\(\int \frac {x^3 (a+b \text {csch}^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\) [158]

Optimal result

Integrand size = 23, antiderivative size = 169 \[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {b c x \sqrt {-1-c^2 x^2}}{3 \left (c^2 d-e\right ) e \sqrt {-c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}-\frac {2 b c x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1-c^2 x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {-c^2 x^2}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.45 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {\frac {b c e \sqrt {1+\frac {1}{c^2 x^2}} x \left (d+e x^2\right )}{c^2 d-e}+a \left (2 d+3 e x^2\right )-\frac {b \sqrt {1+\frac {d}{e x^2}} \left (d+e x^2\right ) \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,-\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )}{c x}+b \left (2 d+3 e x^2\right ) \text {csch}^{-1}(c x)}{3 e^2 \left (d+e x^2\right )^{3/2}} \] Input:

Integrate[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]
 

Output:

-1/3*((b*c*e*Sqrt[1 + 1/(c^2*x^2)]*x*(d + e*x^2))/(c^2*d - e) + a*(2*d + 3 
*e*x^2) - (b*Sqrt[1 + d/(e*x^2)]*(d + e*x^2)*AppellF1[1, 1/2, 1/2, 2, -(1/ 
(c^2*x^2)), -(d/(e*x^2))])/(c*x) + b*(2*d + 3*e*x^2)*ArcCsch[c*x])/(e^2*(d 
 + e*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6856, 27, 435, 169, 27, 104, 217}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 6856

\(\displaystyle -\frac {b c x \int -\frac {3 e x^2+2 d}{3 e^2 x \sqrt {-c^2 x^2-1} \left (e x^2+d\right )^{3/2}}dx}{\sqrt {-c^2 x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c x \int \frac {3 e x^2+2 d}{x \sqrt {-c^2 x^2-1} \left (e x^2+d\right )^{3/2}}dx}{3 e^2 \sqrt {-c^2 x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 435

\(\displaystyle \frac {b c x \int \frac {3 e x^2+2 d}{x^2 \sqrt {-c^2 x^2-1} \left (e x^2+d\right )^{3/2}}dx^2}{6 e^2 \sqrt {-c^2 x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {b c x \left (\frac {2 \int \frac {d \left (c^2 d-e\right )}{x^2 \sqrt {-c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{d \left (c^2 d-e\right )}-\frac {2 e \sqrt {-c^2 x^2-1}}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{6 e^2 \sqrt {-c^2 x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c x \left (2 \int \frac {1}{x^2 \sqrt {-c^2 x^2-1} \sqrt {e x^2+d}}dx^2-\frac {2 e \sqrt {-c^2 x^2-1}}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{6 e^2 \sqrt {-c^2 x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {b c x \left (4 \int \frac {1}{-x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {-c^2 x^2-1}}-\frac {2 e \sqrt {-c^2 x^2-1}}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{6 e^2 \sqrt {-c^2 x^2}}-\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {a+b \text {csch}^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \text {csch}^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac {b c x \left (-\frac {4 \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-c^2 x^2-1}}\right )}{\sqrt {d}}-\frac {2 e \sqrt {-c^2 x^2-1}}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}\right )}{6 e^2 \sqrt {-c^2 x^2}}\)

Input:

Int[(x^3*(a + b*ArcCsch[c*x]))/(d + e*x^2)^(5/2),x]
 

Output:

(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]) + (b*c*x*((-2*e*Sqrt[-1 - c^2*x^2])/((c^2*d - e)*Sqr 
t[d + e*x^2]) - (4*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 - c^2*x^2])])/S 
qrt[d]))/(6*e^2*Sqrt[-(c^2*x^2)])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S 
imp[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 217

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 435

Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^2)^(p_.)*((c_.) + (d_.)*(x_)^2)^(q_.)*(( 
e_.) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2) 
*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^2], x] /; FreeQ[{a, b, c, d, 
 e, f, p, q, r}, x] && IntegerQ[(m - 1)/2]
 

rule 6856

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]}, Si 
mp[(a + b*ArcCsch[c*x])   u, x] - Simp[b*c*(x/Sqrt[(-c^2)*x^2])   Int[Simpl 
ifyIntegrand[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])) || (I 
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))
 

Maple [F]

\[\int \frac {x^{3} \left (a +b \,\operatorname {arccsch}\left (c x \right )\right )}{\left (x^{2} e +d \right )^{\frac {5}{2}}}d x\]

Input:

int(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)
 

Output:

int(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (143) = 286\).

