\(\int \frac {(a+b x^n)^{5/2} (A+B x^n)}{x} \, dx\) [347]

Optimal result

Integrand size = 22, antiderivative size = 111 \[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\frac {2 a^2 A \sqrt {a+b x^n}}{n}+\frac {2 a A \left (a+b x^n\right )^{3/2}}{3 n}+\frac {2 A \left (a+b x^n\right )^{5/2}}{5 n}+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b n}-\frac {2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n} \] Output:

2*a^2*A*(a+b*x^n)^(1/2)/n+2/3*a*A*(a+b*x^n)^(3/2)/n+2/5*A*(a+b*x^n)^(5/2)/ 
n+2/7*B*(a+b*x^n)^(7/2)/b/n-2*a^(5/2)*A*arctanh((a+b*x^n)^(1/2)/a^(1/2))/n
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\frac {2 \left (105 a^2 A b \sqrt {a+b x^n}+35 a A b \left (a+b x^n\right )^{3/2}+21 A b \left (a+b x^n\right )^{5/2}+15 B \left (a+b x^n\right )^{7/2}-105 a^{5/2} A b \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )}{105 b n} \] Input:

Integrate[((a + b*x^n)^(5/2)*(A + B*x^n))/x,x]
 

Output:

(2*(105*a^2*A*b*Sqrt[a + b*x^n] + 35*a*A*b*(a + b*x^n)^(3/2) + 21*A*b*(a + 
 b*x^n)^(5/2) + 15*B*(a + b*x^n)^(7/2) - 105*a^(5/2)*A*b*ArcTanh[Sqrt[a + 
b*x^n]/Sqrt[a]]))/(105*b*n)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {948, 90, 60, 60, 60, 73, 221}

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.

\(\displaystyle \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {\int x^{-n} \left (b x^n+a\right )^{5/2} \left (B x^n+A\right )dx^n}{n}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {A \int x^{-n} \left (b x^n+a\right )^{5/2}dx^n+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b}}{n}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {A \left (a \int x^{-n} \left (b x^n+a\right )^{3/2}dx^n+\frac {2}{5} \left (a+b x^n\right )^{5/2}\right )+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b}}{n}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {A \left (a \left (a \int x^{-n} \sqrt {b x^n+a}dx^n+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}\right )+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b}}{n}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {A \left (a \left (a \left (a \int \frac {x^{-n}}{\sqrt {b x^n+a}}dx^n+2 \sqrt {a+b x^n}\right )+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}\right )+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b}}{n}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {A \left (a \left (a \left (\frac {2 a \int \frac {1}{\frac {x^{2 n}}{b}-\frac {a}{b}}d\sqrt {b x^n+a}}{b}+2 \sqrt {a+b x^n}\right )+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}\right )+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b}}{n}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {A \left (a \left (a \left (2 \sqrt {a+b x^n}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )+\frac {2}{3} \left (a+b x^n\right )^{3/2}\right )+\frac {2}{5} \left (a+b x^n\right )^{5/2}\right )+\frac {2 B \left (a+b x^n\right )^{7/2}}{7 b}}{n}\)

Input:

Int[((a + b*x^n)^(5/2)*(A + B*x^n))/x,x]
 

Output:

((2*B*(a + b*x^n)^(7/2))/(7*b) + A*((2*(a + b*x^n)^(5/2))/5 + a*((2*(a + b 
*x^n)^(3/2))/3 + a*(2*Sqrt[a + b*x^n] - 2*Sqrt[a]*ArcTanh[Sqrt[a + b*x^n]/ 
Sqrt[a]]))))/n
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 90

