3.140 \(\int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x)^2 \, dx\)

Optimal. Leaf size=393 \[ \frac {(b g-a h)^2 (a+b x)^{m+1} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;-n,-p;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+1)}+\frac {2 h (b g-a h) (a+b x)^{m+2} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+2;-n,-p;m+3;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+2)}+\frac {h^2 (a+b x)^{m+3} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+3;-n,-p;m+4;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+3)} \]

[Out]

(-a*h+b*g)^2*(b*x+a)^(1+m)*(d*x+c)^n*(f*x+e)^p*AppellF1(1+m,-n,-p,2+m,-d*(b*x+a)/(-a*d+b*c),-f*(b*x+a)/(-a*f+b
*e))/b^3/(1+m)/((b*(d*x+c)/(-a*d+b*c))^n)/((b*(f*x+e)/(-a*f+b*e))^p)+2*h*(-a*h+b*g)*(b*x+a)^(2+m)*(d*x+c)^n*(f
*x+e)^p*AppellF1(2+m,-n,-p,3+m,-d*(b*x+a)/(-a*d+b*c),-f*(b*x+a)/(-a*f+b*e))/b^3/(2+m)/((b*(d*x+c)/(-a*d+b*c))^
n)/((b*(f*x+e)/(-a*f+b*e))^p)+h^2*(b*x+a)^(3+m)*(d*x+c)^n*(f*x+e)^p*AppellF1(3+m,-n,-p,4+m,-d*(b*x+a)/(-a*d+b*
c),-f*(b*x+a)/(-a*f+b*e))/b^3/(3+m)/((b*(d*x+c)/(-a*d+b*c))^n)/((b*(f*x+e)/(-a*f+b*e))^p)

________________________________________________________________________________________

Rubi [A]  time = 0.49, antiderivative size = 393, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {181, 159, 140, 139, 138} \[ \frac {(b g-a h)^2 (a+b x)^{m+1} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;-n,-p;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+1)}+\frac {2 h (b g-a h) (a+b x)^{m+2} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+2;-n,-p;m+3;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+2)}+\frac {h^2 (a+b x)^{m+3} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+3;-n,-p;m+4;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^2,x]

[Out]

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

Rule 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 139

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rule 159

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Dist[h/b, Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(a + b*x)^m*(
c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n, p}, x] && (SumSimplerQ[m, 1] || ( !SumS
implerQ[n, 1] &&  !SumSimplerQ[p, 1]))

Rule 181

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Dist[h/b, Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*(g + h*x)^(q - 1), x], x] + Dist[(b*g - a*
h)/b, Int[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^(q - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n
, p}, x] && IGtQ[q, 0] && (SumSimplerQ[m, 1] || ( !SumSimplerQ[n, 1] &&  !SumSimplerQ[p, 1]))

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x)^2 \, dx &=\frac {h \int (a+b x)^{1+m} (c+d x)^n (e+f x)^p (g+h x) \, dx}{b}+\frac {(b g-a h) \int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x) \, dx}{b}\\ &=\frac {h^2 \int (a+b x)^{2+m} (c+d x)^n (e+f x)^p \, dx}{b^2}+2 \frac {(h (b g-a h)) \int (a+b x)^{1+m} (c+d x)^n (e+f x)^p \, dx}{b^2}+\frac {(b g-a h)^2 \int (a+b x)^m (c+d x)^n (e+f x)^p \, dx}{b^2}\\ &=\frac {\left (h^2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^{2+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b^2}+2 \frac {\left (h (b g-a h) (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^{1+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b^2}+\frac {\left ((b g-a h)^2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b^2}\\ &=\frac {\left (h^2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^{2+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{b^2}+2 \frac {\left (h (b g-a h) (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^{1+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{b^2}+\frac {\left ((b g-a h)^2 (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{b^2}\\ &=\frac {(b g-a h)^2 (a+b x)^{1+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (1+m;-n,-p;2+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (1+m)}+\frac {2 h (b g-a h) (a+b x)^{2+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (2+m;-n,-p;3+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (2+m)}+\frac {h^2 (a+b x)^{3+m} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (3+m;-n,-p;4+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (3+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 1.53, size = 0, normalized size = 0.00 \[ \int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x)^2 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^2,x]

[Out]

Integrate[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^2, x]

________________________________________________________________________________________

fricas [F]  time = 3.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (h^{2} x^{2} + 2 \, g h x + g^{2}\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \left (h x +g \right )^{2} \left (b x +a \right )^{m} \left (d x +c \right )^{n} \left (f x +e \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^n*(f*x+e)^p*(h*x+g)^2,x)

[Out]

int((b*x+a)^m*(d*x+c)^n*(f*x+e)^p*(h*x+g)^2,x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((h*x + g)^2*(b*x + a)^m*(d*x + c)^n*(f*x + e)^p, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^p*(g + h*x)^2*(a + b*x)^m*(c + d*x)^n,x)

[Out]

int((e + f*x)^p*(g + h*x)^2*(a + b*x)^m*(c + d*x)^n, x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)**n*(f*x+e)**p*(h*x+g)**2,x)

[Out]

Timed out

________________________________________________________________________________________