Time = 0.20 (sec) , antiderivative size = 786, normalized size of antiderivative = 4.65 \[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")
 

Output:

[-1/6*(2*(2*b*c^2*d^3 - 2*b*d^2*e + 3*(b*c^2*d^2*e - b*d*e^2)*x^2)*sqrt(e* 
x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c^2*d^3 + 
 (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*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*(2*a*c^2*d^3 - 2*a*d^2*e + 3*(a*c^2*d^2*e - a*d*e^2 
)*x^2 + (b*c*d*e^2*x^3 + b*c*d^2*e*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*sqrt( 
e*x^2 + d))/(c^2*d^4*e^2 - d^3*e^3 + (c^2*d^2*e^4 - d*e^5)*x^4 + 2*(c^2*d^ 
3*e^3 - d^2*e^4)*x^2), -1/3*((b*c^2*d^3 + (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^ 
2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*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)) + (2*b*c^2*d^3 - 2*b*d^2*e + 3*(b*c^2 
*d^2*e - b*d*e^2)*x^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^ 
2)) + 1)/(c*x)) + (2*a*c^2*d^3 - 2*a*d^2*e + 3*(a*c^2*d^2*e - a*d*e^2)*x^2 
 + (b*c*d*e^2*x^3 + b*c*d^2*e*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))*sqrt(e*x^2 
 + d))/(c^2*d^4*e^2 - d^3*e^3 + (c^2*d^2*e^4 - d*e^5)*x^4 + 2*(c^2*d^3*e^3 
 - d^2*e^4)*x^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(x**3*(a+b*acsch(c*x))/(e*x**2+d)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^3*(a+b*arccsch(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \] Input:

int((x^3*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(5/2),x)
 

Output:

int((x^3*(a + b*asinh(1/(c*x))))/(d + e*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {x^3 \left (a+b \text {csch}^{-1}(c x)\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {-2 \sqrt {e \,x^{2}+d}\, a d -3 \sqrt {e \,x^{2}+d}\, a e \,x^{2}+3 \left (\int \frac {\mathit {acsch} \left (c x \right ) x^{3}}{\sqrt {e \,x^{2}+d}\, d^{2}+2 \sqrt {e \,x^{2}+d}\, d e \,x^{2}+\sqrt {e \,x^{2}+d}\, e^{2} x^{4}}d x \right ) b \,d^{2} e^{2}+6 \left (\int \frac {\mathit {acsch} \left (c x \right ) x^{3}}{\sqrt {e \,x^{2}+d}\, d^{2}+2 \sqrt {e \,x^{2}+d}\, d e \,x^{2}+\sqrt {e \,x^{2}+d}\, e^{2} x^{4}}d x \right ) b d \,e^{3} x^{2}+3 \left (\int \frac {\mathit {acsch} \left (c x \right ) x^{3}}{\sqrt {e \,x^{2}+d}\, d^{2}+2 \sqrt {e \,x^{2}+d}\, d e \,x^{2}+\sqrt {e \,x^{2}+d}\, e^{2} x^{4}}d x \right ) b \,e^{4} x^{4}}{3 e^{2} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:

int(x^3*(a+b*acsch(c*x))/(e*x^2+d)^(5/2),x)
 

Output:

( - 2*sqrt(d + e*x**2)*a*d - 3*sqrt(d + e*x**2)*a*e*x**2 + 3*int((acsch(c* 
x)*x**3)/(sqrt(d + e*x**2)*d**2 + 2*sqrt(d + e*x**2)*d*e*x**2 + sqrt(d + e 
*x**2)*e**2*x**4),x)*b*d**2*e**2 + 6*int((acsch(c*x)*x**3)/(sqrt(d + e*x** 
2)*d**2 + 2*sqrt(d + e*x**2)*d*e*x**2 + sqrt(d + e*x**2)*e**2*x**4),x)*b*d 
*e**3*x**2 + 3*int((acsch(c*x)*x**3)/(sqrt(d + e*x**2)*d**2 + 2*sqrt(d + e 
*x**2)*d*e*x**2 + sqrt(d + e*x**2)*e**2*x**4),x)*b*e**4*x**4)/(3*e**2*(d** 
2 + 2*d*e*x**2 + e**2*x**4))