3.15.50 \(\int (\frac {b e}{2 c}+e x)^m (\frac {b^2}{4 c}+b x+c x^2)^n \, dx\) [1450]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {658, 32} \begin {gather*} \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)} \end {gather*}

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 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]

Rule 658

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]

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*}

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Mathematica [A]
time = 0.03, size = 54, normalized size = 1.08 \begin {gather*} \frac {2^{-1-2 n} (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 (1+m+2 n)} \end {gather*}

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 = 0.86, size = 58, normalized size = 1.16

method	result	size
gosper	\(\frac {\left (2 c x +b \right ) \left (\frac {e \left (2 c x +b \right )}{2 c}\right )^{m} \left (\frac {4 c^{2} x^{2}+4 b c x +b^{2}}{4 c}\right )^{n}}{2 c \left (1+m +2 n \right )}\)	\(58\)
norman	\(\frac {x \,{\mathrm e}^{m \ln \left (\frac {b e}{2 c}+e x \right )} {\mathrm e}^{n \ln \left (\frac {b^{2}}{4 c}+b x +c \,x^{2}\right )}}{1+m +2 n}+\frac {b \,{\mathrm e}^{m \ln \left (\frac {b e}{2 c}+e x \right )} {\mathrm e}^{n \ln \left (\frac {b^{2}}{4 c}+b x +c \,x^{2}\right )}}{2 c \left (1+m +2 n \right )}\)	\(98\)
risch	\(\frac {\left (2 c x +b \right ) {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{c}\right ) \mathrm {csgn}\left (\frac {i e \left (2 c x +b \right )}{c}\right )^{2} m}{2}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{c}\right ) \mathrm {csgn}\left (i e \left (2 c x +b \right )\right ) \mathrm {csgn}\left (\frac {i e \left (2 c x +b \right )}{c}\right ) m}{2}-\frac {i \mathrm {csgn}\left (\frac {i \left (2 c x +b \right )^{2}}{c}\right )^{3} \pi n}{2}-\frac {i \pi \,\mathrm {csgn}\left (i \left (2 c x +b \right )\right ) \mathrm {csgn}\left (i e \left (2 c x +b \right )\right ) \mathrm {csgn}\left (i e \right ) m}{2}-\frac {i \mathrm {csgn}\left (i \left (2 c x +b \right )^{2}\right )^{3} \pi n}{2}-\frac {i \pi \mathrm {csgn}\left (\frac {i e \left (2 c x +b \right )}{c}\right )^{3} m}{2}-\frac {i \pi \mathrm {csgn}\left (i e \left (2 c x +b \right )\right )^{3} m}{2}-2 n \ln \left (2\right )+2 \ln \left (2 c x +b \right ) n -n \ln \left (c \right )+\frac {i \pi \,\mathrm {csgn}\left (i e \left (2 c x +b \right )\right ) \mathrm {csgn}\left (\frac {i e \left (2 c x +b \right )}{c}\right )^{2} m}{2}-\frac {i \mathrm {csgn}\left (i \left (2 c x +b \right )^{2}\right ) \mathrm {csgn}\left (i \left (2 c x +b \right )\right )^{2} \pi n}{2}+\frac {i \pi \,\mathrm {csgn}\left (i \left (2 c x +b \right )\right ) \mathrm {csgn}\left (i e \left (2 c x +b \right )\right )^{2} m}{2}-\frac {i \mathrm {csgn}\left (i \left (2 c x +b \right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (2 c x +b \right )^{2}}{c}\right ) \mathrm {csgn}\left (\frac {i}{c}\right ) \pi n}{2}+\frac {i \mathrm {csgn}\left (\frac {i \left (2 c x +b \right )^{2}}{c}\right )^{2} \mathrm {csgn}\left (\frac {i}{c}\right ) \pi n}{2}+\frac {i \pi \mathrm {csgn}\left (i e \left (2 c x +b \right )\right )^{2} \mathrm {csgn}\left (i e \right ) m}{2}+i \mathrm {csgn}\left (i \left (2 c x +b \right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (2 c x +b \right )\right ) \pi n +\frac {i \mathrm {csgn}\left (i \left (2 c x +b \right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (2 c x +b \right )^{2}}{c}\right )^{2} \pi n}{2}-m \ln \left (2\right )+m \ln \left (2 c x +b \right )+m \ln \left (e \right )-m \ln \left (c \right )}}{2 \left (1+m +2 n \right ) c}\)	\(484\)

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,method=_RETURNVERBOSE)

[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 = 0.51, size = 77, normalized size = 1.54 \begin {gather*} \frac {{\left (2 \, c x e^{m} + 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c*x*e^m + 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 = 2.77, size = 81, normalized size = 1.62 \begin {gather*} \frac {{\left (2 \, c x + b\right )} \left (\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{4 \, c}\right )^{n} e^{\left (\frac {1}{2} \, m \log \left (\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{4 \, c}\right ) + \frac {1}{2} \, m \log \left (\frac {e^{2}}{c}\right )\right )}}{2 \, {\left (c m + 2 \, c n + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \frac {b \left (\frac {b e}{2 c} + e x\right )^{m} \left (\frac {b^{2}}{4 c} + b x + c x^{2}\right )^{n}}{2 c m + 4 c n + 2 c} + \frac {2 c x \left (\frac {b e}{2 c} + e x\right )^{m} \left (\frac {b^{2}}{4 c} + b x + c x^{2}\right )^{n}}{2 c m + 4 c n + 2 c} & \text {for}\: m \neq - 2 n - 1 \\2^{2 n + 1} \cdot 4^{- n} \int \frac {\left (\frac {b^{2}}{c} + 4 b x + 4 c x^{2}\right )^{n}}{\frac {b e \left (\frac {b e}{c} + 2 e x\right )^{2 n}}{c} + 2 e x \left (\frac {b e}{c} + 2 e x\right )^{2 n}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((b*(b*e/(2*c) + e*x)**m*(b**2/(4*c) + b*x + c*x**2)**n/(2*c*m + 4*c*n + 2*c) + 2*c*x*(b*e/(2*c) + e*
x)**m*(b**2/(4*c) + b*x + c*x**2)**n/(2*c*m + 4*c*n + 2*c), Ne(m, -2*n - 1)), (2**(2*n + 1)*Integral((b**2/c +
4*b*x + 4*c*x**2)**n/(b*e*(b*e/c + 2*e*x)**(2*n)/c + 2*e*x*(b*e/c + 2*e*x)**(2*n)), x)/4**n, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (47) = 94\).
time = 1.34, size = 104, normalized size = 2.08 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 0.67, size = 51, normalized size = 1.02 \begin {gather*} \frac {{\left (e\,x+\frac {b\,e}{2\,c}\right )}^m\,\left (b+2\,c\,x\right )\,{\left (b\,x+c\,x^2+\frac {b^2}{4\,c}\right )}^n}{2\,c\,\left (m+2\,n+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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