### 3.1425 $$\int (b d+2 c d x)^m (a+b x+c x^2) \, dx$$

Optimal. Leaf size=65 $\frac{(b d+2 c d x)^{m+3}}{8 c^2 d^3 (m+3)}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{m+1}}{8 c^2 d (m+1)}$

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(1 + m))/(8*c^2*d*(1 + m)) + (b*d + 2*c*d*x)^(3 + m)/(8*c^2*d^3*(3 + m))

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Rubi [A]  time = 0.0284689, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.045, Rules used = {683} $\frac{(b d+2 c d x)^{m+3}}{8 c^2 d^3 (m+3)}-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{m+1}}{8 c^2 d (m+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((b^2 - 4*a*c)*(b*d + 2*c*d*x)^(1 + m))/(8*c^2*d*(1 + m)) + (b*d + 2*c*d*x)^(3 + m)/(8*c^2*d^3*(3 + m))

Rule 683

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] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

\begin{align*} \int (b d+2 c d x)^m \left (a+b x+c x^2\right ) \, dx &=\int \left (\frac{\left (-b^2+4 a c\right ) (b d+2 c d x)^m}{4 c}+\frac{(b d+2 c d x)^{2+m}}{4 c d^2}\right ) \, dx\\ &=-\frac{\left (b^2-4 a c\right ) (b d+2 c d x)^{1+m}}{8 c^2 d (1+m)}+\frac{(b d+2 c d x)^{3+m}}{8 c^2 d^3 (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.0334382, size = 64, normalized size = 0.98 $\frac{(b+2 c x) \left (2 c \left (a (m+3)+c (m+1) x^2\right )-b^2+2 b c (m+1) x\right ) (d (b+2 c x))^m}{4 c^2 (m+1) (m+3)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 76, normalized size = 1.2 \begin{align*}{\frac{ \left ( 2\,cdx+bd \right ) ^{m} \left ( 2\,{c}^{2}m{x}^{2}+2\,bcmx+2\,{c}^{2}{x}^{2}+2\,acm+2\,bcx+6\,ac-{b}^{2} \right ) \left ( 2\,cx+b \right ) }{4\,{c}^{2} \left ({m}^{2}+4\,m+3 \right ) }} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.22908, size = 223, normalized size = 3.43 \begin{align*} \frac{{\left (2 \, a b c m + 4 \,{\left (c^{3} m + c^{3}\right )} x^{3} - b^{3} + 6 \, a b c + 6 \,{\left (b c^{2} m + b c^{2}\right )} x^{2} + 2 \,{\left (6 \, a c^{2} +{\left (b^{2} c + 2 \, a c^{2}\right )} m\right )} x\right )}{\left (2 \, c d x + b d\right )}^{m}}{4 \,{\left (c^{2} m^{2} + 4 \, c^{2} m + 3 \, c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.5383, size = 707, normalized size = 10.88 \begin{align*} \begin{cases} \left (b d\right )^{m} \left (a x + \frac{b x^{2}}{2}\right ) & \text{for}\: c = 0 \\- \frac{4 a c}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac{2 b^{2} \log{\left (\frac{b}{2 c} + x \right )}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac{b^{2}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac{8 b c x \log{\left (\frac{b}{2 c} + x \right )}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac{8 c^{2} x^{2} \log{\left (\frac{b}{2 c} + x \right )}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} & \text{for}\: m = -3 \\\frac{a \log{\left (\frac{b}{2 c} + x \right )}}{2 c d} - \frac{b^{2} \log{\left (\frac{b}{2 c} + x \right )}}{8 c^{2} d} + \frac{b x}{4 c d} + \frac{x^{2}}{4 d} & \text{for}\: m = -1 \\\frac{2 a b c m \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{6 a b c \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{4 a c^{2} m x \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{12 a c^{2} x \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} - \frac{b^{3} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{2 b^{2} c m x \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{6 b c^{2} m x^{2} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{6 b c^{2} x^{2} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{4 c^{3} m x^{3} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac{4 c^{3} x^{3} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((b*d)**m*(a*x + b*x**2/2), Eq(c, 0)), (-4*a*c/(16*b**2*c**2*d**3 + 64*b*c**3*d**3*x + 64*c**4*d**3*
x**2) + 2*b**2*log(b/(2*c) + x)/(16*b**2*c**2*d**3 + 64*b*c**3*d**3*x + 64*c**4*d**3*x**2) + b**2/(16*b**2*c**
2*d**3 + 64*b*c**3*d**3*x + 64*c**4*d**3*x**2) + 8*b*c*x*log(b/(2*c) + x)/(16*b**2*c**2*d**3 + 64*b*c**3*d**3*
x + 64*c**4*d**3*x**2) + 8*c**2*x**2*log(b/(2*c) + x)/(16*b**2*c**2*d**3 + 64*b*c**3*d**3*x + 64*c**4*d**3*x**
2), Eq(m, -3)), (a*log(b/(2*c) + x)/(2*c*d) - b**2*log(b/(2*c) + x)/(8*c**2*d) + b*x/(4*c*d) + x**2/(4*d), Eq(
m, -1)), (2*a*b*c*m*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c**2) + 6*a*b*c*(b*d + 2*c*d*x)**m/(4*c**
2*m**2 + 16*c**2*m + 12*c**2) + 4*a*c**2*m*x*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c**2) + 12*a*c**
2*x*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c**2) - b**3*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m
+ 12*c**2) + 2*b**2*c*m*x*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c**2) + 6*b*c**2*m*x**2*(b*d + 2*c*
d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c**2) + 6*b*c**2*x**2*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c
**2) + 4*c**3*m*x**3*(b*d + 2*c*d*x)**m/(4*c**2*m**2 + 16*c**2*m + 12*c**2) + 4*c**3*x**3*(b*d + 2*c*d*x)**m/(
4*c**2*m**2 + 16*c**2*m + 12*c**2), True))

