3.26.81 \(\int (d+e x)^m (a+b x+c x^2)^{-2-\frac {m}{2}} \, dx\) [2581]

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

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 440, normalized size of antiderivative = 1.00, 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 = {758, 820, 740} \begin {gather*} -\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} \end {gather*}

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 740

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/((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))*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)))], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && Ne
Q[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]

Rule 758

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 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(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[S
implify[m + 2*p + 3], 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*}

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Mathematica [A]
time = 4.62, size = 379, normalized size = 0.86 \begin {gather*} \frac {(d+e x)^{1+m} (a+x (b+c x))^{-2-\frac {m}{2}} \left (2 e \left (c d^2+e (-b d+a e)\right ) (a+x (b+c x))-\frac {e (-2 c d+b e) m (d+e x) (a+x (b+c x))}{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 {2+m}{2}} (d+e x)^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 )}{2 \left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (3+m)}\right )}{2 \left (c d^2+e (-b d+a e)\right )^2 (1+m)} \end {gather*}

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*Hypergeometric2F
1[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 = 0.47, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{-2-\frac {m}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(c*x^2+b*x+a)^(-2-1/2*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 = 0.00, 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)^(-2-1/2*m),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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)^(-2-1/2*m),x, algorithm="giac")

[Out]

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

[Out]

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

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