\(\int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx\) [125]

Optimal result

Integrand size = 29, antiderivative size = 245 \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{2-m} (4 b d (f g+e h)-a d f h (3-m)-b c f h (2+m)+3 b d f h x)}{12 b^2 d^2}+\frac {(b c-a d) \left (a^2 d^2 f h \left (6-5 m+m^2\right )-2 a b d (2-m) (2 d (f g+e h)-c f h (1+m))+b^2 \left (12 d^2 e g-4 c d (f g+e h) (1+m)+c^2 f h \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (-1+m,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{12 b^4 d^2 (1+m)} \]

[Out]

1/12*(b*x+a)^(1+m)*(d*x+c)^(2-m)*(4*b*d*(e*h+f*g)-a*d*f*h*(3-m)-b*c*f*h*(2+m)+3*b*d*f*h*x)/b^2/d^2+1/12*(-a*d+
b*c)*(a^2*d^2*f*h*(m^2-5*m+6)-2*a*b*d*(2-m)*(2*d*(e*h+f*g)-c*f*h*(1+m))+b^2*(12*d^2*e*g-4*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([1+m, -1+m],[2+m],-d*(b*x+a)/(-a*d+b*c)
)/b^4/d^2/(1+m)/((d*x+c)^m)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {152, 72, 71} \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \operatorname {Hypergeometric2F1}\left (m-1,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right ) \left (a^2 d^2 f h \left (m^2-5 m+6\right )-2 a b d (2-m) (2 d (e h+f g)-c f h (m+1))+b^2 \left (c^2 f h \left (m^2+3 m+2\right )-4 c d (m+1) (e h+f g)+12 d^2 e g\right )\right )}{12 b^4 d^2 (m+1)}+\frac {(a+b x)^{m+1} (c+d x)^{2-m} (-a d f h (3-m)-b c f h (m+2)+4 b d (e h+f g)+3 b d f h x)}{12 b^2 d^2} \]

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(2 - m)*(4*b*d*(f*g + e*h) - a*d*f*h*(3 - m) - b*c*f*h*(2 + m) + 3*b*d*f*h*x))/(1
2*b^2*d^2) + ((b*c - a*d)*(a^2*d^2*f*h*(6 - 5*m + m^2) - 2*a*b*d*(2 - m)*(2*d*(f*g + e*h) - c*f*h*(1 + m)) + b
^2*(12*d^2*e*g - 4*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*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(12*b^4*d^2*(1 + 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 152

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*} \text {integral}& = \frac {(a+b x)^{1+m} (c+d x)^{2-m} (4 b d (f g+e h)-a d f h (3-m)-b c f h (2+m)+3 b d f h x)}{12 b^2 d^2}+\frac {\left (a^2 d^2 f h \left (6-5 m+m^2\right )-2 a b d (2-m) (2 d (f g+e h)-c f h (1+m))+b^2 \left (12 d^2 e g-4 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)^{1-m} \, dx}{12 b^2 d^2} \\ & = \frac {(a+b x)^{1+m} (c+d x)^{2-m} (4 b d (f g+e h)-a d f h (3-m)-b c f h (2+m)+3 b d f h x)}{12 b^2 d^2}+\frac {\left ((b c-a d) \left (a^2 d^2 f h \left (6-5 m+m^2\right )-2 a b d (2-m) (2 d (f g+e h)-c f h (1+m))+b^2 \left (12 d^2 e g-4 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 )^{1-m} \, dx}{12 b^3 d^2} \\ & = \frac {(a+b x)^{1+m} (c+d x)^{2-m} (4 b d (f g+e h)-a d f h (3-m)-b c f h (2+m)+3 b d f h x)}{12 b^2 d^2}+\frac {(b c-a d) \left (a^2 d^2 f h \left (6-5 m+m^2\right )-2 a b d (2-m) (2 d (f g+e h)-c f h (1+m))+b^2 \left (12 d^2 e g-4 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 (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{12 b^4 d^2 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.80 \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\frac {(a+b x)^{1+m} (c+d x)^{1-m} \left (\frac {b (c+d x)}{b c-a d}\right )^{-1+m} \left ((b c-a d)^2 f h \operatorname {Hypergeometric2F1}\left (-3+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )+b \left (-\left ((b c-a d) (2 c f h-d (f g+e h)) \operatorname {Hypergeometric2F1}\left (-2+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )+b (d e-c f) (d g-c h) \operatorname {Hypergeometric2F1}\left (-1+m,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{b^3 d^2 (1+m)} \]

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(1 - m)*((b*(c + d*x))/(b*c - a*d))^(-1 + m)*((b*c - a*d)^2*f*h*Hypergeometric2F1
[-3 + 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))*Hypergeometr
ic2F1[-2 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)]) + b*(d*e - c*f)*(d*g - c*h)*Hypergeometric2F1[-1 +
m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)])))/(b^3*d^2*(1 + m))

Maple [F]

\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{1-m} \left (f x +e \right ) \left (h x +g \right )d x\]

[In]

int((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)*(h*x+g),x)

[Out]

int((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)*(h*x+g),x)

Fricas [F]

\[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)*(h*x+g),x, algorithm="fricas")

[Out]

integral((f*h*x^2 + e*g + (f*g + e*h)*x)*(b*x + a)^m*(d*x + c)^(-m + 1), x)

Sympy [F(-2)]

Exception generated. \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

integrate((b*x+a)**m*(d*x+c)**(1-m)*(f*x+e)*(h*x+g),x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

\[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)*(h*x+g),x, algorithm="maxima")

[Out]

integrate((f*x + e)*(h*x + g)*(b*x + a)^m*(d*x + c)^(-m + 1), x)

Giac [F]

\[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\int { {\left (f x + e\right )} {\left (h x + g\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)*(h*x+g),x, algorithm="giac")

[Out]

integrate((f*x + e)*(h*x + g)*(b*x + a)^m*(d*x + c)^(-m + 1), x)

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^{1-m} (e+f x) (g+h x) \, dx=\int \left (e+f\,x\right )\,\left (g+h\,x\right )\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m} \,d x \]

[In]

int((e + f*x)*(g + h*x)*(a + b*x)^m*(c + d*x)^(1 - m),x)

[Out]

int((e + f*x)*(g + h*x)*(a + b*x)^m*(c + d*x)^(1 - m), x)