3.2645 \(\int \frac{(A+B x) (d+e x)^{1+m}}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=212 \[ -\frac{(d+e x)^{m+2} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+2;m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+2) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{(d+e x)^{m+2} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+2;m+3;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+2) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.309465, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {830, 68} \[ -\frac{(d+e x)^{m+2} \left (B-\frac{b B-2 A c}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,m+2;m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+2) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}-\frac{(d+e x)^{m+2} \left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+2;m+3;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+2) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(1 + m))/(a + b*x + c*x^2),x]

[Out]

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

Rule 830

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[m]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.377824, size = 197, normalized size = 0.93 \[ \frac{(d+e x)^{m+2} \left (-\frac{\left (\frac{2 A c-b B}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+2;m+3;\frac{2 c (d+e x)}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}-\frac{\left (\frac{b B-2 A c}{\sqrt{b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+2;m+3;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(1 + m))/(a + b*x + c*x^2),x]

[Out]

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

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Maple [F]  time = 0.112, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( Bx+A \right ) \left ( ex+d \right ) ^{1+m}}{c{x}^{2}+bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + b*x + a), 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 (e x + d\right )}^{m + 1}}{c x^{2} + b x + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x + A)*(e*x + d)^(m + 1)/(c*x^2 + b*x + a), 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)*(e*x+d)**(1+m)/(c*x**2+b*x+a),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 (e x + d\right )}^{m + 1}}{c x^{2} + b x + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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