3.3142 \(\int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx\)

Optimal. Leaf size=144 \[ \frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m;m+1;-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 d m}-\frac {(b c-a d) (a+b x)^m (c+d x)^{-m} \, _2F_1\left (2,m;m+1;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d m} \]

[Out]

-1/4*(-a*d+b*c)*(b*x+a)^m*hypergeom([2, m],[1+m],-d*(b*x+a)/b/(d*x+c))/b^3/d/m/((d*x+c)^m)+1/4*(-a*d+b*c)*(b*x
+a)^m*(b*(d*x+c)/(-a*d+b*c))^m*hypergeom([m, -1+m],[1+m],-d*(b*x+a)/(-a*d+b*c))/b^3/d/m/((d*x+c)^m)

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Rubi [C]  time = 0.06, antiderivative size = 93, normalized size of antiderivative = 0.65, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {137, 136} \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,2;m+2;-\frac {d (a+b x)}{b c-a d},-\frac {2 d (a+b x)}{b c-a d}\right )}{b^3 (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*AppellF1[1 + m, -2 + m, 2, 2 + m, -((d*(a + b*x))/(b*c - a*d)
), (-2*d*(a + b*x))/(b*c - a*d)])/(b^3*(1 + m)*(c + d*x)^m)

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rubi steps

\begin {align*} \int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx &=\frac {\left ((b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int \frac {(a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m}}{(b c+a d+2 b d x)^2} \, dx}{b^2}\\ &=\frac {(a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m F_1\left (1+m;-2+m,2;2+m;-\frac {d (a+b x)}{b c-a d},-\frac {2 d (a+b x)}{b c-a d}\right )}{b^3 (1+m)}\\ \end {align*}

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Mathematica [C]  time = 1.39, size = 308, normalized size = 2.14 \[ \frac {(a+b x)^m (c+d x)^{-m} \left (\frac {\left (\frac {d (a+b x)}{a d-b c}\right )^{-m} \left (4 b (m+1) (c+d x) F_1\left (1-m;-m,1;2-m;\frac {b (c+d x)}{b c-a d},\frac {2 b (c+d x)}{b c-a d}\right )+2 d (m-1) (a+b x) \left (-\frac {b d (a+b x) (c+d x)}{(b c-a d)^2}\right )^m \, _2F_1\left (m,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )\right )}{d (m-1) (m+1)}-\frac {(b c-a d)^2 \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^{1-m} \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^m F_1\left (1;m,-m;2;\frac {a d-b c}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )}{d^2 (a+b x)}\right )}{8 b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x)^m*(-(((b*c - a*d)^2*((d*(a + b*x))/(a*d + b*(c + 2*d*x)))^(1 - m)*((b*(c + d*x))/(a*d + b*(c + 2*d*
x)))^m*AppellF1[1, m, -m, 2, (-(b*c) + a*d)/(a*d + b*(c + 2*d*x)), (b*c - a*d)/(b*c + a*d + 2*b*d*x)])/(d^2*(a
+ b*x))) + (4*b*(1 + m)*(c + d*x)*AppellF1[1 - m, -m, 1, 2 - m, (b*(c + d*x))/(b*c - a*d), (2*b*(c + d*x))/(b
*c - a*d)] + 2*d*(-1 + m)*(a + b*x)*(-((b*d*(a + b*x)*(c + d*x))/(b*c - a*d)^2))^m*Hypergeometric2F1[m, 1 + m,
2 + m, (d*(a + b*x))/(-(b*c) + a*d)])/(d*(-1 + m)*(1 + m)*((d*(a + b*x))/(-(b*c) + a*d))^m)))/(8*b^3*(c + d*x
)^m)

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fricas [F]  time = 1.15, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{4 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 2 \, a b c d + a^{2} d^{2} + 4 \, {\left (b^{2} c d + a b d^{2}\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m +2}}{\left (2 b d x +a d +b c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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