Integrand size = 13, antiderivative size = 48 \[ \int x^m (a+b x)^{5/2} \, dx=\frac {2 x^m \left (-\frac {b x}{a}\right )^{-m} (a+b x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {b x}{a}\right )}{7 b} \]
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int x^m (a+b x)^{5/2} \, dx=\frac {2 x^m \left (-\frac {b x}{a}\right )^{-m} (a+b x)^{7/2} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {b x}{a}\right )}{7 b} \]
Time = 0.16 (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 = {77, 75}
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.
\(\displaystyle \int x^m (a+b x)^{5/2} \, dx\) |
\(\Big \downarrow \) 77 |
\(\displaystyle x^m \left (-\frac {b x}{a}\right )^{-m} \int \left (-\frac {b x}{a}\right )^m (a+b x)^{5/2}dx\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {2 x^m (a+b x)^{7/2} \left (-\frac {b x}{a}\right )^{-m} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},\frac {b x}{a}+1\right )}{7 b}\) |
3.8.7.3.1 Defintions of rubi rules used
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] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-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]
\[\int x^{m} \left (b x +a \right )^{\frac {5}{2}}d x\]
\[ \int x^m (a+b x)^{5/2} \, dx=\int { {\left (b x + a\right )}^{\frac {5}{2}} x^{m} \,d x } \]
Result contains complex when optimal does not.
Time = 7.83 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int x^m (a+b x)^{5/2} \, dx=\frac {a^{\frac {5}{2}} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\Gamma \left (m + 2\right )} \]
a**(5/2)*x**(m + 1)*gamma(m + 1)*hyper((-5/2, m + 1), (m + 2,), b*x*exp_po lar(I*pi)/a)/gamma(m + 2)
\[ \int x^m (a+b x)^{5/2} \, dx=\int { {\left (b x + a\right )}^{\frac {5}{2}} x^{m} \,d x } \]
\[ \int x^m (a+b x)^{5/2} \, dx=\int { {\left (b x + a\right )}^{\frac {5}{2}} x^{m} \,d x } \]
Timed out. \[ \int x^m (a+b x)^{5/2} \, dx=\int x^m\,{\left (a+b\,x\right )}^{5/2} \,d x \]
\[ \int x^m (a+b x)^{5/2} \, dx=\frac {30 x^{m} \sqrt {b x +a}\, a^{3}+16 x^{m} \sqrt {b x +a}\, a^{2} b \,m^{3} x +112 x^{m} \sqrt {b x +a}\, a^{2} b \,m^{2} x +232 x^{m} \sqrt {b x +a}\, a^{2} b m x +90 x^{m} \sqrt {b x +a}\, a^{2} b x +32 x^{m} \sqrt {b x +a}\, a \,b^{2} m^{3} x^{2}+184 x^{m} \sqrt {b x +a}\, a \,b^{2} m^{2} x^{2}+264 x^{m} \sqrt {b x +a}\, a \,b^{2} m \,x^{2}+90 x^{m} \sqrt {b x +a}\, a \,b^{2} x^{2}+16 x^{m} \sqrt {b x +a}\, b^{3} m^{3} x^{3}+72 x^{m} \sqrt {b x +a}\, b^{3} m^{2} x^{3}+92 x^{m} \sqrt {b x +a}\, b^{3} m \,x^{3}+30 x^{m} \sqrt {b x +a}\, b^{3} x^{3}-480 \left (\int \frac {x^{m} \sqrt {b x +a}}{16 b \,m^{4} x^{2}+16 a \,m^{4} x +128 