3.2.69 \(\int \frac {x^3 (a+b \text {sech}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\) [169]
Optimal. Leaf size=179 \[ \frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {2 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {1-c^2 x^2}}\right )}{3 \sqrt {d} e^2} \]
[Out]
1/3*d*(a+b*arcsech(c*x))/e^2/(e*x^2+d)^(3/2)+2/3*b*arctanh((e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2+1)^(1/2))*(1/(c*x
+1))^(1/2)*(c*x+1)^(1/2)/e^2/d^(1/2)+(-a-b*arcsech(c*x))/e^2/(e*x^2+d)^(1/2)+1/3*b*(1/(c*x+1))^(1/2)*(c*x+1)^(
1/2)*(-c^2*x^2+1)^(1/2)/e/(c^2*d+e)/(e*x^2+d)^(1/2)
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Rubi [A]
time = 0.18, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {272, 45, 6436,
12, 587, 157, 95, 213} \begin {gather*} -\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac {2 b \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 )}{3 \sqrt {d} e^2}+\frac {b \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \sqrt {1-c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {d+e x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x^3*(a + b*ArcSech[c*x]))/(d + e*x^2)^(5/2),x]
[Out]
(b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(3*e*(c^2*d + e)*Sqrt[d + e*x^2]) + (d*(a + b*ArcSech
[c*x]))/(3*e^2*(d + e*x^2)^(3/2)) - (a + b*ArcSech[c*x])/(e^2*Sqrt[d + e*x^2]) + (2*b*Sqrt[(1 + c*x)^(-1)]*Sqr
t[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(3*Sqrt[d]*e^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 45
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 95
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 157
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]
Rule 213
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rule 272
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 587
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 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^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-2 d-3 e x^2}{3 e^2 x \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx\\ &=\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {-2 d-3 e x^2}{x \sqrt {1-c^2 x^2} \left (d+e x^2\right )^{3/2}} \, dx}{3 e^2}\\ &=\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {-2 d-3 e x}{x \sqrt {1-c^2 x} (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e^2}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {d \left (c^2 d+e\right )}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 d e^2 \left (c^2 d+e\right )}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {\left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 e^2}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {\left (2 b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {1}{-d+x^2} \, dx,x,\frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}}\right )}{3 e^2}\\ &=\frac {b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \sqrt {1-c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {d+e x^2}}+\frac {d \left (a+b \text {sech}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {sech}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {2 b \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 )}{3 \sqrt {d} e^2}\\ \end {align*}
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Mathematica [A]
time = 10.16, size = 218, normalized size = 1.22 \begin {gather*} \frac {b e \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (d+e x^2\right )-a \left (c^2 d+e\right ) \left (2 d+3 e x^2\right )-b \left (c^2 d+e\right ) \left (2 d+3 e x^2\right ) \text {sech}^{-1}(c x)}{3 e^2 \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\frac {2 b \sqrt {\frac {1-c x}{1+c x}} \sqrt {1-c^2 x^2} \sqrt {-d-e x^2} \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {1-c^2 x^2}}{\sqrt {-d-e x^2}}\right )}{3 \sqrt {d} e^2 (-1+c x) \sqrt {d+e x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x^3*(a + b*ArcSech[c*x]))/(d + e*x^2)^(5/2),x]
[Out]
(b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + e*x^2) - a*(c^2*d + e)*(2*d + 3*e*x^2) - b*(c^2*d + e)*(2*d + 3*
e*x^2)*ArcSech[c*x])/(3*e^2*(c^2*d + e)*(d + e*x^2)^(3/2)) - (2*b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*
Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*Sqrt[1 - c^2*x^2])/Sqrt[-d - e*x^2]])/(3*Sqrt[d]*e^2*(-1 + c*x)*Sqrt[d + e*x^
2])
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Maple [F]
time = 1.24, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(5/2),x)
[Out]
int(x^3*(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^3*(a+b*arcsech(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")
[Out]
-1/3*(3*x^2*e^(-1)/(x^2*e + d)^(3/2) + 2*d*e^(-2)/(x^2*e + d)^(3/2))*a + b*integrate(x^3*log(sqrt(1/(c*x) + 1)
*sqrt(1/(c*x) - 1) + 1/(c*x))/(x^2*e + d)^(5/2), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 898 vs.
\(2 (118) = 236\).
