3.26.77 \(\int (d+e x)^{-3-2 p} (a+b x+c x^2)^p \, dx\) [2577]

Optimal. Leaf size=332 \[ -\frac {e (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right ) (1+p)}+\frac {(2 c d-b e) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, _2F_1\left (-1-2 p,-p;-2 p;-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+2 p)} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {744, 740} \begin {gather*} \frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (2 c d-b e) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \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 )^{-p} \, _2F_1\left (-2 p-1,-p;-2 p;-\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 )}{2 (2 p+1) \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 )}-\frac {e (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) \left (a e^2-b d e+c d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(-3 - 2*p)*(a + b*x + c*x^2)^p,x]

[Out]

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

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 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[(2*c*d - b*e)/(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, 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] && EqQ[m + 2*p + 3, 0]

Rubi steps

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

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Mathematica [A]
time = 0.39, size = 294, normalized size = 0.89 \begin {gather*} \frac {(d+e x)^{-2 (1+p)} (a+x (b+c x))^p \left (-\frac {e (a+x (b+c x))}{1+p}+\frac {(-2 c d+b e) \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 )^{-1-p} (d+e x) \, _2F_1\left (-1-2 p,-p;-2 p;-\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 )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+2 p)}\right )}{2 \left (c d^2+e (-b d+a e)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(-3 - 2*p)*(a + b*x + c*x^2)^p,x]

[Out]

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

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Maple [F]
time = 0.46, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{-3-2 p} \left (c \,x^{2}+b x +a \right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(-3-2*p)*(c*x^2+b*x+a)^p,x)

[Out]

int((e*x+d)^(-3-2*p)*(c*x^2+b*x+a)^p,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)^(-3-2*p)*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p*(x*e + d)^(-2*p - 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)^(-3-2*p)*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p*(x*e + d)^(-2*p - 3), 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)**(-3-2*p)*(c*x**2+b*x+a)**p,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)^(-3-2*p)*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^p*(x*e + d)^(-2*p - 3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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