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

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

[Out]

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

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Rubi [A]  time = 0.0179111, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {73, 14} \[ \frac{a^2 c (e x)^{m+1}}{e (m+1)}-\frac{b^2 c (e x)^{m+3}}{e^3 (m+3)} \]

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 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]

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]]

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*}

Mathematica [A]  time = 0.0178273, size = 31, normalized size = 0.74 \[ c x (e x)^m \left (\frac{a^2}{m+1}-\frac{b^2 x^2}{m+3}\right ) \]

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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Maple [A]  time = 0.002, size = 47, normalized size = 1.1 \begin{align*}{\frac{c \left ( ex \right ) ^{m} \left ( -{b}^{2}m{x}^{2}-{b}^{2}{x}^{2}+{a}^{2}m+3\,{a}^{2} \right ) x}{ \left ( 3+m \right ) \left ( 1+m \right ) }} \end{align*}

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/(3+m)/(1+m)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.85512, size = 100, normalized size = 2.38 \begin{align*} -\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{align*}

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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Sympy [A]  time = 0.535299, size = 141, normalized size = 3.36 \begin{align*} \begin{cases} \frac{- \frac{a^{2} c}{2 x^{2}} - b^{2} c \log{\left (x \right )}}{e^{3}} & \text{for}\: m = -3 \\\frac{a^{2} c \log{\left (x \right )} - \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{align*}

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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Giac [A]  time = 1.24298, size = 88, normalized size = 2.1 \begin{align*} -\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{align*}

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)