3.7.50 \(\int \frac {a+b \text {arcsinh}(c x)}{(d+e x^2)^{5/2}} \, dx\) [650]

3.7.50.1 Optimal result
3.7.50.2 Mathematica [C] (verified)
3.7.50.3 Rubi [A] (verified)
3.7.50.4 Maple [F]
3.7.50.5 Fricas [B] (verification not implemented)
3.7.50.6 Sympy [F]
3.7.50.7 Maxima [F]
3.7.50.8 Giac [F]
3.7.50.9 Mupad [F(-1)]

3.7.50.1 Optimal result

Integrand size = 20, antiderivative size = 146 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {b c \sqrt {1+c^2 x^2}}{3 d \left (c^2 d-e\right ) \sqrt {d+e x^2}}+\frac {x (a+b \text {arcsinh}(c x))}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+e x^2}}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {e} \sqrt {1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{3 d^2 \sqrt {e}} \]

output
1/3*x*(a+b*arcsinh(c*x))/d/(e*x^2+d)^(3/2)-2/3*b*arctanh(e^(1/2)*(c^2*x^2+ 
1)^(1/2)/c/(e*x^2+d)^(1/2))/d^2/e^(1/2)+2/3*x*(a+b*arcsinh(c*x))/d^2/(e*x^ 
2+d)^(1/2)-1/3*b*c*(c^2*x^2+1)^(1/2)/d/(c^2*d-e)/(e*x^2+d)^(1/2)
 
3.7.50.2 Mathematica [C] (verified)

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

Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {-\frac {b c d \sqrt {1+c^2 x^2} \left (d+e x^2\right )}{c^2 d-e}+a x \left (3 d+2 e x^2\right )-b c x^2 \left (d+e x^2\right ) \sqrt {1+\frac {e x^2}{d}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,-c^2 x^2,-\frac {e x^2}{d}\right )+b x \left (3 d+2 e x^2\right ) \text {arcsinh}(c x)}{3 d^2 \left (d+e x^2\right )^{3/2}} \]

input
Integrate[(a + b*ArcSinh[c*x])/(d + e*x^2)^(5/2),x]
 
output
(-((b*c*d*Sqrt[1 + c^2*x^2]*(d + e*x^2))/(c^2*d - e)) + a*x*(3*d + 2*e*x^2 
) - b*c*x^2*(d + e*x^2)*Sqrt[1 + (e*x^2)/d]*AppellF1[1, 1/2, 1/2, 2, -(c^2 
*x^2), -((e*x^2)/d)] + b*x*(3*d + 2*e*x^2)*ArcSinh[c*x])/(3*d^2*(d + e*x^2 
)^(3/2))
 
3.7.50.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6207, 27, 435, 87, 66, 221}

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.

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

\(\Big \downarrow \) 6207

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 435

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

input
Int[(a + b*ArcSinh[c*x])/(d + e*x^2)^(5/2),x]
 
output
(x*(a + b*ArcSinh[c*x]))/(3*d*(d + e*x^2)^(3/2)) + (2*x*(a + b*ArcSinh[c*x 
]))/(3*d^2*Sqrt[d + e*x^2]) - (b*c*((2*d*Sqrt[1 + c^2*x^2])/((c^2*d - e)*S 
qrt[d + e*x^2]) + (4*ArcTanh[(Sqrt[e]*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + e*x^2 
])])/(c*Sqrt[e])))/(6*d^2)
 

3.7.50.3.1 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 66
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 6207
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symb 
ol] :> With[{u = IntHide[(d + e*x^2)^p, x]}, Simp[(a + b*ArcSinh[c*x])   u, 
 x] - Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /; 
FreeQ[{a, b, c, d, e}, x] && NeQ[e, c^2*d] && (IGtQ[p, 0] || ILtQ[p + 1/2, 
0])
 
3.7.50.4 Maple [F]

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

input
int((a+b*arcsinh(c*x))/(e*x^2+d)^(5/2),x)
 
output
int((a+b*arcsinh(c*x))/(e*x^2+d)^(5/2),x)
 
3.7.50.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (122) = 244\).

