3.1450 \(\int (\frac{b e}{2 c}+e x)^m (\frac{b^2}{4 c}+b x+c x^2)^n \, dx\)

Optimal. Leaf size=50 \[ \frac{\left (\frac{b^2}{4 c}+b x+c x^2\right )^n \left (\frac{b e}{2 c}+e x\right )^{m+1}}{e (m+2 n+1)} \]

[Out]

(((b*e)/(2*c) + e*x)^(1 + m)*(b^2/(4*c) + b*x + c*x^2)^n)/(e*(1 + m + 2*n))

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Rubi [A]  time = 0.0205833, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.054, Rules used = {644, 32} \[ \frac{\left (\frac{b^2}{4 c}+b x+c x^2\right )^n \left (\frac{b e}{2 c}+e x\right )^{m+1}}{e (m+2 n+1)} \]

Antiderivative was successfully verified.

[In]

Int[((b*e)/(2*c) + e*x)^m*(b^2/(4*c) + b*x + c*x^2)^n,x]

[Out]

(((b*e)/(2*c) + e*x)^(1 + m)*(b^2/(4*c) + b*x + c*x^2)^n)/(e*(1 + m + 2*n))

Rule 644

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^p/(d
+ e*x)^(2*p), Int[(d + e*x)^(m + 2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !
IntegerQ[p] && EqQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0271691, size = 54, normalized size = 1.08 \[ \frac{2^{-2 n-1} (b+2 c x) \left (\frac{(b+2 c x)^2}{c}\right )^n \left (\frac{b e}{2 c}+e x\right )^m}{c (m+2 n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*e)/(2*c) + e*x)^m*(b^2/(4*c) + b*x + c*x^2)^n,x]

[Out]

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

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Maple [A]  time = 0.044, size = 58, normalized size = 1.2 \begin{align*}{\frac{2\,cx+b}{2\,c \left ( 1+m+2\,n \right ) } \left ({\frac{e \left ( 2\,cx+b \right ) }{2\,c}} \right ) ^{m} \left ({\frac{4\,{c}^{2}{x}^{2}+4\,bcx+{b}^{2}}{4\,c}} \right ) ^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/2*b*e/c+e*x)^m*(1/4*b^2/c+b*x+c*x^2)^n,x)

[Out]

1/2*(2*c*x+b)/c/(1+m+2*n)*(1/2*e*(2*c*x+b)/c)^m*(1/4*(4*c^2*x^2+4*b*c*x+b^2)/c)^n

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Maxima [A]  time = 1.66517, size = 107, normalized size = 2.14 \begin{align*} \frac{{\left (2 \, c e^{m} x + b e^{m}\right )} c^{-m - n - 1} e^{\left (m \log \left (2 \, c x + b\right ) + 2 \, n \log \left (2 \, c x + b\right )\right )}}{{\left (2^{2 \, n + 2} n + 2^{2 \, n + 1}\right )} 2^{m} + 2^{m + 2 \, n + 1} m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2*b*e/c+e*x)^m*(1/4/c*b^2+b*x+c*x^2)^n,x, algorithm="maxima")

[Out]

(2*c*e^m*x + b*e^m)*c^(-m - n - 1)*e^(m*log(2*c*x + b) + 2*n*log(2*c*x + b))/((2^(2*n + 2)*n + 2^(2*n + 1))*2^
m + 2^(m + 2*n + 1)*m)

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Fricas [A]  time = 2.19233, size = 150, normalized size = 3. \begin{align*} \frac{{\left (2 \, c x + b\right )} \left (\frac{2 \, c e x + b e}{2 \, c}\right )^{m} e^{\left (2 \, n \log \left (\frac{2 \, c e x + b e}{2 \, c}\right ) + n \log \left (\frac{c}{e^{2}}\right )\right )}}{2 \,{\left (c m + 2 \, c n + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2*b*e/c+e*x)^m*(1/4/c*b^2+b*x+c*x^2)^n,x, algorithm="fricas")

[Out]

1/2*(2*c*x + b)*(1/2*(2*c*e*x + b*e)/c)^m*e^(2*n*log(1/2*(2*c*e*x + b*e)/c) + n*log(c/e^2))/(c*m + 2*c*n + c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2*b*e/c+e*x)**m*(1/4/c*b**2+b*x+c*x**2)**n,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.18686, size = 140, normalized size = 2.8 \begin{align*} \frac{2 \, c x e^{\left (-m \log \left (2\right ) - 2 \, n \log \left (2\right ) + m \log \left (2 \, c x + b\right ) + 2 \, n \log \left (2 \, c x + b\right ) - m \log \left (c\right ) - n \log \left (c\right ) + m\right )} + b e^{\left (-m \log \left (2\right ) - 2 \, n \log \left (2\right ) + m \log \left (2 \, c x + b\right ) + 2 \, n \log \left (2 \, c x + b\right ) - m \log \left (c\right ) - n \log \left (c\right ) + m\right )}}{2 \,{\left (c m + 2 \, c n + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/2*b*e/c+e*x)^m*(1/4/c*b^2+b*x+c*x^2)^n,x, algorithm="giac")

[Out]

1/2*(2*c*x*e^(-m*log(2) - 2*n*log(2) + m*log(2*c*x + b) + 2*n*log(2*c*x + b) - m*log(c) - n*log(c) + m) + b*e^
(-m*log(2) - 2*n*log(2) + m*log(2*c*x + b) + 2*n*log(2*c*x + b) - m*log(c) - n*log(c) + m))/(c*m + 2*c*n + c)