3.2553 $$\int \frac{(d+e x)^m}{a+b x+c x^2} \, dx$$

Optimal. Leaf size=191 $\frac{2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}$

[Out]

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

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Rubi [A]  time = 0.383565, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {711, 68} $\frac{2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt{b^2-4 a c} \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m/(a + b*x + c*x^2),x]

[Out]

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

Rule 711

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^
m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, 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[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{(d+e x)^m}{a+b x+c x^2} \, dx &=\int \left (\frac{2 c (d+e x)^m}{\sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}+2 c x\right )}-\frac{2 c (d+e x)^m}{\sqrt{b^2-4 a c} \left (b+\sqrt{b^2-4 a c}+2 c x\right )}\right ) \, dx\\ &=\frac{(2 c) \int \frac{(d+e x)^m}{b-\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}-\frac{(2 c) \int \frac{(d+e x)^m}{b+\sqrt{b^2-4 a c}+2 c x} \, dx}{\sqrt{b^2-4 a c}}\\ &=-\frac{2 c (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{\sqrt{b^2-4 a c} \left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) (1+m)}+\frac{2 c (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{\sqrt{b^2-4 a c} \left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.452013, size = 163, normalized size = 0.85 $\frac{2 c (d+e x)^{m+1} \left (\frac{\, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}-\frac{\, _2F_1\left (1,m+1;m+2;\frac{2 c (d+e x)}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}\right )}{(m+1) \sqrt{b^2-4 a c}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 1.28, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{c{x}^{2}+bx+a}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x+d)^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 (e x + d\right )}^{m}}{c x^{2} + b x + a}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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