\(\int \frac {(a+b x^n)^{3/2} (A+B x^n)}{x} \, dx\) [340]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\frac {2 a A \sqrt {a+b x^n}}{n}+\frac {2 A \left (a+b x^n\right )^{3/2}}{3 n}+\frac {2 B \left (a+b x^n\right )^{5/2}}{5 b n}-\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\frac {2 \left (15 a A b \sqrt {a+b x^n}+5 A b \left (a+b x^n\right )^{3/2}+3 B \left (a+b x^n\right )^{5/2}-15 a^{3/2} A b \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )}{15 b n} \] Input:

Integrate[((a + b*x^n)^(3/2)*(A + B*x^n))/x,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {948, 90, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[((a + b*x^n)^(3/2)*(A + B*x^n))/x,x]
 

Output:

((2*B*(a + b*x^n)^(5/2))/(5*b) + 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 = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\frac {2 B \left (a +b \,x^{n}\right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 A a b \sqrt {a +b \,x^{n}}-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{b n}\)	\(69\)
default	\(\frac {\frac {2 B \left (a +b \,x^{n}\right )^{\frac {5}{2}}}{5}+\frac {2 A b \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 A a b \sqrt {a +b \,x^{n}}-2 A \,a^{\frac {3}{2}} b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{b n}\)	\(69\)
risch	\(\frac {2 \left (3 b^{2} B \,{\mathrm e}^{2 n \ln \left (x \right )}+5 A \,b^{2} {\mathrm e}^{n \ln \left (x \right )}+6 B a b \,{\mathrm e}^{n \ln \left (x \right )}+20 a b A +3 a^{2} B \right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{15 b n}-\frac {2 a^{\frac {3}{2}} A \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{\sqrt {a}}\right )}{n}\)	\(92\)

Input:

int((a+b*x^n)^(3/2)*(A+B*x^n)/x,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.03 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\left [\frac {15 \, A a^{\frac {3}{2}} b \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, {\left (3 \, B b^{2} x^{2 \, n} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x^{n}\right )} \sqrt {b x^{n} + a}}{15 \, b n}, \frac {2 \, {\left (15 \, A \sqrt {-a} a b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{n} + a}}\right ) + {\left (3 \, B b^{2} x^{2 \, n} + 3 \, B a^{2} + 20 \, A a b + {\left (6 \, B a b + 5 \, A b^{2}\right )} x^{n}\right )} \sqrt {b x^{n} + a}\right )}}{15 \, b n}\right ] \] Input:

integrate((a+b*x^n)^(3/2)*(A+B*x^n)/x,x, algorithm="fricas")
 

Output:

[1/15*(15*A*a^(3/2)*b*log((b*x^n - 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) + 
 2*(3*B*b^2*x^(2*n) + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x^n)*sqrt(b 
*x^n + a))/(b*n), 2/15*(15*A*sqrt(-a)*a*b*arctan(sqrt(-a)/sqrt(b*x^n + a)) 
 + (3*B*b^2*x^(2*n) + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x^n)*sqrt(b 
*x^n + a))/(b*n)]
                                                                                    
                                                                                    
 

Sympy [A] (verification not implemented)

Time = 20.65 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.72 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\begin {cases} \frac {\begin {cases} \frac {2 A a^{2} \operatorname {atan}{\left (\frac {\sqrt {a + b x^{n}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A a \sqrt {a + b x^{n}} + \frac {2 A \left (a + b x^{n}\right )^{\frac {3}{2}}}{3} + \frac {2 B \left (a + b x^{n}\right )^{\frac {5}{2}}}{5 b} & \text {for}\: b \neq 0 \\A a^{\frac {3}{2}} \log {\left (B a^{\frac {3}{2}} x^{n} \right )} + B a^{\frac {3}{2}} x^{n} & \text {otherwise} \end {cases}}{n} & \text {for}\: n \neq 0 \\\left (A a \sqrt {a + b} + A b \sqrt {a + b} + B a \sqrt {a + b} + B b \sqrt {a + b}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \] Input:

integrate((a+b*x**n)**(3/2)*(A+B*x**n)/x,x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\frac {1}{3} \, {\left (\frac {3 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, {\left ({\left (b x^{n} + a\right )}^{\frac {3}{2}} + 3 \, \sqrt {b x^{n} + a} a\right )}}{n}\right )} A + \frac {2 \, {\left (b x^{n} + a\right )}^{\frac {5}{2}} B}{5 \, b n} \] Input:

integrate((a+b*x^n)^(3/2)*(A+B*x^n)/x,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{\frac {3}{2}}}{x} \,d x } \] Input:

integrate((a+b*x^n)^(3/2)*(A+B*x^n)/x,x, algorithm="giac")
 

Output:

integrate((B*x^n + A)*(b*x^n + a)^(3/2)/x, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\int \frac {\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^{3/2}}{x} \,d x \] Input:

int(((A + B*x^n)*(a + b*x^n)^(3/2))/x,x)
 

Output:

int(((A + B*x^n)*(a + b*x^n)^(3/2))/x, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{x} \, dx=\frac {6 x^{2 n} \sqrt {x^{n} b +a}\, b^{2}+22 x^{n} \sqrt {x^{n} b +a}\, a b +46 \sqrt {x^{n} b +a}\, a^{2}+15 \left (\int \frac {\sqrt {x^{n} b +a}}{x^{n} b x +a x}d x \right ) a^{3} n}{15 n} \] Input:

int((a+b*x^n)^(3/2)*(A+B*x^n)/x,x)
 

Output:

(6*x**(2*n)*sqrt(x**n*b + a)*b**2 + 22*x**n*sqrt(x**n*b + a)*a*b + 46*sqrt 
(x**n*b + a)*a**2 + 15*int(sqrt(x**n*b + a)/(x**n*b*x + a*x),x)*a**3*n)/(1 
5*n)