3.26.73 \(\int \frac {(a+b x+c x^2)^p}{(d+e x)^{3/2}} \, dx\) [2573]

Optimal. Leaf size=183 \[ -\frac {2 \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} F_1\left (-\frac {1}{2};-p,-p;\frac {1}{2};\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e \sqrt {d+e x}} \]

[Out]

-2*(c*x^2+b*x+a)^p*AppellF1(-1/2,-p,-p,1/2,2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*(e*x+d)/(2*c*d-e*(
b+(-4*a*c+b^2)^(1/2))))/e/((1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))))^p)/((1-2*c*(e*x+d)/(2*c*d-e*(b+(-4
*a*c+b^2)^(1/2))))^p)/(e*x+d)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {773, 138} \begin {gather*} -\frac {2 \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} F_1\left (-\frac {1}{2};-p,-p;\frac {1}{2};\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x + c*x^2)^p*AppellF1[-1/2, -p, -p, 1/2, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*
(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(e*Sqrt[d + e*x]*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2
- 4*a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^p)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(a + b*x + c*x^2)^p/(e*(1 - (d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*
c))))^p), Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d - e*((b + q)/(2*c))), x]^p, x],
x, d + e*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] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^p}{(d+e x)^{3/2}} \, dx &=\frac {\left (\left (a+b x+c x^2\right )^p \left (1-\frac {d+e x}{d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{-p} \left (1-\frac {d+e x}{d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c}}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1-\frac {2 c x}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^p \left (1-\frac {2 c x}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^p}{x^{3/2}} \, dx,x,d+e x\right )}{e}\\ &=-\frac {2 \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} F_1\left (-\frac {1}{2};-p,-p;\frac {1}{2};\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.77, size = 209, normalized size = 1.14 \begin {gather*} -\frac {2^{1-2 p} \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{8 c d+4 \left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (a+x (b+c x))^p F_1\left (-\frac {1}{2};-p,-p;\frac {1}{2};\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1 - 2*p)*(a + x*(b + c*x))^p*AppellF1[-1/2, -p, -p, 1/2, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c]
)*e), (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(e*((e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))/(8*c*d +
4*(-b + Sqrt[b^2 - 4*a*c])*e))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e))^p
*Sqrt[d + e*x]))

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Maple [F]
time = 0.42, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (e x +d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral(sqrt(x*e + d)*(c*x^2 + b*x + a)^p/(x^2*e^2 + 2*d*x*e + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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