3.4 \(\int (e x)^m (A+B x^2) (c+d x^2) \, dx\)

Optimal. Leaf size=60 \[ \frac{(e x)^{m+3} (A d+B c)}{e^3 (m+3)}+\frac{A c (e x)^{m+1}}{e (m+1)}+\frac{B d (e x)^{m+5}}{e^5 (m+5)} \]

[Out]

(A*c*(e*x)^(1 + m))/(e*(1 + m)) + ((B*c + A*d)*(e*x)^(3 + m))/(e^3*(3 + m)) + (B*d*(e*x)^(5 + m))/(e^5*(5 + m)
)

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Rubi [A]  time = 0.0314444, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {448} \[ \frac{(e x)^{m+3} (A d+B c)}{e^3 (m+3)}+\frac{A c (e x)^{m+1}}{e (m+1)}+\frac{B d (e x)^{m+5}}{e^5 (m+5)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*(A + B*x^2)*(c + d*x^2),x]

[Out]

(A*c*(e*x)^(1 + m))/(e*(1 + m)) + ((B*c + A*d)*(e*x)^(3 + m))/(e^3*(3 + m)) + (B*d*(e*x)^(5 + m))/(e^5*(5 + m)
)

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int (e x)^m \left (A+B x^2\right ) \left (c+d x^2\right ) \, dx &=\int \left (A c (e x)^m+\frac{(B c+A d) (e x)^{2+m}}{e^2}+\frac{B d (e x)^{4+m}}{e^4}\right ) \, dx\\ &=\frac{A c (e x)^{1+m}}{e (1+m)}+\frac{(B c+A d) (e x)^{3+m}}{e^3 (3+m)}+\frac{B d (e x)^{5+m}}{e^5 (5+m)}\\ \end{align*}

Mathematica [A]  time = 0.0399756, size = 43, normalized size = 0.72 \[ x (e x)^m \left (\frac{x^2 (A d+B c)}{m+3}+\frac{A c}{m+1}+\frac{B d x^4}{m+5}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*(A + B*x^2)*(c + d*x^2),x]

[Out]

x*(e*x)^m*((A*c)/(1 + m) + ((B*c + A*d)*x^2)/(3 + m) + (B*d*x^4)/(5 + m))

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Maple [A]  time = 0.003, size = 111, normalized size = 1.9 \begin{align*}{\frac{ \left ( Bd{m}^{2}{x}^{4}+4\,Bdm{x}^{4}+Ad{m}^{2}{x}^{2}+Bc{m}^{2}{x}^{2}+3\,Bd{x}^{4}+6\,Adm{x}^{2}+6\,Bcm{x}^{2}+Ac{m}^{2}+5\,Ad{x}^{2}+5\,Bc{x}^{2}+8\,Acm+15\,Ac \right ) x \left ( ex \right ) ^{m}}{ \left ( 5+m \right ) \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^2+A)*(d*x^2+c),x)

[Out]

x*(B*d*m^2*x^4+4*B*d*m*x^4+A*d*m^2*x^2+B*c*m^2*x^2+3*B*d*x^4+6*A*d*m*x^2+6*B*c*m*x^2+A*c*m^2+5*A*d*x^2+5*B*c*x
^2+8*A*c*m+15*A*c)*(e*x)^m/(5+m)/(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^2+A)*(d*x^2+c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55228, size = 216, normalized size = 3.6 \begin{align*} \frac{{\left ({\left (B d m^{2} + 4 \, B d m + 3 \, B d\right )} x^{5} +{\left ({\left (B c + A d\right )} m^{2} + 5 \, B c + 5 \, A d + 6 \,{\left (B c + A d\right )} m\right )} x^{3} +{\left (A c m^{2} + 8 \, A c m + 15 \, A c\right )} x\right )} \left (e x\right )^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)*(d*x^2+c),x, algorithm="fricas")

[Out]

((B*d*m^2 + 4*B*d*m + 3*B*d)*x^5 + ((B*c + A*d)*m^2 + 5*B*c + 5*A*d + 6*(B*c + A*d)*m)*x^3 + (A*c*m^2 + 8*A*c*
m + 15*A*c)*x)*(e*x)^m/(m^3 + 9*m^2 + 23*m + 15)

