Integrand size = 20, antiderivative size = 108 \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 f (b x)^{7/2} (c+d x)^{1+n}}{b d (9+2 n)}-\frac {2 (7 c f-d e (9+2 n)) (b x)^{7/2} (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-n,\frac {9}{2},-\frac {d x}{c}\right )}{7 b d (9+2 n)} \] Output:
2*f*(b*x)^(7/2)*(d*x+c)^(1+n)/b/d/(9+2*n)-2/7*(7*c*f-d*e*(9+2*n))*(b*x)^(7 /2)*(d*x+c)^n*hypergeom([7/2, -n],[9/2],-d*x/c)/b/d/(9+2*n)/(((d*x+c)/c)^n )
Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.84 \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 x (b x)^{5/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (7 f (c+d x) \left (1+\frac {d x}{c}\right )^n+(-7 c f+d e (9+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-n,\frac {9}{2},-\frac {d x}{c}\right )\right )}{7 d (9+2 n)} \] Input:
Integrate[(b*x)^(5/2)*(c + d*x)^n*(e + f*x),x]
Output:
(2*x*(b*x)^(5/2)*(c + d*x)^n*(7*f*(c + d*x)*(1 + (d*x)/c)^n + (-7*c*f + d* e*(9 + 2*n))*Hypergeometric2F1[7/2, -n, 9/2, -((d*x)/c)]))/(7*d*(9 + 2*n)* (1 + (d*x)/c)^n)
Time = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {90, 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.
| \(\displaystyle \int (b x)^{5/2} (e+f x) (c+d x)^n \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {2 f (b x)^{7/2} (c+d x)^{n+1}}{b d (2 n+9)}-\frac {(7 c f-d e (2 n+9)) \int (b x)^{5/2} (c+d x)^ndx}{d (2 n+9)}\) |
| \(\Big \downarrow \) 76 |
| \(\displaystyle \frac {2 f (b x)^{7/2} (c+d x)^{n+1}}{b d (2 n+9)}-\frac {(c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} (7 c f-d e (2 n+9)) \int (b x)^{5/2} \left (\frac {d x}{c}+1\right )^ndx}{d (2 n+9)}\) |
| \(\Big \downarrow \) 74 |
| \(\displaystyle \frac {2 f (b x)^{7/2} (c+d x)^{n+1}}{b d (2 n+9)}-\frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} (7 c f-d e (2 n+9)) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-n,\frac {9}{2},-\frac {d x}{c}\right )}{7 b d (2 n+9)}\) |
Input:
Int[(b*x)^(5/2)*(c + d*x)^n*(e + f*x),x]
Output:
(2*f*(b*x)^(7/2)*(c + d*x)^(1 + n))/(b*d*(9 + 2*n)) - (2*(7*c*f - d*e*(9 + 2*n))*(b*x)^(7/2)*(c + d*x)^n*Hypergeometric2F1[7/2, -n, 9/2, -((d*x)/c)] )/(7*b*d*(9 + 2*n)*(1 + (d*x)/c)^n)
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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
\[\int \left (b x \right )^{\frac {5}{2}} \left (x d +c \right )^{n} \left (f x +e \right )d x\]
Input:
int((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x)
Output:
int((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x)
\[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\int { \left (b x\right )^{\frac {5}{2}} {\left (f x + e\right )} {\left (d x + c\right )}^{n} \,d x } \] Input:
integrate((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x, algorithm="fricas")
Output:
integral((b^2*f*x^3 + b^2*e*x^2)*sqrt(b*x)*(d*x + c)^n, x)
Result contains complex when optimal does not.
