\(\int (b x)^{5/2} (c+d x)^n (e+f x) \, dx\) [949]

Optimal result

Integrand size = 20, antiderivative size = 107 \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 f (b x)^{7/2} (c+d x)^{1+n}}{b d (9+2 n)}-\frac {2 (7 c f-d e (9+2 n)) (b x)^{7/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-n,\frac {9}{2},-\frac {d x}{c}\right )}{7 b d (9+2 n)} \]

[Out]

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)/((1+d*x/c)^n)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {81, 68, 66} \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 f (b x)^{7/2} (c+d x)^{n+1}}{b d (2 n+9)}-\frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} (7 c f-d e (2 n+9)) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-n,\frac {9}{2},-\frac {d x}{c}\right )}{7 b d (2 n+9)} \]

[In]

Int[(b*x)^(5/2)*(c + d*x)^n*(e + f*x),x]

[Out]

(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*Hyper
geometric2F1[7/2, -n, 9/2, -((d*x)/c)])/(7*b*d*(9 + 2*n)*(1 + (d*x)/c)^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])

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(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]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.85 \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 x (b x)^{5/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (7 f (c+d x) \left (1+\frac {d x}{c}\right )^n+(-7 c f+d e (9+2 n)) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-n,\frac {9}{2},-\frac {d x}{c}\right )\right )}{7 d (9+2 n)} \]

[In]

Integrate[(b*x)^(5/2)*(c + d*x)^n*(e + f*x),x]

[Out]

(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)

Maple [F]

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

[In]

int((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x)

[Out]

int((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x)

Fricas [F]

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

[In]

integrate((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x, algorithm="fricas")

[Out]

integral((b^2*f*x^3 + b^2*e*x^2)*sqrt(b*x)*(d*x + c)^n, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 65.98 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.65 \[ \int (b x)^{5/2} (c+d x)^n (e+f x) \, dx=\frac {2 b^{\frac {5}{2}} c^{n} e x^{\frac {7}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - n \\ \frac {9}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{7} + \frac {2 b^{\frac {5}{2}} c^{n} f x^{\frac {9}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{2}, - n \\ \frac {11}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{9} \]

[In]

integrate((b*x)**(5/2)*(d*x+c)**n*(f*x+e),x)

[Out]

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)*hype
r((9/2, -n), (11/2,), d*x*exp_polar(I*pi)/c)/9

Maxima [F]

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

[In]

integrate((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x, algorithm="maxima")

[Out]

integrate((b*x)^(5/2)*(f*x + e)*(d*x + c)^n, x)

Giac [F]

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

[In]

integrate((b*x)^(5/2)*(d*x+c)^n*(f*x+e),x, algorithm="giac")

[Out]

integrate((b*x)^(5/2)*(f*x + e)*(d*x + c)^n, x)

Mupad [F(-1)]

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

[In]

int((e + f*x)*(b*x)^(5/2)*(c + d*x)^n,x)

[Out]

int((e + f*x)*(b*x)^(5/2)*(c + d*x)^n, x)