3.10.28 \(\int \frac {(d+e x)^m (a+b x+c x^2)^2}{(f+g x)^2} \, dx\) [928]

Optimal. Leaf size=298 \[ \frac {\left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^{1+m}}{e^3 g^4 (1+m)}-\frac {2 c (c e f+c d g-b e g) (d+e x)^{2+m}}{e^3 g^3 (2+m)}+\frac {c^2 (d+e x)^{3+m}}{e^3 g^2 (3+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}+\frac {\left (c f^2-b f g+a g^2\right ) (c f (4 d g-e f (4+m))-g (a e g m+b (2 d g-e f (2+m)))) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{g^4 (e f-d g)^2 (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.73, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {963, 1634, 70} \begin {gather*} \frac {(d+e x)^{m+1} \left (2 c e g (a e g-b (d g+2 e f))+b^2 e^2 g^2+c^2 \left (d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{e^3 g^4 (m+1)}-\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) \, _2F_1\left (1,m+1;m+2;-\frac {g (d+e x)}{e f-d g}\right ) (g (a e g m+2 b d g-b e f (m+2))-c f (4 d g-e f (m+4)))}{g^4 (m+1) (e f-d g)^2}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2}{g^4 (f+g x) (e f-d g)}-\frac {2 c (d+e x)^{m+2} (-b e g+c d g+c e f)}{e^3 g^3 (m+2)}+\frac {c^2 (d+e x)^{m+3}}{e^3 g^2 (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*e^2*g^2 + c^2*(3*e^2*f^2 + 2*d*e*f*g + d^2*g^2) + 2*c*e*g*(a*e*g - b*(2*e*f + d*g)))*(d + e*x)^(1 + m))/
(e^3*g^4*(1 + m)) - (2*c*(c*e*f + c*d*g - b*e*g)*(d + e*x)^(2 + m))/(e^3*g^3*(2 + m)) + (c^2*(d + e*x)^(3 + m)
)/(e^3*g^2*(3 + m)) + ((c*f^2 - b*f*g + a*g^2)^2*(d + e*x)^(1 + m))/(g^4*(e*f - d*g)*(f + g*x)) - ((c*f^2 - b*
f*g + a*g^2)*(g*(2*b*d*g + a*e*g*m - b*e*f*(2 + m)) - c*f*(4*d*g - e*f*(4 + m)))*(d + e*x)^(1 + m)*Hypergeomet
ric2F1[1, 1 + m, 2 + m, -((g*(d + e*x))/(e*f - d*g))])/(g^4*(e*f - d*g)^2*(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 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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[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)^2} \, dx &=\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}+\frac {\int \frac {(d+e x)^m \left (\frac {c^2 f^3 (d g-e f (1+m))-2 c f g (b f-a g) (d g-e f (1+m))-g^2 \left (a^2 e g^2 m-b^2 f (d g-e f (1+m))+2 a b g (d g-e f (1+m))\right )}{g^4}+\frac {(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 {c (c f-2 b g) (e f-d g) x^2}{g^2}-c^2 \left (d-\frac {e f}{g}\right ) x^3\right )}{f+g x} \, dx}{e f-d g}\\ &=\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}+\frac {\int \left (\frac {(e f-d g) \left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^m}{e^2 g^4}-\frac {2 c (e f-d g) (c e f+c d g-b e g) (d+e x)^{1+m}}{e^2 g^3}+\frac {c^2 (e f-d g) (d+e x)^{2+m}}{e^2 g^2}+\frac {\left (c f^2-b f g+a g^2\right ) (-g (2 b d g+a e g m-b e f (2+m))+c f (4 d g-e f (4+m))) (d+e x)^m}{g^4 (f+g x)}\right ) \, dx}{e f-d g}\\ &=\frac {\left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^{1+m}}{e^3 g^4 (1+m)}-\frac {2 c (c e f+c d g-b e g) (d+e x)^{2+m}}{e^3 g^3 (2+m)}+\frac {c^2 (d+e x)^{3+m}}{e^3 g^2 (3+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}-\frac {\left (\left (c f^2-b f g+a g^2\right ) (g (2 b d g+a e g m-b e f (2+m))-c f (4 d g-e f (4+m)))\right ) \int \frac {(d+e x)^m}{f+g x} \, dx}{g^4 (e f-d g)}\\ &=\frac {\left (b^2 e^2 g^2+c^2 \left (3 e^2 f^2+2 d e f g+d^2 g^2\right )+2 c e g (a e g-b (2 e f+d g))\right ) (d+e x)^{1+m}}{e^3 g^4 (1+m)}-\frac {2 c (c e f+c d g-b e g) (d+e x)^{2+m}}{e^3 g^3 (2+m)}+\frac {c^2 (d+e x)^{3+m}}{e^3 g^2 (3+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m}}{g^4 (e f-d g) (f+g x)}-\frac {\left (c f^2-b f g+a g^2\right ) (g (2 b d g+a e g m-b e f (2+m))-c f (4 d g-e f (4+m))) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{g^4 (e f-d g)^2 (1+m)}\\ \end {align*}

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Mathematica [F]
time = 0.99, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^m \left (a+b x+c x^2\right )^2}{(f+g x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{2}}{\left (g x +f \right )^{2}}\, dx\]

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)^2,x)

[Out]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \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)**2,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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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)^2,x, algorithm="giac")

[Out]

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

[Out]

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

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