\(\int \frac {x^m (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\) [207]

Optimal result

Integrand size = 26, antiderivative size = 268 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {x^{1+m} (a+b \text {arcsinh}(c x))}{d \sqrt {d+c^2 d x^2}}-\frac {m x^{1+m} \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{d (1+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{d (2+m) \sqrt {d+c^2 d x^2}}+\frac {b c m x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{d \left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.77 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{3/2}} \, dx=\frac {x^{1+m} \left (-m (2+m) \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )+(1+m) \left ((2+m) (a+b \text {arcsinh}(c x))-b c x \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )\right )+b c m x \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )}{d (1+m) (2+m) \sqrt {d+c^2 d x^2}} \] Input:

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

Output:

(x^(1 + m)*(-(m*(2 + m)*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Hypergeomet 
ric2F1[1/2, (1 + m)/2, (3 + m)/2, -(c^2*x^2)]) + (1 + m)*((2 + m)*(a + b*A 
rcSinh[c*x]) - b*c*x*Sqrt[1 + c^2*x^2]*Hypergeometric2F1[1, 1 + m/2, 2 + m 
/2, -(c^2*x^2)]) + b*c*m*x*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)]))/(d*(1 + m)*(2 + m)*Sqrt[d 
+ c^2*d*x^2])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6226, 278, 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.

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

\(\Big \downarrow \) 6226

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

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 6232

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 6226

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/(2*d*f*(p + 1))), x] + (Simp[(m + 2*p + 3)/(2*d*(p + 1 
))   Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Simp[ 
b*c*(n/(2*f*(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] && LtQ[p, -1] &&  !G 
tQ[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**m*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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