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

Optimal. Leaf size=75 \[ \frac {d (c d-b e) (d+e x)^{m+1}}{e^3 (m+1)}-\frac {(2 c d-b e) (d+e x)^{m+2}}{e^3 (m+2)}+\frac {c (d+e x)^{m+3}}{e^3 (m+3)} \]

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Rubi [A]  time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {698} \begin {gather*} \frac {d (c d-b e) (d+e x)^{m+1}}{e^3 (m+1)}-\frac {(2 c d-b e) (d+e x)^{m+2}}{e^3 (m+2)}+\frac {c (d+e x)^{m+3}}{e^3 (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m*(b*x + c*x^2),x]

[Out]

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

Rule 698

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 75, normalized size = 1.00 \begin {gather*} \frac {d (c d-b e) (d+e x)^{m+1}}{e^3 (m+1)}-\frac {(2 c d-b e) (d+e x)^{m+2}}{e^3 (m+2)}+\frac {c (d+e x)^{m+3}}{e^3 (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.05, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m \left (b x+c x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 160, normalized size = 2.13 \begin {gather*} -\frac {{\left (b d^{2} e m - 2 \, c d^{3} + 3 \, b d^{2} e - {\left (c e^{3} m^{2} + 3 \, c e^{3} m + 2 \, c e^{3}\right )} x^{3} - {\left (3 \, b e^{3} + {\left (c d e^{2} + b e^{3}\right )} m^{2} + {\left (c d e^{2} + 4 \, b e^{3}\right )} m\right )} x^{2} - {\left (b d e^{2} m^{2} - {\left (2 \, c d^{2} e - 3 \, b d e^{2}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.18, size = 263, normalized size = 3.51 \begin {gather*} \frac {{\left (x e + d\right )}^{m} c m^{2} x^{3} e^{3} + {\left (x e + d\right )}^{m} c d m^{2} x^{2} e^{2} + {\left (x e + d\right )}^{m} b m^{2} x^{2} e^{3} + 3 \, {\left (x e + d\right )}^{m} c m x^{3} e^{3} + {\left (x e + d\right )}^{m} b d m^{2} x e^{2} + {\left (x e + d\right )}^{m} c d m x^{2} e^{2} - 2 \, {\left (x e + d\right )}^{m} c d^{2} m x e + 4 \, {\left (x e + d\right )}^{m} b m x^{2} e^{3} + 2 \, {\left (x e + d\right )}^{m} c x^{3} e^{3} + 3 \, {\left (x e + d\right )}^{m} b d m x e^{2} - {\left (x e + d\right )}^{m} b d^{2} m e + 2 \, {\left (x e + d\right )}^{m} c d^{3} + 3 \, {\left (x e + d\right )}^{m} b x^{2} e^{3} - 3 \, {\left (x e + d\right )}^{m} b d^{2} e}{m^{3} e^{3} + 6 \, m^{2} e^{3} + 11 \, m e^{3} + 6 \, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*e + d)^m*c*m^2*x^3*e^3 + (x*e + d)^m*c*d*m^2*x^2*e^2 + (x*e + d)^m*b*m^2*x^2*e^3 + 3*(x*e + d)^m*c*m*x^3*e
^3 + (x*e + d)^m*b*d*m^2*x*e^2 + (x*e + d)^m*c*d*m*x^2*e^2 - 2*(x*e + d)^m*c*d^2*m*x*e + 4*(x*e + d)^m*b*m*x^2
*e^3 + 2*(x*e + d)^m*c*x^3*e^3 + 3*(x*e + d)^m*b*d*m*x*e^2 - (x*e + d)^m*b*d^2*m*e + 2*(x*e + d)^m*c*d^3 + 3*(
x*e + d)^m*b*x^2*e^3 - 3*(x*e + d)^m*b*d^2*e)/(m^3*e^3 + 6*m^2*e^3 + 11*m*e^3 + 6*e^3)

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maple [A]  time = 0.04, size = 116, normalized size = 1.55 \begin {gather*} -\frac {\left (-c \,e^{2} m^{2} x^{2}-b \,e^{2} m^{2} x -3 c \,e^{2} m \,x^{2}-4 b \,e^{2} m x +2 c d e m x -2 c \,e^{2} x^{2}+b d e m -3 b \,e^{2} x +2 c d e x +3 b d e -2 c \,d^{2}\right ) \left (e x +d \right )^{m +1}}{\left (m^{3}+6 m^{2}+11 m +6\right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e*x+d)^(m+1)*(-c*e^2*m^2*x^2-b*e^2*m^2*x-3*c*e^2*m*x^2-4*b*e^2*m*x+2*c*d*e*m*x-2*c*e^2*x^2+b*d*e*m-3*b*e^2*x
+2*c*d*e*x+3*b*d*e-2*c*d^2)/e^3/(m^3+6*m^2+11*m+6)

