3.2.75 \(\int \frac {x^2 (a+b \text {sech}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\) [175]

Optimal. Leaf size=246 \[ -\frac {b x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b c \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2} E\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{3 d e \left (c^2 d+e\right ) \sqrt {1+\frac {e x^2}{d}}}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{3 c d e \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {270, 6436, 12, 482, 434, 437, 435, 432, 430} \begin {gather*} \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {\frac {e x^2}{d}+1} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{3 c d e \sqrt {d+e x^2}}-\frac {b c \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {d+e x^2} E\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{3 d e \left (c^2 d+e\right ) \sqrt {\frac {e x^2}{d}+1}}-\frac {b x \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*x*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(d*(c^2*d + e)*Sqrt[d + e*x^2]) + (x^3*(a + b*
ArcSech[c*x]))/(3*d*(d + e*x^2)^(3/2)) - (b*c*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[d + e*x^2]*EllipticE[Arc
Sin[c*x], -(e/(c^2*d))])/(3*d*e*(c^2*d + e)*Sqrt[1 + (e*x^2)/d]) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[
1 + (e*x^2)/d]*EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(3*c*d*e*Sqrt[d + e*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 270

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

Rule 430

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]
))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && Gt
Q[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])

Rule 432

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

Rule 434

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

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 437

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2]
, Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 482

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

Rule 6436

Int[((a_.) + ArcSech[(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*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)],
Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), 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 \frac {x^2 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^2}{3 d \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx\\ &=\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 d}\\ &=-\frac {b x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {1-c^2 x^2}}{\sqrt {d+e x^2}} \, dx}{3 d \left (c^2 d+e\right )}\\ &=-\frac {b x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \, dx}{3 d e}-\frac {\left (b c^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}} \, dx}{3 d e \left (c^2 d+e\right )}\\ &=-\frac {b x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {\left (b c^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2}\right ) \int \frac {\sqrt {1+\frac {e x^2}{d}}}{\sqrt {1-c^2 x^2}} \, dx}{3 d e \left (c^2 d+e\right ) \sqrt {1+\frac {e x^2}{d}}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{3 d e \sqrt {d+e x^2}}\\ &=-\frac {b x \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 d \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b c \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 d e \left (c^2 d+e\right ) \sqrt {1+\frac {e x^2}{d}}}+\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1+\frac {e x^2}{d}} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{3 c d e \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.79, size = 488, normalized size = 1.98 \begin {gather*} \frac {a x^3-\frac {b \sqrt {\frac {1-c x}{1+c x}} (-c d+e x) \left (d+e x^2\right )}{e \left (c^2 d+e\right )}+b x^3 \text {sech}^{-1}(c x)+\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \sqrt {\frac {c \left (\sqrt {d}+i \sqrt {e} x\right )}{\left (c \sqrt {d}+i \sqrt {e}\right ) (1+c x)}} \sqrt {\frac {c \left (i \sqrt {d}+\sqrt {e} x\right )}{\left (i c \sqrt {d}+\sqrt {e}\right ) (1+c x)}} \left (d+e x^2\right ) \left (\left (i c \sqrt {d}+\sqrt {e}\right ) E\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (c^2 d+e\right ) (1-c x)}{\left (c \sqrt {d}+i \sqrt {e}\right )^2 (1+c x)}}\right )|\frac {\left (c \sqrt {d}+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )-2 \sqrt {e} F\left (i \sinh ^{-1}\left (\sqrt {\frac {\left (c^2 d+e\right ) (1-c x)}{\left (c \sqrt {d}+i \sqrt {e}\right )^2 (1+c x)}}\right )|\frac {\left (c \sqrt {d}+i \sqrt {e}\right )^2}{\left (c \sqrt {d}-i \sqrt {e}\right )^2}\right )\right )}{c \left (c \sqrt {d}+i \sqrt {e}\right ) e \sqrt {\frac {\left (i c \sqrt {d}+\sqrt {e}\right ) (-1+c x)}{\left (-i c \sqrt {d}+\sqrt {e}\right ) (1+c x)}}}}{3 d \left (d+e x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 1.12, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [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^2*(a+b*arcsech(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [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^2*(a+b*arcsech(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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