3.3090 \(\int (a+b x)^m (c+d x)^{-3-m} (e+f x)^2 \, dx\)

Optimal. Leaf size=205 \[ \frac{(a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-2}}{d^2 (m+2) (b c-a d)}-\frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1} (2 a d f (m+2)-b (c f (2 m+3)+d e))}{d^2 (m+1) (m+2) (b c-a d)^2}-\frac{f^2 (a+b x)^m (c+d x)^{-m} \left (-\frac{d (a+b x)}{b c-a d}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{d^3 m} \]

[Out]

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

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Rubi [A]  time = 0.190818, antiderivative size = 202, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {89, 79, 70, 69} \[ \frac{(a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-2}}{d^2 (m+2) (b c-a d)}+\frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1} (-2 a d f (m+2)+b c f (2 m+3)+b d e)}{d^2 (m+1) (m+2) (b c-a d)^2}-\frac{f^2 (a+b x)^m (c+d x)^{-m} \left (-\frac{d (a+b x)}{b c-a d}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{d^3 m} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-3 - m)*(e + f*x)^2,x]

[Out]

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

Rule 89

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

Rule 79

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

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

Mathematica [A]  time = 0.486068, size = 184, normalized size = 0.9 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (\frac{(a+b x) (d e-c f) (-2 a d f (m+2)+b c f (2 m+3)+b d e)}{(m+1) (c+d x) (b c-a d)}+\frac{(a+b x) (d e-c f)^2}{(c+d x)^2}+\frac{f^2 (m+2) (a+b x) \left (\frac{d (a+b x)}{a d-b c}\right )^{-m-1} \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{m}\right )}{d^2 (m+2) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^(-3 - m)*(e + f*x)^2,x]

[Out]

((a + b*x)^m*(((d*e - c*f)^2*(a + b*x))/(c + d*x)^2 + ((d*e - c*f)*(b*d*e - 2*a*d*f*(2 + m) + b*c*f*(3 + 2*m))
*(a + b*x))/((b*c - a*d)*(1 + m)*(c + d*x)) + (f^2*(2 + m)*(a + b*x)*((d*(a + b*x))/(-(b*c) + a*d))^(-1 - m)*H
ypergeometric2F1[-m, -m, 1 - m, (b*(c + d*x))/(b*c - a*d)])/m))/(d^2*(b*c - a*d)*(2 + m)*(c + d*x)^m)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-3-m} \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)^(-3-m)*(f*x+e)^2,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-3-m)*(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 )}^{-m - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)^2,x, algorithm="maxima")

[Out]

integrate((f*x + e)^2*(b*x + a)^m*(d*x + c)^(-m - 3), 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 )}^{-m - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(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)^(-m - 3), 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+a)**m*(d*x+c)**(-3-m)*(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 (f x + e\right )}^{2}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e)^2,x, algorithm="giac")

[Out]

integrate((f*x + e)^2*(b*x + a)^m*(d*x + c)^(-m - 3), x)