3.3160 \(\int (a+b x)^m (c+d x)^n (e+f x)^2 \, dx\)

Optimal. Leaf size=259 \[ \frac{(a+b x)^{m+1} (c+d x)^{n+1} (f (a d (n+1)+b c (m+1)) (b d e (m+n+4)-f (a d (n+2)+b c (m+2)))+b d (m+n+2) (a f (c f+d e (n+1))+b e (c f (m+1)-d e (m+n+3)))) \, _2F_1\left (1,m+n+2;n+2;\frac{b (c+d x)}{b c-a d}\right )}{b^2 d^2 (n+1) (m+n+2) (m+n+3) (b c-a d)}+\frac{f (a+b x)^{m+1} (c+d x)^{n+1} (b d e (m+n+4)-f (a d (n+2)+b c (m+2)))}{b^2 d^2 (m+n+2) (m+n+3)}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{n+1}}{b d (m+n+3)} \]

[Out]

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

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Rubi [A]  time = 0.340053, antiderivative size = 272, normalized size of antiderivative = 1.05, 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, 70, 69} \[ \frac{(a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (b d (m+n+2) \left (b d e^2 (m+n+3)-f (a c f+a d e (n+1)+b c e (m+1))\right )-f (a d (n+1)+b c (m+1)) (b d e (m+n+4)-f (a d (n+2)+b c (m+2)))\right ) \, _2F_1\left (m+1,-n;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b^3 d^2 (m+1) (m+n+2) (m+n+3)}+\frac{f (a+b x)^{m+1} (c+d x)^{n+1} (b d e (m+n+4)-f (a d (n+2)+b c (m+2)))}{b^2 d^2 (m+n+2) (m+n+3)}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{n+1}}{b d (m+n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(b*d*e*(4 + m + n) - f*(b*c*(2 + m) + a*d*(2 + n)))*(a + b*x)^(1 + m)*(c + d*x)^(1 + n))/(b^2*d^2*(2 + m +
n)*(3 + m + n)) + (f*(a + b*x)^(1 + m)*(c + d*x)^(1 + n)*(e + f*x))/(b*d*(3 + m + n)) + ((b*d*(2 + m + n)*(b*d
*e^2*(3 + m + n) - f*(a*c*f + b*c*e*(1 + m) + a*d*e*(1 + n))) - f*(b*c*(1 + m) + a*d*(1 + n))*(b*d*e*(4 + m +
n) - f*(b*c*(2 + m) + a*d*(2 + n))))*(a + b*x)^(1 + m)*(c + d*x)^n*Hypergeometric2F1[1 + m, -n, 2 + m, -((d*(a
+ b*x))/(b*c - a*d))])/(b^3*d^2*(1 + m)*(2 + m + n)*(3 + m + n)*((b*(c + d*x))/(b*c - a*d))^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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.416482, size = 253, normalized size = 0.98 \[ \frac{(a+b x)^{m+1} (c+d x)^n \left (\frac{\left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (a^2 d^2 f^2 \left (n^2+3 n+2\right )-2 a b d f (n+1) (d e (m+n+3)-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-2 c d e f (m+1) (m+n+3)+d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right ) \, _2F_1\left (m+1,-n;m+2;\frac{d (a+b x)}{a d-b c}\right )}{b^2 d (m+1) (m+n+2)}+\frac{f (c+d x) (b d e (m+n+4)-f (a d (n+2)+b c (m+2)))}{b d (m+n+2)}+f (c+d x) (e+f x)\right )}{b d (m+n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \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+a)^m*(d*x+c)^n*(f*x+e)^2,x)

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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