3.16.54 \(\int (a+b x)^m (a+b (2+m) x) \, dx\)

Optimal. Leaf size=11 \[ x (a+b x)^{m+1} \]

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Rubi [A]  time = 0.00, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {34} \begin {gather*} x (a+b x)^{m+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(a + b*(2 + m)*x),x]

[Out]

x*(a + b*x)^(1 + m)

Rule 34

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_)), x_Symbol] :> Simp[(d*x*(a + b*x)^(m + 1))/(b*(m + 2)), x] /
; FreeQ[{a, b, c, d, m}, x] && EqQ[a*d - b*c*(m + 2), 0]

Rubi steps

\begin {align*} \int (a+b x)^m (a+b (2+m) x) \, dx &=x (a+b x)^{1+m}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 1.00 \begin {gather*} x (a+b x)^{m+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(a + b*(2 + m)*x),x]

[Out]

x*(a + b*x)^(1 + m)

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^m*(a + b*(2 + m)*x),x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x)^m*(a + b*(2 + m)*x), x]

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fricas [A]  time = 1.19, size = 17, normalized size = 1.55 \begin {gather*} {\left (b x^{2} + a x\right )} {\left (b x + a\right )}^{m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(a+b*(2+m)*x),x, algorithm="fricas")

[Out]

(b*x^2 + a*x)*(b*x + a)^m

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giac [B]  time = 1.00, size = 23, normalized size = 2.09 \begin {gather*} {\left (b x + a\right )}^{m} b x^{2} + {\left (b x + a\right )}^{m} a x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(a+b*(2+m)*x),x, algorithm="giac")

[Out]

(b*x + a)^m*b*x^2 + (b*x + a)^m*a*x

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maple [A]  time = 0.00, size = 12, normalized size = 1.09 \begin {gather*} x \left (b x +a \right )^{m +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(a+b*(m+2)*x),x)

[Out]

x*(b*x+a)^(m+1)

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maxima [B]  time = 1.12, size = 106, normalized size = 9.64 \begin {gather*} \frac {{\left (b^{2} {\left (m + 1\right )} x^{2} + a b m x - a^{2}\right )} {\left (b x + a\right )}^{m} m}{{\left (m^{2} + 3 \, m + 2\right )} b} + \frac {2 \, {\left (b^{2} {\left (m + 1\right )} x^{2} + a b m x - a^{2}\right )} {\left (b x + a\right )}^{m}}{{\left (m^{2} + 3 \, m + 2\right )} b} + \frac {{\left (b x + a\right )}^{m + 1} a}{b {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(a+b*(2+m)*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.46, size = 11, normalized size = 1.00 \begin {gather*} x\,{\left (a+b\,x\right )}^{m+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x*(m + 2))*(a + b*x)^m,x)

[Out]

x*(a + b*x)^(m + 1)

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sympy [B]  time = 0.27, size = 20, normalized size = 1.82 \begin {gather*} a x \left (a + b x\right )^{m} + b x^{2} \left (a + b x\right )^{m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(a+b*(2+m)*x),x)

[Out]

a*x*(a + b*x)**m + b*x**2*(a + b*x)**m

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