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

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Rubi [A]  time = 0.38, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 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]

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]

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*}

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Mathematica [A]  time = 0.43, 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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fricas [F]  time = 1.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{m}}{c x^{2} + b x + a}, x\right ) \]

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

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)

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maple [F]  time = 1.40, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{c \,x^{2}+b x +a}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{c x^{2} + b x + a}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{c\,x^2+b\,x+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(a + b*x + c*x^2),x)

[Out]

int((d + e*x)^m/(a + b*x + c*x^2), x)

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

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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