3.32.42 \(\int \frac {(a+b x)^m (c+d x)^{2-m}}{(b c+a d+2 b d x)^2} \, dx\) [3142]

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

[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 [A]
time = 0.07, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {130, 72, 71, 133} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 130

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

Rule 133

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*c - a
*d)^n*((a + b*x)^(m + 1)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

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] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 1.77, size = 308, normalized size = 2.14 \begin {gather*} \frac {(a+b x)^m (c+d x)^{-m} \left (-\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 {-b c+a d}{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 {\left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \left (4 b (1+m) (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 (-1+m) (a+b x) \left (-\frac {b d (a+b x) (c+d x)}{(b c-a d)^2}\right )^m \, _2F_1\left (m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )\right )}{d (-1+m) (1+m)}\right )}{8 b^3} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{\left (2 b d x +a d +b c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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