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

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ 
p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ 
[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {\frac {2 B \left (a +b \,x^{n}\right )^{\frac {7}{2}}}{7}+\frac {2 A b \left (a +b \,x^{n}\right )^{\frac {5}{2}}}{5}+\frac {2 A a b \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 A \,a^{2} b \sqrt {a +b \,x^{n}}-2 A \,a^{\frac {5}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{b n}\)	\(85\)
default	\(\frac {\frac {2 B \left (a +b \,x^{n}\right )^{\frac {7}{2}}}{7}+\frac {2 A b \left (a +b \,x^{n}\right )^{\frac {5}{2}}}{5}+\frac {2 A a b \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 A \,a^{2} b \sqrt {a +b \,x^{n}}-2 A \,a^{\frac {5}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{b n}\)	\(85\)
risch	\(\frac {2 \left (15 B \,b^{3} {\mathrm e}^{3 n \ln \left (x \right )}+21 A \,b^{3} {\mathrm e}^{2 n \ln \left (x \right )}+45 B a \,b^{2} {\mathrm e}^{2 n \ln \left (x \right )}+77 A a \,b^{2} {\mathrm e}^{n \ln \left (x \right )}+45 B \,a^{2} b \,{\mathrm e}^{n \ln \left (x \right )}+161 A \,a^{2} b +15 B \,a^{3}\right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{105 b n}-\frac {2 A \,a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{\sqrt {a}}\right )}{n}\)	\(124\)

Input:

int((a+b*x^n)^(5/2)*(A+B*x^n)/x,x,method=_RETURNVERBOSE)
 

Output:

2/n/b*(1/7*B*(a+b*x^n)^(7/2)+1/5*A*b*(a+b*x^n)^(5/2)+1/3*A*a*b*(a+b*x^n)^( 
3/2)+A*a^2*b*(a+b*x^n)^(1/2)-A*a^(5/2)*b*arctanh((a+b*x^n)^(1/2)/a^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.12 \[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\left [\frac {105 \, A a^{\frac {5}{2}} b \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, {\left (15 \, B b^{3} x^{3 \, n} + 15 \, B a^{3} + 161 \, A a^{2} b + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2 \, n} + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x^{n}\right )} \sqrt {b x^{n} + a}}{105 \, b n}, \frac {2 \, {\left (105 \, A \sqrt {-a} a^{2} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{n} + a}}\right ) + {\left (15 \, B b^{3} x^{3 \, n} + 15 \, B a^{3} + 161 \, A a^{2} b + 3 \, {\left (15 \, B a b^{2} + 7 \, A b^{3}\right )} x^{2 \, n} + {\left (45 \, B a^{2} b + 77 \, A a b^{2}\right )} x^{n}\right )} \sqrt {b x^{n} + a}\right )}}{105 \, b n}\right ] \] Input:

integrate((a+b*x^n)^(5/2)*(A+B*x^n)/x,x, algorithm="fricas")
 

Output:

[1/105*(105*A*a^(5/2)*b*log((b*x^n - 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) 
 + 2*(15*B*b^3*x^(3*n) + 15*B*a^3 + 161*A*a^2*b + 3*(15*B*a*b^2 + 7*A*b^3) 
*x^(2*n) + (45*B*a^2*b + 77*A*a*b^2)*x^n)*sqrt(b*x^n + a))/(b*n), 2/105*(1 
05*A*sqrt(-a)*a^2*b*arctan(sqrt(-a)/sqrt(b*x^n + a)) + (15*B*b^3*x^(3*n) + 
 15*B*a^3 + 161*A*a^2*b + 3*(15*B*a*b^2 + 7*A*b^3)*x^(2*n) + (45*B*a^2*b + 
 77*A*a*b^2)*x^n)*sqrt(b*x^n + a))/(b*n)]
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (99) = 198\).

Time = 28.37 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.86 \[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\begin {cases} \frac {\begin {cases} \frac {2 A a^{3} \operatorname {atan}{\left (\frac {\sqrt {a + b x^{n}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a^{2} \sqrt {a + b x^{n}} + \frac {2 A a \left (a + b x^{n}\right )^{\frac {3}{2}}}{3} + \frac {2 A \left (a + b x^{n}\right )^{\frac {5}{2}}}{5} + \frac {2 B \left (a + b x^{n}\right )^{\frac {7}{2}}}{7 b} & \text {for}\: b \neq 0 \\A a^{\frac {5}{2}} \log {\left (B a^{\frac {5}{2}} x^{n} \right )} + B a^{\frac {5}{2}} x^{n} & \text {otherwise} \end {cases}}{n} & \text {for}\: n \neq 0 \\\left (A a^{2} \sqrt {a + b} + 2 A a b \sqrt {a + b} + A b^{2} \sqrt {a + b} + B a^{2} \sqrt {a + b} + 2 B a b \sqrt {a + b} + B b^{2} \sqrt {a + b}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \] Input:

integrate((a+b*x**n)**(5/2)*(A+B*x**n)/x,x)
 

