3.1.65 \(\int (e x)^m (a+b x) (a c-b c x) \, dx\)

Optimal. Leaf size=42 \[ \frac {a^2 c (e x)^{m+1}}{e (m+1)}-\frac {b^2 c (e x)^{m+3}}{e^3 (m+3)} \]

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {73, 14} \begin {gather*} \frac {a^2 c (e x)^{m+1}}{e (m+1)}-\frac {b^2 c (e x)^{m+3}}{e^3 (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 73

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.74 \begin {gather*} c x (e x)^m \left (\frac {a^2}{m+1}-\frac {b^2 x^2}{m+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int (e x)^m (a+b x) (a c-b c x) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.49, size = 50, normalized size = 1.19 \begin {gather*} -\frac {{\left ({\left (b^{2} c m + b^{2} c\right )} x^{3} - {\left (a^{2} c m + 3 \, a^{2} c\right )} x\right )} \left (e x\right )^{m}}{m^{2} + 4 \, m + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.24, size = 65, normalized size = 1.55 \begin {gather*} -\frac {b^{2} c m x^{3} x^{m} e^{m} + b^{2} c x^{3} x^{m} e^{m} - a^{2} c m x x^{m} e^{m} - 3 \, a^{2} c x x^{m} e^{m}}{m^{2} + 4 \, m + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 47, normalized size = 1.12 \begin {gather*} \frac {\left (-b^{2} m \,x^{2}-b^{2} x^{2}+a^{2} m +3 a^{2}\right ) c x \left (e x \right )^{m}}{\left (m +3\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)*(-b*c*x+a*c),x)

[Out]

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

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maxima [A]  time = 1.00, size = 41, normalized size = 0.98 \begin {gather*} -\frac {b^{2} c e^{m} x^{3} x^{m}}{m + 3} + \frac {\left (e x\right )^{m + 1} a^{2} c}{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)*(-b*c*x+a*c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.37, size = 46, normalized size = 1.10 \begin {gather*} \frac {c\,x\,{\left (e\,x\right )}^m\,\left (a^2\,m+3\,a^2-b^2\,x^2-b^2\,m\,x^2\right )}{m^2+4\,m+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.72, size = 141, normalized size = 3.36 \begin {gather*} \begin {cases} \frac {- \frac {a^{2} c}{2 x^{2}} - b^{2} c \log {\relax (x )}}{e^{3}} & \text {for}\: m = -3 \\\frac {a^{2} c \log {\relax (x )} - \frac {b^{2} c x^{2}}{2}}{e} & \text {for}\: m = -1 \\\frac {a^{2} c e^{m} m x x^{m}}{m^{2} + 4 m + 3} + \frac {3 a^{2} c e^{m} x x^{m}}{m^{2} + 4 m + 3} - \frac {b^{2} c e^{m} m x^{3} x^{m}}{m^{2} + 4 m + 3} - \frac {b^{2} c e^{m} x^{3} x^{m}}{m^{2} + 4 m + 3} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-a**2*c/(2*x**2) - b**2*c*log(x))/e**3, Eq(m, -3)), ((a**2*c*log(x) - b**2*c*x**2/2)/e, Eq(m, -1))
, (a**2*c*e**m*m*x*x**m/(m**2 + 4*m + 3) + 3*a**2*c*e**m*x*x**m/(m**2 + 4*m + 3) - b**2*c*e**m*m*x**3*x**m/(m*
*2 + 4*m + 3) - b**2*c*e**m*x**3*x**m/(m**2 + 4*m + 3), True))

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