3.10.29 \(\int \frac {(d+e x)^m (a+b x+c x^2)^2}{(f+g x)^3} \, dx\) [929]

Optimal. Leaf size=461 \[ -\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}+\frac {\left (c f^2-b f g+a g^2\right ) (c f (8 d g-e f (7+m))+g (a e g (1-m)-b (4 d g-e f (3+m)))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{2 g^4 (e f-d g)^3 (1+m)} \]

[Out]

-c*(-2*b*e*g+c*d*g+3*c*e*f)*(e*x+d)^(1+m)/e^2/g^4/(1+m)+c^2*(e*x+d)^(2+m)/e^2/g^3/(2+m)+1/2*(a*g^2-b*f*g+c*f^2
)^2*(e*x+d)^(1+m)/g^4/(-d*g+e*f)/(g*x+f)^2+1/2*(a*g^2-b*f*g+c*f^2)*(c*f*(8*d*g-e*f*(7+m))+g*(a*e*g*(1-m)-b*(4*
d*g-e*f*(3+m))))*(e*x+d)^(1+m)/g^4/(-d*g+e*f)^2/(g*x+f)+1/2*(c^2*f^2*(12*d^2*g^2-8*d*e*f*g*(3+m)+e^2*f^2*(m^2+
7*m+12))-g^2*(a^2*e^2*g^2*(1-m)*m-2*a*b*e*g*m*(2*d*g-e*f*(1+m))-b^2*(2*d^2*g^2-4*d*e*f*g*(1+m)+e^2*f^2*(m^2+3*
m+2)))+2*c*g*(a*g*(2*d^2*g^2-4*d*e*f*g*(1+m)+e^2*f^2*(m^2+3*m+2))-b*f*(6*d^2*g^2-6*d*e*f*g*(2+m)+e^2*f^2*(m^2+
5*m+6))))*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],-g*(e*x+d)/(-d*g+e*f))/g^4/(-d*g+e*f)^3/(1+m)

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Rubi [A]
time = 0.94, antiderivative size = 461, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {963, 1635, 965, 81, 70} \begin {gather*} \frac {(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {g (d+e x)}{e f-d g}\right ) \left (-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (m+1))-\left (b^2 \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (m+1)+e^2 f^2 \left (m^2+3 m+2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (m+2)+e^2 f^2 \left (m^2+5 m+6\right )\right )\right )+c^2 f^2 \left (12 d^2 g^2-8 d e f g (m+3)+e^2 f^2 \left (m^2+7 m+12\right )\right )\right )}{2 g^4 (m+1) (e f-d g)^3}-\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) (g (-a e g (1-m)+4 b d g-b e f (m+3))-c f (8 d g-e f (m+7)))}{2 g^4 (f+g x) (e f-d g)^2}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{2 g^4 (f+g x)^2 (e f-d g)}-\frac {c (d+e x)^{m+1} (-2 b e g+c d g+3 c e f)}{e^2 g^4 (m+1)}+\frac {c^2 (d+e x)^{m+2}}{e^2 g^3 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(3*c*e*f + c*d*g - 2*b*e*g)*(d + e*x)^(1 + m))/(e^2*g^4*(1 + m))) + (c^2*(d + e*x)^(2 + m))/(e^2*g^3*(2 +
m)) + ((c*f^2 - b*f*g + a*g^2)^2*(d + e*x)^(1 + m))/(2*g^4*(e*f - d*g)*(f + g*x)^2) - ((c*f^2 - b*f*g + a*g^2
)*(g*(4*b*d*g - a*e*g*(1 - m) - b*e*f*(3 + m)) - c*f*(8*d*g - e*f*(7 + m)))*(d + e*x)^(1 + m))/(2*g^4*(e*f - d
*g)^2*(f + g*x)) + ((c^2*f^2*(12*d^2*g^2 - 8*d*e*f*g*(3 + m) + e^2*f^2*(12 + 7*m + m^2)) - g^2*(a^2*e^2*g^2*(1
- m)*m - 2*a*b*e*g*m*(2*d*g - e*f*(1 + m)) - b^2*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2))) +
2*c*g*(a*g*(2*d^2*g^2 - 4*d*e*f*g*(1 + m) + e^2*f^2*(2 + 3*m + m^2)) - b*f*(6*d^2*g^2 - 6*d*e*f*g*(2 + m) + e
^2*f^2*(6 + 5*m + m^2))))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/
(2*g^4*(e*f - d*g)^3*(1 + m))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 81

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 963

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g))), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rule 965

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

Rule 1635

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(
b*c - a*d))), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rubi steps

