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

Optimal. Leaf size=19 \[ \frac {c (d+e x)^{m+3}}{e (m+3)} \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {27, 12, 32} \begin {gather*} \frac {c (d+e x)^{m+3}}{e (m+3)} \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),x]

[Out]

(c*(d + e*x)^(3 + m))/(e*(3 + m))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[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]

Rubi steps

\begin {align*} \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx &=\int c (d+e x)^{2+m} \, dx\\ &=c \int (d+e x)^{2+m} \, dx\\ &=\frac {c (d+e x)^{3+m}}{e (3+m)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.05 \begin {gather*} \frac {c (d+e x)^{m+3}}{e m+3 e} \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),x]

[Out]

(c*(d + e*x)^(3 + m))/(3*e + e*m)

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IntegrateAlgebraic [F]  time = 0.03, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \left (c d^2+2 c d e x+c e^2 x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[IntegrateAlgebraic][(d + e*x)^m*(c*d^2 + 2*c*d*e*x + c*e^2*x^2), x]

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fricas [B]  time = 0.42, size = 49, normalized size = 2.58 \begin {gather*} \frac {{\left (c e^{3} x^{3} + 3 \, c d e^{2} x^{2} + 3 \, c d^{2} e x + c d^{3}\right )} {\left (e x + d\right )}^{m}}{e m + 3 \, 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),x, algorithm="fricas")

[Out]

(c*e^3*x^3 + 3*c*d*e^2*x^2 + 3*c*d^2*e*x + c*d^3)*(e*x + d)^m/(e*m + 3*e)

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giac [B]  time = 0.18, size = 75, normalized size = 3.95 \begin {gather*} \frac {{\left (x e + d\right )}^{m} c x^{3} e^{3} + 3 \, {\left (x e + d\right )}^{m} c d x^{2} e^{2} + 3 \, {\left (x e + d\right )}^{m} c d^{2} x e + {\left (x e + d\right )}^{m} c d^{3}}{m e + 3 \, 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),x, algorithm="giac")

[Out]

((x*e + d)^m*c*x^3*e^3 + 3*(x*e + d)^m*c*d*x^2*e^2 + 3*(x*e + d)^m*c*d^2*x*e + (x*e + d)^m*c*d^3)/(m*e + 3*e)

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maple [A]  time = 0.05, size = 36, normalized size = 1.89 \begin {gather*} \frac {\left (e^{2} x^{2}+2 d x e +d^{2}\right ) c \left (e x +d \right )^{m +1}}{\left (m +3\right ) e} \end {gather*}

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),x)

[Out]

(e*x+d)^(m+1)*c*(e^2*x^2+2*d*e*x+d^2)/e/(m+3)

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maxima [B]  time = 1.46, size = 137, normalized size = 7.21 \begin {gather*} \frac {2 \, {\left (e^{2} {\left (m + 1\right )} x^{2} + d e m x - d^{2}\right )} {\left (e x + d\right )}^{m} c d}{{\left (m^{2} + 3 \, m + 2\right )} e} + \frac {{\left (e x + d\right )}^{m + 1} c d^{2}}{e {\left (m + 1\right )}} + \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} e^{3} x^{3} + {\left (m^{2} + m\right )} d e^{2} x^{2} - 2 \, d^{2} e m x + 2 \, d^{3}\right )} {\left (e x + d\right )}^{m} c}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} 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),x, algorithm="maxima")

[Out]

2*(e^2*(m + 1)*x^2 + d*e*m*x - d^2)*(e*x + d)^m*c*d/((m^2 + 3*m + 2)*e) + (e*x + d)^(m + 1)*c*d^2/(e*(m + 1))
+ ((m^2 + 3*m + 2)*e^3*x^3 + (m^2 + m)*d*e^2*x^2 - 2*d^2*e*m*x + 2*d^3)*(e*x + d)^m*c/((m^3 + 6*m^2 + 11*m + 6
)*e)

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mupad [B]  time = 0.46, size = 60, normalized size = 3.16 \begin {gather*} {\left (d+e\,x\right )}^m\,\left (\frac {3\,c\,d^2\,x}{m+3}+\frac {c\,d^3}{e\,\left (m+3\right )}+\frac {c\,e^2\,x^3}{m+3}+\frac {3\,c\,d\,e\,x^2}{m+3}\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),x)

[Out]

(d + e*x)^m*((3*c*d^2*x)/(m + 3) + (c*d^3)/(e*(m + 3)) + (c*e^2*x^3)/(m + 3) + (3*c*d*e*x^2)/(m + 3))

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sympy [A]  time = 0.76, size = 116, normalized size = 6.11 \begin {gather*} \begin {cases} \frac {c x}{d} & \text {for}\: e = 0 \wedge m = -3 \\c d^{2} d^{m} x & \text {for}\: e = 0 \\\frac {c \log {\left (\frac {d}{e} + x \right )}}{e} & \text {for}\: m = -3 \\\frac {c d^{3} \left (d + e x\right )^{m}}{e m + 3 e} + \frac {3 c d^{2} e x \left (d + e x\right )^{m}}{e m + 3 e} + \frac {3 c d e^{2} x^{2} \left (d + e x\right )^{m}}{e m + 3 e} + \frac {c e^{3} x^{3} \left (d + e x\right )^{m}}{e m + 3 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),x)

[Out]

Piecewise((c*x/d, Eq(e, 0) & Eq(m, -3)), (c*d**2*d**m*x, Eq(e, 0)), (c*log(d/e + x)/e, Eq(m, -3)), (c*d**3*(d
+ e*x)**m/(e*m + 3*e) + 3*c*d**2*e*x*(d + e*x)**m/(e*m + 3*e) + 3*c*d*e**2*x**2*(d + e*x)**m/(e*m + 3*e) + c*e
**3*x**3*(d + e*x)**m/(e*m + 3*e), True))

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