3.31.81 \(\int (a+b x)^m (c+d x)^{-2-m} (e+f x)^2 \, dx\) [3081]

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

[Out]

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

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Rubi [A]
time = 0.12, 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 = {92, 80, 72, 71} \begin {gather*} -\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} \end {gather*}

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 71

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

Rule 72

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 80

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 92

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]

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.42, size = 163, normalized size = 0.80 \begin {gather*} \frac {1}{3} (a+b x)^m (c+d x)^{-m} \left (\frac {3 e^2 (a+b x)}{(b c-a d) (1+m) (c+d x)}+\frac {3 e f x^2 \left (1+\frac {b x}{a}\right )^{-m} \left (1+\frac {d x}{c}\right )^m F_1\left (2;-m,2+m;3;-\frac {b x}{a},-\frac {d x}{c}\right )}{c^2}+\frac {f^2 x^3 \left (1+\frac {b x}{a}\right )^{-m} \left (1+\frac {d x}{c}\right )^m F_1\left (3;-m,2+m;4;-\frac {b x}{a},-\frac {d x}{c}\right )}{c^2}\right ) \end {gather*}

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*((3*e^2*(a + b*x))/((b*c - a*d)*(1 + m)*(c + d*x)) + (3*e*f*x^2*(1 + (d*x)/c)^m*AppellF1[2, -m, 2
+ m, 3, -((b*x)/a), -((d*x)/c)])/(c^2*(1 + (b*x)/a)^m) + (f^2*x^3*(1 + (d*x)/c)^m*AppellF1[3, -m, 2 + m, 4, -
((b*x)/a), -((d*x)/c)])/(c^2*(1 + (b*x)/a)^m)))/(3*(c + d*x)^m)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-2-m} \left (f x +e \right )^{2}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^2\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)^2*(a + b*x)^m)/(c + d*x)^(m + 2),x)

[Out]

int(((e + f*x)^2*(a + b*x)^m)/(c + d*x)^(m + 2), x)

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