\begin {align*} \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^3} \, dx &=\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}+\frac {\int \frac {(d+e x)^m \left (\frac {c^2 f^3 (2 d g-e f (1+m))-2 c f g (b f-a g) (2 d g-e f (1+m))+g^2 \left (a^2 e g^2 (1-m)+b^2 f (2 d g-e f (1+m))-2 a b g (2 d g-e f (1+m))\right )}{g^4}+\frac {2 (e f-d g) \left (c^2 f^2+b^2 g^2-2 c g (b f-a g)\right ) x}{g^3}-\frac {2 c (c f-2 b g) (e f-d g) x^2}{g^2}-2 c^2 \left (d-\frac {e f}{g}\right ) x^3\right )}{(f+g x)^2} \, dx}{2 (e f-d g)}\\ &=\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\int \frac {(d+e x)^m \left (\frac {c^2 f^2 \left (6 d^2 g^2-4 d e f g (3+2 m)+e^2 f^2 \left (6+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (4 d^2 g^2-2 d e f g (4+3 m)+e^2 f^2 \left (4+5 m+m^2\right )\right )\right )}{g^4}-\frac {4 c (c f-b g) (e f-d g)^2 x}{g^3}+\frac {2 c^2 (e f-d g)^2 x^2}{g^2}\right )}{f+g x} \, dx}{2 (e f-d g)^2}\\ &=\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\int \frac {(d+e x)^m \left (\frac {e (2+m) \left (c^2 f \left (10 d^2 e f g^2-2 d^3 g^3-2 d e^2 f^2 g (7+4 m)+e^3 f^3 \left (6+7 m+m^2\right )\right )-e g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c e g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (4 d^2 g^2-2 d e f g (4+3 m)+e^2 f^2 \left (4+5 m+m^2\right )\right )\right )\right )}{g^3}-\frac {2 c e (e f-d g)^2 (3 c e f+c d g-2 b e g) (2+m) x}{g^2}\right )}{f+g x} \, dx}{2 e^2 g (e f-d g)^2 (2+m)}\\ &=-\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) \int \frac {(d+e x)^m}{f+g x} \, dx}{2 g^4 (e f-d g)^2}\\ &=-\frac {c (3 c e f+c d g-2 b e g) (d+e x)^{1+m}}{e^2 g^4 (1+m)}+\frac {c^2 (d+e x)^{2+m}}{e^2 g^3 (2+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{2 g^4 (e f-d g) (f+g x)^2}-\frac {\left (c f^2-b f g+a g^2\right ) (g (4 b d g-a e g (1-m)-b e f (3+m))-c f (8 d g-e f (7+m))) (d+e x)^{1+m}}{2 g^4 (e f-d g)^2 (f+g x)}+\frac {\left (c^2 f^2 \left (12 d^2 g^2-8 d e f g (3+m)+e^2 f^2 \left (12+7 m+m^2\right )\right )-g^2 \left (a^2 e^2 g^2 (1-m) m-2 a b e g m (2 d g-e f (1+m))-b^2 \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )\right )+2 c g \left (a g \left (2 d^2 g^2-4 d e f g (1+m)+e^2 f^2 \left (2+3 m+m^2\right )\right )-b f \left (6 d^2 g^2-6 d e f g (2+m)+e^2 f^2 \left (6+5 m+m^2\right )\right )\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{2 g^4 (e f-d g)^3 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 2.12, size = 281, normalized size = 0.61 \begin {gather*} \frac {e \left (c f^2+g (-b f+a g)\right ) (d+e x)^{1+m} \left (-2 (2 c f-b g) (e f-d g) \, _2F_1\left (2,1+m;2+m;\frac {g (d+e x)}{-e f+d g}\right )+e \left (c f^2+g (-b f+a g)\right ) \, _2F_1\left (3,1+m;2+m;\frac {g (d+e x)}{-e f+d g}\right )\right )}{g^4 (e f-d g)^3 (1+m)}+\frac {(d+e x)^m \left (\frac {c g (d+e x) (2 b e g (2+m)+c (-d g-3 e f (2+m)+e g (1+m) x))}{e^2 (1+m) (2+m)}+\frac {\left (6 c^2 f^2+b^2 g^2+2 c g (-3 b f+a g)\right ) \left (\frac {g (d+e x)}{e (f+g x)}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac {e f-d g}{e f+e g x}\right )}{m}\right )}{g^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(c*f^2 + g*(-(b*f) + a*g))*(d + e*x)^(1 + m)*(-2*(2*c*f - b*g)*(e*f - d*g)*Hypergeometric2F1[2, 1 + m, 2 +
m, (g*(d + e*x))/(-(e*f) + d*g)] + e*(c*f^2 + g*(-(b*f) + a*g))*Hypergeometric2F1[3, 1 + m, 2 + m, (g*(d + e*x
))/(-(e*f) + d*g)]))/(g^4*(e*f - d*g)^3*(1 + m)) + ((d + e*x)^m*((c*g*(d + e*x)*(2*b*e*g*(2 + m) + c*(-(d*g) -
3*e*f*(2 + m) + e*g*(1 + m)*x)))/(e^2*(1 + m)*(2 + m)) + ((6*c^2*f^2 + b^2*g^2 + 2*c*g*(-3*b*f + a*g))*Hyperg
eometric2F1[-m, -m, 1 - m, (e*f - d*g)/(e*f + e*g*x)])/(m*((g*(d + e*x))/(e*(f + g*x)))^m)))/g^5

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{2}}{\left (g x +f \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f)^3,x)

[Out]

int((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f)^3,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f)^3,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^2*(x*e + d)^m/(g*x + f)^3, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f)^3,x, algorithm="fricas")

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*(x*e + d)^m/(g^3*x^3 + 3*f*g^2*x^2 + 3*f^2*
g*x + f^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )^{2}}{\left (f + g x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(c*x**2+b*x+a)**2/(g*x+f)**3,x)

[Out]

Integral((d + e*x)**m*(a + b*x + c*x**2)**2/(f + g*x)**3, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*x^2+b*x+a)^2/(g*x+f)^3,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^2*(x*e + d)^m/(g*x + f)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^2}{{\left (f+g\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x)^m*(a + b*x + c*x^2)^2)/(f + g*x)^3,x)

[Out]

int(((d + e*x)^m*(a + b*x + c*x^2)^2)/(f + g*x)^3, x)

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