3.2643 \(\int \frac {(A+B x) (d+e x)^m}{a+b x+c x^2} \, dx\)

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

[Out]

-(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))))*(B+(2*A*c-B*b)/(-4*a*c+b
^2)^(1/2))/(1+m)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))-(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))))*(B+(-2*A*c+B*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 = 0.46, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {830, 68} \[ -\frac {(d+e x)^{m+1} \left (B-\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {(d+e x)^{m+1} \left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((B - (b*B - 2*A*c)/Sqrt[b^2 - 4*a*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))) - ((B + (b*B - 2*A*c)/Sqr
t[b^2 - 4*a*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 {(A+B x) (d+e x)^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)^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)^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)^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)^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)^{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 {\left (B+\frac {b B-2 A c}{\sqrt {b^2-4 a c}}\right ) (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 = 0.40, size = 197, normalized size = 0.93 \[ \frac {(d+e x)^{m+1} \left (-\frac {\left (\frac {2 A c-b B}{\sqrt {b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )}{e \left (\sqrt {b^2-4 a c}-b\right )+2 c d}-\frac {\left (\frac {b B-2 A c}{\sqrt {b^2-4 a c}}+B\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(-(((B + (-(b*B) + 2*A*c)/Sqrt[b^2 - 4*a*c])*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)) - ((B + (b*B - 2*A*c)/Sqrt[b
^2 - 4*a*c])*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 = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{c \,x^{2}+b x +a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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