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

Optimal. Leaf size=525 \[ \frac {(d+e x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (e g^2 (m+1) (b d-a e)+c \left (2 d^2 g^2 (p+1)-2 d e f g (m+2 p+3)+e^2 f^2 (m+2 p+3)\right )\right )}{c e^3 (m+1) (m+2 p+3)}-\frac {g (d+e x)^{m+2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} (b e g (m+p+2)+2 c (d g (p+1)-e f (m+2 p+3))) F_1\left (m+2;-p,-p;m+3;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (m+2) (m+2 p+3)}+\frac {g^2 (d+e x)^{m+1} \left (a+b x+c x^2\right )^{p+1}}{c e (m+2 p+3)} \]

[Out]

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

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Rubi [A]  time = 0.70, antiderivative size = 523, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1653, 843, 759, 133} \[ \frac {(d+e x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (g^2 (b d-a e)+\frac {c \left (2 d^2 g^2 (p+1)-2 d e f g (m+2 p+3)+e^2 f^2 (m+2 p+3)\right )}{e (m+1)}\right )}{c e^2 (m+2 p+3)}-\frac {g (d+e x)^{m+2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} (b e g (m+p+2)+2 c d g (p+1)-2 c e f (m+2 p+3)) F_1\left (m+2;-p,-p;m+3;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (m+2) (m+2 p+3)}+\frac {g^2 (d+e x)^{m+1} \left (a+b x+c x^2\right )^{p+1}}{c e (m+2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g^2*(d + e*x)^(1 + m)*(a + b*x + c*x^2)^(1 + p))/(c*e*(3 + m + 2*p)) + (((b*d - a*e)*g^2 + (c*(2*d^2*g^2*(1 +
p) + e^2*f^2*(3 + m + 2*p) - 2*d*e*f*g*(3 + m + 2*p)))/(e*(1 + m)))*(d + e*x)^(1 + m)*(a + b*x + c*x^2)^p*App
ellF1[1 + m, -p, -p, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b +
Sqrt[b^2 - 4*a*c])*e)])/(c*e^2*(3 + m + 2*p)*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))^p*(1 -
(2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^p) - (g*(2*c*d*g*(1 + p) + b*e*g*(2 + m + p) - 2*c*e*f*(
3 + m + 2*p))*(d + e*x)^(2 + m)*(a + b*x + c*x^2)^p*AppellF1[2 + m, -p, -p, 3 + m, (2*c*(d + e*x))/(2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(c*e^3*(2 + m)*(3 + m + 2*p)*(
1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c
])*e))^p)

Rule 133

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

Rule 759

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(a + b*x + c*x^2)^p/(e*(1 - (d + e*x)/(d - (e*(b - q))/(2*c)))^p*(1 - (d + e*x)/(d - (e*(b + q))/(2
*c)))^p), Subst[Int[x^m*Simp[1 - x/(d - (e*(b - q))/(2*c)), x]^p*Simp[1 - x/(d - (e*(b + q))/(2*c)), x]^p, x],
x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

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

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Mathematica [F]  time = 2.24, size = 0, normalized size = 0.00 \[ \int (d+e x)^m (f+g x)^2 \left (a+b x+c x^2\right )^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (g^{2} x^{2} + 2 \, f g x + f^{2}\right )} {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (g x +f \right )^{2} \left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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