3.33.1 \(\int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx\) [3201]

Optimal. Leaf size=281 \[ \frac {f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac {d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac {d^2 (2 a d f-b (c f (2-m)+d e m)) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^2 (d e-c f)^3 (1+m)}-\frac {f^2 (2 a d f-b (d e (2-m)+c f m)) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^3 (1+m)} \]

[Out]

f*(-2*a*d*f+b*c*f+b*d*e)*(b*x+a)^(1+m)/(-a*d+b*c)/(-a*f+b*e)/(-c*f+d*e)^2/(f*x+e)+d*(b*x+a)^(1+m)/(-a*d+b*c)/(
-c*f+d*e)/(d*x+c)/(f*x+e)+d^2*(2*a*d*f-b*(c*f*(2-m)+d*e*m))*(b*x+a)^(1+m)*hypergeom([1, 1+m],[2+m],-d*(b*x+a)/
(-a*d+b*c))/(-a*d+b*c)^2/(-c*f+d*e)^3/(1+m)-f^2*(2*a*d*f-b*(d*e*(2-m)+c*f*m))*(b*x+a)^(1+m)*hypergeom([1, 1+m]
,[2+m],-f*(b*x+a)/(-a*f+b*e))/(-a*f+b*e)^2/(-c*f+d*e)^3/(1+m)

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Rubi [A]
time = 0.28, antiderivative size = 281, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {105, 156, 162, 70} \begin {gather*} \frac {d^2 (a+b x)^{m+1} (2 a d f-b c f (2-m)-b d e m) \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)^2 (d e-c f)^3}-\frac {f^2 (a+b x)^{m+1} (2 a d f-b c f m-b d e (2-m)) \, _2F_1\left (1,m+1;m+2;-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2 (d e-c f)^3}+\frac {f (a+b x)^{m+1} (-2 a d f+b c f+b d e)}{(e+f x) (b c-a d) (b e-a f) (d e-c f)^2}+\frac {d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(b*d*e + b*c*f - 2*a*d*f)*(a + b*x)^(1 + m))/((b*c - a*d)*(b*e - a*f)*(d*e - c*f)^2*(e + f*x)) + (d*(a + b*
x)^(1 + m))/((b*c - a*d)*(d*e - c*f)*(c + d*x)*(e + f*x)) + (d^2*(2*a*d*f - b*c*f*(2 - m) - b*d*e*m)*(a + b*x)
^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/((b*c - a*d)^2*(d*e - c*f)^3*(1 + m
)) - (f^2*(2*a*d*f - b*d*e*(2 - m) - b*c*f*m)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((f*(a + b
*x))/(b*e - a*f))])/((b*e - a*f)^2*(d*e - c*f)^3*(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 105

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

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rubi steps

\begin {align*} \int \frac {(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx &=\frac {d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac {\int \frac {(a+b x)^m (2 a d f-b (c f+d e m)+b d f (1-m) x)}{(c+d x) (e+f x)^2} \, dx}{(b c-a d) (d e-c f)}\\ &=\frac {f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac {d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}-\frac {\int \frac {(a+b x)^m \left (2 a^2 d^2 f^2-a b d f (d e+c f) (2+m)+b^2 \left (2 c d e f+d^2 e^2 m+c^2 f^2 m\right )+b d f (b d e+b c f-2 a d f) m x\right )}{(c+d x) (e+f x)} \, dx}{(b c-a d) (b e-a f) (d e-c f)^2}\\ &=\frac {f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac {d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac {\left (d^2 (2 a d f-b c f (2-m)-b d e m)\right ) \int \frac {(a+b x)^m}{c+d x} \, dx}{(b c-a d) (d e-c f)^3}-\frac {\left (f^2 (2 a d f-b d e (2-m)-b c f m)\right ) \int \frac {(a+b x)^m}{e+f x} \, dx}{(b e-a f) (d e-c f)^3}\\ &=\frac {f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac {d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac {d^2 (2 a d f-b c f (2-m)-b d e m) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d)^2 (d e-c f)^3 (1+m)}-\frac {f^2 (2 a d f-b d e (2-m)-b c f m) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^3 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.56, size = 247, normalized size = 0.88 \begin {gather*} \frac {(a+b x)^{1+m} \left (-\frac {f (b d e+b c f-2 a d f)}{(b e-a f) (d e-c f) (e+f x)}-\frac {d}{(c+d x) (e+f x)}-\frac {-d^2 (b e-a f)^2 (-2 a d f-b c f (-2+m)+b d e m) \, _2F_1\left (1,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )+(b c-a d)^2 f^2 (-2 a d f-b d e (-2+m)+b c f m) \, _2F_1\left (1,1+m;2+m;\frac {f (a+b x)}{-b e+a f}\right )}{(b c-a d) (b e-a f)^2 (d e-c f)^2 (1+m)}\right )}{(b c-a d) (-d e+c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + m)*(-((f*(b*d*e + b*c*f - 2*a*d*f))/((b*e - a*f)*(d*e - c*f)*(e + f*x))) - d/((c + d*x)*(e + f
*x)) - (-(d^2*(b*e - a*f)^2*(-2*a*d*f - b*c*f*(-2 + m) + b*d*e*m)*Hypergeometric2F1[1, 1 + m, 2 + m, (d*(a + b
*x))/(-(b*c) + a*d)]) + (b*c - a*d)^2*f^2*(-2*a*d*f - b*d*e*(-2 + m) + b*c*f*m)*Hypergeometric2F1[1, 1 + m, 2
+ m, (f*(a + b*x))/(-(b*e) + a*f)])/((b*c - a*d)*(b*e - a*f)^2*(d*e - c*f)^2*(1 + m))))/((b*c - a*d)*(-(d*e) +
c*f))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right )^{2} \left (f x +e \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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