3.1657 \(\int \frac {(b+2 c x) (d+e x)^m}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=167 \[ -\frac {2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )} \]

[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)))-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)))

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Rubi [A]  time = 0.30, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {830, 68} \[ -\frac {2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {2 c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(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)*(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)])/((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 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]

Rubi steps

\begin {align*} \int \frac {(b+2 c x) (d+e x)^m}{a+b x+c x^2} \, dx &=\int \left (\frac {2 c (d+e x)^m}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {2 c (d+e x)^m}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx\\ &=(2 c) \int \frac {(d+e x)^m}{b-\sqrt {b^2-4 a c}+2 c x} \, dx+(2 c) \int \frac {(d+e x)^m}{b+\sqrt {b^2-4 a c}+2 c x} \, dx\\ &=-\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 )}{\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 )}{\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 = 0.24, size = 152, normalized size = 0.91 \[ \frac {2 c (d+e x)^{m+1} \left (-\frac {\, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c d}-\frac {\, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(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)))/(1 + m)

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{m}}{c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((2*c*x + b)*(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 \[ \int \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{m}}{c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.42, size = 0, normalized size = 0.00 \[ \int \frac {\left (2 c x +b \right ) \left (e x +d \right )^{m}}{c \,x^{2}+b x +a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((2*c*x+b)*(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 \[ \int \frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{m}}{c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*c*x + b)*(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 \[ \int \frac {\left (b+2\,c\,x\right )\,{\left (d+e\,x\right )}^m}{c\,x^2+b\,x+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((b + 2*c*x)*(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 \[ \int \frac {\left (b + 2 c x\right ) \left (d + e x\right )^{m}}{a + b x + c x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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