3.1.61 \(\int (e x)^m (a+b x)^2 (a d-b d x) \, dx\)

Optimal. Leaf size=85 \[ \frac {a^3 d (e x)^{m+1}}{e (m+1)}+\frac {a^2 b d (e x)^{m+2}}{e^2 (m+2)}-\frac {a b^2 d (e x)^{m+3}}{e^3 (m+3)}-\frac {b^3 d (e x)^{m+4}}{e^4 (m+4)} \]

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {75} \begin {gather*} \frac {a^2 b d (e x)^{m+2}}{e^2 (m+2)}+\frac {a^3 d (e x)^{m+1}}{e (m+1)}-\frac {a b^2 d (e x)^{m+3}}{e^3 (m+3)}-\frac {b^3 d (e x)^{m+4}}{e^4 (m+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 87, normalized size = 1.02 \begin {gather*} \frac {d (e x)^m \left (\frac {a (2 m+5) x \left (a^2 \left (m^2+5 m+6\right )+2 a b \left (m^2+4 m+3\right ) x+b^2 \left (m^2+3 m+2\right ) x^2\right )}{(m+1) (m+2) (m+3)}-x (a+b x)^3\right )}{m+4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(e*x)^m*(-(x*(a + b*x)^3) + (a*(5 + 2*m)*x*(a^2*(6 + 5*m + m^2) + 2*a*b*(3 + 4*m + m^2)*x + b^2*(2 + 3*m +
m^2)*x^2))/((1 + m)*(2 + m)*(3 + m))))/(4 + m)

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IntegrateAlgebraic [F]  time = 0.10, size = 0, normalized size = 0.00 \begin {gather*} \int (e x)^m (a+b x)^2 (a d-b d x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(e*x)^m*(a + b*x)^2*(a*d - b*d*x),x]

[Out]

Defer[IntegrateAlgebraic][(e*x)^m*(a + b*x)^2*(a*d - b*d*x), x]

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fricas [B]  time = 1.23, size = 176, normalized size = 2.07 \begin {gather*} -\frac {{\left ({\left (b^{3} d m^{3} + 6 \, b^{3} d m^{2} + 11 \, b^{3} d m + 6 \, b^{3} d\right )} x^{4} + {\left (a b^{2} d m^{3} + 7 \, a b^{2} d m^{2} + 14 \, a b^{2} d m + 8 \, a b^{2} d\right )} x^{3} - {\left (a^{2} b d m^{3} + 8 \, a^{2} b d m^{2} + 19 \, a^{2} b d m + 12 \, a^{2} b d\right )} x^{2} - {\left (a^{3} d m^{3} + 9 \, a^{3} d m^{2} + 26 \, a^{3} d m + 24 \, a^{3} d\right )} x\right )} \left (e x\right )^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^3*d*m^3 + 6*b^3*d*m^2 + 11*b^3*d*m + 6*b^3*d)*x^4 + (a*b^2*d*m^3 + 7*a*b^2*d*m^2 + 14*a*b^2*d*m + 8*a*b^2
*d)*x^3 - (a^2*b*d*m^3 + 8*a^2*b*d*m^2 + 19*a^2*b*d*m + 12*a^2*b*d)*x^2 - (a^3*d*m^3 + 9*a^3*d*m^2 + 26*a^3*d*
m + 24*a^3*d)*x)*(e*x)^m/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)

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giac [B]  time = 1.23, size = 273, normalized size = 3.21 \begin {gather*} -\frac {b^{3} d m^{3} x^{4} x^{m} e^{m} + a b^{2} d m^{3} x^{3} x^{m} e^{m} + 6 \, b^{3} d m^{2} x^{4} x^{m} e^{m} - a^{2} b d m^{3} x^{2} x^{m} e^{m} + 7 \, a b^{2} d m^{2} x^{3} x^{m} e^{m} + 11 \, b^{3} d m x^{4} x^{m} e^{m} - a^{3} d m^{3} x x^{m} e^{m} - 8 \, a^{2} b d m^{2} x^{2} x^{m} e^{m} + 14 \, a b^{2} d m x^{3} x^{m} e^{m} + 6 \, b^{3} d x^{4} x^{m} e^{m} - 9 \, a^{3} d m^{2} x x^{m} e^{m} - 19 \, a^{2} b d m x^{2} x^{m} e^{m} + 8 \, a b^{2} d x^{3} x^{m} e^{m} - 26 \, a^{3} d m x x^{m} e^{m} - 12 \, a^{2} b d x^{2} x^{m} e^{m} - 24 \, a^{3} d x x^{m} e^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x+a)^2*(-b*d*x+a*d),x, algorithm="giac")

