\(\int (-\frac {b n x^{-1+m+n}}{2 (a+b x^n)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}) \, dx\) [2686]

Optimal result

Integrand size = 42, antiderivative size = 15 \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\frac {x^m}{\sqrt {a+b x^n}} \]

[Out]

x^m/(a+b*x^n)^(1/2)

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.07 (sec) , antiderivative size = 126, normalized size of antiderivative = 8.40, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {372, 371} \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\frac {x^m \sqrt {\frac {b x^n}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{n},\frac {m+n}{n},-\frac {b x^n}{a}\right )}{\sqrt {a+b x^n}}-\frac {b n x^{m+n} \sqrt {\frac {b x^n}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+n}{n},\frac {m}{n}+2,-\frac {b x^n}{a}\right )}{2 a (m+n) \sqrt {a+b x^n}} \]

[In]

Int[-1/2*(b*n*x^(-1 + m + n))/(a + b*x^n)^(3/2) + (m*x^(-1 + m))/Sqrt[a + b*x^n],x]

[Out]

(x^m*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, m/n, (m + n)/n, -((b*x^n)/a)])/Sqrt[a + b*x^n] - (b*n*x^(m + n
)*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[3/2, (m + n)/n, 2 + m/n, -((b*x^n)/a)])/(2*a*(m + n)*Sqrt[a + b*x^n])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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

Rubi steps \begin{align*} \text {integral}& = m \int \frac {x^{-1+m}}{\sqrt {a+b x^n}} \, dx-\frac {1}{2} (b n) \int \frac {x^{-1+m+n}}{\left (a+b x^n\right )^{3/2}} \, dx \\ & = \frac {\left (m \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {x^{-1+m}}{\sqrt {1+\frac {b x^n}{a}}} \, dx}{\sqrt {a+b x^n}}-\frac {\left (b n \sqrt {1+\frac {b x^n}{a}}\right ) \int \frac {x^{-1+m+n}}{\left (1+\frac {b x^n}{a}\right )^{3/2}} \, dx}{2 a \sqrt {a+b x^n}} \\ & = \frac {x^m \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {1}{2},\frac {m}{n};\frac {m+n}{n};-\frac {b x^n}{a}\right )}{\sqrt {a+b x^n}}-\frac {b n x^{m+n} \sqrt {1+\frac {b x^n}{a}} \, _2F_1\left (\frac {3}{2},\frac {m+n}{n};2+\frac {m}{n};-\frac {b x^n}{a}\right )}{2 a (m+n) \sqrt {a+b x^n}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 7.40 \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\frac {x^m \sqrt {1+\frac {b x^n}{a}} \left (2 a (m+n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m}{n},\frac {m+n}{n},-\frac {b x^n}{a}\right )+b (2 m-n) x^n \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+n}{n},2+\frac {m}{n},-\frac {b x^n}{a}\right )\right )}{2 a (m+n) \sqrt {a+b x^n}} \]

[In]

Integrate[-1/2*(b*n*x^(-1 + m + n))/(a + b*x^n)^(3/2) + (m*x^(-1 + m))/Sqrt[a + b*x^n],x]

[Out]

(x^m*Sqrt[1 + (b*x^n)/a]*(2*a*(m + n)*Hypergeometric2F1[3/2, m/n, (m + n)/n, -((b*x^n)/a)] + b*(2*m - n)*x^n*H
ypergeometric2F1[3/2, (m + n)/n, 2 + m/n, -((b*x^n)/a)]))/(2*a*(m + n)*Sqrt[a + b*x^n])

Maple [F]

\[\int \left (-\frac {b n \,x^{-1+m +n}}{2 \left (a +b \,x^{n}\right )^{\frac {3}{2}}}+\frac {m \,x^{-1+m}}{\sqrt {a +b \,x^{n}}}\right )d x\]

[In]

int(-1/2*b*n*x^(-1+m+n)/(a+b*x^n)^(3/2)+m*x^(-1+m)/(a+b*x^n)^(1/2),x)

[Out]

int(-1/2*b*n*x^(-1+m+n)/(a+b*x^n)^(3/2)+m*x^(-1+m)/(a+b*x^n)^(1/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(-1/2*b*n*x^(-1+m+n)/(a+b*x^n)^(3/2)+m*x^(-1+m)/(a+b*x^n)^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.79 (sec) , antiderivative size = 114, normalized size of antiderivative = 7.60 \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\frac {a^{\frac {m}{n}} a^{- \frac {m}{n} - \frac {1}{2}} m x^{m} \Gamma \left (\frac {m}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{n} \\ \frac {m}{n} + 1 \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (\frac {m}{n} + 1\right )} - \frac {a^{- \frac {m}{n} - \frac {5}{2}} a^{\frac {m}{n} + 1} b x^{m + n} \Gamma \left (\frac {m}{n} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{n} + 1 \\ \frac {m}{n} + 2 \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{n} + 2\right )} \]

[In]

integrate(-1/2*b*n*x**(-1+m+n)/(a+b*x**n)**(3/2)+m*x**(-1+m)/(a+b*x**n)**(1/2),x)

[Out]

a**(m/n)*a**(-m/n - 1/2)*m*x**m*gamma(m/n)*hyper((1/2, m/n), (m/n + 1,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(m/
n + 1)) - a**(-m/n - 5/2)*a**(m/n + 1)*b*x**(m + n)*gamma(m/n + 1)*hyper((3/2, m/n + 1), (m/n + 2,), b*x**n*ex
p_polar(I*pi)/a)/(2*gamma(m/n + 2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\frac {x^{m}}{\sqrt {b x^{n} + a}} \]

[In]

integrate(-1/2*b*n*x^(-1+m+n)/(a+b*x^n)^(3/2)+m*x^(-1+m)/(a+b*x^n)^(1/2),x, algorithm="maxima")

[Out]

x^m/sqrt(b*x^n + a)

Giac [F]

\[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\int { -\frac {b n x^{m + n - 1}}{2 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}}} + \frac {m x^{m - 1}}{\sqrt {b x^{n} + a}} \,d x } \]

[In]

integrate(-1/2*b*n*x^(-1+m+n)/(a+b*x^n)^(3/2)+m*x^(-1+m)/(a+b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(-1/2*b*n*x^(m + n - 1)/(b*x^n + a)^(3/2) + m*x^(m - 1)/sqrt(b*x^n + a), x)

Mupad [F(-1)]

Timed out. \[ \int \left (-\frac {b n x^{-1+m+n}}{2 \left (a+b x^n\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^n}}\right ) \, dx=\int \frac {m\,x^{m-1}}{\sqrt {a+b\,x^n}}-\frac {b\,n\,x^{m+n-1}}{2\,{\left (a+b\,x^n\right )}^{3/2}} \,d x \]

[In]

int((m*x^(m - 1))/(a + b*x^n)^(1/2) - (b*n*x^(m + n - 1))/(2*(a + b*x^n)^(3/2)),x)

[Out]

int((m*x^(m - 1))/(a + b*x^n)^(1/2) - (b*n*x^(m + n - 1))/(2*(a + b*x^n)^(3/2)), x)