Integrand size = 22, antiderivative size = 69 \[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\frac {2 A \sqrt {a+b x^n}}{n}+\frac {2 B \left (a+b x^n\right )^{3/2}}{3 b n}-\frac {2 \sqrt {a} A \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{n} \] Output:
2*A*(a+b*x^n)^(1/2)/n+2/3*B*(a+b*x^n)^(3/2)/b/n-2*a^(1/2)*A*arctanh((a+b*x ^n)^(1/2)/a^(1/2))/n
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\frac {2 \sqrt {a+b x^n} \left (3 A b+B \left (a+b x^n\right )\right )-6 \sqrt {a} A b \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )}{3 b n} \] Input:
Integrate[(Sqrt[a + b*x^n]*(A + B*x^n))/x,x]
Output:
(2*Sqrt[a + b*x^n]*(3*A*b + B*(a + b*x^n)) - 6*Sqrt[a]*A*b*ArcTanh[Sqrt[a + b*x^n]/Sqrt[a]])/(3*b*n)
Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {948, 90, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {\int x^{-n} \sqrt {b x^n+a} \left (B x^n+A\right )dx^n}{n}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {A \int x^{-n} \sqrt {b x^n+a}dx^n+\frac {2 B \left (a+b x^n\right )^{3/2}}{3 b}}{n}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {A \left (a \int \frac {x^{-n}}{\sqrt {b x^n+a}}dx^n+2 \sqrt {a+b x^n}\right )+\frac {2 B \left (a+b x^n\right )^{3/2}}{3 b}}{n}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {A \left (\frac {2 a \int \frac {1}{\frac {x^{2 n}}{b}-\frac {a}{b}}d\sqrt {b x^n+a}}{b}+2 \sqrt {a+b x^n}\right )+\frac {2 B \left (a+b x^n\right )^{3/2}}{3 b}}{n}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {A \left (2 \sqrt {a+b x^n}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b x^n}}{\sqrt {a}}\right )\right )+\frac {2 B \left (a+b x^n\right )^{3/2}}{3 b}}{n}\) |
Input:
Int[(Sqrt[a + b*x^n]*(A + B*x^n))/x,x]
Output:
((2*B*(a + b*x^n)^(3/2))/(3*b) + A*(2*Sqrt[a + b*x^n] - 2*Sqrt[a]*ArcTanh[ Sqrt[a + b*x^n]/Sqrt[a]]))/n
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]
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]
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]
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]
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]]
Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {\frac {2 B \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 A b \sqrt {a +b \,x^{n}}-2 A \sqrt {a}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{b n}\) | \(55\) |
default | \(\frac {\frac {2 B \left (a +b \,x^{n}\right )^{\frac {3}{2}}}{3}+2 A b \sqrt {a +b \,x^{n}}-2 A \sqrt {a}\, b \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \,x^{n}}}{\sqrt {a}}\right )}{b n}\) | \(55\) |
risch | \(\frac {2 \left (B b \,{\mathrm e}^{n \ln \left (x \right )}+3 A b +B a \right ) \sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{3 b n}-\frac {2 A \sqrt {a}\, \operatorname {arctanh}\left (\frac {\sqrt {a +b \,{\mathrm e}^{n \ln \left (x \right )}}}{\sqrt {a}}\right )}{n}\) | \(62\) |
Input:
int((a+b*x^n)^(1/2)*(A+B*x^n)/x,x,method=_RETURNVERBOSE)
Output:
2/n/b*(1/3*B*(a+b*x^n)^(3/2)+A*b*(a+b*x^n)^(1/2)-A*a^(1/2)*b*arctanh((a+b* x^n)^(1/2)/a^(1/2)))
Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.88 \[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\left [\frac {3 \, A \sqrt {a} b \log \left (\frac {b x^{n} - 2 \, \sqrt {b x^{n} + a} \sqrt {a} + 2 \, a}{x^{n}}\right ) + 2 \, {\left (B b x^{n} + B a + 3 \, A b\right )} \sqrt {b x^{n} + a}}{3 \, b n}, \frac {2 \, {\left (3 \, A \sqrt {-a} b \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{n} + a}}\right ) + {\left (B b x^{n} + B a + 3 \, A b\right )} \sqrt {b x^{n} + a}\right )}}{3 \, b n}\right ] \] Input:
integrate((a+b*x^n)^(1/2)*(A+B*x^n)/x,x, algorithm="fricas")
Output:
[1/3*(3*A*sqrt(a)*b*log((b*x^n - 2*sqrt(b*x^n + a)*sqrt(a) + 2*a)/x^n) + 2 *(B*b*x^n + B*a + 3*A*b)*sqrt(b*x^n + a))/(b*n), 2/3*(3*A*sqrt(-a)*b*arcta n(sqrt(-a)/sqrt(b*x^n + a)) + (B*b*x^n + B*a + 3*A*b)*sqrt(b*x^n + a))/(b* n)]
Time = 13.57 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\begin {cases} \frac {\begin {cases} \frac {2 A a \operatorname {atan}{\left (\frac {\sqrt {a + b x^{n}}}{\sqrt {- a}} \right )}}{\sqrt {- a}} + 2 A \sqrt {a + b x^{n}} + \frac {2 B \left (a + b x^{n}\right )^{\frac {3}{2}}}{3 b} & \text {for}\: b \neq 0 \\A \sqrt {a} \log {\left (B \sqrt {a} x^{n} \right )} + B \sqrt {a} x^{n} & \text {otherwise} \end {cases}}{n} & \text {for}\: n \neq 0 \\\left (A \sqrt {a + b} + B \sqrt {a + b}\right ) \log {\left (x \right )} & \text {otherwise} \end {cases} \] Input:
integrate((a+b*x**n)**(1/2)*(A+B*x**n)/x,x)
Output:
Piecewise((Piecewise((2*A*a*atan(sqrt(a + b*x**n)/sqrt(-a))/sqrt(-a) + 2*A *sqrt(a + b*x**n) + 2*B*(a + b*x**n)**(3/2)/(3*b), Ne(b, 0)), (A*sqrt(a)*l og(B*sqrt(a)*x**n) + B*sqrt(a)*x**n, True))/n, Ne(n, 0)), ((A*sqrt(a + b) + B*sqrt(a + b))*log(x), True))
Time = 0.12 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=A {\left (\frac {\sqrt {a} \log \left (\frac {\sqrt {b x^{n} + a} - \sqrt {a}}{\sqrt {b x^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, \sqrt {b x^{n} + a}}{n}\right )} + \frac {2 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}} B}{3 \, b n} \] Input:
integrate((a+b*x^n)^(1/2)*(A+B*x^n)/x,x, algorithm="maxima")
Output:
A*(sqrt(a)*log((sqrt(b*x^n + a) - sqrt(a))/(sqrt(b*x^n + a) + sqrt(a)))/n + 2*sqrt(b*x^n + a)/n) + 2/3*(b*x^n + a)^(3/2)*B/(b*n)
\[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\int { \frac {{\left (B x^{n} + A\right )} \sqrt {b x^{n} + a}}{x} \,d x } \] Input:
integrate((a+b*x^n)^(1/2)*(A+B*x^n)/x,x, algorithm="giac")
Output:
integrate((B*x^n + A)*sqrt(b*x^n + a)/x, x)
Timed out. \[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\int \frac {\left (A+B\,x^n\right )\,\sqrt {a+b\,x^n}}{x} \,d x \] Input:
int(((A + B*x^n)*(a + b*x^n)^(1/2))/x,x)
Output:
int(((A + B*x^n)*(a + b*x^n)^(1/2))/x, x)
\[ \int \frac {\sqrt {a+b x^n} \left (A+B x^n\right )}{x} \, dx=\frac {2 x^{n} \sqrt {x^{n} b +a}\, b +8 \sqrt {x^{n} b +a}\, a +3 \left (\int \frac {\sqrt {x^{n} b +a}}{x^{n} b x +a x}d x \right ) a^{2} n}{3 n} \] Input:
int((a+b*x^n)^(1/2)*(A+B*x^n)/x,x)
Output:
(2*x**n*sqrt(x**n*b + a)*b + 8*sqrt(x**n*b + a)*a + 3*int(sqrt(x**n*b + a) /(x**n*b*x + a*x),x)*a**2*n)/(3*n)