\(\int \frac {x^2 (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [177]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {b}{6 c^3 d^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}-\frac {b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 c^3 d^2 \sqrt {d+c^2 d x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {\sqrt {d+c^2 d x^2} \left (b+b c^2 x^2-2 a c^3 x^3 \sqrt {1+c^2 x^2}-2 b c^3 x^3 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+b \left (1+c^2 x^2\right )^2 \log \left (1+c^2 x^2\right )\right )}{6 c^3 d^3 \left (1+c^2 x^2\right )^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6215, 243, 49, 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.

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

\(\Big \downarrow \) 6215

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

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

rule 6215

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_ 
.)*(x_)^2)^(p_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + 
b*ArcSinh[c*x])^n/(d*f*(m + 1))), x] - Simp[b*c*(n/(f*(m + 1)))*Simp[(d + e 
*x^2)^p/(1 + c^2*x^2)^p]   Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b 
*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ 
[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(103)=206\).

Time = 1.01 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.47

method	result	size
default	\(a \left (-\frac {x}{2 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {\frac {x}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {c^{2} d \,x^{2}+d}}}{2 c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (x^{3} c^{3}+x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+\sqrt {c^{2} x^{2}+1}\right ) \left (-2 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{6} c^{6}+2 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{5} c^{5}+6 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-6 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}+2 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{3} c^{3}-c^{4} x^{4}+\sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}+6 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-6 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-2 c^{2} x^{2}+2 \,\operatorname {arcsinh}\left (x c \right )-2 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )-1\right )}{6 \left (3 c^{8} x^{8}+9 c^{6} x^{6}+10 c^{4} x^{4}+5 c^{2} x^{2}+1\right ) d^{3} c^{3}}\)	\(413\)
parts	\(a \left (-\frac {x}{2 c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {\frac {x}{3 d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{2} \sqrt {c^{2} d \,x^{2}+d}}}{2 c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (x^{3} c^{3}+x^{2} c^{2} \sqrt {c^{2} x^{2}+1}+\sqrt {c^{2} x^{2}+1}\right ) \left (-2 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{6} c^{6}+2 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{5} c^{5}+6 \,\operatorname {arcsinh}\left (x c \right ) c^{4} x^{4}-6 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{4} c^{4}+2 \sqrt {c^{2} x^{2}+1}\, \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{3} c^{3}-c^{4} x^{4}+\sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}+6 \,\operatorname {arcsinh}\left (x c \right ) c^{2} x^{2}-6 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right ) x^{2} c^{2}-2 c^{2} x^{2}+2 \,\operatorname {arcsinh}\left (x c \right )-2 \ln \left (1+\left (x c +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )-1\right )}{6 \left (3 c^{8} x^{8}+9 c^{6} x^{6}+10 c^{4} x^{4}+5 c^{2} x^{2}+1\right ) d^{3} c^{3}}\)	\(413\)

Input:

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

Output:

a*(-1/2*x/c^2/d/(c^2*d*x^2+d)^(3/2)+1/2/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)))+1/6*b*(d*(c^2*x^2+1))^(1/2)*(x^3*c^3+x^2*c^ 
2*(c^2*x^2+1)^(1/2)+(c^2*x^2+1)^(1/2))*(-2*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2) 
*x^6*c^6+2*(c^2*x^2+1)^(1/2)*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)*x^5*c^5+6*arc 
sinh(x*c)*c^4*x^4-6*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)*x^4*c^4+2*(c^2*x^2+1)^ 
(1/2)*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)*x^3*c^3-c^4*x^4+(c^2*x^2+1)^(1/2)*c^ 
3*x^3+6*arcsinh(x*c)*c^2*x^2-6*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)*x^2*c^2-2*c 
^2*x^2+2*arcsinh(x*c)-2*ln(1+(x*c+(c^2*x^2+1)^(1/2))^2)-1)/(3*c^8*x^8+9*c^ 
6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3/c^3
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=-\frac {1}{6} \, b c {\left (\frac {1}{c^{6} d^{\frac {5}{2}} x^{2} + c^{4} d^{\frac {5}{2}}} + \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4} d^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {x}{\sqrt {c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {x}{\sqrt {c^{2} d x^{2} + d} c^{2} d^{2}} - \frac {x}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} c^{2} d}\right )} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^{2} x^{2}+1}\, a \,c^{3} x^{3}+3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{2}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{7} x^{4}+6 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{2}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{5} x^{2}+3 \left (\int \frac {\mathit {asinh} \left (c x \right ) x^{2}}{\sqrt {c^{2} x^{2}+1}\, c^{4} x^{4}+2 \sqrt {c^{2} x^{2}+1}\, c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}}d x \right ) b \,c^{3}+a \,c^{4} x^{4}+2 a \,c^{2} x^{2}+a}{3 \sqrt {d}\, c^{3} d^{2} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )} \] Input:

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

Output:

(sqrt(c**2*x**2 + 1)*a*c**3*x**3 + 3*int((asinh(c*x)*x**2)/(sqrt(c**2*x**2 
 + 1)*c**4*x**4 + 2*sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x 
)*b*c**7*x**4 + 6*int((asinh(c*x)*x**2)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2 
*sqrt(c**2*x**2 + 1)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**5*x**2 + 3*i 
nt((asinh(c*x)*x**2)/(sqrt(c**2*x**2 + 1)*c**4*x**4 + 2*sqrt(c**2*x**2 + 1 
)*c**2*x**2 + sqrt(c**2*x**2 + 1)),x)*b*c**3 + a*c**4*x**4 + 2*a*c**2*x**2 
 + a)/(3*sqrt(d)*c**3*d**2*(c**4*x**4 + 2*c**2*x**2 + 1))