2} \sqrt {\frac {d \left (1+c^2 x^2\right )}{d+e x^2}} \sqrt {d+e x^2}}-\frac {b \left (3 c^2 d-2 e\right ) x \sqrt {d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac {e}{c^2 d}\right )}{3 d^3 \left (c^2 d-e\right ) \sqrt {-c^2 x^2} \sqrt {-1-c^2 x^2} \sqrt {\frac {d+e x^2}{d \left (1+c^2 x^2\right )}}}\\ \end {align*}

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Mathematica [C]  time = 0.56, size = 248, normalized size = 0.89 \[ \frac {x \left (a \left (c^2 d-e\right ) \left (3 d+2 e x^2\right )-b c e x \sqrt {\frac {1}{c^2 x^2}+1} \left (d+e x^2\right )+b \left (c^2 d-e\right ) \text {csch}^{-1}(c x) \left (3 d+2 e x^2\right )\right )}{3 d^2 \left (c^2 d-e\right ) \left (d+e x^2\right )^{3/2}}-\frac {i b c x \sqrt {\frac {1}{c^2 x^2}+1} \sqrt {\frac {e x^2}{d}+1} \left (2 \left (c^2 d-e\right ) F\left (i \sinh ^{-1}\left (\sqrt {c^2} x\right )|\frac {e}{c^2 d}\right )+c^2 d E\left (i \sinh ^{-1}\left (\sqrt {c^2} x\right )|\frac {e}{c^2 d}\right )\right )}{3 \sqrt {c^2} d^2 \sqrt {c^2 x^2+1} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/(d + e*x^2)^(5/2),x]

[Out]

(x*(-(b*c*e*Sqrt[1 + 1/(c^2*x^2)]*x*(d + e*x^2)) + a*(c^2*d - e)*(3*d + 2*e*x^2) + b*(c^2*d - e)*(3*d + 2*e*x^
2)*ArcCsch[c*x]))/(3*d^2*(c^2*d - e)*(d + e*x^2)^(3/2)) - ((I/3)*b*c*Sqrt[1 + 1/(c^2*x^2)]*x*Sqrt[1 + (e*x^2)/
d]*(c^2*d*EllipticE[I*ArcSinh[Sqrt[c^2]*x], e/(c^2*d)] + 2*(c^2*d - e)*EllipticF[I*ArcSinh[Sqrt[c^2]*x], e/(c^
2*d)]))/(Sqrt[c^2]*d^2*(c^2*d - e)*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )}}{e^{3} x^{6} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x^2 + d)*(b*arccsch(c*x) + a)/(e^3*x^6 + 3*d*e^2*x^4 + 3*d^2*e*x^2 + d^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arccsch}\left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int((a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a {\left (\frac {2 \, x}{\sqrt {e x^{2} + d} d^{2}} + \frac {x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d}\right )} + b \int \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + \frac {1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/3*a*(2*x/(sqrt(e*x^2 + d)*d^2) + x/((e*x^2 + d)^(3/2)*d)) + b*integrate(log(sqrt(1/(c^2*x^2) + 1) + 1/(c*x))
/(e*x^2 + d)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(c*x))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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