[Out]

-(b^3*d*m^3*x^4*x^m*e^m + a*b^2*d*m^3*x^3*x^m*e^m + 6*b^3*d*m^2*x^4*x^m*e^m - a^2*b*d*m^3*x^2*x^m*e^m + 7*a*b^
2*d*m^2*x^3*x^m*e^m + 11*b^3*d*m*x^4*x^m*e^m - a^3*d*m^3*x*x^m*e^m - 8*a^2*b*d*m^2*x^2*x^m*e^m + 14*a*b^2*d*m*
x^3*x^m*e^m + 6*b^3*d*x^4*x^m*e^m - 9*a^3*d*m^2*x*x^m*e^m - 19*a^2*b*d*m*x^2*x^m*e^m + 8*a*b^2*d*x^3*x^m*e^m -
26*a^3*d*m*x*x^m*e^m - 12*a^2*b*d*x^2*x^m*e^m - 24*a^3*d*x*x^m*e^m)/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)

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maple [B]  time = 0.01, size = 172, normalized size = 2.02 \begin {gather*} \frac {\left (-b^{3} m^{3} x^{3}-a \,b^{2} m^{3} x^{2}-6 b^{3} m^{2} x^{3}+a^{2} b \,m^{3} x -7 a \,b^{2} m^{2} x^{2}-11 b^{3} m \,x^{3}+a^{3} m^{3}+8 a^{2} b \,m^{2} x -14 a \,b^{2} m \,x^{2}-6 b^{3} x^{3}+9 a^{3} m^{2}+19 a^{2} b m x -8 a \,b^{2} x^{2}+26 a^{3} m +12 a^{2} b x +24 a^{3}\right ) d x \left (e x \right )^{m}}{\left (m +4\right ) \left (m +3\right ) \left (m +2\right ) \left (m +1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(e*x)^m*(-b^3*m^3*x^3-a*b^2*m^3*x^2-6*b^3*m^2*x^3+a^2*b*m^3*x-7*a*b^2*m^2*x^2-11*b^3*m*x^3+a^3*m^3+8*a^2*b*m
^2*x-14*a*b^2*m*x^2-6*b^3*x^3+9*a^3*m^2+19*a^2*b*m*x-8*a*b^2*x^2+26*a^3*m+12*a^2*b*x+24*a^3)*x/(m+4)/(m+3)/(m+
2)/(m+1)

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maxima [A]  time = 1.23, size = 82, normalized size = 0.96 \begin {gather*} -\frac {b^{3} d e^{m} x^{4} x^{m}}{m + 4} - \frac {a b^{2} d e^{m} x^{3} x^{m}}{m + 3} + \frac {a^{2} b d e^{m} x^{2} x^{m}}{m + 2} + \frac {\left (e x\right )^{m + 1} a^{3} d}{e {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x+a)^2*(-b*d*x+a*d),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.42, size = 174, normalized size = 2.05 \begin {gather*} -{\left (e\,x\right )}^m\,\left (\frac {b^3\,d\,x^4\,\left (m^3+6\,m^2+11\,m+6\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}-\frac {a^3\,d\,x\,\left (m^3+9\,m^2+26\,m+24\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}+\frac {a\,b^2\,d\,x^3\,\left (m^3+7\,m^2+14\,m+8\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}-\frac {a^2\,b\,d\,x^2\,\left (m^3+8\,m^2+19\,m+12\right )}{m^4+10\,m^3+35\,m^2+50\,m+24}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e*x)^m*((b^3*d*x^4*(11*m + 6*m^2 + m^3 + 6))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) - (a^3*d*x*(26*m + 9*m^2 +
m^3 + 24))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24) + (a*b^2*d*x^3*(14*m + 7*m^2 + m^3 + 8))/(50*m + 35*m^2 + 10*m^
3 + m^4 + 24) - (a^2*b*d*x^2*(19*m + 8*m^2 + m^3 + 12))/(50*m + 35*m^2 + 10*m^3 + m^4 + 24))