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Giac [B]  time = 1.16209, size = 282, normalized size = 4.34 \begin{align*} \frac{4 \,{\left (2 \, c d x + b d\right )}^{m} c^{3} m x^{3} + 6 \,{\left (2 \, c d x + b d\right )}^{m} b c^{2} m x^{2} + 4 \,{\left (2 \, c d x + b d\right )}^{m} c^{3} x^{3} + 2 \,{\left (2 \, c d x + b d\right )}^{m} b^{2} c m x + 4 \,{\left (2 \, c d x + b d\right )}^{m} a c^{2} m x + 6 \,{\left (2 \, c d x + b d\right )}^{m} b c^{2} x^{2} + 2 \,{\left (2 \, c d x + b d\right )}^{m} a b c m + 12 \,{\left (2 \, c d x + b d\right )}^{m} a c^{2} x -{\left (2 \, c d x + b d\right )}^{m} b^{3} + 6 \,{\left (2 \, c d x + b d\right )}^{m} a b c}{4 \,{\left (c^{2} m^{2} + 4 \, c^{2} m + 3 \, c^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(2*c*d*x + b*d)^m*c^3*m*x^3 + 6*(2*c*d*x + b*d)^m*b*c^2*m*x^2 + 4*(2*c*d*x + b*d)^m*c^3*x^3 + 2*(2*c*d*
x + b*d)^m*b^2*c*m*x + 4*(2*c*d*x + b*d)^m*a*c^2*m*x + 6*(2*c*d*x + b*d)^m*b*c^2*x^2 + 2*(2*c*d*x + b*d)^m*a*b
*c*m + 12*(2*c*d*x + b*d)^m*a*c^2*x - (2*c*d*x + b*d)^m*b^3 + 6*(2*c*d*x + b*d)^m*a*b*c)/(c^2*m^2 + 4*c^2*m +
3*c^2)