3.10.48 \(\int (b x)^{5/2} (c+d x)^n (e+f x)^2 \, dx\) [948]

Optimal. Leaf size=195 \[ -\frac {2 f (9 c f-d e (13+2 n)) (b x)^{7/2} (c+d x)^{1+n}}{b d^2 (9+2 n) (11+2 n)}+\frac {2 f (b x)^{7/2} (c+d x)^{1+n} (e+f x)}{b d (11+2 n)}+\frac {2 \left (63 c^2 f^2-14 c d e f (11+2 n)+d^2 e^2 \left (99+40 n+4 n^2\right )\right ) (b x)^{7/2} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \, _2F_1\left (\frac {7}{2},-n;\frac {9}{2};-\frac {d x}{c}\right )}{7 b d^2 (9+2 n) (11+2 n)} \]

[Out]

-2*f*(9*c*f-d*e*(13+2*n))*(b*x)^(7/2)*(d*x+c)^(1+n)/b/d^2/(9+2*n)/(11+2*n)+2*f*(b*x)^(7/2)*(d*x+c)^(1+n)*(f*x+
e)/b/d/(11+2*n)+2/7*(63*c^2*f^2-14*c*d*e*f*(11+2*n)+d^2*e^2*(4*n^2+40*n+99))*(b*x)^(7/2)*(d*x+c)^n*hypergeom([
7/2, -n],[9/2],-d*x/c)/b/d^2/(9+2*n)/(11+2*n)/((1+d*x/c)^n)

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Rubi [A]
time = 0.09, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {92, 81, 68, 66} \begin {gather*} \frac {2 (b x)^{7/2} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} \left (63 c^2 f^2-14 c d e f (2 n+11)+d^2 e^2 \left (4 n^2+40 n+99\right )\right ) \, _2F_1\left (\frac {7}{2},-n;\frac {9}{2};-\frac {d x}{c}\right )}{7 b d^2 (2 n+9) (2 n+11)}-\frac {2 f (b x)^{7/2} (c+d x)^{n+1} (9 c f-d e (2 n+13))}{b d^2 (2 n+9) (2 n+11)}+\frac {2 f (b x)^{7/2} (e+f x) (c+d x)^{n+1}}{b d (2 n+11)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*f*(9*c*f - d*e*(13 + 2*n))*(b*x)^(7/2)*(c + d*x)^(1 + n))/(b*d^2*(9 + 2*n)*(11 + 2*n)) + (2*f*(b*x)^(7/2)*
(c + d*x)^(1 + n)*(e + f*x))/(b*d*(11 + 2*n)) + (2*(63*c^2*f^2 - 14*c*d*e*f*(11 + 2*n) + d^2*e^2*(99 + 40*n +
4*n^2))*(b*x)^(7/2)*(c + d*x)^n*Hypergeometric2F1[7/2, -n, 9/2, -((d*x)/c)])/(7*b*d^2*(9 + 2*n)*(11 + 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]

Rule 92

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a + b*x
)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rubi steps

\begin {align*} \int (b x)^{5/2} (c+d x)^n (e+f x)^2 \, dx &=\frac {2 f (b x)^{7/2} (c+d x)^{1+n} (e+f x)}{b d (11+2 n)}+\frac {2 \int (b x)^{5/2} (c+d x)^n \left (-\frac {1}{2} b e \left (7 c f-2 d e \left (\frac {11}{2}+n\right )\right )-\frac {1}{2} b f (9 c f-d e (13+2 n)) x\right ) \, dx}{b d (11+2 n)}\\ &=-\frac {2 f (9 c f-d e (13+2 n)) (b x)^{7/2} (c+d x)^{1+n}}{b d^2 (9+2 n) (11+2 n)}+\frac {2 f (b x)^{7/2} (c+d x)^{1+n} (e+f x)}{b d (11+2 n)}+\frac {\left (63 c^2 f^2-14 c d e f (11+2 n)+d^2 e^2 \left (99+40 n+4 n^2\right )\right ) \int (b x)^{5/2} (c+d x)^n \, dx}{d^2 (9+2 n) (11+2 n)}\\ &=-\frac {2 f (9 c f-d e (13+2 n)) (b x)^{7/2} (c+d x)^{1+n}}{b d^2 (9+2 n) (11+2 n)}+\frac {2 f (b x)^{7/2} (c+d x)^{1+n} (e+f x)}{b d (11+2 n)}+\frac {\left (\left (63 c^2 f^2-14 c d e f (11+2 n)+d^2 e^2 \left (99+40 n+4 n^2\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}{d^2 (9+2 n) (11+2 n)}\\ &=-\frac {2 f (9 c f-d e (13+2 n)) (b x)^{7/2} (c+d x)^{1+n}}{b d^2 (9+2 n) (11+2 n)}+\frac {2 f (b x)^{7/2} (c+d x)^{1+n} (e+f x)}{b d (11+2 n)}+\frac {2 \left (63 c^2 f^2-14 c d e f (11+2 n)+d^2 e^2 \left (99+40 n+4 n^2\right )\right ) (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^2 (9+2 n) (11+2 n)}\\ \end {align*}

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Mathematica [A]
time = 0.25, size = 146, normalized size = 0.75 \begin {gather*} \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 (9 c f-d (e (22+4 n)+f (9+2 n) x))+\left (63 c^2 f^2-14 c d e f (11+2 n)+d^2 e^2 \left (99+40 n+4 n^2\right )\right ) \, _2F_1\left (\frac {7}{2},-n;\frac {9}{2};-\frac {d x}{c}\right )\right )}{7 d^2 (9+2 n) (11+2 n)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(b*x)^(5/2)*(c + d*x)^n*(-7*f*(c + d*x)*(1 + (d*x)/c)^n*(9*c*f - d*(e*(22 + 4*n) + f*(9 + 2*n)*x)) + (63*
c^2*f^2 - 14*c*d*e*f*(11 + 2*n) + d^2*e^2*(99 + 40*n + 4*n^2))*Hypergeometric2F1[7/2, -n, 9/2, -((d*x)/c)]))/(
7*d^2*(9 + 2*n)*(11 + 2*n)*(1 + (d*x)/c)^n)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (b x \right )^{\frac {5}{2}} \left (d x +c \right )^{n} \left (f x +e \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*f^2*x^4 + 2*b^2*f*x^3*e + b^2*x^2*e^2)*sqrt(b*x)*(d*x + c)^n, x)

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Sympy [C] Result contains complex when optimal does not.
time = 106.14, size = 110, normalized size = 0.56 \begin {gather*} \frac {2 b^{\frac {5}{2}} c^{n} e^{2} 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 {4 b^{\frac {5}{2}} c^{n} e 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} + \frac {2 b^{\frac {5}{2}} c^{n} f^{2} x^{\frac {11}{2}} {{}_{2}F_{1}\left (\begin {matrix} \frac {11}{2}, - n \\ \frac {13}{2} \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b**(5/2)*c**n*e**2*x**(7/2)*hyper((7/2, -n), (9/2,), d*x*exp_polar(I*pi)/c)/7 + 4*b**(5/2)*c**n*e*f*x**(9/2)
*hyper((9/2, -n), (11/2,), d*x*exp_polar(I*pi)/c)/9 + 2*b**(5/2)*c**n*f**2*x**(11/2)*hyper((11/2, -n), (13/2,)
, d*x*exp_polar(I*pi)/c)/11

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (e+f\,x\right )}^2\,{\left (b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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