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

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

[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)

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Rubi [A]  time = 0.124683, antiderivative size = 195, normalized size of antiderivative = 1., 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 = {90, 80, 66, 64} \[ \frac{2 (b x)^{7/2} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} \left (63 c^2 f^2-14 c d e f (2 n+11)+d^2 e^2 \left (4 n^2+40 n+99\right )\right ) \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{d x}{c}\right )}{7 b d^2 (2 n+9) (2 n+11)}-\frac{2 f (b x)^{7/2} (c+d x)^{n+1} (9 c f-d e (2 n+13))}{b d^2 (2 n+9) (2 n+11)}+\frac{2 f (b x)^{7/2} (e+f x) (c+d x)^{n+1}}{b d (2 n+11)} \]

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 90

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]

Rule 80

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 66

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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])))

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*}

Mathematica [A]  time = 0.14591, size = 146, normalized size = 0.75 \[ \frac{2 x (b x)^{5/2} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} \left (\left (63 c^2 f^2-14 c d e f (2 n+11)+d^2 e^2 \left (4 n^2+40 n+99\right )\right ) \, _2F_1\left (\frac{7}{2},-n;\frac{9}{2};-\frac{d x}{c}\right )-7 f (c+d x) \left (\frac{d x}{c}+1\right )^n (9 c f-d (e (4 n+22)+f (2 n+9) x))\right )}{7 d^2 (2 n+9) (2 n+11)} \]

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 = 0.029, size = 0, normalized size = 0. \begin{align*} \int \left ( bx \right ) ^{{\frac{5}{2}}} \left ( dx+c \right ) ^{n} \left ( fx+e \right ) ^{2}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{\frac{5}{2}}{\left (f x + e\right )}^{2}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} f^{2} x^{4} + 2 \, b^{2} e f x^{3} + b^{2} e^{2} x^{2}\right )} \sqrt{b x}{\left (d x + c\right )}^{n}, x\right ) \end{align*}

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{\frac{5}{2}}{\left (f x + e\right )}^{2}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

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)