Output:

Piecewise((Piecewise((2*A*a**3*atan(sqrt(a + b*x**n)/sqrt(-a))/sqrt(-a) + 
2*A*a**2*sqrt(a + b*x**n) + 2*A*a*(a + b*x**n)**(3/2)/3 + 2*A*(a + b*x**n) 
**(5/2)/5 + 2*B*(a + b*x**n)**(7/2)/(7*b), Ne(b, 0)), (A*a**(5/2)*log(B*a* 
*(5/2)*x**n) + B*a**(5/2)*x**n, True))/n, Ne(n, 0)), ((A*a**2*sqrt(a + b) 
+ 2*A*a*b*sqrt(a + b) + A*b**2*sqrt(a + b) + B*a**2*sqrt(a + b) + 2*B*a*b* 
sqrt(a + b) + B*b**2*sqrt(a + b))*log(x), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\frac {2 \, {\left (b x^{n} + a\right )}^{\frac {7}{2}} B}{7 \, b n} + \frac {1}{15} \, {\left (\frac {15 \, a^{\frac {5}{2}} \log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, {\left (3 \, {\left (b x^{n} + a\right )}^{\frac {5}{2}} + 5 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x^{n} + a} a^{2}\right )}}{n}\right )} A \] Input:

integrate((a+b*x^n)^(5/2)*(A+B*x^n)/x,x, algorithm="maxima")
 

Output:

2/7*(b*x^n + a)^(7/2)*B/(b*n) + 1/15*(15*a^(5/2)*log((sqrt(b*x^n + a) - sq 
rt(a))/(sqrt(b*x^n + a) + sqrt(a)))/n + 2*(3*(b*x^n + a)^(5/2) + 5*(b*x^n 
+ a)^(3/2)*a + 15*sqrt(b*x^n + a)*a^2)/n)*A
 

Giac [F]

\[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{\frac {5}{2}}}{x} \,d x } \] Input:

integrate((a+b*x^n)^(5/2)*(A+B*x^n)/x,x, algorithm="giac")
 

Output:

integrate((B*x^n + A)*(b*x^n + a)^(5/2)/x, x)
 

Mupad [F(-1)]

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

int(((A + B*x^n)*(a + b*x^n)^(5/2))/x,x)
 

Output:

int(((A + B*x^n)*(a + b*x^n)^(5/2))/x, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^n\right )^{5/2} \left (A+B x^n\right )}{x} \, dx=\frac {30 x^{3 n} \sqrt {x^{n} b +a}\, b^{3}+132 x^{2 n} \sqrt {x^{n} b +a}\, a \,b^{2}+244 x^{n} \sqrt {x^{n} b +a}\, a^{2} b +352 \sqrt {x^{n} b +a}\, a^{3}+105 \left (\int \frac {\sqrt {x^{n} b +a}}{x^{n} b x +a x}d x \right ) a^{4} n}{105 n} \] Input:

int((a+b*x^n)^(5/2)*(A+B*x^n)/x,x)
 

Output:

(30*x**(3*n)*sqrt(x**n*b + a)*b**3 + 132*x**(2*n)*sqrt(x**n*b + a)*a*b**2 
+ 244*x**n*sqrt(x**n*b + a)*a**2*b + 352*sqrt(x**n*b + a)*a**3 + 105*int(s 
qrt(x**n*b + a)/(x**n*b*x + a*x),x)*a**4*n)/(105*n)