\(\int \frac {a+b \text {arcsinh}(c x)}{x^2 (d+c^2 d x^2)^{5/2}} \, dx\) [181]

Optimal result

Integrand size = 26, antiderivative size = 214 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {b c}{6 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {a+b \text {arcsinh}(c x)}{3 d x \left (d+c^2 d x^2\right )^{3/2}}+\frac {4 (a+b \text {arcsinh}(c x))}{3 d^2 x \sqrt {d+c^2 d x^2}}-\frac {8 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{3 d^3 x}+\frac {b c \sqrt {1+c^2 x^2} \log (x)}{d^2 \sqrt {d+c^2 d x^2}}+\frac {5 b c \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 d^2 \sqrt {d+c^2 d x^2}} \] Output:

-1/6*b*c/d^2/(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)+1/3*(a+b*arcsinh(c*x))/ 
d/x/(c^2*d*x^2+d)^(3/2)+4/3*(a+b*arcsinh(c*x))/d^2/x/(c^2*d*x^2+d)^(1/2)-8 
/3*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(c*x))/d^3/x+b*c*(c^2*x^2+1)^(1/2)*ln(x 
)/d^2/(c^2*d*x^2+d)^(1/2)+5/6*b*c*(c^2*x^2+1)^(1/2)*ln(c^2*x^2+1)/d^2/(c^2 
*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.06 \[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d+c^2 d x^2} \left (b c x+b c^3 x^3+6 a \sqrt {1+c^2 x^2}+24 a c^2 x^2 \sqrt {1+c^2 x^2}+16 a c^4 x^4 \sqrt {1+c^2 x^2}+2 b \sqrt {1+c^2 x^2} \left (3+12 c^2 x^2+8 c^4 x^4\right ) \text {arcsinh}(c x)+3 b c x \left (1+c^2 x^2\right )^2 \log \left (1+\frac {1}{c^2 x^2}\right )-8 b c x \log \left (1+c^2 x^2\right )-16 b c^3 x^3 \log \left (1+c^2 x^2\right )-8 b c^5 x^5 \log \left (1+c^2 x^2\right )\right )}{6 d^3 x \left (1+c^2 x^2\right )^{5/2}} \] Input:

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

Output:

-1/6*(Sqrt[d + c^2*d*x^2]*(b*c*x + b*c^3*x^3 + 6*a*Sqrt[1 + c^2*x^2] + 24* 
a*c^2*x^2*Sqrt[1 + c^2*x^2] + 16*a*c^4*x^4*Sqrt[1 + c^2*x^2] + 2*b*Sqrt[1 
+ c^2*x^2]*(3 + 12*c^2*x^2 + 8*c^4*x^4)*ArcSinh[c*x] + 3*b*c*x*(1 + c^2*x^ 
2)^2*Log[1 + 1/(c^2*x^2)] - 8*b*c*x*Log[1 + c^2*x^2] - 16*b*c^3*x^3*Log[1 
+ c^2*x^2] - 8*b*c^5*x^5*Log[1 + c^2*x^2]))/(d^3*x*(1 + c^2*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6219, 27, 1578, 1195, 2009}

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.

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

\(\Big \downarrow \) 6219

\(\displaystyle -\frac {b c \sqrt {c^2 d x^2+d} \int -\frac {8 c^4 x^4+12 c^2 x^2+3}{3 d^3 x \left (c^2 x^2+1\right )^2}dx}{\sqrt {c^2 x^2+1}}-\frac {8 c^2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 c^2 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b c \sqrt {c^2 d x^2+d} \int \frac {8 c^4 x^4+12 c^2 x^2+3}{x \left (c^2 x^2+1\right )^2}dx}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {8 c^2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 c^2 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 1578

\(\displaystyle \frac {b c \sqrt {c^2 d x^2+d} \int \frac {8 c^4 x^4+12 c^2 x^2+3}{x^2 \left (c^2 x^2+1\right )^2}dx^2}{6 d^3 \sqrt {c^2 x^2+1}}-\frac {8 c^2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 c^2 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 1195

\(\displaystyle \frac {b c \sqrt {c^2 d x^2+d} \int \left (\frac {5 c^2}{c^2 x^2+1}+\frac {c^2}{\left (c^2 x^2+1\right )^2}+\frac {3}{x^2}\right )dx^2}{6 d^3 \sqrt {c^2 x^2+1}}-\frac {8 c^2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 c^2 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {8 c^2 x (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {c^2 d x^2+d}}-\frac {4 c^2 x (a+b \text {arcsinh}(c x))}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {a+b \text {arcsinh}(c x)}{d x \left (c^2 d x^2+d\right )^{3/2}}+\frac {b c \sqrt {c^2 d x^2+d} \left (-\frac {1}{c^2 x^2+1}+5 \log \left (c^2 x^2+1\right )+3 \log \left (x^2\right )\right )}{6 d^3 \sqrt {c^2 x^2+1}}\)

Input:

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

Output:

-((a + b*ArcSinh[c*x])/(d*x*(d + c^2*d*x^2)^(3/2))) - (4*c^2*x*(a + b*ArcS 
inh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) - (8*c^2*x*(a + b*ArcSinh[c*x]))/(3 
*d^2*Sqrt[d + c^2*d*x^2]) + (b*c*Sqrt[d + c^2*d*x^2]*(-(1 + c^2*x^2)^(-1) 
+ 3*Log[x^2] + 5*Log[1 + c^2*x^2]))/(6*d^3*Sqrt[1 + 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 1195

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x 
_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + 
 g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && IGtQ[p, 0]
 

rule 1578

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6219

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_ 
), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSi 
nh[c*x])   u, x] - Simp[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   Int[S 
implifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x 
] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1) 
/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1256\) vs. \(2(186)=372\).

