3.26.36 \(\int (a+b x)^m (c+d x)^{-3-m} (e+f x) \, dx\)

Optimal. Leaf size=114 \[ \frac {(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-2}}{d (m+2) (b c-a d)}-\frac {(a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b (c f (m+1)+d e))}{d (m+1) (m+2) (b c-a d)^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 112, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {79, 37} \begin {gather*} \frac {(a+b x)^{m+1} (d e-c f) (c+d x)^{-m-2}}{d (m+2) (b c-a d)}+\frac {(a+b x)^{m+1} (c+d x)^{-m-1} (-a d f (m+2)+b c f (m+1)+b d e)}{d (m+1) (m+2) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d*e - c*f)*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(d*(b*c - a*d)*(2 + m)) + ((b*d*e + b*c*f*(1 + m) - a*d*f*(
2 + m))*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d*(b*c - a*d)^2*(1 + m)*(2 + m))

Rule 37

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

Rule 79

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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 82, normalized size = 0.72 \begin {gather*} \frac {(a+b x)^{m+1} (c+d x)^{-m-2} (b (c e (m+2)+c f (m+1) x+d e x)-a (c f+d e (m+1)+d f (m+2) x))}{(m+1) (m+2) (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + m)*(c + d*x)^(-2 - m)*(b*(c*e*(2 + m) + d*e*x + c*f*(1 + m)*x) - a*(c*f + d*e*(1 + m) + d*f*(2
+ m)*x)))/((b*c - a*d)^2*(1 + m)*(2 + m))

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IntegrateAlgebraic [F]  time = 0.07, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x)^m (c+d x)^{-3-m} (e+f x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^m*(c + d*x)^(-3 - m)*(e + f*x),x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x)^m*(c + d*x)^(-3 - m)*(e + f*x), x]

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fricas [B]  time = 1.24, size = 336, normalized size = 2.95 \begin {gather*} -\frac {{\left (a^{2} c^{2} f - {\left (b^{2} d^{2} e + {\left (b^{2} c d - a b d^{2}\right )} f m + {\left (b^{2} c d - 2 \, a b d^{2}\right )} f\right )} x^{3} - {\left (a b c^{2} - a^{2} c d\right )} e m - {\left (3 \, b^{2} c d e + {\left (b^{2} c^{2} - 2 \, a b c d - 2 \, a^{2} d^{2}\right )} f + {\left ({\left (b^{2} c d - a b d^{2}\right )} e + {\left (b^{2} c^{2} - a^{2} d^{2}\right )} f\right )} m\right )} x^{2} - {\left (2 \, a b c^{2} - a^{2} c d\right )} e + {\left (3 \, a^{2} c d f - {\left (2 \, b^{2} c^{2} + 2 \, a b c d - a^{2} d^{2}\right )} e - {\left ({\left (b^{2} c^{2} - a^{2} d^{2}\right )} e + {\left (a b c^{2} - a^{2} c d\right )} f\right )} m\right )} x\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}}{2 \, b^{2} c^{2} - 4 \, a b c d + 2 \, a^{2} d^{2} + {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m^{2} + 3 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (f x + e\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e),x, algorithm="giac")

[Out]

integrate((f*x + e)*(b*x + a)^m*(d*x + c)^(-m - 3), x)

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maple [A]  time = 0.01, size = 158, normalized size = 1.39 \begin {gather*} -\frac {\left (a d f m x -b c f m x +a d e m +2 a d f x -b c e m -b c f x -b d e x +a c f +a d e -2 b c e \right ) \left (b x +a \right )^{m +1} \left (d x +c \right )^{-m -2}}{a^{2} d^{2} m^{2}-2 a b c d \,m^{2}+b^{2} c^{2} m^{2}+3 a^{2} d^{2} m -6 a b c d m +3 b^{2} c^{2} m +2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-m-3)*(f*x+e),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (f x + e\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-3-m)*(f*x+e),x, algorithm="maxima")

[Out]

integrate((f*x + e)*(b*x + a)^m*(d*x + c)^(-m - 3), x)

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mupad [B]  time = 2.80, size = 360, normalized size = 3.16 \begin {gather*} \frac {b\,d\,x^3\,{\left (a+b\,x\right )}^m\,\left (b\,c\,f-2\,a\,d\,f+b\,d\,e-a\,d\,f\,m+b\,c\,f\,m\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{m+3}\,\left (m^2+3\,m+2\right )}-\frac {x\,{\left (a+b\,x\right )}^m\,\left (a^2\,d^2\,e-2\,b^2\,c^2\,e+3\,a^2\,c\,d\,f+a^2\,d^2\,e\,m-b^2\,c^2\,e\,m-2\,a\,b\,c\,d\,e-a\,b\,c^2\,f\,m+a^2\,c\,d\,f\,m\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{m+3}\,\left (m^2+3\,m+2\right )}-\frac {a\,c\,{\left (a+b\,x\right )}^m\,\left (a\,c\,f+a\,d\,e-2\,b\,c\,e+a\,d\,e\,m-b\,c\,e\,m\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{m+3}\,\left (m^2+3\,m+2\right )}-\frac {x^2\,{\left (a+b\,x\right )}^m\,\left (2\,a^2\,d^2\,f-b^2\,c^2\,f-3\,b^2\,c\,d\,e+a^2\,d^2\,f\,m-b^2\,c^2\,f\,m+2\,a\,b\,c\,d\,f+a\,b\,d^2\,e\,m-b^2\,c\,d\,e\,m\right )}{{\left (a\,d-b\,c\right )}^2\,{\left (c+d\,x\right )}^{m+3}\,\left (m^2+3\,m+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e + f*x)*(a + b*x)^m)/(c + d*x)^(m + 3),x)

[Out]

(b*d*x^3*(a + b*x)^m*(b*c*f - 2*a*d*f + b*d*e - a*d*f*m + b*c*f*m))/((a*d - b*c)^2*(c + d*x)^(m + 3)*(3*m + m^
2 + 2)) - (x*(a + b*x)^m*(a^2*d^2*e - 2*b^2*c^2*e + 3*a^2*c*d*f + a^2*d^2*e*m - b^2*c^2*e*m - 2*a*b*c*d*e - a*
b*c^2*f*m + a^2*c*d*f*m))/((a*d - b*c)^2*(c + d*x)^(m + 3)*(3*m + m^2 + 2)) - (a*c*(a + b*x)^m*(a*c*f + a*d*e
- 2*b*c*e + a*d*e*m - b*c*e*m))/((a*d - b*c)^2*(c + d*x)^(m + 3)*(3*m + m^2 + 2)) - (x^2*(a + b*x)^m*(2*a^2*d^
2*f - b^2*c^2*f - 3*b^2*c*d*e + a^2*d^2*f*m - b^2*c^2*f*m + 2*a*b*c*d*f + a*b*d^2*e*m - b^2*c*d*e*m))/((a*d -
b*c)^2*(c + d*x)^(m + 3)*(3*m + m^2 + 2))

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sympy [F(-1)]  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)**(-3-m)*(f*x+e),x)

[Out]

Timed out

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