3.32.27 \(\int \frac {(a+b x)^m (c+d x)^{2-m}}{e+f x} \, dx\) [3127]

Optimal. Leaf size=370 \[ -\frac {d \left (2 a b c d f^2 m-a^2 d^2 f^2 m-b^2 \left (2 d^2 e^2-4 c d e f+c^2 f^2 (2+m)\right )\right ) (a+b x)^{1+m} (c+d x)^{-m}}{2 b^2 (b c-a d) f^3 m}+\frac {d^2 (a+b x)^{2+m} (c+d x)^{-m}}{2 b^2 f}+\frac {(d e-c f)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{f^3 m}+\frac {d \left (2 a b d f (d e-c f (2-m)) m+a^2 d^2 f^2 (1-m) m-b^2 \left (2 d^2 e^2-2 c d e f (2-m)+c^2 f^2 \left (2-3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 (b c-a d) f^3 m (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.29, antiderivative size = 370, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {135, 133, 965, 80, 72, 71} \begin {gather*} \frac {d (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 (1-m) m+2 a b d f m (d e-c f (2-m))-\left (b^2 \left (c^2 f^2 \left (m^2-3 m+2\right )-2 c d e f (2-m)+2 d^2 e^2\right )\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{2 b^2 f^3 m (m+1) (b c-a d)}-\frac {d (a+b x)^{m+1} (c+d x)^{-m} \left (-a^2 d^2 f^2 m+2 a b c d f^2 m-\left (b^2 \left (c^2 f^2 (m+2)-4 c d e f+2 d^2 e^2\right )\right )\right )}{2 b^2 f^3 m (b c-a d)}+\frac {d^2 (a+b x)^{m+2} (c+d x)^{-m}}{2 b^2 f}+\frac {(a+b x)^m (d e-c f)^2 (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;\frac {(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{f^3 m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d*(2*a*b*c*d*f^2*m - a^2*d^2*f^2*m - b^2*(2*d^2*e^2 - 4*c*d*e*f + c^2*f^2*(2 + m)))*(a + b*x)^(1 + m))/(
b^2*(b*c - a*d)*f^3*m*(c + d*x)^m) + (d^2*(a + b*x)^(2 + m))/(2*b^2*f*(c + d*x)^m) + ((d*e - c*f)^2*(a + b*x)^
m*Hypergeometric2F1[1, -m, 1 - m, ((b*e - a*f)*(c + d*x))/((d*e - c*f)*(a + b*x))])/(f^3*m*(c + d*x)^m) + (d*(
2*a*b*d*f*(d*e - c*f*(2 - m))*m + a^2*d^2*f^2*(1 - m)*m - b^2*(2*d^2*e^2 - 2*c*d*e*f*(2 - m) + c^2*f^2*(2 - 3*
m + m^2)))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, 1 + m, 2 + m, -((d*(a + b*x))/
(b*c - a*d))])/(2*b^2*(b*c - a*d)*f^3*m*(1 + 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 80

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

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]

Rule 135

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

Rule 965

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x)^(n + 1)/(g*e^(2*p)*(m + n + 2*p + 1))), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rubi steps

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

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Mathematica [A]
time = 0.26, size = 258, normalized size = 0.70 \begin {gather*} \frac {(a+b x)^m (c+d x)^{-m} \left (-d (-b c+a d) f^2 m (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )-b (d e-c f) \left (b (d e-c f) (1+m) \left (\, _2F_1\left (1,m;1+m;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )-\left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;1+m;\frac {d (a+b x)}{-b c+a d}\right )\right )+d f m (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;\frac {d (a+b x)}{-b c+a d}\right )\right )\right )}{b^2 f^3 m (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{2-m}}{f x +e}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

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

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