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maxima [A]  time = 1.41, size = 113, normalized size = 1.51 \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} b}{{\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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.37, size = 146, normalized size = 1.95 \begin {gather*} {\left (d+e\,x\right )}^m\,\left (\frac {c\,x^3\,\left (m^2+3\,m+2\right )}{m^3+6\,m^2+11\,m+6}-\frac {d^2\,\left (3\,b\,e-2\,c\,d+b\,e\,m\right )}{e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {x^2\,\left (m+1\right )\,\left (3\,b\,e+b\,e\,m+c\,d\,m\right )}{e\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {d\,m\,x\,\left (3\,b\,e-2\,c\,d+b\,e\,m\right )}{e^2\,\left (m^3+6\,m^2+11\,m+6\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)*(d + e*x)^m,x)

[Out]

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

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sympy [A]  time = 1.72, size = 1095, normalized size = 14.60 \begin {gather*} \begin {cases} d^{m} \left (\frac {b x^{2}}{2} + \frac {c x^{3}}{3}\right ) & \text {for}\: e = 0 \\- \frac {b d e}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} - \frac {2 b e^{2} x}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {2 c d^{2} \log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {3 c d^{2}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {4 c d e x \log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {4 c d e x}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} + \frac {2 c e^{2} x^{2} \log {\left (\frac {d}{e} + x \right )}}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} & \text {for}\: m = -3 \\\frac {b d e \log {\left (\frac {d}{e} + x \right )}}{d e^{3} + e^{4} x} + \frac {b d e}{d e^{3} + e^{4} x} + \frac {b e^{2} x \log {\left (\frac {d}{e} + x \right )}}{d e^{3} + e^{4} x} - \frac {2 c d^{2} \log {\left (\frac {d}{e} + x \right )}}{d e^{3} + e^{4} x} - \frac {2 c d^{2}}{d e^{3} + e^{4} x} - \frac {2 c d e x \log {\left (\frac {d}{e} + x \right )}}{d e^{3} + e^{4} x} + \frac {c e^{2} x^{2}}{d e^{3} + e^{4} x} & \text {for}\: m = -2 \\- \frac {b d \log {\left (\frac {d}{e} + x \right )}}{e^{2}} + \frac {b x}{e} + \frac {c d^{2} \log {\left (\frac {d}{e} + x \right )}}{e^{3}} - \frac {c d x}{e^{2}} + \frac {c x^{2}}{2 e} & \text {for}\: m = -1 \\- \frac {b d^{2} e m \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} - \frac {3 b d^{2} e \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {b d e^{2} m^{2} x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {3 b d e^{2} m x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {b e^{3} m^{2} x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {4 b e^{3} m x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {3 b e^{3} x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {2 c d^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} - \frac {2 c d^{2} e m x \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {c d e^{2} m^{2} x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {c d e^{2} m x^{2} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {c e^{3} m^{2} x^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {3 c e^{3} m x^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} + \frac {2 c e^{3} x^{3} \left (d + e x\right )^{m}}{e^{3} m^{3} + 6 e^{3} m^{2} + 11 e^{3} m + 6 e^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((d**m*(b*x**2/2 + c*x**3/3), Eq(e, 0)), (-b*d*e/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) - 2*b*e**2*
x/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 2*c*d**2*log(d/e + x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) +
3*c*d**2/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 4*c*d*e*x*log(d/e + x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*
x**2) + 4*c*d*e*x/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2) + 2*c*e**2*x**2*log(d/e + x)/(2*d**2*e**3 + 4*d*e**
4*x + 2*e**5*x**2), Eq(m, -3)), (b*d*e*log(d/e + x)/(d*e**3 + e**4*x) + b*d*e/(d*e**3 + e**4*x) + b*e**2*x*log
(d/e + x)/(d*e**3 + e**4*x) - 2*c*d**2*log(d/e + x)/(d*e**3 + e**4*x) - 2*c*d**2/(d*e**3 + e**4*x) - 2*c*d*e*x
*log(d/e + x)/(d*e**3 + e**4*x) + c*e**2*x**2/(d*e**3 + e**4*x), Eq(m, -2)), (-b*d*log(d/e + x)/e**2 + b*x/e +
c*d**2*log(d/e + x)/e**3 - c*d*x/e**2 + c*x**2/(2*e), Eq(m, -1)), (-b*d**2*e*m*(d + e*x)**m/(e**3*m**3 + 6*e*
*3*m**2 + 11*e**3*m + 6*e**3) - 3*b*d**2*e*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + b*d*e
**2*m**2*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 3*b*d*e**2*m*x*(d + e*x)**m/(e**3*m**
3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + b*e**3*m**2*x**2*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6
*e**3) + 4*b*e**3*m*x**2*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 3*b*e**3*x**2*(d + e*x)
**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 2*c*d**3*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*
m + 6*e**3) - 2*c*d**2*e*m*x*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + c*d*e**2*m**2*x**2*
(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + c*d*e**2*m*x**2*(d + e*x)**m/(e**3*m**3 + 6*e**3
*m**2 + 11*e**3*m + 6*e**3) + c*e**3*m**2*x**3*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 3
*c*e**3*m*x**3*(d + e*x)**m/(e**3*m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3) + 2*c*e**3*x**3*(d + e*x)**m/(e**3*
m**3 + 6*e**3*m**2 + 11*e**3*m + 6*e**3), True))

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