3.3119 \(\int \frac{(a+b x)^m (c+d x)^{1-m}}{(e+f x)^4} \, dx\)

Optimal. Leaf size=176 \[ \frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m-1} (b (3 d e-c f (2-m))-a d f (m+1)) \, _2F_1\left (3,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{3 (m+1) (b e-a f)^4 (d e-c f)}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.0760859, antiderivative size = 175, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {96, 131} \[ \frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m-1} (-a d f (m+1)-b c f (2-m)+3 b d e) \, _2F_1\left (3,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{3 (m+1) (b e-a f)^4 (d e-c f)}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m}}{3 (e+f x)^3 (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^m*(c + d*x)^(1 - m))/(e + f*x)^4,x]

[Out]

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

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 131

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)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin{align*} \int \frac{(a+b x)^m (c+d x)^{1-m}}{(e+f x)^4} \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{2-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}-\frac{(-3 b d e+b c f (2-m)+a d f (1+m)) \int \frac{(a+b x)^m (c+d x)^{1-m}}{(e+f x)^3} \, dx}{3 (-b e+a f) (-d e+c f)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{2-m}}{3 (b e-a f) (d e-c f) (e+f x)^3}+\frac{(b c-a d)^2 (3 b d e-b c f (2-m)-a d f (1+m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (3,1+m;2+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{3 (b e-a f)^4 (d e-c f) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.143275, size = 148, normalized size = 0.84 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (\frac{(b c-a d)^2 (-a d f (m+1)+b c f (m-2)+3 b d e) \, _2F_1\left (3,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^3}-\frac{f (c+d x)^3}{(e+f x)^3}\right )}{3 (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(c + d*x)^(1 - m))/(e + f*x)^4,x]

[Out]

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

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Maple [F]  time = 0.124, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{1-m}}{ \left ( fx+e \right ) ^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^4,x)

[Out]

int((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^4,x)

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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 + 1}}{{\left (f x + e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 1)/(f*x + e)^4, x)

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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 + 1}}{f^{4} x^{4} + 4 \, e f^{3} x^{3} + 6 \, e^{2} f^{2} x^{2} + 4 \, e^{3} f x + e^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 1)/(f^4*x^4 + 4*e*f^3*x^3 + 6*e^2*f^2*x^2 + 4*e^3*f*x + e^4), x)

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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)**(1-m)/(f*x+e)**4,x)

[Out]

Timed out

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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 + 1}}{{\left (f x + e\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)/(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 1)/(f*x + e)^4, x)