\(\int x^m (a+b x)^{5/2} \, dx\) [707]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [C] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 13, antiderivative size = 48 \[ \int x^m (a+b x)^{5/2} \, dx=\frac {2 x^m \left (-\frac {b x}{a}\right )^{-m} (a+b x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {b x}{a}\right )}{7 b} \]

[Out]

2/7*x^m*(b*x+a)^(7/2)*hypergeom([7/2, -m],[9/2],1+b*x/a)/b/((-b*x/a)^m)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, 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 = {69, 67} \[ \int x^m (a+b x)^{5/2} \, dx=\frac {2 x^m (a+b x)^{7/2} \left (-\frac {b x}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},\frac {b x}{a}+1\right )}{7 b} \]

[In]

Int[x^m*(a + b*x)^(5/2),x]

[Out]

(2*x^m*(a + b*x)^(7/2)*Hypergeometric2F1[7/2, -m, 9/2, 1 + (b*x)/a])/(7*b*(-((b*x)/a))^m)

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

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

Rubi steps \begin{align*} \text {integral}& = \left (x^m \left (-\frac {b x}{a}\right )^{-m}\right ) \int \left (-\frac {b x}{a}\right )^m (a+b x)^{5/2} \, dx \\ & = \frac {2 x^m \left (-\frac {b x}{a}\right )^{-m} (a+b x)^{7/2} \, _2F_1\left (\frac {7}{2},-m;\frac {9}{2};1+\frac {b x}{a}\right )}{7 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int x^m (a+b x)^{5/2} \, dx=\frac {2 x^m \left (-\frac {b x}{a}\right )^{-m} (a+b x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {b x}{a}\right )}{7 b} \]

[In]

Integrate[x^m*(a + b*x)^(5/2),x]

[Out]

(2*x^m*(a + b*x)^(7/2)*Hypergeometric2F1[7/2, -m, 9/2, 1 + (b*x)/a])/(7*b*(-((b*x)/a))^m)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((b^2*x^2 + 2*a*b*x + a^2)*sqrt(b*x + a)*x^m, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.83 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int x^m (a+b x)^{5/2} \, dx=\frac {a^{\frac {5}{2}} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\Gamma \left (m + 2\right )} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((b*x + a)^(5/2)*x^m, x)

Giac [F]

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

[In]

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

[Out]

integrate((b*x + a)^(5/2)*x^m, x)

Mupad [F(-1)]

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

[In]

int(x^m*(a + b*x)^(5/2),x)

[Out]

int(x^m*(a + b*x)^(5/2), x)