### 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]

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

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Rubi [A]  time = 0.0177609, antiderivative size = 43, normalized size of antiderivative = 1., 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 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 (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*}

Mathematica [A]  time = 0.0169445, 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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Maple [A]  time = 0.041, size = 44, normalized size = 1. \begin{align*}{\frac{ \left ( ex+d \right ) ^{1+m} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{p}}{e \left ( 1+m+2\,p \right ) }} \end{align*}

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)^(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 = 1.09992, size = 58, normalized size = 1.35 \begin{align*} \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 )}} \end{align*}

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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Fricas [A]  time = 2.43167, size = 99, normalized size = 2.3 \begin{align*} \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} \end{align*}

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.24254, size = 93, normalized size = 2.16 \begin{align*} \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{align*}

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)