3.1658 \(\int \frac {(b+2 c x) (d+e x)^m}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=358 \[ -\frac {c e m (d+e x)^{m+1} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt {b^2-4 a c} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac {c e m (d+e x)^{m+1} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt {b^2-4 a c} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac {(d+e x)^{m+1} \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.57, antiderivative size = 358, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {822, 830, 68} \[ -\frac {c e m (d+e x)^{m+1} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt {b^2-4 a c} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac {c e m (d+e x)^{m+1} \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \sqrt {b^2-4 a c} \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac {(d+e x)^{m+1} \left (\left (b^2-4 a c\right ) (c d-b e)-c e x \left (b^2-4 a c\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((d + e*x)^(1 + m)*((b^2 - 4*a*c)*(c*d - b*e) - c*(b^2 - 4*a*c)*e*x))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)
*(a + b*x + c*x^2))) - (c*e*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*m*(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)*(c*d^2 - b*d*e + a*e^2)*(1 + m)) + (c*e*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*m*(d + e*x)^(1 + m)*Hyperg
eometric2F1[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)*(c*d^2 - b*d*e + a*e^2)*(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 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

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}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {(d+e x)^{1+m} \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {\int \frac {(d+e x)^m \left (-\left (b^2-4 a c\right ) e (c d-b e) m+c \left (b^2-4 a c\right ) e^2 m x\right )}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {(d+e x)^{1+m} \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {\left (c \left (b^2-4 a c\right ) e^2 m+c \sqrt {b^2-4 a c} e (-2 c d+b e) m\right ) (d+e x)^m}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (c \left (b^2-4 a c\right ) e^2 m-c \sqrt {b^2-4 a c} e (-2 c d+b e) m\right ) (d+e x)^m}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {(d+e x)^{1+m} \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {\left (c e \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) m\right ) \int \frac {(d+e x)^m}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {\left (c e \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) m\right ) \int \frac {(d+e x)^m}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}\\ &=-\frac {(d+e x)^{1+m} \left (\left (b^2-4 a c\right ) (c d-b e)-c \left (b^2-4 a c\right ) e x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {c e \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) m (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 ) \left (c d^2-b d e+a e^2\right ) (1+m)}+\frac {c e \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) m (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 ) \left (c d^2-b d e+a e^2\right ) (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.59, size = 287, normalized size = 0.80 \[ \frac {(d+e x)^{m+1} \left (\frac {c e m \sqrt {b^2-4 a c} \left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d+\left (\sqrt {b^2-4 a c}-b\right ) e}\right )}{(m+1) \left (e \left (\sqrt {b^2-4 a c}-b\right )+2 c d\right )}-\frac {c e m \sqrt {b^2-4 a c} \left (e \left (\sqrt {b^2-4 a c}-b\right )+2 c d\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (e \left (\sqrt {b^2-4 a c}+b\right )-2 c d\right )}+\frac {\left (b^2-4 a c\right ) (b e-c d+c e x)}{a+x (b+c x)}\right )}{\left (b^2-4 a c\right ) \left (e (a e-b d)+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (2 \, c x + b\right )} {\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}}, x\right ) \]

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)^2,x, algorithm="fricas")

[Out]

integral((2*c*x + b)*(e*x + d)^m/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2), x)

________________________________________________________________________________________

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

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)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.66, size = 0, normalized size = 0.00 \[ \int \frac {\left (2 c x +b \right ) \left (e x +d \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{2}}\, dx \]

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)^2,x)

[Out]

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

________________________________________________________________________________________

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

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)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

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)^2,x)

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**2,x)

[Out]

Timed out

________________________________________________________________________________________