b \,m^{3} x^{2}+128 a \,m^{3} x +344 b \,m^{2} x^{2}+344 a \,m^{2} x +352 b m \,x^{2}+352 a m x +105 b \,x^{2}+105 a x}d x \right ) a^{4} m^{5}-3840 \left (\int \frac {x^{m} \sqrt {b x +a}}{16 b \,m^{4} x^{2}+16 a \,m^{4} x +128 b \,m^{3} x^{2}+128 a \,m^{3} x +344 b \,m^{2} x^{2}+344 a \,m^{2} x +352 b m \,x^{2}+352 a m x +105 b \,x^{2}+105 a x}d x \right ) a^{4} m^{4}-10320 \left (\int \frac {x^{m} \sqrt {b x +a}}{16 b \,m^{4} x^{2}+16 a \,m^{4} x +128 b \,m^{3} x^{2}+128 a \,m^{3} x +344 b \,m^{2} x^{2}+344 a \,m^{2} x +352 b m \,x^{2}+352 a m x +105 b \,x^{2}+105 a x}d x \right ) a^{4} m^{3}-10560 \left (\int \frac {x^{m} \sqrt {b x +a}}{16 b \,m^{4} x^{2}+16 a \,m^{4} x +128 b \,m^{3} x^{2}+128 a \,m^{3} x +344 b \,m^{2} x^{2}+344 a \,m^{2} x +352 b m \,x^{2}+352 a m x +105 b \,x^{2}+105 a x}d x \right ) a^{4} m^{2}-3150 \left (\int \frac {x^{m} \sqrt {b x +a}}{16 b \,m^{4} x^{2}+16 a \,m^{4} x +128 b \,m^{3} x^{2}+128 a \,m^{3} x +344 b \,m^{2} x^{2}+344 a \,m^{2} x +352 b m \,x^{2}+352 a m x +105 b \,x^{2}+105 a x}d x \right ) a^{4} m}{b \left (16 m^{4}+128 m^{3}+344 m^{2}+352 m +105\right )} \]
(2*(15*x**m*sqrt(a + b*x)*a**3 + 8*x**m*sqrt(a + b*x)*a**2*b*m**3*x + 56*x **m*sqrt(a + b*x)*a**2*b*m**2*x + 116*x**m*sqrt(a + b*x)*a**2*b*m*x + 45*x **m*sqrt(a + b*x)*a**2*b*x + 16*x**m*sqrt(a + b*x)*a*b**2*m**3*x**2 + 92*x **m*sqrt(a + b*x)*a*b**2*m**2*x**2 + 132*x**m*sqrt(a + b*x)*a*b**2*m*x**2 + 45*x**m*sqrt(a + b*x)*a*b**2*x**2 + 8*x**m*sqrt(a + b*x)*b**3*m**3*x**3 + 36*x**m*sqrt(a + b*x)*b**3*m**2*x**3 + 46*x**m*sqrt(a + b*x)*b**3*m*x**3 + 15*x**m*sqrt(a + b*x)*b**3*x**3 - 240*int((x**m*sqrt(a + b*x))/(16*a*m* *4*x + 128*a*m**3*x + 344*a*m**2*x + 352*a*m*x + 105*a*x + 16*b*m**4*x**2 + 128*b*m**3*x**2 + 344*b*m**2*x**2 + 352*b*m*x**2 + 105*b*x**2),x)*a**4*m **5 - 1920*int((x**m*sqrt(a + b*x))/(16*a*m**4*x + 128*a*m**3*x + 344*a*m* *2*x + 352*a*m*x + 105*a*x + 16*b*m**4*x**2 + 128*b*m**3*x**2 + 344*b*m**2 *x**2 + 352*b*m*x**2 + 105*b*x**2),x)*a**4*m**4 - 5160*int((x**m*sqrt(a + b*x))/(16*a*m**4*x + 128*a*m**3*x + 344*a*m**2*x + 352*a*m*x + 105*a*x + 1 6*b*m**4*x**2 + 128*b*m**3*x**2 + 344*b*m**2*x**2 + 352*b*m*x**2 + 105*b*x **2),x)*a**4*m**3 - 5280*int((x**m*sqrt(a + b*x))/(16*a*m**4*x + 128*a*m** 3*x + 344*a*m**2*x + 352*a*m*x + 105*a*x + 16*b*m**4*x**2 + 128*b*m**3*x** 2 + 344*b*m**2*x**2 + 352*b*m*x**2 + 105*b*x**2),x)*a**4*m**2 - 1575*int(( x**m*sqrt(a + b*x))/(16*a*m**4*x + 128*a*m**3*x + 344*a*m**2*x + 352*a*m*x + 105*a*x + 16*b*m**4*x**2 + 128*b*m**3*x**2 + 344*b*m**2*x**2 + 352*b*m* x**2 + 105*b*x**2),x)*a**4*m))/(b*(16*m**4 + 128*m**3 + 344*m**2 + 352*...