3.2167 \(\int (a c+b c x)^m (f+g x) \left (a^2+2 a b x+b^2 x^2\right )^p \, dx\)

Optimal. Leaf size=100 \[ \frac{g \left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{m+2}}{b^2 c^2 (m+2 p+2)}+\frac{(b f-a g) \left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{m+1}}{b^2 c (m+2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.175202, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088 \[ \frac{g \left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{m+2}}{b^2 c^2 (m+2 p+2)}+\frac{(b f-a g) \left (a^2+2 a b x+b^2 x^2\right )^p (a c+b c x)^{m+1}}{b^2 c (m+2 p+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 43.9654, size = 94, normalized size = 0.94 \[ - \frac{\left (a c + b c x\right )^{m + 1} \left (a g - b f\right ) \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{b^{2} c \left (m + 2 p + 1\right )} + \frac{g \left (a c + b c x\right )^{m + 2} \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{p}}{b^{2} c^{2} \left (m + 2 p + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*c*x+a*c)**m*(g*x+f)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

-(a*c + b*c*x)**(m + 1)*(a*g - b*f)*(a**2 + 2*a*b*x + b**2*x**2)**p/(b**2*c*(m +
2*p + 1)) + g*(a*c + b*c*x)**(m + 2)*(a**2 + 2*a*b*x + b**2*x**2)**p/(b**2*c**2
*(m + 2*p + 2))

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Mathematica [A]  time = 0.0813778, size = 67, normalized size = 0.67 \[ \frac{(a+b x) \left ((a+b x)^2\right )^p (c (a+b x))^m (-a g+b f (m+2 p+2)+b g x (m+2 p+1))}{b^2 (m+2 p+1) (m+2 p+2)} \]

Antiderivative was successfully verified.

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

[Out]

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

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Maple [A]  time = 0.008, size = 96, normalized size = 1. \[ -{\frac{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p} \left ( bxc+ac \right ) ^{m} \left ( -bgmx-2\,bgpx-bfm-2\,bfp-bgx+ag-2\,bf \right ) \left ( bx+a \right ) }{{b}^{2} \left ({m}^{2}+4\,mp+4\,{p}^{2}+3\,m+6\,p+2 \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*c*x+a*c)^m*(g*x+f)*(b^2*x^2+2*a*b*x+a^2)^p,x)

[Out]

-(b^2*x^2+2*a*b*x+a^2)^p*(b*c*x+a*c)^m*(-b*g*m*x-2*b*g*p*x-b*f*m-2*b*f*p-b*g*x+a
*g-2*b*f)*(b*x+a)/b^2/(m^2+4*m*p+4*p^2+3*m+6*p+2)

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Maxima [A]  time = 0.727056, size = 173, normalized size = 1.73 \[ \frac{{\left (b c^{m} x + a c^{m}\right )} f e^{\left (m \log \left (b x + a\right ) + 2 \, p \log \left (b x + a\right )\right )}}{b{\left (m + 2 \, p + 1\right )}} + \frac{{\left (b^{2} c^{m}{\left (m + 2 \, p + 1\right )} x^{2} + a b c^{m}{\left (m + 2 \, p\right )} x - a^{2} c^{m}\right )} g e^{\left (m \log \left (b x + a\right ) + 2 \, p \log \left (b x + a\right )\right )}}{{\left (m^{2} + m{\left (4 \, p + 3\right )} + 4 \, p^{2} + 6 \, p + 2\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((g*x + f)*(b^2*x^2 + 2*a*b*x + a^2)^p*(b*c*x + a*c)^m,x, algorithm="maxima")

[Out]

(b*c^m*x + a*c^m)*f*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)*g*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^2)

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Fricas [A]  time = 0.306593, size = 209, normalized size = 2.09 \[ \frac{{\left (a b f m + 2 \, a b f p + 2 \, a b f - a^{2} g +{\left (b^{2} g m + 2 \, b^{2} g p + b^{2} g\right )} x^{2} +{\left (2 \, b^{2} f +{\left (b^{2} f + a b g\right )} m + 2 \,{\left (b^{2} f + a b g\right )} p\right )} x\right )}{\left (b c x + a c\right )}^{m} e^{\left (2 \, p \log \left (b c x + a c\right ) + p \log \left (\frac{1}{c^{2}}\right )\right )}}{b^{2} m^{2} + 4 \, b^{2} p^{2} + 3 \, b^{2} m + 2 \, b^{2} + 2 \,{\left (2 \, b^{2} m + 3 \, b^{2}\right )} p} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((g*x + f)*(b^2*x^2 + 2*a*b*x + a^2)^p*(b*c*x + a*c)^m,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*c*x+a*c)**m*(g*x+f)*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Timed out

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GIAC/XCAS [A]  time = 0.290422, size = 545, normalized size = 5.45 \[ \frac{b^{2} g m x^{2} e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + 2 \, b^{2} g p x^{2} e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + b^{2} f m x e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + a b g m x e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + 2 \, b^{2} f p x e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + 2 \, a b g p x e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + b^{2} g x^{2} e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + a b f m e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + 2 \, a b f p e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + 2 \, b^{2} f x e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} + 2 \, a b f e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )} - a^{2} g e^{\left (m{\rm ln}\left (b x + a\right ) + 2 \, p{\rm ln}\left (b x + a\right ) + m{\rm ln}\left (c\right )\right )}}{b^{2} m^{2} + 4 \, b^{2} m p + 4 \, b^{2} p^{2} + 3 \, b^{2} m + 6 \, b^{2} p + 2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((g*x + f)*(b^2*x^2 + 2*a*b*x + a^2)^p*(b*c*x + a*c)^m,x, algorithm="giac")

[Out]

(b^2*g*m*x^2*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + 2*b^2*g*p*x^2*e^(m*
ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + b^2*f*m*x*e^(m*ln(b*x + a) + 2*p*ln(b
*x + a) + m*ln(c)) + a*b*g*m*x*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + 2
*b^2*f*p*x*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + 2*a*b*g*p*x*e^(m*ln(b
*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + b^2*g*x^2*e^(m*ln(b*x + a) + 2*p*ln(b*x +
a) + m*ln(c)) + a*b*f*m*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + 2*a*b*f
*p*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)) + 2*b^2*f*x*e^(m*ln(b*x + a) +
2*p*ln(b*x + a) + m*ln(c)) + 2*a*b*f*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c
)) - a^2*g*e^(m*ln(b*x + a) + 2*p*ln(b*x + a) + m*ln(c)))/(b^2*m^2 + 4*b^2*m*p +
4*b^2*p^2 + 3*b^2*m + 6*b^2*p + 2*b^2)