\(\int \frac {x^m (a+b \text {arcsinh}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [208]

Optimal result

Integrand size = 26, antiderivative size = 402 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {x^{1+m} (a+b \text {arcsinh}(c x))}{3 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {(2-m) x^{1+m} (a+b \text {arcsinh}(c x))}{3 d^2 \sqrt {d+c^2 d x^2}}-\frac {(2-m) m x^{1+m} \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )}{3 d^2 (1+m) \sqrt {d+c^2 d x^2}}-\frac {b c (2-m) x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}-\frac {b c x^{2+m} \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-c^2 x^2\right )}{3 d^2 (2+m) \sqrt {d+c^2 d x^2}}+\frac {b c (2-m) m x^{2+m} \sqrt {1+c^2 x^2} \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )}{3 d^2 \left (2+3 m+m^2\right ) \sqrt {d+c^2 d x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.71 \[ \int \frac {x^m (a+b \text {arcsinh}(c x))}{\left (d+c^2 d x^2\right )^{5/2}} \, dx=\frac {x^{1+m} \left ((1+m) (2+m) (a+b \text {arcsinh}(c x))-b c (1+m) x \left (1+c^2 x^2\right )^{3/2} \operatorname {Hypergeometric2F1}\left (2,1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )+(2-m) \left (1+c^2 x^2\right ) \left ((1+m) (2+m) (a+b \text {arcsinh}(c x))-b c (1+m) x \sqrt {1+c^2 x^2} \operatorname {Hypergeometric2F1}\left (1,1+\frac {m}{2},2+\frac {m}{2},-c^2 x^2\right )-m \sqrt {1+c^2 x^2} \left ((2+m) (a+b \text {arcsinh}(c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},-c^2 x^2\right )-b c x \, _3F_2\left (1,1+\frac {m}{2},1+\frac {m}{2};\frac {3}{2}+\frac {m}{2},2+\frac {m}{2};-c^2 x^2\right )\right )\right )\right )}{3 d^2 (1+m) (2+m) \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 384, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {6226, 278, 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.

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

\(\Big \downarrow \) 6226

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

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 6226

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

\(\Big \downarrow \) 278

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

\(\Big \downarrow \) 6232

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

Input:

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

Output:

(x^(1 + m)*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2)) - (b*c*x^(2 + 
 m)*Sqrt[1 + c^2*x^2]*Hypergeometric2F1[2, (2 + m)/2, (4 + m)/2, -(c^2*x^2 
)])/(3*d^2*(2 + m)*Sqrt[d + c^2*d*x^2]) + ((2 - m)*((x^(1 + m)*(a + b*ArcS 
inh[c*x]))/(d*Sqrt[d + c^2*d*x^2]) - (b*c*x^(2 + m)*Sqrt[1 + c^2*x^2]*Hype 
rgeometric2F1[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])*Hypergeo 
metric2F1[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])))/d))/(3*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]

\[\int \frac {x^{m} \left (a +b \,\operatorname {arcsinh}\left (x c \right )\right )}{\left (c^{2} d \,x^{2}+d \right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(x^m*(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*arcsinh(c*x) + a)*x^m/(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(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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