3.3143 \(\int \frac{(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^3} \, dx\)

Optimal. Leaf size=261 \[ -\frac{\left (2 m^2-4 m+1\right ) (a+b x)^{m-1} (c+d x)^{1-m} \, _2F_1\left (1,m-1;m;-\frac{d (a+b x)}{b (c+d x)}\right )}{8 b^2 d^2 (1-m)}-\frac{(b c-a d) (a+b x)^{m-1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m-1;m;-\frac{d (a+b x)}{b c-a d}\right )}{8 b^3 d^2 (1-m)}+\frac{(1-2 m) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)}+\frac{(b c-a d) (a+b x)^{m-1} (c+d x)^{2-m}}{8 b d^2 (a d+b c+2 b d x)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [C]  time = 0.0659831, antiderivative size = 103, normalized size of antiderivative = 0.39, 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} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,3;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{2 d (a+b x)}{b c-a d}\right )}{b^3 (m+1) (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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]

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])

Rubi steps

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

Mathematica [C]  time = 1.2903, size = 373, normalized size = 1.43 \[ \frac{(a+b x)^m (c+d x)^{-m} \left (\frac{4 (a d-b c) \left (\frac{b (c+d x)}{a d+b (c+2 d x)}\right )^m \left (\frac{d (a+b x)}{a d+b (c+2 d x)}\right )^{1-m} F_1\left (1;m,-m;2;\frac{a d-b c}{a d+b (c+2 d x)},\frac{b c-a d}{b c+a d+2 b d x}\right )}{d^2 (a+b x)}+\frac{\frac{4 b (c+d x) \left (\frac{d (a+b x)}{a d-b c}\right )^{-m} F_1\left (1-m;-m,1;2-m;\frac{b (c+d x)}{b c-a d},\frac{2 b (c+d x)}{b c-a d}\right )}{m-1}-\frac{(b c-a d)^3 \left (\frac{d (a+b x)}{a d+b (c+2 d x)}\right )^{-m} \left (\frac{b (c+d x)}{a d+b (c+2 d x)}\right )^m F_1\left (2;m,-m;3;\frac{a d-b c}{a d+b (c+2 d x)},\frac{b c-a d}{b c+a d+2 b d x}\right )}{(a d+b (c+2 d x))^2}}{d (b c-a d)}\right )}{16 b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x)^m*((4*(-(b*c) + a*d)*((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)) + (-(((b*c - a*d)^3*((b*(c + d*x))/(a*d + b*(c + 2*d*x)))^m*AppellF1[2, m, -m, 3, (-(b*c) + a*d)/(a*
d + b*(c + 2*d*x)), (b*c - a*d)/(b*c + a*d + 2*b*d*x)])/(((d*(a + b*x))/(a*d + b*(c + 2*d*x)))^m*(a*d + b*(c +
2*d*x))^2)) + (4*b*(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)])/((-1 + m)*((d*(a + b*x))/(-(b*c) + a*d))^m))/(d*(b*c - a*d))))/(16*b^3*(c + d*x)^m)

________________________________________________________________________________________

Maple [F]  time = 0.084, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{2-m}}{ \left ( 2\,bdx+ad+bc \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^3,x)

[Out]

int((b*x+a)^m*(d*x+c)^(2-m)/(2*b*d*x+a*d+b*c)^3,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{3}}\,{d x} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2)/(2*b*d*x + b*c + a*d)^3, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 2}}{8 \, b^{3} d^{3} x^{3} + b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + a^{3} d^{3} + 12 \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} + 6 \,{\left (b^{3} c^{2} d + 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x}, x\right ) \end{align*}

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)^3,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 2)/(8*b^3*d^3*x^3 + b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + a^3*d^3 + 1
2*(b^3*c*d^2 + a*b^2*d^3)*x^2 + 6*(b^3*c^2*d + 2*a*b^2*c*d^2 + a^2*b*d^3)*x), x)

________________________________________________________________________________________

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

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)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{3}}\,{d x} \end{align*}

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)^3,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2)/(2*b*d*x + b*c + a*d)^3, x)