\(\int x^m (d+c^2 d x^2)^{3/2} (a+b \text {arcsinh}(c x)) \, dx\) [204]

Optimal result

Integrand size = 26, antiderivative size = 390 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {3 b c d x^{2+m} \sqrt {d+c^2 d x^2}}{(2+m)^2 (4+m) \sqrt {1+c^2 x^2}}-\frac {b c d x^{2+m} \sqrt {d+c^2 d x^2}}{\left (8+6 m+m^2\right ) \sqrt {1+c^2 x^2}}-\frac {b c^3 d x^{4+m} \sqrt {d+c^2 d x^2}}{(4+m)^2 \sqrt {1+c^2 x^2}}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{8+6 m+m^2}+\frac {x^{1+m} \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{4+m}+\frac {3 d x^{1+m} \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{\left (8+14 m+7 m^2+m^3\right ) \sqrt {1+c^2 x^2}}-\frac {3 b c d x^{2+m} \sqrt {d+c^2 d x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{(1+m) (2+m)^2 (4+m) \sqrt {1+c^2 x^2}} \] Output:

-3*b*c*d*x^(2+m)*(c^2*d*x^2+d)^(1/2)/(2+m)^2/(4+m)/(c^2*x^2+1)^(1/2)-b*c*d 
*x^(2+m)*(c^2*d*x^2+d)^(1/2)/(m^2+6*m+8)/(c^2*x^2+1)^(1/2)-b*c^3*d*x^(4+m) 
*(c^2*d*x^2+d)^(1/2)/(4+m)^2/(c^2*x^2+1)^(1/2)+3*d*x^(1+m)*(c^2*d*x^2+d)^( 
1/2)*(a+b*arcsinh(c*x))/(m^2+6*m+8)+x^(1+m)*(c^2*d*x^2+d)^(3/2)*(a+b*arcsi 
nh(c*x))/(4+m)+3*d*x^(1+m)*(c^2*d*x^2+d)^(1/2)*(a+b*arcsinh(c*x))*hypergeo 
m([1/2, 1/2+1/2*m],[3/2+1/2*m],-c^2*x^2)/(m^3+7*m^2+14*m+8)/(c^2*x^2+1)^(1 
/2)-3*b*c*d*x^(2+m)*(c^2*d*x^2+d)^(1/2)*hypergeom([1, 1+1/2*m, 1+1/2*m],[2 
+1/2*m, 3/2+1/2*m],-c^2*x^2)/(1+m)/(2+m)^2/(4+m)/(c^2*x^2+1)^(1/2)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.60 \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {d x^{1+m} \sqrt {d+c^2 d x^2} \left (-\frac {b c x \left (4+m+c^2 (2+m) x^2\right )}{(2+m) (4+m) \sqrt {1+c^2 x^2}}+\left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))-\frac {3 \left (b c (1+m) x-(1+m) (2+m) \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))-(2+m) (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )+b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )}{(1+m) (2+m)^2 \sqrt {1+c^2 x^2}}\right )}{4+m} \] Input:

Integrate[x^m*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]),x]
 

Output:

(d*x^(1 + m)*Sqrt[d + c^2*d*x^2]*(-((b*c*x*(4 + m + c^2*(2 + m)*x^2))/((2 
+ m)*(4 + m)*Sqrt[1 + c^2*x^2])) + (1 + c^2*x^2)*(a + b*ArcSinh[c*x]) - (3 
*(b*c*(1 + m)*x - (1 + m)*(2 + m)*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]) - 
 (2 + m)*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, 
 -(c^2*x^2)] + b*c*x*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 
2 + m/2}, -(c^2*x^2)]))/((1 + m)*(2 + m)^2*Sqrt[1 + c^2*x^2])))/(4 + m)
 

Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 320, normalized size of antiderivative = 0.82, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6223, 244, 2009, 6221, 15, 6232}

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.