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sympy [A]  time = 1.15, size = 768, normalized size = 9.04 \begin {gather*} \begin {cases} \frac {- \frac {a^{3} d}{3 x^{3}} - \frac {a^{2} b d}{2 x^{2}} + \frac {a b^{2} d}{x} - b^{3} d \log {\relax (x )}}{e^{4}} & \text {for}\: m = -4 \\\frac {- \frac {a^{3} d}{2 x^{2}} - \frac {a^{2} b d}{x} - a b^{2} d \log {\relax (x )} - b^{3} d x}{e^{3}} & \text {for}\: m = -3 \\\frac {- \frac {a^{3} d}{x} + a^{2} b d \log {\relax (x )} - a b^{2} d x - \frac {b^{3} d x^{2}}{2}}{e^{2}} & \text {for}\: m = -2 \\\frac {a^{3} d \log {\relax (x )} + a^{2} b d x - \frac {a b^{2} d x^{2}}{2} - \frac {b^{3} d x^{3}}{3}}{e} & \text {for}\: m = -1 \\\frac {a^{3} d e^{m} m^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {9 a^{3} d e^{m} m^{2} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {26 a^{3} d e^{m} m x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {24 a^{3} d e^{m} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {a^{2} b d e^{m} m^{3} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {8 a^{2} b d e^{m} m^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {19 a^{2} b d e^{m} m x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac {12 a^{2} b d e^{m} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {a b^{2} d e^{m} m^{3} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {7 a b^{2} d e^{m} m^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {14 a b^{2} d e^{m} m x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {8 a b^{2} d e^{m} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {b^{3} d e^{m} m^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {6 b^{3} d e^{m} m^{2} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {11 b^{3} d e^{m} m x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac {6 b^{3} d e^{m} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(b*x+a)**2*(-b*d*x+a*d),x)

[Out]

Piecewise(((-a**3*d/(3*x**3) - a**2*b*d/(2*x**2) + a*b**2*d/x - b**3*d*log(x))/e**4, Eq(m, -4)), ((-a**3*d/(2*
x**2) - a**2*b*d/x - a*b**2*d*log(x) - b**3*d*x)/e**3, Eq(m, -3)), ((-a**3*d/x + a**2*b*d*log(x) - a*b**2*d*x
- b**3*d*x**2/2)/e**2, Eq(m, -2)), ((a**3*d*log(x) + a**2*b*d*x - a*b**2*d*x**2/2 - b**3*d*x**3/3)/e, Eq(m, -1
)), (a**3*d*e**m*m**3*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 9*a**3*d*e**m*m**2*x*x**m/(m**4 + 10*m**
3 + 35*m**2 + 50*m + 24) + 26*a**3*d*e**m*m*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 24*a**3*d*e**m*x*x
**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + a**2*b*d*e**m*m**3*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 2
4) + 8*a**2*b*d*e**m*m**2*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 19*a**2*b*d*e**m*m*x**2*x**m/(m**
4 + 10*m**3 + 35*m**2 + 50*m + 24) + 12*a**2*b*d*e**m*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - a*b**
2*d*e**m*m**3*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 7*a*b**2*d*e**m*m**2*x**3*x**m/(m**4 + 10*m**
3 + 35*m**2 + 50*m + 24) - 14*a*b**2*d*e**m*m*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 8*a*b**2*d*e*
*m*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - b**3*d*e**m*m**3*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 5
0*m + 24) - 6*b**3*d*e**m*m**2*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 11*b**3*d*e**m*m*x**4*x**m/(
m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 6*b**3*d*e**m*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))

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