3.32.100 \(\int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx\) [3200]

Optimal. Leaf size=187 \[ -\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {d^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (d e-c f)^2 (1+m)}+\frac {f (a d f-b (d e (1-m)+c f m)) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^2 (1+m)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {105, 162, 70} \begin {gather*} \frac {d^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f)^2}+\frac {f (a+b x)^{m+1} (a d f-b c f m-b d e (1-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)^2}-\frac {f (a+b x)^{m+1}}{(e+f x) (b e-a f) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((f*(a + b*x)^(1 + m))/((b*e - a*f)*(d*e - c*f)*(e + f*x))) + (d^2*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 +
m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/((b*c - a*d)*(d*e - c*f)^2*(1 + m)) + (f*(a*d*f - b*d*e*(1 - 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)^2*(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 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) (e+f x)^2} \, dx &=-\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}-\frac {\int \frac {(a+b x)^m (a d f-b (d e+c f m)-b d f m x)}{(c+d x) (e+f x)} \, dx}{(b e-a f) (d e-c f)}\\ &=-\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {d^2 \int \frac {(a+b x)^m}{c+d x} \, dx}{(d e-c f)^2}+\frac {(f (a d f-b d e (1-m)-b c f m)) \int \frac {(a+b x)^m}{e+f x} \, dx}{(b e-a f) (d e-c f)^2}\\ &=-\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {d^2 (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) (d e-c f)^2 (1+m)}+\frac {f (a d f-b d e (1-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)^2 (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.22, size = 174, normalized size = 0.93 \begin {gather*} \frac {(a+b x)^{1+m} \left (-\frac {f}{e+f x}-\frac {d^2 (b e-a f) \, _2F_1\left (1,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )}{(b c-a d) (-d e+c f) (1+m)}+\frac {f (a d f+b d e (-1+m)-b c f m) \, _2F_1\left (1,1+m;2+m;\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f) (d e-c f) (1+m)}\right )}{(b e-a f) (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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)/(f*x+e)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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)/(f*x+e)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)/(f*x+e)**2,x)

[Out]

Timed out

________________________________________________________________________________________

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)/(f*x+e)^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2\,\left (c+d\,x\right )} \,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)),x)

[Out]

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

________________________________________________________________________________________