3.31.90 \(\int (a+b x)^m (c+d x)^{-3-m} (e+f x)^2 \, dx\) [3090]

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

[Out]

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

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Rubi [A]
time = 0.13, 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 = {91, 80, 72, 71} \begin {gather*} -\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}+\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} \end {gather*}

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 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 91

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])))

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.77, size = 360, normalized size = 1.76 \begin {gather*} \frac {1}{3} (a+b x)^m (c+d x)^{-m} \left (\frac {6 e f \left (1+\frac {b x}{a}\right )^{-m} \left (1+\frac {d x}{c}\right )^m \left (b^2 c^2 (1+m) x^2 \left (\frac {c (a+b x)}{a (c+d x)}\right )^m-a b c x \left (\frac {c (a+b x)}{a (c+d x)}\right )^m (-c m+d (2+m) x)+a^2 \left (d^2 x^2-c^2 \left (-1+\left (\frac {c (a+b x)}{a (c+d x)}\right )^m\right )-c d x \left (-2+2 \left (\frac {c (a+b x)}{a (c+d x)}\right )^m+m \left (\frac {c (a+b x)}{a (c+d x)}\right )^m\right )\right )\right )}{c (b c-a d)^2 (1+m) (2+m) (c+d x)^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,3+m;4;-\frac {b x}{a},-\frac {d x}{c}\right )}{c^3}-\frac {3 e^2 \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \, _2F_1\left (-2-m,-m;-1-m;\frac {b (c+d x)}{b c-a d}\right )}{d (2+m) (c+d x)^2}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x)^m*((6*e*f*(1 + (d*x)/c)^m*(b^2*c^2*(1 + m)*x^2*((c*(a + b*x))/(a*(c + d*x)))^m - a*b*c*x*((c*(a + b
*x))/(a*(c + d*x)))^m*(-(c*m) + d*(2 + m)*x) + a^2*(d^2*x^2 - c^2*(-1 + ((c*(a + b*x))/(a*(c + d*x)))^m) - c*d
*x*(-2 + 2*((c*(a + b*x))/(a*(c + d*x)))^m + m*((c*(a + b*x))/(a*(c + d*x)))^m))))/(c*(b*c - a*d)^2*(1 + m)*(2
+ m)*(1 + (b*x)/a)^m*(c + d*x)^2) + (f^2*x^3*(1 + (d*x)/c)^m*AppellF1[3, -m, 3 + m, 4, -((b*x)/a), -((d*x)/c)
])/(c^3*(1 + (b*x)/a)^m) - (3*e^2*Hypergeometric2F1[-2 - m, -m, -1 - m, (b*(c + d*x))/(b*c - a*d)])/(d*(2 + m)
*((d*(a + b*x))/(-(b*c) + a*d))^m*(c + d*x)^2)))/(3*(c + d*x)^m)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-3-m} \left (f x +e \right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-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 = 0.00, 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)^(-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 = 0.00, 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)^(-3-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 - 3), 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)**(-3-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)^(-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)

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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+3}} \,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 + 3),x)

[Out]

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

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