Time = 0.32 (sec) , antiderivative size = 738, normalized size of antiderivative = 5.05 \[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx=\left [\frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {e} \log \left (8 \, c^{4} e^{2} x^{4} + c^{4} d^{2} + 6 \, c^{2} d e + 8 \, {\left (c^{4} d e + c^{2} e^{2}\right )} x^{2} - 4 \, {\left (2 \, c^{3} e x^{2} + c^{3} d + c e\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {e x^{2} + d} \sqrt {e} + e^{2}\right ) + 2 \, {\left (2 \, {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{3} + 3 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x\right )} \sqrt {e x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (2 \, {\left (a c^{2} d e^{2} - a e^{3}\right )} x^{3} + 3 \, {\left (a c^{2} d^{2} e - a d e^{2}\right )} x - {\left (b c d e^{2} x^{2} + b c d^{2} e\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{2} d^{5} e - d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} - d^{3} e^{3}\right )} x^{2}\right )}}, \frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt {-e} \arctan \left (\frac {{\left (2 \, c^{2} e x^{2} + c^{2} d + e\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {e x^{2} + d} \sqrt {-e}}{2 \, {\left (c^{3} e^{2} x^{4} + c d e + {\left (c^{3} d e + c e^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, {\left (b c^{2} d e^{2} - b e^{3}\right )} x^{3} + 3 \, {\left (b c^{2} d^{2} e - b d e^{2}\right )} x\right )} \sqrt {e x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (2 \, {\left (a c^{2} d e^{2} - a e^{3}\right )} x^{3} + 3 \, {\left (a c^{2} d^{2} e - a d e^{2}\right )} x - {\left (b c d e^{2} x^{2} + b c d^{2} e\right )} \sqrt {c^{2} x^{2} + 1}\right )} \sqrt {e x^{2} + d}}{3 \, {\left (c^{2} d^{5} e - d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} - d^{3} e^{3}\right )} x^{2}\right )}}\right ] \]

input
integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")
 
output
[1/6*((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(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^3*e*x^2 + c^3*d + c*e)*sqrt(c^2*x^2 + 1)*sqrt(e*x 
^2 + d)*sqrt(e) + e^2) + 2*(2*(b*c^2*d*e^2 - b*e^3)*x^3 + 3*(b*c^2*d^2*e - 
 b*d*e^2)*x)*sqrt(e*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(2*(a*c^2*d* 
e^2 - a*e^3)*x^3 + 3*(a*c^2*d^2*e - a*d*e^2)*x - (b*c*d*e^2*x^2 + b*c*d^2* 
e)*sqrt(c^2*x^2 + 1))*sqrt(e*x^2 + d))/(c^2*d^5*e - d^4*e^2 + (c^2*d^3*e^3 
 - d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 - d^3*e^3)*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(-e)*ar 
ctan(1/2*(2*c^2*e*x^2 + c^2*d + e)*sqrt(c^2*x^2 + 1)*sqrt(e*x^2 + d)*sqrt( 
-e)/(c^3*e^2*x^4 + c*d*e + (c^3*d*e + c*e^2)*x^2)) + (2*(b*c^2*d*e^2 - b*e 
^3)*x^3 + 3*(b*c^2*d^2*e - b*d*e^2)*x)*sqrt(e*x^2 + d)*log(c*x + sqrt(c^2* 
x^2 + 1)) + (2*(a*c^2*d*e^2 - a*e^3)*x^3 + 3*(a*c^2*d^2*e - a*d*e^2)*x - ( 
b*c*d*e^2*x^2 + b*c*d^2*e)*sqrt(c^2*x^2 + 1))*sqrt(e*x^2 + d))/(c^2*d^5*e 
- d^4*e^2 + (c^2*d^3*e^3 - d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 - d^3*e^3)*x^2)]
 
3.7.50.6 Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \]

input
integrate((a+b*asinh(c*x))/(e*x**2+d)**(5/2),x)
 
output
Integral((a + b*asinh(c*x))/(d + e*x**2)**(5/2), x)
 
3.7.50.7 Maxima [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

input
integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")
 
output
1/3*a*(2*x/(sqrt(e*x^2 + d)*d^2) + x/((e*x^2 + d)^(3/2)*d)) + b*integrate( 
log(c*x + sqrt(c^2*x^2 + 1))/(e*x^2 + d)^(5/2), x)
 
3.7.50.8 Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

input
integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")
 
output
integrate((b*arcsinh(c*x) + a)/(e*x^2 + d)^(5/2), x)
 
3.7.50.9 Mupad [F(-1)]

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

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