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

Optimal. Leaf size=41 \[ \frac {c (d+e x)^{m+3}}{e^2 (m+3)}-\frac {c d (d+e x)^{m+2}}{e^2 (m+2)} \]

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Rubi [A]  time = 0.02, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {626, 12, 43} \begin {gather*} \frac {c (d+e x)^{m+3}}{e^2 (m+3)}-\frac {c d (d+e x)^{m+2}}{e^2 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m*(c*d*x + c*e*x^2),x]

[Out]

-((c*d*(d + e*x)^(2 + m))/(e^2*(2 + m))) + (c*(d + e*x)^(3 + m))/(e^2*(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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 626

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

Rubi steps

\begin {align*} \int (d+e x)^m \left (c d x+c e x^2\right ) \, dx &=\int c x (d+e x)^{1+m} \, dx\\ &=c \int x (d+e x)^{1+m} \, dx\\ &=c \int \left (-\frac {d (d+e x)^{1+m}}{e}+\frac {(d+e x)^{2+m}}{e}\right ) \, dx\\ &=-\frac {c d (d+e x)^{2+m}}{e^2 (2+m)}+\frac {c (d+e x)^{3+m}}{e^2 (3+m)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.83 \begin {gather*} \frac {c (d+e x)^{m+2} (e (m+2) x-d)}{e^2 (m+2) (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.08, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \left (c d x+c e x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 81, normalized size = 1.98 \begin {gather*} \frac {{\left (c d^{2} e m x - c d^{3} + {\left (c e^{3} m + 2 \, c e^{3}\right )} x^{3} + {\left (2 \, c d e^{2} m + 3 \, c d e^{2}\right )} x^{2}\right )} {\left (e x + d\right )}^{m}}{e^{2} m^{2} + 5 \, e^{2} m + 6 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*d^2*e*m*x - c*d^3 + (c*e^3*m + 2*c*e^3)*x^3 + (2*c*d*e^2*m + 3*c*d*e^2)*x^2)*(e*x + d)^m/(e^2*m^2 + 5*e^2*m
+ 6*e^2)

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giac [B]  time = 0.18, size = 118, normalized size = 2.88 \begin {gather*} \frac {{\left (x e + d\right )}^{m} c m x^{3} e^{3} + 2 \, {\left (x e + d\right )}^{m} c d m x^{2} e^{2} + {\left (x e + d\right )}^{m} c d^{2} m x e + 2 \, {\left (x e + d\right )}^{m} c x^{3} e^{3} + 3 \, {\left (x e + d\right )}^{m} c d x^{2} e^{2} - {\left (x e + d\right )}^{m} c d^{3}}{m^{2} e^{2} + 5 \, m e^{2} + 6 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*e + d)^m*c*m*x^3*e^3 + 2*(x*e + d)^m*c*d*m*x^2*e^2 + (x*e + d)^m*c*d^2*m*x*e + 2*(x*e + d)^m*c*x^3*e^3 + 3
*(x*e + d)^m*c*d*x^2*e^2 - (x*e + d)^m*c*d^3)/(m^2*e^2 + 5*m*e^2 + 6*e^2)

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maple [A]  time = 0.04, size = 37, normalized size = 0.90 \begin {gather*} -\frac {\left (-m e x -2 e x +d \right ) c \left (e x +d \right )^{m +2}}{\left (m^{2}+5 m +6\right ) e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*(e*x+d)^(m+2)*(-e*m*x-2*e*x+d)/e^2/(m^2+5*m+6)

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maxima [B]  time = 1.41, size = 114, normalized size = 2.78 \begin {gather*} \frac {{\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^{2}} + \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^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.33, size = 88, normalized size = 2.15 \begin {gather*} {\left (d+e\,x\right )}^m\,\left (\frac {c\,e\,x^3\,\left (m+2\right )}{m^2+5\,m+6}-\frac {c\,d^3}{e^2\,\left (m^2+5\,m+6\right )}+\frac {c\,d\,x^2\,\left (2\,m+3\right )}{m^2+5\,m+6}+\frac {c\,d^2\,m\,x}{e\,\left (m^2+5\,m+6\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x + c*e*x^2)*(d + e*x)^m,x)

[Out]

(d + e*x)^m*((c*e*x^3*(m + 2))/(5*m + m^2 + 6) - (c*d^3)/(e^2*(5*m + m^2 + 6)) + (c*d*x^2*(2*m + 3))/(5*m + m^
2 + 6) + (c*d^2*m*x)/(e*(5*m + m^2 + 6)))

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sympy [A]  time = 1.17, size = 299, normalized size = 7.29 \begin {gather*} \begin {cases} \frac {c d d^{m} x^{2}}{2} & \text {for}\: e = 0 \\\frac {c d \log {\left (\frac {d}{e} + x \right )}}{d e^{2} + e^{3} x} + \frac {c d}{d e^{2} + e^{3} x} + \frac {c e x \log {\left (\frac {d}{e} + x \right )}}{d e^{2} + e^{3} x} & \text {for}\: m = -3 \\- \frac {c d \log {\left (\frac {d}{e} + x \right )}}{e^{2}} + \frac {c x}{e} & \text {for}\: m = -2 \\- \frac {c d^{3} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {c d^{2} e m x \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {2 c d e^{2} m x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {3 c d e^{2} x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {c e^{3} m x^{3} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {2 c e^{3} x^{3} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c*d*d**m*x**2/2, Eq(e, 0)), (c*d*log(d/e + x)/(d*e**2 + e**3*x) + c*d/(d*e**2 + e**3*x) + c*e*x*log
(d/e + x)/(d*e**2 + e**3*x), Eq(m, -3)), (-c*d*log(d/e + x)/e**2 + c*x/e, Eq(m, -2)), (-c*d**3*(d + e*x)**m/(e
**2*m**2 + 5*e**2*m + 6*e**2) + c*d**2*e*m*x*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 2*c*d*e**2*m*x**2*
(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 3*c*d*e**2*x**2*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) +
c*e**3*m*x**3*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 2*c*e**3*x**3*(d + e*x)**m/(e**2*m**2 + 5*e**2*m
+ 6*e**2), True))

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