### 3.2581 $$\int (d+e x)^m (a+b x+c x^2)^{-2-\frac{m}{2}} \, dx$$

Optimal. Leaf size=440 $-\frac{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) (d+e x)^{m+3} \left (a+b x+c x^2\right )^{-\frac{m}{2}-2} \left (4 c e (a e-b d (m+1))+b^2 e^2 m+4 c^2 d^2 (m+1)\right ) \left (\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}\right )^{\frac{m+4}{2}} \, _2F_1\left (m+3,\frac{m+4}{2};m+4;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt{b^2-4 a c}\right )}\right )}{4 (m+1) (m+3) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac{e (d+e x)^{m+1} \left (a+b x+c x^2\right )^{-\frac{m}{2}-1}}{(m+1) \left (a e^2-b d e+c d^2\right )}+\frac{e m (2 c d-b e) (d+e x)^{m+2} \left (a+b x+c x^2\right )^{-\frac{m}{2}-1}}{2 (m+1) (m+2) \left (a e^2-b d e+c d^2\right )^2}$

[Out]

(e*(d + e*x)^(1 + m)*(a + b*x + c*x^2)^(-1 - m/2))/((c*d^2 - b*d*e + a*e^2)*(1 + m)) + (e*(2*c*d - b*e)*m*(d +
e*x)^(2 + m)*(a + b*x + c*x^2)^(-1 - m/2))/(2*(c*d^2 - b*d*e + a*e^2)^2*(1 + m)*(2 + m)) - ((b^2*e^2*m + 4*c^
2*d^2*(1 + m) + 4*c*e*(a*e - b*d*(1 + m)))*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d - (b - Sqrt[b^2 - 4*a*c])*
e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)))^((4
+ m)/2)*(d + e*x)^(3 + m)*(a + b*x + c*x^2)^(-2 - m/2)*Hypergeometric2F1[3 + m, (4 + m)/2, 4 + m, (-4*c*Sqrt[
b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))])/(4*(2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e)*(c*d^2 - b*d*e + a*e^2)^2*(1 + m)*(3 + m))

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Rubi [A]  time = 0.385077, antiderivative size = 440, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {744, 806, 726} $-\frac{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) (d+e x)^{m+3} \left (a+b x+c x^2\right )^{-\frac{m}{2}-2} \left (4 c e (a e-b d (m+1))+b^2 e^2 m+4 c^2 d^2 (m+1)\right ) \left (\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}{\left (-\sqrt{b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}\right )^{\frac{m+4}{2}} \, _2F_1\left (m+3,\frac{m+4}{2};m+4;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt{b^2-4 a c}\right )}\right )}{4 (m+1) (m+3) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )^2}+\frac{e (d+e x)^{m+1} \left (a+b x+c x^2\right )^{-\frac{m}{2}-1}}{(m+1) \left (a e^2-b d e+c d^2\right )}+\frac{e m (2 c d-b e) (d+e x)^{m+2} \left (a+b x+c x^2\right )^{-\frac{m}{2}-1}}{2 (m+1) (m+2) \left (a e^2-b d e+c d^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(e*(d + e*x)^(1 + m)*(a + b*x + c*x^2)^(-1 - m/2))/((c*d^2 - b*d*e + a*e^2)*(1 + m)) + (e*(2*c*d - b*e)*m*(d +
e*x)^(2 + m)*(a + b*x + c*x^2)^(-1 - m/2))/(2*(c*d^2 - b*d*e + a*e^2)^2*(1 + m)*(2 + m)) - ((b^2*e^2*m + 4*c^
2*d^2*(1 + m) + 4*c*e*(a*e - b*d*(1 + m)))*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d - (b - Sqrt[b^2 - 4*a*c])*
e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)))^((4
+ m)/2)*(d + e*x)^(3 + m)*(a + b*x + c*x^2)^(-2 - m/2)*Hypergeometric2F1[3 + m, (4 + m)/2, 4 + m, (-4*c*Sqrt[
b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))])/(4*(2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e)*(c*d^2 - b*d*e + a*e^2)^2*(1 + m)*(3 + m))

Rule 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(a + b*x + c*x^2)^p, 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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 726

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b - Rt[b^2 - 4*a*
c, 2] + 2*c*x)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Hypergeometric2F1[m + 1, -p, m + 2, (-4*c*Rt[b^2 - 4*a*c,
2]*(d + e*x))/((2*c*d - b*e - e*Rt[b^2 - 4*a*c, 2])*(b - Rt[b^2 - 4*a*c, 2] + 2*c*x))])/((m + 1)*(2*c*d - b*e
+ e*Rt[b^2 - 4*a*c, 2])*(((2*c*d - b*e + e*Rt[b^2 - 4*a*c, 2])*(b + Rt[b^2 - 4*a*c, 2] + 2*c*x))/((2*c*d - b*
e - e*Rt[b^2 - 4*a*c, 2])*(b - Rt[b^2 - 4*a*c, 2] + 2*c*x)))^p), 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] && EqQ[m + 2*p + 2, 0
]

