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

Optimal. Leaf size=52 \[ \frac {c d (d+e x)^{m+3}}{e^2 (m+3)}-\frac {\left (c d^2-a e^2\right ) (d+e x)^{m+2}}{e^2 (m+2)} \]

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Rubi [A]  time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {626, 43} \begin {gather*} \frac {c d (d+e x)^{m+3}}{e^2 (m+3)}-\frac {\left (c d^2-a e^2\right ) (d+e x)^{m+2}}{e^2 (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 45, normalized size = 0.87 \begin {gather*} \frac {(d+e x)^{m+2} \left (a e^2 (m+3)+c d (e (m+2) x-d)\right )}{e^2 (m+2) (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 136, normalized size = 2.62 \begin {gather*} \frac {{\left (a d^{2} e^{2} m - c d^{4} + 3 \, a d^{2} e^{2} + {\left (c d e^{3} m + 2 \, c d e^{3}\right )} x^{3} + {\left (3 \, c d^{2} e^{2} + 3 \, a e^{4} + {\left (2 \, c d^{2} e^{2} + a e^{4}\right )} m\right )} x^{2} + {\left (6 \, a d e^{3} + {\left (c d^{3} e + 2 \, a d e^{3}\right )} m\right )} x\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*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="fricas")

[Out]

(a*d^2*e^2*m - c*d^4 + 3*a*d^2*e^2 + (c*d*e^3*m + 2*c*d*e^3)*x^3 + (3*c*d^2*e^2 + 3*a*e^4 + (2*c*d^2*e^2 + a*e
^4)*m)*x^2 + (6*a*d*e^3 + (c*d^3*e + 2*a*d*e^3)*m)*x)*(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 = 219, normalized size = 4.21 \begin {gather*} \frac {{\left (x e + d\right )}^{m} c d m x^{3} e^{3} + 2 \, {\left (x e + d\right )}^{m} c d^{2} m x^{2} e^{2} + {\left (x e + d\right )}^{m} c d^{3} m x e + 2 \, {\left (x e + d\right )}^{m} c d x^{3} e^{3} + 3 \, {\left (x e + d\right )}^{m} c d^{2} x^{2} e^{2} - {\left (x e + d\right )}^{m} c d^{4} + {\left (x e + d\right )}^{m} a m x^{2} e^{4} + 2 \, {\left (x e + d\right )}^{m} a d m x e^{3} + {\left (x e + d\right )}^{m} a d^{2} m e^{2} + 3 \, {\left (x e + d\right )}^{m} a x^{2} e^{4} + 6 \, {\left (x e + d\right )}^{m} a d x e^{3} + 3 \, {\left (x e + d\right )}^{m} a d^{2} e^{2}}{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*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="giac")

[Out]

((x*e + d)^m*c*d*m*x^3*e^3 + 2*(x*e + d)^m*c*d^2*m*x^2*e^2 + (x*e + d)^m*c*d^3*m*x*e + 2*(x*e + d)^m*c*d*x^3*e
^3 + 3*(x*e + d)^m*c*d^2*x^2*e^2 - (x*e + d)^m*c*d^4 + (x*e + d)^m*a*m*x^2*e^4 + 2*(x*e + d)^m*a*d*m*x*e^3 + (
x*e + d)^m*a*d^2*m*e^2 + 3*(x*e + d)^m*a*x^2*e^4 + 6*(x*e + d)^m*a*d*x*e^3 + 3*(x*e + d)^m*a*d^2*e^2)/(m^2*e^2
+ 5*m*e^2 + 6*e^2)

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maple [A]  time = 0.05, size = 55, normalized size = 1.06 \begin {gather*} \frac {\left (c d e m x +a \,e^{2} m +2 c d e x +3 a \,e^{2}-c \,d^{2}\right ) \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*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x),x)

[Out]

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

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maxima [B]  time = 1.20, size = 174, normalized size = 3.35 \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} a}{m^{2} + 3 \, m + 2} + \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^{2}}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} + \frac {{\left (e x + d\right )}^{m + 1} a d}{m + 1} + \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 d}{{\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*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.74, size = 141, normalized size = 2.71 \begin {gather*} {\left (d+e\,x\right )}^m\,\left (\frac {x^2\,\left (3\,a\,e^2+3\,c\,d^2+a\,e^2\,m+2\,c\,d^2\,m\right )}{m^2+5\,m+6}+\frac {d^2\,\left (3\,a\,e^2-c\,d^2+a\,e^2\,m\right )}{e^2\,\left (m^2+5\,m+6\right )}+\frac {d\,x\,\left (6\,a\,e^2+2\,a\,e^2\,m+c\,d^2\,m\right )}{e\,\left (m^2+5\,m+6\right )}+\frac {c\,d\,e\,x^3\,\left (m+2\right )}{m^2+5\,m+6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.34, size = 556, normalized size = 10.69 \begin {gather*} \begin {cases} \frac {c d^{2} d^{m} x^{2}}{2} & \text {for}\: e = 0 \\- \frac {a e^{2}}{d e^{2} + e^{3} x} + \frac {c d^{2} \log {\left (\frac {d}{e} + x \right )}}{d e^{2} + e^{3} x} + \frac {c d^{2}}{d e^{2} + e^{3} x} + \frac {c d e x \log {\left (\frac {d}{e} + x \right )}}{d e^{2} + e^{3} x} & \text {for}\: m = -3 \\a \log {\left (\frac {d}{e} + x \right )} - \frac {c d^{2} \log {\left (\frac {d}{e} + x \right )}}{e^{2}} + \frac {c d x}{e} & \text {for}\: m = -2 \\\frac {a d^{2} e^{2} m \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {3 a d^{2} e^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {2 a d e^{3} m x \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {6 a d e^{3} x \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {a e^{4} m x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {3 a e^{4} x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} - \frac {c d^{4} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {c d^{3} e m x \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {2 c d^{2} 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^{2} e^{2} x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 5 e^{2} m + 6 e^{2}} + \frac {c d 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 d 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*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)

[Out]

Piecewise((c*d**2*d**m*x**2/2, Eq(e, 0)), (-a*e**2/(d*e**2 + e**3*x) + c*d**2*log(d/e + x)/(d*e**2 + e**3*x) +
c*d**2/(d*e**2 + e**3*x) + c*d*e*x*log(d/e + x)/(d*e**2 + e**3*x), Eq(m, -3)), (a*log(d/e + x) - c*d**2*log(d
/e + x)/e**2 + c*d*x/e, Eq(m, -2)), (a*d**2*e**2*m*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 3*a*d**2*e**
2*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 2*a*d*e**3*m*x*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) +
6*a*d*e**3*x*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + a*e**4*m*x**2*(d + e*x)**m/(e**2*m**2 + 5*e**2*m
+ 6*e**2) + 3*a*e**4*x**2*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) - c*d**4*(d + e*x)**m/(e**2*m**2 + 5*e*
*2*m + 6*e**2) + c*d**3*e*m*x*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 2*c*d**2*e**2*m*x**2*(d + e*x)**m
/(e**2*m**2 + 5*e**2*m + 6*e**2) + 3*c*d**2*e**2*x**2*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + c*d*e**3*
m*x**3*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e**2) + 2*c*d*e**3*x**3*(d + e*x)**m/(e**2*m**2 + 5*e**2*m + 6*e
**2), True))

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