3.952 \(\int (b x)^m (c+d x)^n (e+f x)^2 \, dx\)

Optimal. Leaf size=209 \[ \frac{(b x)^{m+1} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} \left (c^2 f^2 \left (m^2+3 m+2\right )-2 c d e f (m+1) (m+n+3)+d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right ) \, _2F_1\left (m+1,-n;m+2;-\frac{d x}{c}\right )}{b d^2 (m+1) (m+n+2) (m+n+3)}-\frac{f (b x)^{m+1} (c+d x)^{n+1} (c f (m+2)-d e (m+n+4))}{b d^2 (m+n+2) (m+n+3)}+\frac{f (b x)^{m+1} (e+f x) (c+d x)^{n+1}}{b d (m+n+3)} \]

[Out]

-((f*(c*f*(2 + m) - d*e*(4 + m + n))*(b*x)^(1 + m)*(c + d*x)^(1 + n))/(b*d^2*(2 + m + n)*(3 + m + n))) + (f*(b
*x)^(1 + m)*(c + d*x)^(1 + n)*(e + f*x))/(b*d*(3 + m + n)) + ((c^2*f^2*(2 + 3*m + m^2) - 2*c*d*e*f*(1 + m)*(3
+ m + n) + d^2*e^2*(6 + m^2 + 5*n + n^2 + m*(5 + 2*n)))*(b*x)^(1 + m)*(c + d*x)^n*Hypergeometric2F1[1 + m, -n,
2 + m, -((d*x)/c)])/(b*d^2*(1 + m)*(2 + m + n)*(3 + m + n)*(1 + (d*x)/c)^n)

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Rubi [A]  time = 0.18416, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {90, 80, 66, 64} \[ \frac{(b x)^{m+1} (c+d x)^n \left (\frac{d x}{c}+1\right )^{-n} \left (c^2 f^2 \left (m^2+3 m+2\right )-2 c d e f (m+1) (m+n+3)+d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right ) \, _2F_1\left (m+1,-n;m+2;-\frac{d x}{c}\right )}{b d^2 (m+1) (m+n+2) (m+n+3)}-\frac{f (b x)^{m+1} (c+d x)^{n+1} (c f (m+2)-d e (m+n+4))}{b d^2 (m+n+2) (m+n+3)}+\frac{f (b x)^{m+1} (e+f x) (c+d x)^{n+1}}{b d (m+n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((f*(c*f*(2 + m) - d*e*(4 + m + n))*(b*x)^(1 + m)*(c + d*x)^(1 + n))/(b*d^2*(2 + m + n)*(3 + m + n))) + (f*(b
*x)^(1 + m)*(c + d*x)^(1 + n)*(e + f*x))/(b*d*(3 + m + n)) + ((c^2*f^2*(2 + 3*m + m^2) - 2*c*d*e*f*(1 + m)*(3
+ m + n) + d^2*e^2*(6 + m^2 + 5*n + n^2 + m*(5 + 2*n)))*(b*x)^(1 + m)*(c + d*x)^n*Hypergeometric2F1[1 + m, -n,
2 + m, -((d*x)/c)])/(b*d^2*(1 + m)*(2 + m + n)*(3 + m + 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)^m (c+d x)^n (e+f x)^2 \, dx &=\frac{f (b x)^{1+m} (c+d x)^{1+n} (e+f x)}{b d (3+m+n)}+\frac{\int (b x)^m (c+d x)^n (-b e (c f (1+m)-d e (3+m+n))-b f (c f (2+m)-d e (4+m+n)) x) \, dx}{b d (3+m+n)}\\ &=-\frac{f (c f (2+m)-d e (4+m+n)) (b x)^{1+m} (c+d x)^{1+n}}{b d^2 (2+m+n) (3+m+n)}+\frac{f (b x)^{1+m} (c+d x)^{1+n} (e+f x)}{b d (3+m+n)}+\frac{\left (c^2 f^2 \left (2+3 m+m^2\right )-2 c d e f (1+m) (3+m+n)+d^2 e^2 \left (6+m^2+5 n+n^2+m (5+2 n)\right )\right ) \int (b x)^m (c+d x)^n \, dx}{d^2 (2+m+n) (3+m+n)}\\ &=-\frac{f (c f (2+m)-d e (4+m+n)) (b x)^{1+m} (c+d x)^{1+n}}{b d^2 (2+m+n) (3+m+n)}+\frac{f (b x)^{1+m} (c+d x)^{1+n} (e+f x)}{b d (3+m+n)}+\frac{\left (\left (c^2 f^2 \left (2+3 m+m^2\right )-2 c d e f (1+m) (3+m+n)+d^2 e^2 \left (6+m^2+5 n+n^2+m (5+2 n)\right )\right ) (c+d x)^n \left (1+\frac{d x}{c}\right )^{-n}\right ) \int (b x)^m \left (1+\frac{d x}{c}\right )^n \, dx}{d^2 (2+m+n) (3+m+n)}\\ &=-\frac{f (c f (2+m)-d e (4+m+n)) (b x)^{1+m} (c+d x)^{1+n}}{b d^2 (2+m+n) (3+m+n)}+\frac{f (b x)^{1+m} (c+d x)^{1+n} (e+f x)}{b d (3+m+n)}+\frac{\left (c^2 f^2 \left (2+3 m+m^2\right )-2 c d e f (1+m) (3+m+n)+d^2 e^2 \left (6+m^2+5 n+n^2+m (5+2 n)\right )\right ) (b x)^{1+m} (c+d x)^n \left (1+\frac{d x}{c}\right )^{-n} \, _2F_1\left (1+m,-n;2+m;-\frac{d x}{c}\right )}{b d^2 (1+m) (2+m+n) (3+m+n)}\\ \end{align*}

