3.1098 \(\int (d+e x)^m (c d^2+2 c d e x+c e^2 x^2)^p \, dx\)

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

[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 = 0.02, 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 = {644, 32} \[ \frac {(d+e x)^{m+1} \left (c d^2+2 c d e x+c e^2 x^2\right )^p}{e (m+2 p+1)} \]

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 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]

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 \[ \frac {(d+e x)^{m+1} \left (c (d+e x)^2\right )^p}{e m+2 e p+e} \]

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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fricas [A]  time = 1.19, size = 39, normalized size = 0.91 \[ \frac {{\left (e x + d\right )} {\left (e x + d\right )}^{m} e^{\left (2 \, p \log \left (e x + d\right ) + p \log \relax (c)\right )}}{e m + 2 \, e p + e} \]

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]

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

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giac [A]  time = 0.21, size = 69, normalized size = 1.60 \[ \frac {{\left (x e + d\right )}^{m} x e^{\left (2 \, p \log \left (x e + d\right ) + p \log \relax (c) + 1\right )} + {\left (x e + d\right )}^{m} d e^{\left (2 \, p \log \left (x e + d\right ) + p \log \relax (c)\right )}}{m e + 2 \, p e + e} \]

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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maple [A]  time = 0.04, size = 44, normalized size = 1.02 \[ \frac {\left (e x +d \right )^{m +1} \left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{p}}{\left (m +2 p +1\right ) e} \]

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)

[Out]

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

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maxima [A]  time = 1.53, size = 43, normalized size = 1.00 \[ \frac {{\left (c^{p} e x + c^{p} d\right )} e^{\left (m \log \left (e x + d\right ) + 2 \, p \log \left (e x + d\right )\right )}}{e {\left (m + 2 \, p + 1\right )}} \]

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

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

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

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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