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

Optimal result

Integrand size = 30, antiderivative size = 43 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x)^{1+m} \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (1+m+2 p)} \]

[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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {658, 32} \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x)^{m+1} \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (m+2 p+1)} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {(d+e x)^{1+m} \left (c (d+e x)^2\right )^p}{e (1+m+2 p)} \]

[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*(1 + m + 2*p))

Maple [A] (verified)

Time = 2.48 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02

method	result	size
gosper	\(\frac {\left (e x +d \right )^{1+m} \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{p}}{e \left (1+m +2 p \right )}\)	\(44\)
parallelrisch	\(\frac {x \left (e x +d \right )^{m} {\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} e +{\left (c \left (x^{2} e^{2}+2 d e x +d^{2}\right )\right )}^{p} d \left (e x +d \right )^{m}}{e \left (1+m +2 p \right )}\)	\(73\)
norman	\(\frac {x \,{\mathrm e}^{m \ln \left (e x +d \right )} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{1+m +2 p}+\frac {d \,{\mathrm e}^{m \ln \left (e x +d \right )} {\mathrm e}^{p \ln \left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )}}{e \left (1+m +2 p \right )}\)	\(91\)
risch	\(\frac {\left (e x +d \right ) \left (e x +d \right )^{m} c^{p} \left (e x +d \right )^{2 p} {\mathrm e}^{-\frac {i \pi p \left (\operatorname {csgn}\left (i \left (e x +d \right )^{2}\right )^{3}-2 \operatorname {csgn}\left (i \left (e x +d \right )^{2}\right )^{2} \operatorname {csgn}\left (i \left (e x +d \right )\right )+\operatorname {csgn}\left (i \left (e x +d \right )^{2}\right ) \operatorname {csgn}\left (i \left (e x +d \right )\right )^{2}-\operatorname {csgn}\left (i \left (e x +d \right )^{2}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{2}\right )^{2}+\operatorname {csgn}\left (i \left (e x +d \right )^{2}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{2}\right ) \operatorname {csgn}\left (i c \right )+\operatorname {csgn}\left (i c \left (e x +d \right )^{2}\right )^{3}-\operatorname {csgn}\left (i c \left (e x +d \right )^{2}\right )^{2} \operatorname {csgn}\left (i c \right )\right )}{2}}}{e \left (1+m +2 p \right )}\)	\(195\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (e x + d\right )} {\left (e x + d\right )}^{m} e^{\left (2 \, p \log \left (e x + d\right ) + p \log \left (c\right )\right )}}{e m + 2 \, e p + e} \]

[In]

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

[Out]

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

Sympy [F]

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

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (c^{p} e x + c^{p} d\right )} e^{\left (m \log \left (e x + d\right ) + 2 \, p \log \left (e x + d\right )\right )}}{e {\left (m + 2 \, p + 1\right )}} \]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.44 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (e x + d\right )}^{m} e x e^{\left (2 \, p \log \left (e x + d\right ) + p \log \left (c\right )\right )} + {\left (e x + d\right )}^{m} d e^{\left (2 \, p \log \left (e x + d\right ) + p \log \left (c\right )\right )}}{e m + 2 \, e p + e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right )^p \, dx=\frac {{\left (d+e\,x\right )}^{m+1}\,{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^p}{e\,\left (m+2\,p+1\right )} \]

[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))