Mathematica [A]  time = 0.231982, size = 153, normalized size = 0.73 \[ \frac{x (b x)^m (c+d x)^n \left (f (c+d x) (e+f x)-\frac{\left (\frac{d x}{c}+1\right )^{-n} (d e (m+n+2) (c f (m+1)-d e (m+n+3))-c f (m+1) (c f (m+2)-d e (m+n+4))) \, _2F_1\left (m+1,-n;m+2;-\frac{d x}{c}\right )+f (m+1) (c+d x) (c f (m+2)-d e (m+n+4))}{d (m+1) (m+n+2)}\right )}{d (m+n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b*x)^m*(c + d*x)^n*(f*(c + d*x)*(e + f*x) - (f*(1 + m)*(c*f*(2 + m) - d*e*(4 + m + n))*(c + d*x) + ((d*e*(
2 + m + n)*(c*f*(1 + m) - d*e*(3 + m + n)) - c*f*(1 + m)*(c*f*(2 + m) - d*e*(4 + m + n)))*Hypergeometric2F1[1
+ m, -n, 2 + m, -((d*x)/c)])/(1 + (d*x)/c)^n)/(d*(1 + m)*(2 + m + n))))/(d*(3 + m + n))

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \begin{align*} \int \left ( bx \right ) ^{m} \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)^m*(d*x+c)^n*(f*x+e)^2,x)

[Out]

int((b*x)^m*(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 (f x + e\right )}^{2} \left (b x\right )^{m}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (f^{2} x^{2} + 2 \, e f x + e^{2}\right )} \left (b x\right )^{m}{\left (d x + c\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 16.892, size = 131, normalized size = 0.63 \begin{align*} \frac{b^{m} c^{n} e^{2} x x^{m} \Gamma \left (m + 1\right ){{}_{2}F_{1}\left (\begin{matrix} - n, m + 1 \\ m + 2 \end{matrix}\middle |{\frac{d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 2\right )} + \frac{2 b^{m} c^{n} e f x^{2} x^{m} \Gamma \left (m + 2\right ){{}_{2}F_{1}\left (\begin{matrix} - n, m + 2 \\ m + 3 \end{matrix}\middle |{\frac{d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 3\right )} + \frac{b^{m} c^{n} f^{2} x^{3} x^{m} \Gamma \left (m + 3\right ){{}_{2}F_{1}\left (\begin{matrix} - n, m + 3 \\ m + 4 \end{matrix}\middle |{\frac{d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**m*c**n*e**2*x*x**m*gamma(m + 1)*hyper((-n, m + 1), (m + 2,), d*x*exp_polar(I*pi)/c)/gamma(m + 2) + 2*b**m*c
**n*e*f*x**2*x**m*gamma(m + 2)*hyper((-n, m + 2), (m + 3,), d*x*exp_polar(I*pi)/c)/gamma(m + 3) + b**m*c**n*f*
*2*x**3*x**m*gamma(m + 3)*hyper((-n, m + 3), (m + 4,), d*x*exp_polar(I*pi)/c)/gamma(m + 4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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