3.126 \(\int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx\)

Optimal. Leaf size=235 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right ) \left (a^2 d^2 f h \left (m^2-3 m+2\right )-a b d (1-m) (3 d (e h+f g)-2 c f h (m+1))+b^2 \left (c^2 f h \left (m^2+3 m+2\right )-3 c d (m+1) (e h+f g)+6 d^2 e g\right )\right )}{6 b^3 d^2 (m+1)}+\frac {(a+b x)^{m+1} (c+d x)^{1-m} (-a d f h (2-m)-b c f h (m+2)+3 b d (e h+f g)+2 b d f h x)}{6 b^2 d^2} \]

[Out]

1/6*(b*x+a)^(1+m)*(d*x+c)^(1-m)*(3*b*d*(e*h+f*g)-a*d*f*h*(2-m)-b*c*f*h*(2+m)+2*b*d*f*h*x)/b^2/d^2+1/6*(a^2*d^2
*f*h*(m^2-3*m+2)-a*b*d*(1-m)*(3*d*(e*h+f*g)-2*c*f*h*(1+m))+b^2*(6*d^2*e*g-3*c*d*(e*h+f*g)*(1+m)+c^2*f*h*(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 = 0.14, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {147, 70, 69} \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right ) \left (a^2 d^2 f h \left (m^2-3 m+2\right )-a b d (1-m) (3 d (e h+f g)-2 c f h (m+1))+b^2 \left (c^2 f h \left (m^2+3 m+2\right )-3 c d (m+1) (e h+f g)+6 d^2 e g\right )\right )}{6 b^3 d^2 (m+1)}+\frac {(a+b x)^{m+1} (c+d x)^{1-m} (-a d f h (2-m)-b c f h (m+2)+3 b d (e h+f g)+2 b d f h x)}{6 b^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(1 - m)*(3*b*d*(f*g + e*h) - a*d*f*h*(2 - m) - b*c*f*h*(2 + m) + 2*b*d*f*h*x))/(6
*b^2*d^2) + ((a^2*d^2*f*h*(2 - 3*m + m^2) - a*b*d*(1 - m)*(3*d*(f*g + e*h) - 2*c*f*h*(1 + m)) + b^2*(6*d^2*e*g
- 3*c*d*(f*g + e*h)*(1 + m) + c^2*f*h*(2 + 3*m + m^2)))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hyper
geometric2F1[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 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^{-m} (e+f x) (g+h x) \, dx &=\frac {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \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 {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \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 {(a+b x)^{1+m} (c+d x)^{1-m} (3 b d (f g+e h)-a d f h (2-m)-b c f h (2+m)+2 b d f h x)}{6 b^2 d^2}+\frac {\left (a^2 d^2 f h \left (2-3 m+m^2\right )-a b d (1-m) (3 d (f g+e h)-2 c f h (1+m))+b^2 \left (6 d^2 e g-3 c d (f g+e h) (1+m)+c^2 f h \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 = 0.22, size = 189, normalized size = 0.80 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (b \left (b (d e-c f) (d g-c h) \, _2F_1\left (m,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )-(b c-a d) (2 c f h-d (e h+f g)) \, _2F_1\left (m-1,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )\right )+f h (b c-a d)^2 \, _2F_1\left (m-2,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )\right )}{b^3 d^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f h x^{2} + e g + {\left (f g + e h\right )} x\right )} {\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((e + f*x)*(g + h*x)*(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)*(h*x+g)/((d*x+c)**m),x)

[Out]

Exception raised: HeuristicGCDFailed

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