time = 0.64, size = 1834, normalized size = 10.25 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsech(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")
[Out]
[-1/6*(2*(2*b*c^2*d^3 + 3*b*d*x^2*cosh(1)^2 + 3*b*d*x^2*sinh(1)^2 + (3*b*c^2*d^2*x^2 + 2*b*d^2)*cosh(1) + (3*b
*c^2*d^2*x^2 + 6*b*d*x^2*cosh(1) + 2*b*d^2)*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*log((c*x*sqrt(-(c^2*x
^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (b*x^4*cosh(1)^3 + b*x^4*sinh(1)^3 + b*c^2*d^3 + (b*c^2*d*x^4 + 2*b*d*x^2)*co
sh(1)^2 + (b*c^2*d*x^4 + 3*b*x^4*cosh(1) + 2*b*d*x^2)*sinh(1)^2 + (2*b*c^2*d^2*x^2 + b*d^2)*cosh(1) + (2*b*c^2
*d^2*x^2 + 3*b*x^4*cosh(1)^2 + b*d^2 + 2*(b*c^2*d*x^4 + 2*b*d*x^2)*cosh(1))*sinh(1))*sqrt(d)*log((c^4*d^2*x^4
- 8*c^2*d^2*x^2 + x^4*cosh(1)^2 + x^4*sinh(1)^2 - 4*(c^3*d*x^3 - c*x^3*cosh(1) - c*x^3*sinh(1) - 2*c*d*x)*sqrt
(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(d)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 8*d^2 - 2*(3*c^2*d*x^4 - 4*d*x^2)*cos
h(1) - 2*(3*c^2*d*x^4 - x^4*cosh(1) - 4*d*x^2)*sinh(1))/x^4) + 2*(2*a*c^2*d^3 + 3*a*d*x^2*cosh(1)^2 + 3*a*d*x^
2*sinh(1)^2 + (3*a*c^2*d^2*x^2 + 2*a*d^2)*cosh(1) + (3*a*c^2*d^2*x^2 + 6*a*d*x^2*cosh(1) + 2*a*d^2)*sinh(1) -
(b*c*d*x^3*cosh(1)^2 + b*c*d*x^3*sinh(1)^2 + b*c*d^2*x*cosh(1) + (2*b*c*d*x^3*cosh(1) + b*c*d^2*x)*sinh(1))*sq
rt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d))/(d*x^4*cosh(1)^5 + d*x^4*sinh(1)^5 + c^2*d^
4*cosh(1)^2 + (c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1)^4 + (c^2*d^2*x^4 + 5*d*x^4*cosh(1) + 2*d^2*x^2)*sinh(1)^4 + (2
*c^2*d^3*x^2 + d^3)*cosh(1)^3 + (2*c^2*d^3*x^2 + 10*d*x^4*cosh(1)^2 + d^3 + 4*(c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1
))*sinh(1)^3 + (10*d*x^4*cosh(1)^3 + c^2*d^4 + 6*(c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1)^2 + 3*(2*c^2*d^3*x^2 + d^3)
*cosh(1))*sinh(1)^2 + (5*d*x^4*cosh(1)^4 + 2*c^2*d^4*cosh(1) + 4*(c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1)^3 + 3*(2*c^
2*d^3*x^2 + d^3)*cosh(1)^2)*sinh(1)), 1/3*((b*x^4*cosh(1)^3 + b*x^4*sinh(1)^3 + b*c^2*d^3 + (b*c^2*d*x^4 + 2*b
*d*x^2)*cosh(1)^2 + (b*c^2*d*x^4 + 3*b*x^4*cosh(1) + 2*b*d*x^2)*sinh(1)^2 + (2*b*c^2*d^2*x^2 + b*d^2)*cosh(1)
+ (2*b*c^2*d^2*x^2 + 3*b*x^4*cosh(1)^2 + b*d^2 + 2*(b*c^2*d*x^4 + 2*b*d*x^2)*cosh(1))*sinh(1))*sqrt(-d)*arctan
(-1/2*(c^3*d*x^3 - c*x^3*cosh(1) - c*x^3*sinh(1) - 2*c*d*x)*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*sqrt(-d)*sqrt(
-(c^2*x^2 - 1)/(c^2*x^2))/(c^2*d^2*x^2 - d^2 + (c^2*d*x^4 - d*x^2)*cosh(1) + (c^2*d*x^4 - d*x^2)*sinh(1))) - (
2*b*c^2*d^3 + 3*b*d*x^2*cosh(1)^2 + 3*b*d*x^2*sinh(1)^2 + (3*b*c^2*d^2*x^2 + 2*b*d^2)*cosh(1) + (3*b*c^2*d^2*x
^2 + 6*b*d*x^2*cosh(1) + 2*b*d^2)*sinh(1))*sqrt(x^2*cosh(1) + x^2*sinh(1) + d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c
^2*x^2)) + 1)/(c*x)) - (2*a*c^2*d^3 + 3*a*d*x^2*cosh(1)^2 + 3*a*d*x^2*sinh(1)^2 + (3*a*c^2*d^2*x^2 + 2*a*d^2)*
cosh(1) + (3*a*c^2*d^2*x^2 + 6*a*d*x^2*cosh(1) + 2*a*d^2)*sinh(1) - (b*c*d*x^3*cosh(1)^2 + b*c*d*x^3*sinh(1)^2
+ b*c*d^2*x*cosh(1) + (2*b*c*d*x^3*cosh(1) + b*c*d^2*x)*sinh(1))*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sqrt(x^2*cos
h(1) + x^2*sinh(1) + d))/(d*x^4*cosh(1)^5 + d*x^4*sinh(1)^5 + c^2*d^4*cosh(1)^2 + (c^2*d^2*x^4 + 2*d^2*x^2)*co
sh(1)^4 + (c^2*d^2*x^4 + 5*d*x^4*cosh(1) + 2*d^2*x^2)*sinh(1)^4 + (2*c^2*d^3*x^2 + d^3)*cosh(1)^3 + (2*c^2*d^3
*x^2 + 10*d*x^4*cosh(1)^2 + d^3 + 4*(c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1))*sinh(1)^3 + (10*d*x^4*cosh(1)^3 + c^2*d
^4 + 6*(c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1)^2 + 3*(2*c^2*d^3*x^2 + d^3)*cosh(1))*sinh(1)^2 + (5*d*x^4*cosh(1)^4 +
2*c^2*d^4*cosh(1) + 4*(c^2*d^2*x^4 + 2*d^2*x^2)*cosh(1)^3 + 3*(2*c^2*d^3*x^2 + d^3)*cosh(1)^2)*sinh(1))]
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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**3*(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^3*(a+b*arcsech(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")
[Out]
integrate((b*arcsech(c*x) + a)*x^3/(e*x^2 + d)^(5/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\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^3*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(5/2),x)
[Out]
int((x^3*(a + b*acosh(1/(c*x))))/(d + e*x^2)^(5/2), x)
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