3.2169 \(\int (a+b x) (a c+b c x)^m (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=45 \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{m+2}}{b c^2 (m+2 p+2)} \]

[Out]

((a*c + b*c*x)^(2 + m)*(a^2 + 2*a*b*x + b^2*x^2)^p)/(b*c^2*(2 + m + 2*p))

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Rubi [A]  time = 0.0278098, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {21, 644, 32} \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{m+2}}{b c^2 (m+2 p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*(a*c + b*c*x)^m*(a^2 + 2*a*b*x + b^2*x^2)^p,x]

[Out]

((a*c + b*c*x)^(2 + m)*(a^2 + 2*a*b*x + b^2*x^2)^p)/(b*c^2*(2 + m + 2*p))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 644

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.021459, size = 32, normalized size = 0.71 \[ \frac{\left ((a+b x)^2\right )^{p+1} (c (a+b x))^m}{b (m+2 p+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c*(a + b*x))^m*((a + b*x)^2)^(1 + p))/(b*(2 + m + 2*p))

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Maple [A]  time = 0.003, size = 48, normalized size = 1.1 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2} \left ( bcx+ac \right ) ^{m} \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p}}{b \left ( 2+m+2\,p \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)^2/b/(2+m+2*p)*(b*c*x+a*c)^m*(b^2*x^2+2*a*b*x+a^2)^p

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Maxima [B]  time = 1.16281, size = 171, normalized size = 3.8 \begin{align*} \frac{{\left (b c^{m} x + a c^{m}\right )} a e^{\left (m \log \left (b x + a\right ) + 2 \, p \log \left (b x + a\right )\right )}}{b{\left (m + 2 \, p + 1\right )}} + \frac{{\left (b^{2} c^{m}{\left (m + 2 \, p + 1\right )} x^{2} + a b c^{m}{\left (m + 2 \, p\right )} x - a^{2} c^{m}\right )} e^{\left (m \log \left (b x + a\right ) + 2 \, p \log \left (b x + a\right )\right )}}{{\left (m^{2} + m{\left (4 \, p + 3\right )} + 4 \, p^{2} + 6 \, p + 2\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c^m*x + a*c^m)*a*e^(m*log(b*x + a) + 2*p*log(b*x + a))/(b*(m + 2*p + 1)) + (b^2*c^m*(m + 2*p + 1)*x^2 + a*b
*c^m*(m + 2*p)*x - a^2*c^m)*e^(m*log(b*x + a) + 2*p*log(b*x + a))/((m^2 + m*(4*p + 3) + 4*p^2 + 6*p + 2)*b)

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Fricas [A]  time = 1.57082, size = 140, normalized size = 3.11 \begin{align*} \frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}{\left (b c x + a c\right )}^{m} e^{\left (2 \, p \log \left (b c x + a c\right ) + p \log \left (\frac{1}{c^{2}}\right )\right )}}{b m + 2 \, b p + 2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*x^2 + 2*a*b*x + a^2)*(b*c*x + a*c)^m*e^(2*p*log(b*c*x + a*c) + p*log(c^(-2)))/(b*m + 2*b*p + 2*b)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.19921, size = 135, normalized size = 3. \begin{align*} \frac{{\left (b x + a\right )}^{2 \, p} b^{2} x^{2} e^{\left (m \log \left (b x + a\right ) + m \log \left (c\right )\right )} + 2 \,{\left (b x + a\right )}^{2 \, p} a b x e^{\left (m \log \left (b x + a\right ) + m \log \left (c\right )\right )} +{\left (b x + a\right )}^{2 \, p} a^{2} e^{\left (m \log \left (b x + a\right ) + m \log \left (c\right )\right )}}{b m + 2 \, b p + 2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x + a)^(2*p)*b^2*x^2*e^(m*log(b*x + a) + m*log(c)) + 2*(b*x + a)^(2*p)*a*b*x*e^(m*log(b*x + a) + m*log(c))
+ (b*x + a)^(2*p)*a^2*e^(m*log(b*x + a) + m*log(c)))/(b*m + 2*b*p + 2*b)