3.1101 \(\int \frac{(A+B x) (a+b x+c x^2)^p}{x^3} \, dx\)

Optimal. Leaf size=376 \[ \frac{4^{p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \left (A \left (2 a c-b^2 (1-p)\right )+2 a b B\right ) F_1\left (-2 p;-p,-p;1-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x},-\frac{b+\sqrt{b^2-4 a c}}{2 c x}\right )}{a^2}-\frac{c 2^p (2 p+1) (2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{a^2 (p+1) \sqrt{b^2-4 a c}}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{2 a^2 x}-\frac{A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2} \]

[Out]

-(A*(a + b*x + c*x^2)^(1 + p))/(2*a*x^2) - ((2*a*B - A*b*(1 - p))*(a + b*x + c*x^2)^(1 + p))/(2*a^2*x) + (4^(-
1 + p)*(2*a*b*B + A*(2*a*c - b^2*(1 - p)))*(a + b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -(b - Sqrt[b^2
- 4*a*c])/(2*c*x), -(b + Sqrt[b^2 - 4*a*c])/(2*c*x)])/(a^2*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x))^p*((b + Sqr
t[b^2 - 4*a*c] + 2*c*x)/(c*x))^p) - (2^p*c*(2*a*B - A*b*(1 - p))*(1 + 2*p)*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/
Sqrt[b^2 - 4*a*c]))^(-1 - p)*(a + b*x + c*x^2)^(1 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a
*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(a^2*Sqrt[b^2 - 4*a*c]*(1 + p))

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Rubi [A]  time = 0.355723, antiderivative size = 375, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {834, 843, 624, 758, 133} \[ \frac{4^{p-1} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \left (2 a A c+2 a b B-A b^2 (1-p)\right ) F_1\left (-2 p;-p,-p;1-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x},-\frac{b+\sqrt{b^2-4 a c}}{2 c x}\right )}{a^2}-\frac{c 2^p (2 p+1) (2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1} \left (-\frac{-\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}\right )^{-p-1} \, _2F_1\left (-p,p+1;p+2;\frac{b+2 c x+\sqrt{b^2-4 a c}}{2 \sqrt{b^2-4 a c}}\right )}{a^2 (p+1) \sqrt{b^2-4 a c}}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{p+1}}{2 a^2 x}-\frac{A \left (a+b x+c x^2\right )^{p+1}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*(a + b*x + c*x^2)^(1 + p))/(2*a*x^2) - ((2*a*B - A*b*(1 - p))*(a + b*x + c*x^2)^(1 + p))/(2*a^2*x) + (4^(-
1 + p)*(2*a*b*B + 2*a*A*c - A*b^2*(1 - p))*(a + b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 2*p, -(b - Sqrt[b^2
- 4*a*c])/(2*c*x), -(b + Sqrt[b^2 - 4*a*c])/(2*c*x)])/(a^2*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x))^p*((b + Sqr
t[b^2 - 4*a*c] + 2*c*x)/(c*x))^p) - (2^p*c*(2*a*B - A*b*(1 - p))*(1 + 2*p)*(-((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/
Sqrt[b^2 - 4*a*c]))^(-1 - p)*(a + b*x + c*x^2)^(1 + p)*Hypergeometric2F1[-p, 1 + p, 2 + p, (b + Sqrt[b^2 - 4*a
*c] + 2*c*x)/(2*Sqrt[b^2 - 4*a*c])])/(a^2*Sqrt[b^2 - 4*a*c]*(1 + p))

Rule 834

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))/((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)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 843

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

Rule 624

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, -Simp[((a + b*x + c*
x^2)^(p + 1)*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q)])/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p
+ 1)), x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, -Dist[((1/(d + e*x))^(2*p)*(a + b*x + c*x^2)^p)/(e*((e*(b - q + 2*c*x))/(2*c*(d + e*x)))^p*((e*(b + q +
2*c*x))/(2*c*(d + e*x)))^p), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - (e*(b - q))/(2*c))*x, x]^p*Simp[1 - (d
- (e*(b + q))/(2*c))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, 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] && ILtQ[m, 0]

