3.26.51 \(\int (A+B x) (d+e x)^m \, dx\)

Optimal. Leaf size=47 \[ \frac {B (d+e x)^{m+2}}{e^2 (m+2)}-\frac {(B d-A e) (d+e x)^{m+1}}{e^2 (m+1)} \]

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {43} \begin {gather*} \frac {B (d+e x)^{m+2}}{e^2 (m+2)}-\frac {(B d-A e) (d+e x)^{m+1}}{e^2 (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(d + e*x)^m,x]

[Out]

-(((B*d - A*e)*(d + e*x)^(1 + m))/(e^2*(1 + m))) + (B*(d + e*x)^(2 + m))/(e^2*(2 + 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])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.87 \begin {gather*} \frac {(d+e x)^{m+1} (A e (m+2)-B d+B e (m+1) x)}{e^2 (m+1) (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(d + e*x)^m,x]

[Out]

((d + e*x)^(1 + m)*(-(B*d) + A*e*(2 + m) + B*e*(1 + m)*x))/(e^2*(1 + m)*(2 + m))

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IntegrateAlgebraic [F]  time = 0.04, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) (d+e x)^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)*(d + e*x)^m,x]

[Out]

Defer[IntegrateAlgebraic][(A + B*x)*(d + e*x)^m, x]

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fricas [A]  time = 1.08, size = 83, normalized size = 1.77 \begin {gather*} \frac {{\left (A d e m - B d^{2} + 2 \, A d e + {\left (B e^{2} m + B e^{2}\right )} x^{2} + {\left (2 \, A e^{2} + {\left (B d e + A e^{2}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{2} m^{2} + 3 \, e^{2} m + 2 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^m,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.03, size = 136, normalized size = 2.89 \begin {gather*} \frac {{\left (x e + d\right )}^{m} B m x^{2} e^{2} + {\left (x e + d\right )}^{m} B d m x e + {\left (x e + d\right )}^{m} A m x e^{2} + {\left (x e + d\right )}^{m} B x^{2} e^{2} + {\left (x e + d\right )}^{m} A d m e - {\left (x e + d\right )}^{m} B d^{2} + 2 \, {\left (x e + d\right )}^{m} A x e^{2} + 2 \, {\left (x e + d\right )}^{m} A d e}{m^{2} e^{2} + 3 \, m e^{2} + 2 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^m,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 46, normalized size = 0.98 \begin {gather*} \frac {\left (B e m x +A e m +B e x +2 A e -B d \right ) \left (e x +d \right )^{m +1}}{\left (m^{2}+3 m +2\right ) e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^m,x)

[Out]

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

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maxima [A]  time = 0.47, size = 63, normalized size = 1.34 \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 (e x + d\right )}^{m + 1} A}{e {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^m,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) + (e*x + d)^(m + 1)*A/(e*(m + 1))

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mupad [B]  time = 2.60, size = 88, normalized size = 1.87 \begin {gather*} {\left (d+e\,x\right )}^m\,\left (\frac {x\,\left (2\,A\,e^2+A\,e^2\,m+B\,d\,e\,m\right )}{e^2\,\left (m^2+3\,m+2\right )}+\frac {B\,x^2\,\left (m+1\right )}{m^2+3\,m+2}+\frac {d\,\left (2\,A\,e-B\,d+A\,e\,m\right )}{e^2\,\left (m^2+3\,m+2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)*(d + e*x)^m,x)

[Out]

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

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sympy [A]  time = 0.87, size = 377, normalized size = 8.02 \begin {gather*} \begin {cases} d^{m} \left (A x + \frac {B x^{2}}{2}\right ) & \text {for}\: e = 0 \\- \frac {A e}{d e^{2} + e^{3} x} + \frac {B d \log {\left (\frac {d}{e} + x \right )}}{d e^{2} + e^{3} x} + \frac {B d}{d e^{2} + e^{3} x} + \frac {B e x \log {\left (\frac {d}{e} + x \right )}}{d e^{2} + e^{3} x} & \text {for}\: m = -2 \\\frac {A \log {\left (\frac {d}{e} + x \right )}}{e} - \frac {B d \log {\left (\frac {d}{e} + x \right )}}{e^{2}} + \frac {B x}{e} & \text {for}\: m = -1 \\\frac {A d e m \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} + \frac {2 A d e \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} + \frac {A e^{2} m x \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} + \frac {2 A e^{2} x \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} - \frac {B d^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} + \frac {B d e m x \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} + \frac {B e^{2} m x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} + \frac {B e^{2} x^{2} \left (d + e x\right )^{m}}{e^{2} m^{2} + 3 e^{2} m + 2 e^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**m,x)

[Out]

Piecewise((d**m*(A*x + B*x**2/2), Eq(e, 0)), (-A*e/(d*e**2 + e**3*x) + B*d*log(d/e + x)/(d*e**2 + e**3*x) + B*
d/(d*e**2 + e**3*x) + B*e*x*log(d/e + x)/(d*e**2 + e**3*x), Eq(m, -2)), (A*log(d/e + x)/e - B*d*log(d/e + x)/e
**2 + B*x/e, Eq(m, -1)), (A*d*e*m*(d + e*x)**m/(e**2*m**2 + 3*e**2*m + 2*e**2) + 2*A*d*e*(d + e*x)**m/(e**2*m*
*2 + 3*e**2*m + 2*e**2) + A*e**2*m*x*(d + e*x)**m/(e**2*m**2 + 3*e**2*m + 2*e**2) + 2*A*e**2*x*(d + e*x)**m/(e
**2*m**2 + 3*e**2*m + 2*e**2) - B*d**2*(d + e*x)**m/(e**2*m**2 + 3*e**2*m + 2*e**2) + B*d*e*m*x*(d + e*x)**m/(
e**2*m**2 + 3*e**2*m + 2*e**2) + B*e**2*m*x**2*(d + e*x)**m/(e**2*m**2 + 3*e**2*m + 2*e**2) + B*e**2*x**2*(d +
e*x)**m/(e**2*m**2 + 3*e**2*m + 2*e**2), True))

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