\(\displaystyle \int x^m \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx\)

\(\Big \downarrow \) 6223

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

\(\Big \downarrow \) 244

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

\(\Big \downarrow \) 2009

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

\(\Big \downarrow \) 6221

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

\(\Big \downarrow \) 15

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

\(\Big \downarrow \) 6232

\(\displaystyle \frac {3 d \left (\frac {\sqrt {c^2 d x^2+d} \left (\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},-c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{m+1}-\frac {b c x^{m+2} \, _3F_2\left (1,\frac {m}{2}+1,\frac {m}{2}+1;\frac {m}{2}+\frac {3}{2},\frac {m}{2}+2;-c^2 x^2\right )}{m^2+3 m+2}\right )}{(m+2) \sqrt {c^2 x^2+1}}+\frac {x^{m+1} \sqrt {c^2 d x^2+d} (a+b \text {arcsinh}(c x))}{m+2}-\frac {b c x^{m+2} \sqrt {c^2 d x^2+d}}{(m+2)^2 \sqrt {c^2 x^2+1}}\right )}{m+4}+\frac {x^{m+1} \left (c^2 d x^2+d\right )^{3/2} (a+b \text {arcsinh}(c x))}{m+4}-\frac {b c d \sqrt {c^2 d x^2+d} \left (\frac {c^2 x^{m+4}}{m+4}+\frac {x^{m+2}}{m+2}\right )}{(m+4) \sqrt {c^2 x^2+1}}\)

Input:

Int[x^m*(d + c^2*d*x^2)^(3/2)*(a + b*ArcSinh[c*x]),x]
 

Output:

-((b*c*d*Sqrt[d + c^2*d*x^2]*(x^(2 + m)/(2 + m) + (c^2*x^(4 + m))/(4 + m)) 
)/((4 + m)*Sqrt[1 + c^2*x^2])) + (x^(1 + m)*(d + c^2*d*x^2)^(3/2)*(a + b*A 
rcSinh[c*x]))/(4 + m) + (3*d*(-((b*c*x^(2 + m)*Sqrt[d + c^2*d*x^2])/((2 + 
m)^2*Sqrt[1 + c^2*x^2])) + (x^(1 + m)*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c 
*x]))/(2 + m) + (Sqrt[d + c^2*d*x^2]*((x^(1 + m)*(a + b*ArcSinh[c*x])*Hype 
rgeometric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)])/(1 + m) - (b*c*x^(2 
+ m)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, -(c^2* 
x^2)])/(2 + 3*m + m^2)))/((2 + m)*Sqrt[1 + c^2*x^2])))/(4 + m)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 244

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand 
Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p 
, 0]
 

rule 2009

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

rule 6221

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + 
 (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*Arc 
Sinh[c*x])^n/(f*(m + 2))), x] + (Simp[(1/(m + 2))*Simp[Sqrt[d + e*x^2]/Sqrt 
[1 + c^2*x^2]]   Int[(f*x)^m*((a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2]), x] 
, x] - Simp[b*c*(n/(f*(m + 2)))*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]]   I 
nt[(f*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d 
, e, f, m}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && (IGtQ[m, -2] || EqQ[n, 1])
 

rule 6223

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*((a + b*Arc 
Sinh[c*x])^n/(f*(m + 2*p + 1))), x] + (Simp[2*d*(p/(m + 2*p + 1))   Int[(f* 
x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^n, x], x] - Simp[b*c*(n/(f*(m 
 + 2*p + 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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
 

rule 6232

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_ 
.)*(x_)^2], x_Symbol] :> Simp[((f*x)^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 + c^2 
*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1 + m)/ 
2, (3 + m)/2, (-c^2)*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2 
)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 
1 + m/2}, {3/2 + m/2, 2 + m/2}, (-c^2)*x^2], x] /; FreeQ[{a, b, c, d, e, f, 
 m}, x] && EqQ[e, c^2*d] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

int(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c)),x)
 

Output:

int(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(x*c)),x)
 

Fricas [F]

\[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas" 
)
 

Output:

integral((a*c^2*d*x^2 + a*d + (b*c^2*d*x^2 + b*d)*arcsinh(c*x))*sqrt(c^2*d 
*x^2 + d)*x^m, x)
 

Sympy [F(-1)]

Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Timed out} \] Input:

integrate(x**m*(c**2*d*x**2+d)**(3/2)*(a+b*asinh(c*x)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{m} \,d x } \] Input:

integrate(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="maxima" 
)
 

Output:

integrate((c^2*d*x^2 + d)^(3/2)*(b*arcsinh(c*x) + a)*x^m, x)
 

Giac [F(-2)]

Exception generated. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^m*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int x^m \left (d+c^2 d x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int x^m\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(d)*d*(int(x**m*sqrt(c**2*x**2 + 1)*asinh(c*x)*x**2,x)*b*c**2 + int(x* 
*m*sqrt(c**2*x**2 + 1)*asinh(c*x),x)*b + int(x**m*sqrt(c**2*x**2 + 1)*x**2 
,x)*a*c**2 + int(x**m*sqrt(c**2*x**2 + 1),x)*a)