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Sympy [A]  time = 0.906236, size = 459, normalized size = 7.65 \begin{align*} \begin{cases} \frac{- \frac{A c}{4 x^{4}} - \frac{A d}{2 x^{2}} - \frac{B c}{2 x^{2}} + B d \log{\left (x \right )}}{e^{5}} & \text{for}\: m = -5 \\\frac{- \frac{A c}{2 x^{2}} + A d \log{\left (x \right )} + B c \log{\left (x \right )} + \frac{B d x^{2}}{2}}{e^{3}} & \text{for}\: m = -3 \\\frac{A c \log{\left (x \right )} + \frac{A d x^{2}}{2} + \frac{B c x^{2}}{2} + \frac{B d x^{4}}{4}}{e} & \text{for}\: m = -1 \\\frac{A c e^{m} m^{2} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{8 A c e^{m} m x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{15 A c e^{m} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{A d e^{m} m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{6 A d e^{m} m x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{5 A d e^{m} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{B c e^{m} m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{6 B c e^{m} m x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{5 B c e^{m} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{B d e^{m} m^{2} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{4 B d e^{m} m x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac{3 B d e^{m} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x**2+A)*(d*x**2+c),x)

[Out]

Piecewise(((-A*c/(4*x**4) - A*d/(2*x**2) - B*c/(2*x**2) + B*d*log(x))/e**5, Eq(m, -5)), ((-A*c/(2*x**2) + A*d*
log(x) + B*c*log(x) + B*d*x**2/2)/e**3, Eq(m, -3)), ((A*c*log(x) + A*d*x**2/2 + B*c*x**2/2 + B*d*x**4/4)/e, Eq
(m, -1)), (A*c*e**m*m**2*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + 8*A*c*e**m*m*x*x**m/(m**3 + 9*m**2 + 23*m + 15)
+ 15*A*c*e**m*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + A*d*e**m*m**2*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 6*A*d
*e**m*m*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 5*A*d*e**m*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + B*c*e**m*m*
*2*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 6*B*c*e**m*m*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 5*B*c*e**m*x**
3*x**m/(m**3 + 9*m**2 + 23*m + 15) + B*d*e**m*m**2*x**5*x**m/(m**3 + 9*m**2 + 23*m + 15) + 4*B*d*e**m*m*x**5*x
**m/(m**3 + 9*m**2 + 23*m + 15) + 3*B*d*e**m*x**5*x**m/(m**3 + 9*m**2 + 23*m + 15), True))

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Giac [B]  time = 1.20939, size = 225, normalized size = 3.75 \begin{align*} \frac{B d m^{2} x^{5} x^{m} e^{m} + 4 \, B d m x^{5} x^{m} e^{m} + B c m^{2} x^{3} x^{m} e^{m} + A d m^{2} x^{3} x^{m} e^{m} + 3 \, B d x^{5} x^{m} e^{m} + 6 \, B c m x^{3} x^{m} e^{m} + 6 \, A d m x^{3} x^{m} e^{m} + A c m^{2} x x^{m} e^{m} + 5 \, B c x^{3} x^{m} e^{m} + 5 \, A d x^{3} x^{m} e^{m} + 8 \, A c m x x^{m} e^{m} + 15 \, A c x x^{m} e^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(B*x^2+A)*(d*x^2+c),x, algorithm="giac")

[Out]

(B*d*m^2*x^5*x^m*e^m + 4*B*d*m*x^5*x^m*e^m + B*c*m^2*x^3*x^m*e^m + A*d*m^2*x^3*x^m*e^m + 3*B*d*x^5*x^m*e^m + 6
*B*c*m*x^3*x^m*e^m + 6*A*d*m*x^3*x^m*e^m + A*c*m^2*x*x^m*e^m + 5*B*c*x^3*x^m*e^m + 5*A*d*x^3*x^m*e^m + 8*A*c*m
*x*x^m*e^m + 15*A*c*x*x^m*e^m)/(m^3 + 9*m^2 + 23*m + 15)