3.3081 \(\int (a+b x)^m (c+d x)^{-2-m} (e+f x)^2 \, dx\)

Optimal. Leaf size=204 \[ -\frac{f (a+b x)^m (c+d x)^{-m} \left (-\frac{d (a+b x)}{b c-a d}\right )^{-m} (a d f m+b (2 d e-c f (m+2))) \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{b d^3 m}+\frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1} (a d f (m+1)+b (d e-c f (m+2)))}{b d^2 (m+1) (b c-a d)}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-1}}{b d} \]

[Out]

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

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Rubi [A]  time = 0.187174, 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 = {90, 79, 70, 69} \[ -\frac{f (a+b x)^m (c+d x)^{-m} \left (-\frac{d (a+b x)}{b c-a d}\right )^{-m} (a d f m-b c f (m+2)+2 b d e) \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{b d^3 m}+\frac{(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1} (a d f (m+1)-b c f (m+2)+b d e)}{b d^2 (m+1) (b c-a d)}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{-m-1}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.429895, size = 179, normalized size = 0.88 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (\frac{f \left (\frac{d (a+b x)}{a d-b c}\right )^{-m} (-a d f m+b c f (m+2)-2 b d e) \, _2F_1\left (-m,-m;1-m;\frac{b (c+d x)}{b c-a d}\right )}{d^2 m}+\frac{(a+b x) (d e-c f) (a d f (m+1)-b c f (m+2)+b d e)}{d (m+1) (c+d x) (b c-a d)}+\frac{f (a+b x) (e+f x)}{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.058, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-2-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)^(-2-m)*(f*x+e)^2,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-2-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 - 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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