3.3062 \(\int (a+b x)^m (c+d x)^{-m} (e+f x)^2 \, dx\)

Optimal. Leaf size=250 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (1-m) (3 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (m+1)}-\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (a d f (2-m)-b (4 d e-c f (m+2)))}{6 b^2 d^2}+\frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{1-m}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {90, 80, 70, 69} \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (1-m) (3 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (m+1)}+\frac {f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (2-m)-b c f (m+2)+4 b d e)}{6 b^2 d^2}+\frac {f (e+f x) (a+b x)^{m+1} (c+d x)^{1-m}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

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 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 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]

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^{-m} (e+f x)^2 \, dx &=\frac {f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac {\int (a+b x)^m (c+d x)^{-m} \left (3 b d e^2-f (a c f+a d e (1-m)+b c e (1+m))+f (4 b d e-a d f (2-m)-b c f (2+m)) x\right ) \, dx}{3 b d}\\ &=\frac {f (4 b d e-a d f (2-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 b^2 d^2}+\frac {f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac {\left (a^2 d^2 f^2 \left (2-3 m+m^2\right )-2 a b d f (1-m) (3 d e-c f (1+m))+b^2 \left (6 d^2 e^2-6 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{6 b^2 d^2}\\ &=\frac {f (4 b d e-a d f (2-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 b^2 d^2}+\frac {f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac {\left (\left (a^2 d^2 f^2 \left (2-3 m+m^2\right )-2 a b d f (1-m) (3 d e-c f (1+m))+b^2 \left (6 d^2 e^2-6 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m} \, dx}{6 b^2 d^2}\\ &=\frac {f (4 b d e-a d f (2-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 b^2 d^2}+\frac {f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac {\left (a^2 d^2 f^2 \left (2-3 m+m^2\right )-2 a b d f (1-m) (3 d e-c f (1+m))+b^2 \left (6 d^2 e^2-6 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 201, normalized size = 0.80 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {\left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (m-1) (c f (m+1)-3 d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )}{m+1}+b f (c+d x) (a d f (m-2)-b c f (m+2)+4 b d e)+2 b^2 d f (c+d x) (e+f x)\right )}{6 b^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.16, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f^{2} x^{2} + 2 \, e f x + e^{2}\right )} {\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (f x +e \right )^{2} \left (b x +a \right )^{m} \left (d x +c \right )^{-m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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