Rule 133

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

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^p}{x^3} \, dx &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{\int \frac{(-2 a B+A (b-b p)-2 A c p x) \left (a+b x+c x^2\right )^p}{x^2} \, dx}{2 a}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac{\int \frac{\left (\left (2 a b B+2 a A c-A b^2 (1-p)\right ) p+c (1+2 p) (2 a B-A (b-b p)) x\right ) \left (a+b x+c x^2\right )^p}{x} \, dx}{2 a^2}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac{\left (\left (2 a b B+2 a A c-A b^2 (1-p)\right ) p\right ) \int \frac{\left (a+b x+c x^2\right )^p}{x} \, dx}{2 a^2}+\frac{(c (2 a B-A b (1-p)) (1+2 p)) \int \left (a+b x+c x^2\right )^p \, dx}{2 a^2}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}-\frac{2^p c (2 a B-A b (1-p)) (1+2 p) \left (-\frac{b-\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac{b+\sqrt{b^2-4 a c}+2 c x}{2 \sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} (1+p)}-\frac{\left (2^{-1+2 p} \left (2 a b B+2 a A c-A b^2 (1-p)\right ) p \left (\frac{1}{x}\right )^{2 p} \left (\frac{b-\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac{b+\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p\right ) \operatorname{Subst}\left (\int x^{1-2 (1+p)} \left (1+\frac{\left (b-\sqrt{b^2-4 a c}\right ) x}{2 c}\right )^p \left (1+\frac{\left (b+\sqrt{b^2-4 a c}\right ) x}{2 c}\right )^p \, dx,x,\frac{1}{x}\right )}{a^2}\\ &=-\frac{A \left (a+b x+c x^2\right )^{1+p}}{2 a x^2}-\frac{(2 a B-A b (1-p)) \left (a+b x+c x^2\right )^{1+p}}{2 a^2 x}+\frac{4^{-1+p} \left (2 a b B+2 a A c-A b^2 (1-p)\right ) \left (\frac{b-\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac{b+\sqrt{b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (-2 p;-p,-p;1-2 p;-\frac{b-\sqrt{b^2-4 a c}}{2 c x},-\frac{b+\sqrt{b^2-4 a c}}{2 c x}\right )}{a^2}-\frac{2^p c (2 a B-A b (1-p)) (1+2 p) \left (-\frac{b-\sqrt{b^2-4 a c}+2 c x}{\sqrt{b^2-4 a c}}\right )^{-1-p} \left (a+b x+c x^2\right )^{1+p} \, _2F_1\left (-p,1+p;2+p;\frac{b+\sqrt{b^2-4 a c}+2 c x}{2 \sqrt{b^2-4 a c}}\right )}{a^2 \sqrt{b^2-4 a c} (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.316734, size = 295, normalized size = 0.78 \[ \frac{\left (\frac{b-\sqrt{b^2-4 a c}}{2 c x}+1\right )^{-p} \left (\frac{b-\sqrt{b^2-4 a c}}{2 c}+x\right )^{-p} \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{c}\right )^p \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{c x}\right )^{-p} (a+x (b+c x))^p \left (A (2 p-1) F_1\left (2-2 p;-p,-p;3-2 p;-\frac{b+\sqrt{b^2-4 a c}}{2 c x},\frac{\sqrt{b^2-4 a c}-b}{2 c x}\right )+2 B (p-1) x F_1\left (1-2 p;-p,-p;2-2 p;-\frac{b+\sqrt{b^2-4 a c}}{2 c x},\frac{\sqrt{b^2-4 a c}-b}{2 c x}\right )\right )}{2 (p-1) (2 p-1) x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/c)^p*(a + x*(b + c*x))^p*(2*B*(-1 + p)*x*AppellF1[1 - 2*p, -p, -p, 2 - 2*p,
-(b + Sqrt[b^2 - 4*a*c])/(2*c*x), (-b + Sqrt[b^2 - 4*a*c])/(2*c*x)] + A*(-1 + 2*p)*AppellF1[2 - 2*p, -p, -p, 3
- 2*p, -(b + Sqrt[b^2 - 4*a*c])/(2*c*x), (-b + Sqrt[b^2 - 4*a*c])/(2*c*x)]))/(2*(-1 + p)*(-1 + 2*p)*(1 + (b -
Sqrt[b^2 - 4*a*c])/(2*c*x))^p*x^2*((b - Sqrt[b^2 - 4*a*c])/(2*c) + x)^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(c*x
))^p)

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Maple [F]  time = 0.086, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( c{x}^{2}+bx+a \right ) ^{p}}{{x}^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^p/x^3,x)

[Out]

int((B*x+A)*(c*x^2+b*x+a)^p/x^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (c x^{2} + b x + a\right )}^{p}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^p/x^3,x, algorithm="maxima")

[Out]

integrate((B*x + A)*(c*x^2 + b*x + a)^p/x^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x + A\right )}{\left (c x^{2} + b x + a\right )}^{p}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x + A)*(c*x^2 + b*x + a)^p/x^3, 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((B*x+A)*(c*x**2+b*x+a)**p/x**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x + A\right )}{\left (c x^{2} + b x + a\right )}^{p}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^p/x^3,x, algorithm="giac")

[Out]

integrate((B*x + A)*(c*x^2 + b*x + a)^p/x^3, x)