Time = 1.00 (sec) , antiderivative size = 1257, normalized size of antiderivative = 5.87

method	result	size
default	\(\text {Expression too large to display}\)	\(1257\)
parts	\(\text {Expression too large to display}\)	\(1257\)

Input:

int((a+b*arcsinh(x*c))/x^2/(c^2*d*x^2+d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

a*(-1/d/x/(c^2*d*x^2+d)^(3/2)-4*c^2*(1/3*x/d/(c^2*d*x^2+d)^(3/2)+2/3/d^2*x 
/(c^2*d*x^2+d)^(1/2)))-16/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d^3* 
arcsinh(x*c)*c-32/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x 
^2+9)/d^3*x^9*c^10+32/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c 
^2*x^2+9)/d^3*x^7*(c^2*x^2+1)*c^8-112/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6 
+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^7*c^8+80/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6* 
x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^5*(c^2*x^2+1)*c^6-64/3*b*(d*(c^2*x^2+1) 
)^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^5*arcsinh(x*c)*c^6+64/3* 
b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^4*(c^2*x 
^2+1)^(1/2)*arcsinh(x*c)*c^5-140/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c 
^4*x^4+26*c^2*x^2+9)/d^3*x^5*c^6+20*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25* 
c^4*x^4+26*c^2*x^2+9)/d^3*x^3*(c^2*x^2+1)*c^4-56*b*(d*(c^2*x^2+1))^(1/2)/( 
8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^3*arcsinh(x*c)*c^4+136/3*b*(d*(c^ 
2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x^2*(c^2*x^2+1)^(1 
/2)*arcsinh(x*c)*c^3-24*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c 
^2*x^2+9)/d^3*x^3*c^4-4/3*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26 
*c^2*x^2+9)/d^3*x^2*(c^2*x^2+1)^(1/2)*c^3+4*b*(d*(c^2*x^2+1))^(1/2)/(8*c^6 
*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x*(c^2*x^2+1)*c^2-44*b*(d*(c^2*x^2+1))^( 
1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*x*arcsinh(x*c)*c^2+24*b*(d*(c 
^2*x^2+1))^(1/2)/(8*c^6*x^6+25*c^4*x^4+26*c^2*x^2+9)/d^3*(c^2*x^2+1)^(1...
 

Fricas [F]

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

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

Output:

integral(sqrt(c^2*d*x^2 + d)*(b*arcsinh(c*x) + a)/(c^6*d^3*x^8 + 3*c^4*d^3 
*x^6 + 3*c^2*d^3*x^4 + d^3*x^2), x)
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate((a+b*asinh(c*x))/x**2/(c**2*d*x**2+d)**(5/2),x)
 

Output:

Integral((a + b*asinh(c*x))/(x**2*(d*(c**2*x**2 + 1))**(5/2)), x)
 

Maxima [F]

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

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

Output:

-1/3*a*(8*c^2*x/(sqrt(c^2*d*x^2 + d)*d^2) + 4*c^2*x/((c^2*d*x^2 + d)^(3/2) 
*d) + 3/((c^2*d*x^2 + d)^(3/2)*d*x)) + b*integrate(log(c*x + sqrt(c^2*x^2 
+ 1))/((c^2*d*x^2 + d)^(5/2)*x^2), x)
 

Giac [F]

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

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

Output:

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^(5/2)*x^2), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{x^2 \left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {-8 \sqrt {c^{2} x^{2}+1}\, a \,c^{4} x^{4}-12 \sqrt {c^{2} x^{2}+1}\, a \,c^{2} x^{2}-3 \sqrt {c^{2} x^{2}+1}\, a +3 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{6}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{4}+\sqrt {c^{2} x^{2}+1}\, x^{2}}d x \right ) b \,c^{4} x^{5}+6 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{6}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{4}+\sqrt {c^{2} x^{2}+1}\, x^{2}}d x \right ) b \,c^{2} x^{3}+3 \left (\int \frac {\mathit {asinh} \left (c x \right )}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{6}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{4}+\sqrt {c^{2} x^{2}+1}\, x^{2}}d x \right ) b x +8 a \,c^{5} x^{5}+16 a \,c^{3} x^{3}+8 a c x}{3 \sqrt {d}\, d^{2} x \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )} \] Input:

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

Output:

( - 8*sqrt(c**2*x**2 + 1)*a*c**4*x**4 - 12*sqrt(c**2*x**2 + 1)*a*c**2*x**2 
 - 3*sqrt(c**2*x**2 + 1)*a + 3*int(asinh(c*x)/(sqrt(c**2*x**2 + 1)*c**4*x* 
*6 + 2*sqrt(c**2*x**2 + 1)*c**2*x**4 + sqrt(c**2*x**2 + 1)*x**2),x)*b*c**4 
*x**5 + 6*int(asinh(c*x)/(sqrt(c**2*x**2 + 1)*c**4*x**6 + 2*sqrt(c**2*x**2 
 + 1)*c**2*x**4 + sqrt(c**2*x**2 + 1)*x**2),x)*b*c**2*x**3 + 3*int(asinh(c 
*x)/(sqrt(c**2*x**2 + 1)*c**4*x**6 + 2*sqrt(c**2*x**2 + 1)*c**2*x**4 + sqr 
t(c**2*x**2 + 1)*x**2),x)*b*x + 8*a*c**5*x**5 + 16*a*c**3*x**3 + 8*a*c*x)/ 
(3*sqrt(d)*d**2*x*(c**4*x**4 + 2*c**2*x**2 + 1))