3.11.98 \(\int (d+e x)^m (c d^2+2 c d e x+c e^2 x^2)^p \, dx\) [1098]

Optimal. Leaf size=43 \[ \frac {(d+e x)^{1+m} \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (1+m+2 p)} \]

[Out]

(e*x+d)^(1+m)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^p/e/(1+m+2*p)

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {658, 32} \begin {gather*} \frac {(d+e x)^{m+1} \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (m+2 p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^p)/(e*(1 + m + 2*p))

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

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Mathematica [A]
time = 0.02, size = 32, normalized size = 0.74 \begin {gather*} \frac {(d+e x)^{1+m} \left (c (d+e x)^2\right )^p}{e+e m+2 e p} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + m)*(c*(d + e*x)^2)^p)/(e + e*m + 2*e*p)

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Maple [A]
time = 0.70, size = 44, normalized size = 1.02

method	result	size
gosper	\(\frac {\left (e x +d \right )^{1+m} \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{p}}{e \left (1+m +2 p \right )}\)	\(44\)
norman	\(\frac {x \,{\mathrm e}^{m \ln \left (e x +d \right )} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{1+m +2 p}+\frac {d \,{\mathrm e}^{m \ln \left (e x +d \right )} {\mathrm e}^{p \ln \left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )}}{e \left (1+m +2 p \right )}\)	\(91\)
risch	\(\frac {\left (e x +d \right ) \left (e x +d \right )^{m} {\mathrm e}^{\frac {p \left (-i \mathrm {csgn}\left (i \left (e x +d \right )^{2}\right )^{3} \pi +2 i \mathrm {csgn}\left (i \left (e x +d \right )^{2}\right )^{2} \mathrm {csgn}\left (i \left (e x +d \right )\right ) \pi -i \mathrm {csgn}\left (i \left (e x +d \right )^{2}\right ) \mathrm {csgn}\left (i \left (e x +d \right )\right )^{2} \pi +i \mathrm {csgn}\left (i \left (e x +d \right )^{2}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{2}\right )^{2} \pi -i \mathrm {csgn}\left (i \left (e x +d \right )^{2}\right ) \mathrm {csgn}\left (i c \left (e x +d \right )^{2}\right ) \mathrm {csgn}\left (i c \right ) \pi -i \mathrm {csgn}\left (i c \left (e x +d \right )^{2}\right )^{3} \pi +i \mathrm {csgn}\left (i c \left (e x +d \right )^{2}\right )^{2} \mathrm {csgn}\left (i c \right ) \pi +4 \ln \left (e x +d \right )+2 \ln \left (c \right )\right )}{2}}}{e \left (1+m +2 p \right )}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(c*e^2*x^2+2*c*d*e*x+c*d^2)^p,x,method=_RETURNVERBOSE)

[Out]

(e*x+d)^(1+m)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^p/e/(1+m+2*p)

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Maxima [A]
time = 0.28, size = 44, normalized size = 1.02 \begin {gather*} \frac {{\left (c^{p} x e + c^{p} d\right )} e^{\left (m \log \left (x e + d\right ) + 2 \, p \log \left (x e + d\right ) - 1\right )}}{m + 2 \, p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(c*e^2*x^2+2*c*d*e*x+c*d^2)^p,x, algorithm="maxima")

[Out]

(c^p*x*e + c^p*d)*e^(m*log(x*e + d) + 2*p*log(x*e + d) - 1)/(m + 2*p + 1)

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Fricas [A]
time = 1.73, size = 40, normalized size = 0.93 \begin {gather*} \frac {{\left (x e + d\right )} {\left (x e + d\right )}^{m} e^{\left (2 \, p \log \left (x e + d\right ) + p \log \left (c\right ) - 1\right )}}{m + 2 \, p + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*e + d)*(x*e + d)^m*e^(2*p*log(x*e + d) + p*log(c) - 1)/(m + 2*p + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((d**(-2*p - 1)*x*(c*d**2)**p, Eq(e, 0) & Eq(m, -2*p - 1)), (d**m*x*(c*d**2)**p, Eq(e, 0)), (Integral
((c*(d + e*x)**2)**p*(d + e*x)**(-2*p - 1), x), Eq(m, -2*p - 1)), (d*(d + e*x)**m*(c*d**2 + 2*c*d*e*x + c*e**2
*x**2)**p/(e*m + 2*e*p + e) + e*x*(d + e*x)**m*(c*d**2 + 2*c*d*e*x + c*e**2*x**2)**p/(e*m + 2*e*p + e), True))

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Giac [A]
time = 4.73, size = 69, normalized size = 1.60 \begin {gather*} \frac {{\left (x e + d\right )}^{m} x e^{\left (2 \, p \log \left (x e + d\right ) + p \log \left (c\right ) + 1\right )} + {\left (x e + d\right )}^{m} d e^{\left (2 \, p \log \left (x e + d\right ) + p \log \left (c\right )\right )}}{m e + 2 \, p e + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*e + d)^m*x*e^(2*p*log(x*e + d) + p*log(c) + 1) + (x*e + d)^m*d*e^(2*p*log(x*e + d) + p*log(c)))/(m*e + 2*p
*e + e)

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Mupad [B]
time = 0.48, size = 43, normalized size = 1.00 \begin {gather*} \frac {{\left (d+e\,x\right )}^{m+1}\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p}{e\,\left (m+2\,p+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^p,x)

[Out]

((d + e*x)^(m + 1)*(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^p)/(e*(m + 2*p + 1))

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