\(\int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx\) [206]

Optimal result

Integrand size = 26, antiderivative size = 161 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {x^{1+m} \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{(1+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{\left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \] Output:

x^(1+m)*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))*hypergeom([1/2, 1/2+1/2*m],[3 
/2+1/2*m],-c^2*x^2)/(1+m)/(c^2*d*x^2+d)^(1/2)-b*c*x^(2+m)*(c^2*x^2+1)^(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)/(m^2+3*m+ 
2)/(c^2*d*x^2+d)^(1/2)
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.80 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {x^{1+m} \sqrt {1+c^2 x^2} \left ((2+m) (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )-b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )}{(1+m) (2+m) \sqrt {d+c^2 d x^2}} \] Input:

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

Output:

(x^(1 + m)*Sqrt[1 + c^2*x^2]*((2 + m)*(a + b*ArcSinh[c*x])*Hypergeometric2 
F1[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)*Sqrt 
[d + c^2*d*x^2])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {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.

\(\displaystyle \int \frac {x^m (a+b \text {arcsinh}(c x))}{\sqrt {c^2 d x^2+d}} \, dx\)

\(\Big \downarrow \) 6232

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

Input:

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

Output:

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

Defintions of rubi rules used

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]

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

Input:

int(x^m*(a+b*arcsinh(x*c))/(c^2*d*x^2+d)^(1/2),x)
 

Output:

int(x^m*(a+b*arcsinh(x*c))/(c^2*d*x^2+d)^(1/2),x)
 

Fricas [F]

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

integrate(x^m*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas" 
)
 

Output:

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

Sympy [F]

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

integrate(x**m*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(x^m*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima" 
)
 

Output:

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

Giac [F]

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

integrate(x^m*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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