3.160 \(\int \frac{x^3 (a+b \text{sech}^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\)
Optimal. Leaf size=177 \[ \frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{c e^{3/2}}-\frac{2 b \sqrt{d} \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{e^2} \]
[Out]
(d*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*ArcSech[c*x]))/e^2 - (b*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(c*e^(3/2)) - (2*b*Sqrt[d]*Sqrt
[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/e^2
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Rubi [A] time = 0.272205, antiderivative size = 177, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used
= {266, 43, 6301, 12, 573, 157, 63, 217, 203, 93, 207} \[ \frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{c e^{3/2}}-\frac{2 b \sqrt{d} \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{e^2} \]
Antiderivative was successfully verified.
[In]
Int[(x^3*(a + b*ArcSech[c*x]))/(d + e*x^2)^(3/2),x]
[Out]
(d*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*ArcSech[c*x]))/e^2 - (b*Sqrt[(1 + c*x
)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(c*e^(3/2)) - (2*b*Sqrt[d]*Sqrt
[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/e^2
Rule 266
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rule 6301
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]))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 573
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n],
x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 157
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
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 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 93
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]
Rule 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{x^3 \left (a+b \text{sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{2 d+e x^2}{e^2 x \sqrt{1-c^2 x^2} \sqrt{d+e x^2}} \, dx\\ &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{2 d+e x^2}{x \sqrt{1-c^2 x^2} \sqrt{d+e x^2}} \, dx}{e^2}\\ &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{2 d+e x}{x \sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{2 e^2}\\ &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\frac{\left (b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{e^2}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{2 e}\\ &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}+\frac{\left (2 b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{-d+x^2} \, dx,x,\frac{\sqrt{d+e x^2}}{\sqrt{1-c^2 x^2}}\right )}{e^2}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d+\frac{e}{c^2}-\frac{e x^2}{c^2}}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{c^2 e}\\ &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}-\frac{2 b \sqrt{d} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{e^2}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{e x^2}{c^2}} \, dx,x,\frac{\sqrt{1-c^2 x^2}}{\sqrt{d+e x^2}}\right )}{c^2 e}\\ &=\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{\sqrt{d+e x^2} \left (a+b \text{sech}^{-1}(c x)\right )}{e^2}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{c e^{3/2}}-\frac{2 b \sqrt{d} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{e^2}\\ \end{align*}
Mathematica [A] time = 1.6662, size = 249, normalized size = 1.41 \[ \frac{\left (2 d+e x^2\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{e^2 \sqrt{d+e x^2}}+\frac{b \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \left (\sqrt{-c^2} \sqrt{e} \sqrt{c^2 (-d)-e} \sqrt{\frac{c^2 \left (d+e x^2\right )}{c^2 d+e}} \sin ^{-1}\left (\frac{c \sqrt{e} \sqrt{1-c^2 x^2}}{\sqrt{-c^2} \sqrt{c^2 (-d)-e}}\right )+2 c^3 \sqrt{d} \sqrt{-d-e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-c^2 x^2}}{\sqrt{-d-e x^2}}\right )\right )}{c^3 e^2 (c x-1) \sqrt{d+e x^2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(x^3*(a + b*ArcSech[c*x]))/(d + e*x^2)^(3/2),x]
[Out]
((2*d + e*x^2)*(a + b*ArcSech[c*x]))/(e^2*Sqrt[d + e*x^2]) + (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*(S
qrt[-c^2]*Sqrt[-(c^2*d) - e]*Sqrt[e]*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSin[(c*Sqrt[e]*Sqrt[1 - c^2*x^2])/
(Sqrt[-c^2]*Sqrt[-(c^2*d) - e])] + 2*c^3*Sqrt[d]*Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d -
e*x^2]]))/(c^3*e^2*(-1 + c*x)*Sqrt[d + e*x^2])
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Maple [F] time = 1.487, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x)
[Out]
int(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.88903, size = 2867, normalized size = 16.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")
[Out]
[-1/4*((b*e*x^2 + b*d)*sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4
*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + e^2) - 4*(b*c*e*x^2 + 2*
b*c*d)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(b*c*e*x^2 + b*c*d)*sqrt(d)*log
(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sq
rt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2)/x^4) - 4*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d
*e^2), -1/2*((b*e*x^2 + b*d)*sqrt(e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt(-(c
^2*x^2 - 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)) - 2*(b*c*e*x^2 + 2*b*c*d)*sqrt(e*x^2 + d)*lo
g((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c*e*x^2 + b*c*d)*sqrt(d)*log(((c^4*d^2 - 6*c^2*d*e + e^
2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt(-(c^2*x^2 - 1)/(
c^2*x^2)) + 8*d^2)/x^4) - 2*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2), -1/4*(4*(b*c*e*x^2 +
b*c*d)*sqrt(-d)*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt(-(c^2*x^2 - 1)/(c^2*x
^2))/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (b*e*x^2 + b*d)*sqrt(-e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2
*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^4*e*x^3 + (c^4*d - c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(-e)*sqrt(-(c^2*x^2
- 1)/(c^2*x^2)) + e^2) - 4*(b*c*e*x^2 + 2*b*c*d)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)
/(c*x)) - 4*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 + d))/(c*e^3*x^2 + c*d*e^2), -1/2*(2*(b*c*e*x^2 + b*c*d)*sqrt(-d)
*arctan(-1/2*((c^3*d - c*e)*x^3 - 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d*e*x^
4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (b*e*x^2 + b*d)*sqrt(e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d - e)*x)*sqrt(e*x^2
+ d)*sqrt(e)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)) - 2*(b*c*e*x^2 + 2*b*c*
d)*sqrt(e*x^2 + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(a*c*e*x^2 + 2*a*c*d)*sqrt(e*x^2 +
d))/(c*e^3*x^2 + c*d*e^2)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**3*(a+b*asech(c*x))/(e*x**2+d)**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsech}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")
[Out]
integrate((b*arcsech(c*x) + a)*x^3/(e*x^2 + d)^(3/2), x)