Time = 61.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.65 \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 b^{\frac {5}{2}} c^{n} e x^{\frac {7}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - n \\ \frac {9}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{7} + \frac {2 b^{\frac {5}{2}} c^{n} f x^{\frac {9}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{2}, - n \\ \frac {11}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{9} \] Input:
integrate((b*x)**(5/2)*(d*x+c)**n*(f*x+e),x)
Output:
2*b**(5/2)*c**n*e*x**(7/2)*hyper((7/2, -n), (9/2,), d*x*exp_polar(I*pi)/c) /7 + 2*b**(5/2)*c**n*f*x**(9/2)*hyper((9/2, -n), (11/2,), d*x*exp_polar(I* pi)/c)/9
\[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\int { \left (b x\right )^{\frac {5}{2}} {\left (f x + e\right )} {\left (d x + c\right )}^{n} \,d x } \] Input:
integrate((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x, algorithm="maxima")
Output:
integrate((b*x)^(5/2)*(f*x + e)*(d*x + c)^n, x)
\[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\int { \left (b x\right )^{\frac {5}{2}} {\left (f x + e\right )} {\left (d x + c\right )}^{n} \,d x } \] Input:
integrate((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x, algorithm="giac")
Output:
integrate((b*x)^(5/2)*(f*x + e)*(d*x + c)^n, x)
Timed out. \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\int \left (e+f\,x\right )\,{\left (b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^n \,d x \] Input:
int((e + f*x)*(b*x)^(5/2)*(c + d*x)^n,x)
Output:
int((e + f*x)*(b*x)^(5/2)*(c + d*x)^n, x)
\[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\text {too large to display} \] Input:
int((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x)
Output:
(2*sqrt(b)*b**2*( - 210*sqrt(x)*(c + d*x)**n*c**4*f*n + 60*sqrt(x)*(c + d* x)**n*c**3*d*e*n**2 + 270*sqrt(x)*(c + d*x)**n*c**3*d*e*n + 140*sqrt(x)*(c + d*x)**n*c**3*d*f*n**2*x + 70*sqrt(x)*(c + d*x)**n*c**3*d*f*n*x - 40*sqr t(x)*(c + d*x)**n*c**2*d**2*e*n**3*x - 200*sqrt(x)*(c + d*x)**n*c**2*d**2* e*n**2*x - 90*sqrt(x)*(c + d*x)**n*c**2*d**2*e*n*x - 56*sqrt(x)*(c + d*x)* *n*c**2*d**2*f*n**3*x**2 - 112*sqrt(x)*(c + d*x)**n*c**2*d**2*f*n**2*x**2 - 42*sqrt(x)*(c + d*x)**n*c**2*d**2*f*n*x**2 + 16*sqrt(x)*(c + d*x)**n*c*d **3*e*n**4*x**2 + 104*sqrt(x)*(c + d*x)**n*c*d**3*e*n**3*x**2 + 156*sqrt(x )*(c + d*x)**n*c*d**3*e*n**2*x**2 + 54*sqrt(x)*(c + d*x)**n*c*d**3*e*n*x** 2 + 16*sqrt(x)*(c + d*x)**n*c*d**3*f*n**4*x**3 + 72*sqrt(x)*(c + d*x)**n*c *d**3*f*n**3*x**3 + 92*sqrt(x)*(c + d*x)**n*c*d**3*f*n**2*x**3 + 30*sqrt(x )*(c + d*x)**n*c*d**3*f*n*x**3 + 16*sqrt(x)*(c + d*x)**n*d**4*e*n**4*x**3 + 144*sqrt(x)*(c + d*x)**n*d**4*e*n**3*x**3 + 416*sqrt(x)*(c + d*x)**n*d** 4*e*n**2*x**3 + 444*sqrt(x)*(c + d*x)**n*d**4*e*n*x**3 + 135*sqrt(x)*(c + d*x)**n*d**4*e*x**3 + 16*sqrt(x)*(c + d*x)**n*d**4*f*n**4*x**4 + 128*sqrt( x)*(c + d*x)**n*d**4*f*n**3*x**4 + 344*sqrt(x)*(c + d*x)**n*d**4*f*n**2*x* *4 + 352*sqrt(x)*(c + d*x)**n*d**4*f*n*x**4 + 105*sqrt(x)*(c + d*x)**n*d** 4*f*x**4 + 3360*int((sqrt(x)*(c + d*x)**n)/(32*c*n**5*x + 400*c*n**4*x + 1 840*c*n**3*x + 3800*c*n**2*x + 3378*c*n*x + 945*c*x + 32*d*n**5*x**2 + 400 *d*n**4*x**2 + 1840*d*n**3*x**2 + 3800*d*n**2*x**2 + 3378*d*n*x**2 + 94...