3.2.57 \(\int \frac {x^3 (a+b \text {csch}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\) [157]
Optimal. Leaf size=169 \[ -\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {2 b c x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}} \]
[Out]
1/3*d*(a+b*arccsch(c*x))/e^2/(e*x^2+d)^(3/2)-2/3*b*c*x*arctan((e*x^2+d)^(1/2)/d^(1/2)/(-c^2*x^2-1)^(1/2))/e^2/
d^(1/2)/(-c^2*x^2)^(1/2)+(-a-b*arccsch(c*x))/e^2/(e*x^2+d)^(1/2)-1/3*b*c*x*(-c^2*x^2-1)^(1/2)/(c^2*d-e)/e/(-c^
2*x^2)^(1/2)/(e*x^2+d)^(1/2)
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Rubi [A]
time = 0.17, antiderivative size = 169, 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, 6437,
12, 587, 157, 95, 210} \begin {gather*} -\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {2 b c x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}}-\frac {b c x \sqrt {-c^2 x^2-1}}{3 e \sqrt {-c^2 x^2} \left (c^2 d-e\right ) \sqrt {d+e x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]
[Out]
-1/3*(b*c*x*Sqrt[-1 - c^2*x^2])/((c^2*d - e)*e*Sqrt[-(c^2*x^2)]*Sqrt[d + e*x^2]) + (d*(a + b*ArcCsch[c*x]))/(3
*e^2*(d + e*x^2)^(3/2)) - (a + b*ArcCsch[c*x])/(e^2*Sqrt[d + e*x^2]) - (2*b*c*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d
]*Sqrt[-1 - c^2*x^2])])/(3*Sqrt[d]*e^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 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 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[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 6437
Int[((a_.) + ArcCsch[(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*ArcCsch[c*x], u, x] - Dist[b*c*(x/Sqrt[(-c^2)*x^2]), Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), 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 {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {(b c x) \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}{\sqrt {-c^2 x^2}}\\ &=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {(b c x) \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 \sqrt {-c^2 x^2}}\\ &=\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {(b c x) \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 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {(b c x) \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 \left (c^2 d-e\right ) e^2 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {(b c x) \text {Subst}\left (\int \frac {1}{x \sqrt {-1-c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{3 e^2 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {(2 b c x) \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 \sqrt {-c^2 x^2}}\\ &=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {2 b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}}\\ \end {align*}
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Mathematica [A]
time = 0.16, size = 173, normalized size = 1.02 \begin {gather*} \frac {b c e \sqrt {1+\frac {1}{c^2 x^2}} 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 {csch}^{-1}(c x)}{3 e^2 \left (-c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}+\frac {2 b c \sqrt {1+\frac {1}{c^2 x^2}} x \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {1+c^2 x^2}}{\sqrt {d+e x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {1+c^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]
[Out]
(b*c*e*Sqrt[1 + 1/(c^2*x^2)]*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)*Arc
Csch[c*x])/(3*e^2*(-(c^2*d) + e)*(d + e*x^2)^(3/2)) + (2*b*c*Sqrt[1 + 1/(c^2*x^2)]*x*ArcTanh[(Sqrt[d]*Sqrt[1 +
c^2*x^2])/Sqrt[d + e*x^2]])/(3*Sqrt[d]*e^2*Sqrt[1 + c^2*x^2])
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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \,\mathrm {arccsch}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)
[Out]
int(x^3*(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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arccsch(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^2*x^2)
+ 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. 909 vs.
\(2 (144) = 288\).
time = 0.56, size = 1858, normalized size = 10.99 \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*arccsch(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)*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)*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)*cosh(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*s
inh(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*cosh(1) + x^2*sinh(1) + d))/(d*x^4*cosh(1)^5 + d*x^4*sinh(1)^5 - c^2*d^4*co
sh(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))*s
inh(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)*cos
h(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*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))]
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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*acsch(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*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")
[Out]
integrate((b*arccsch(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 {asinh}\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*asinh(1/(c*x))))/(d + e*x^2)^(5/2),x)
[Out]
int((x^3*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(5/2), x)
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