Integrand size = 13, antiderivative size = 46 \[ \int x^{3/2} (a+b x)^p \, dx=\frac {2}{5} x^{5/2} (a+b x)^p \left (\frac {a+b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x}{a}\right ) \] Output:
2/5*x^(5/2)*(b*x+a)^p*hypergeom([5/2, -p],[7/2],-b*x/a)/(((b*x+a)/a)^p)
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int x^{3/2} (a+b x)^p \, dx=\frac {2}{5} x^{5/2} (a+b x)^p \left (1+\frac {b x}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x}{a}\right ) \] Input:
Integrate[x^(3/2)*(a + b*x)^p,x]
Output:
(2*x^(5/2)*(a + b*x)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x)/a)])/(5*(1 + (b*x)/a)^p)
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {76, 74}
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.
\(\displaystyle \int x^{3/2} (a+b x)^p \, dx\) |
\(\Big \downarrow \) 76 |
\(\displaystyle (a+b x)^p \left (\frac {b x}{a}+1\right )^{-p} \int x^{3/2} \left (\frac {b x}{a}+1\right )^pdx\) |
\(\Big \downarrow \) 74 |
\(\displaystyle \frac {2}{5} x^{5/2} (a+b x)^p \left (\frac {b x}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x}{a}\right )\) |
Input:
Int[x^(3/2)*(a + b*x)^p,x]
Output:
(2*x^(5/2)*(a + b*x)^p*Hypergeometric2F1[5/2, -p, 7/2, -((b*x)/a)])/(5*(1 + (b*x)/a)^p)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-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])))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[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] && !Integer Q[n] && !GtQ[c, 0] && !GtQ[-d/(b*c), 0] && ((RationalQ[m] && !(EqQ[n, -2 ^(-1)] && EqQ[c^2 - d^2, 0])) || !RationalQ[n])
\[\int x^{\frac {3}{2}} \left (b x +a \right )^{p}d x\]
Input:
int(x^(3/2)*(b*x+a)^p,x)
Output:
int(x^(3/2)*(b*x+a)^p,x)
\[ \int x^{3/2} (a+b x)^p \, dx=\int { {\left (b x + a\right )}^{p} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(b*x+a)^p,x, algorithm="fricas")
Output:
integral((b*x + a)^p*x^(3/2), x)
Result contains complex when optimal does not.
Time = 68.84 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.59 \[ \int x^{3/2} (a+b x)^p \, dx=\frac {2 a^{p} x^{\frac {5}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{5} \] Input:
integrate(x**(3/2)*(b*x+a)**p,x)
Output:
2*a**p*x**(5/2)*hyper((5/2, -p), (7/2,), b*x*exp_polar(I*pi)/a)/5
\[ \int x^{3/2} (a+b x)^p \, dx=\int { {\left (b x + a\right )}^{p} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(b*x+a)^p,x, algorithm="maxima")
Output:
integrate((b*x + a)^p*x^(3/2), x)
\[ \int x^{3/2} (a+b x)^p \, dx=\int { {\left (b x + a\right )}^{p} x^{\frac {3}{2}} \,d x } \] Input:
integrate(x^(3/2)*(b*x+a)^p,x, algorithm="giac")
Output:
integrate((b*x + a)^p*x^(3/2), x)
Timed out. \[ \int x^{3/2} (a+b x)^p \, dx=\int x^{3/2}\,{\left (a+b\,x\right )}^p \,d x \] Input:
int(x^(3/2)*(a + b*x)^p,x)
Output:
int(x^(3/2)*(a + b*x)^p, x)
\[ \int x^{3/2} (a+b x)^p \, dx=\frac {-12 \sqrt {x}\, \left (b x +a \right )^{p} a^{2} p +8 \sqrt {x}\, \left (b x +a \right )^{p} a b \,p^{2} x +4 \sqrt {x}\, \left (b x +a \right )^{p} a b p x +8 \sqrt {x}\, \left (b x +a \right )^{p} b^{2} p^{2} x^{2}+16 \sqrt {x}\, \left (b x +a \right )^{p} b^{2} p \,x^{2}+6 \sqrt {x}\, \left (b x +a \right )^{p} b^{2} x^{2}+48 \left (\int \frac {\sqrt {x}\, \left (b x +a \right )^{p}}{8 b \,p^{3} x^{2}+8 a \,p^{3} x +36 b \,p^{2} x^{2}+36 a \,p^{2} x +46 b p \,x^{2}+46 a p x +15 b \,x^{2}+15 a x}d x \right ) a^{3} p^{4}+216 \left (\int \frac {\sqrt {x}\, \left (b x +a \right )^{p}}{8 b \,p^{3} x^{2}+8 a \,p^{3} x +36 b \,p^{2} x^{2}+36 a \,p^{2} x +46 b p \,x^{2}+46 a p x +15 b \,x^{2}+15 a x}d x \right ) a^{3} p^{3}+276 \left (\int \frac {\sqrt {x}\, \left (b x +a \right )^{p}}{8 b \,p^{3} x^{2}+8 a \,p^{3} x +36 b \,p^{2} x^{2}+36 a \,p^{2} x +46 b p \,x^{2}+46 a p x +15 b \,x^{2}+15 a x}d x \right ) a^{3} p^{2}+90 \left (\int \frac {\sqrt {x}\, \left (b x +a \right )^{p}}{8 b \,p^{3} x^{2}+8 a \,p^{3} x +36 b \,p^{2} x^{2}+36 a \,p^{2} x +46 b p \,x^{2}+46 a p x +15 b \,x^{2}+15 a x}d x \right ) a^{3} p}{b^{2} \left (8 p^{3}+36 p^{2}+46 p +15\right )} \] Input:
int(x^(3/2)*(b*x+a)^p,x)
Output:
(2*( - 6*sqrt(x)*(a + b*x)**p*a**2*p + 4*sqrt(x)*(a + b*x)**p*a*b*p**2*x + 2*sqrt(x)*(a + b*x)**p*a*b*p*x + 4*sqrt(x)*(a + b*x)**p*b**2*p**2*x**2 + 8*sqrt(x)*(a + b*x)**p*b**2*p*x**2 + 3*sqrt(x)*(a + b*x)**p*b**2*x**2 + 24 *int((sqrt(x)*(a + b*x)**p)/(8*a*p**3*x + 36*a*p**2*x + 46*a*p*x + 15*a*x + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*p**4 + 108*int((sqrt(x)*(a + b*x)**p)/(8*a*p**3*x + 36*a*p**2*x + 46*a*p*x + 15* a*x + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a**3*p* *3 + 138*int((sqrt(x)*(a + b*x)**p)/(8*a*p**3*x + 36*a*p**2*x + 46*a*p*x + 15*a*x + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)*a** 3*p**2 + 45*int((sqrt(x)*(a + b*x)**p)/(8*a*p**3*x + 36*a*p**2*x + 46*a*p* x + 15*a*x + 8*b*p**3*x**2 + 36*b*p**2*x**2 + 46*b*p*x**2 + 15*b*x**2),x)* a**3*p))/(b**2*(8*p**3 + 36*p**2 + 46*p + 15))