\(\int \frac {(a+b x)^n}{x^{5/2}} \, dx\) [749]

Optimal result

Integrand size = 13, antiderivative size = 45 \[ \int \frac {(a+b x)^n}{x^{5/2}} \, dx=-\frac {2 (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {b x}{a}\right )}{3 x^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {68, 66} \[ \int \frac {(a+b x)^n}{x^{5/2}} \, dx=-\frac {2 (a+b x)^n \left (\frac {b x}{a}+1\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {b x}{a}\right )}{3 x^{3/2}} \]

[In]

Int[(a + b*x)^n/x^(5/2),x]

[Out]

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

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(
x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rubi steps \begin{align*} \text {integral}& = \left ((a+b x)^n \left (1+\frac {b x}{a}\right )^{-n}\right ) \int \frac {\left (1+\frac {b x}{a}\right )^n}{x^{5/2}} \, dx \\ & = -\frac {2 (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \, _2F_1\left (-\frac {3}{2},-n;-\frac {1}{2};-\frac {b x}{a}\right )}{3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^n}{x^{5/2}} \, dx=-\frac {2 (a+b x)^n \left (1+\frac {b x}{a}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-n,-\frac {1}{2},-\frac {b x}{a}\right )}{3 x^{3/2}} \]

[In]

Integrate[(a + b*x)^n/x^(5/2),x]

[Out]

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

Maple [F]

\[\int \frac {\left (b x +a \right )^{n}}{x^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b*x + a)^n/x^(5/2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 157.98 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71 \[ \int \frac {(a+b x)^n}{x^{5/2}} \, dx=- \frac {2 a^{n} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - n \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{3 x^{\frac {3}{2}}} \]

[In]

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

[Out]

-2*a**n*hyper((-3/2, -n), (-1/2,), b*x*exp_polar(I*pi)/a)/(3*x**(3/2))

Maxima [F]

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

[In]

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

[Out]

integrate((b*x + a)^n/x^(5/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x + a)^n/x^(5/2), x)

Mupad [F(-1)]

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

[In]

int((a + b*x)^n/x^(5/2),x)

[Out]

int((a + b*x)^n/x^(5/2), x)