Rubi steps

\begin{align*} \int (d+e x)^m \left (a+b x+c x^2\right )^{-2-\frac{m}{2}} \, dx &=\frac{e (d+e x)^{1+m} \left (a+b x+c x^2\right )^{-1-\frac{m}{2}}}{\left (c d^2-b d e+a e^2\right ) (1+m)}+\frac{\int (d+e x)^{1+m} \left (\frac{1}{2} (-b e m+2 c d (1+m))+c e x\right ) \left (a+b x+c x^2\right )^{-2-\frac{m}{2}} \, dx}{\left (c d^2-b d e+a e^2\right ) (1+m)}\\ &=\frac{e (d+e x)^{1+m} \left (a+b x+c x^2\right )^{-1-\frac{m}{2}}}{\left (c d^2-b d e+a e^2\right ) (1+m)}+\frac{e (2 c d-b e) m (d+e x)^{2+m} \left (a+b x+c x^2\right )^{-1-\frac{m}{2}}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+m) (2+m)}+\frac{\left (b^2 e^2 m+4 c^2 d^2 (1+m)+4 c e (a e-b d (1+m))\right ) \int (d+e x)^{2+m} \left (a+b x+c x^2\right )^{-2-\frac{m}{2}} \, dx}{4 \left (c d^2-b d e+a e^2\right )^2 (1+m)}\\ &=\frac{e (d+e x)^{1+m} \left (a+b x+c x^2\right )^{-1-\frac{m}{2}}}{\left (c d^2-b d e+a e^2\right ) (1+m)}+\frac{e (2 c d-b e) m (d+e x)^{2+m} \left (a+b x+c x^2\right )^{-1-\frac{m}{2}}}{2 \left (c d^2-b d e+a e^2\right )^2 (1+m) (2+m)}-\frac{\left (b^2 e^2 m+4 c^2 d^2 (1+m)+4 c e (a e-b d (1+m))\right ) \left (b-\sqrt{b^2-4 a c}+2 c x\right ) \left (\frac{\left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (b-\sqrt{b^2-4 a c}+2 c x\right )}\right )^{\frac{4+m}{2}} (d+e x)^{3+m} \left (a+b x+c x^2\right )^{-2-\frac{m}{2}} \, _2F_1\left (3+m,\frac{4+m}{2};4+m;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) \left (b-\sqrt{b^2-4 a c}+2 c x\right )}\right )}{4 \left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right )^2 (1+m) (3+m)}\\ \end{align*}

Mathematica [A]  time = 3.78267, size = 379, normalized size = 0.86 $\frac{(d+e x)^{m+1} (a+x (b+c x))^{-\frac{m}{2}-2} \left (\frac{(d+e x)^2 \left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (-4 c e (b d (m+1)-a e)+b^2 e^2 m+4 c^2 d^2 (m+1)\right ) \left (\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )}{\left (\sqrt{b^2-4 a c}-b-2 c x\right ) \left (e \left (\sqrt{b^2-4 a c}+b\right )-2 c d\right )}\right )^{\frac{m+2}{2}} \, _2F_1\left (m+3,\frac{m+4}{2};m+4;-\frac{4 c \sqrt{b^2-4 a c} (d+e x)}{\left (\left (b+\sqrt{b^2-4 a c}\right ) e-2 c d\right ) \left (-b-2 c x+\sqrt{b^2-4 a c}\right )}\right )}{2 (m+3) \left (e \left (\sqrt{b^2-4 a c}+b\right )-2 c d\right )}+2 e (a+x (b+c x)) \left (e (a e-b d)+c d^2\right )+\frac{e m (d+e x) (a+x (b+c x)) (2 c d-b e)}{m+2}\right )}{2 (m+1) \left (e (a e-b d)+c d^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d + e*x)^(1 + m)*(a + x*(b + c*x))^(-2 - m/2)*(2*e*(c*d^2 + e*(-(b*d) + a*e))*(a + x*(b + c*x)) + (e*(2*c*d
- b*e)*m*(d + e*x)*(a + x*(b + c*x)))/(2 + m) + ((b^2*e^2*m + 4*c^2*d^2*(1 + m) - 4*c*e*(-(a*e) + b*d*(1 + m))
)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((-2
*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)))^((2 + m)/2)*(d + e*x)^2*Hypergeometric2F1
[3 + m, (4 + m)/2, 4 + m, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[
b^2 - 4*a*c] - 2*c*x))])/(2*(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(3 + m))))/(2*(c*d^2 + e*(-(b*d) + a*e))^2*(1
+ m))

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Maple [F]  time = 1.258, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) ^{-2-{\frac{m}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(c*x^2+b*x+a)^(-2-1/2*m),x)

[Out]

int((e*x+d)^m*(c*x^2+b*x+a)^(-2-1/2*m),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{-\frac{1}{2} \, m - 2}{\left (e x + d\right )}^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*x^2+b*x+a)^(-2-1/2*m),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(-1/2*m - 2)*(e*x + d)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{-\frac{1}{2} \, m - 2}{\left (e x + d\right )}^{m}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*x^2+b*x+a)^(-2-1/2*m),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(c*x**2+b*x+a)**(-2-1/2*m),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*x^2+b*x+a)^(-2-1/2*m),x, algorithm="giac")

[Out]

Exception raised: TypeError