\(\int \frac {(d+e x)^m}{a+b x+c x^2} \, dx\) [2553]

Optimal result

Integrand size = 20, antiderivative size = 191 \[ \int \frac {(d+e x)^m}{a+b x+c x^2} \, dx=-\frac {2 c (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}+\frac {2 c (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)} \]

[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)))

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {725, 70} \[ \int \frac {(d+e x)^m}{a+b x+c x^2} \, dx=\frac {2 c (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt {b^2-4 a c} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {2 c (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt {b^2-4 a c} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )} \]

[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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 725

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*} \text {integral}& = \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] (verified)

Time = 0.41 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^m}{a+b x+c x^2} \, dx=\frac {2 c (d+e x)^{1+m} \left (-\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\sqrt {b^2-4 a c} (1+m)} \]

[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))

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[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)