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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, 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 = 0.04, 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))

________________________________________________________________________________________

fricas [A]  time = 1.11, size = 106, normalized size = 1.63 \[ \frac {{\left (2 \, a b c m + 4 \, {\left (c^{3} m + c^{3}\right )} x^{3} - b^{3} + 6 \, a b c + 6 \, {\left (b c^{2} m + b c^{2}\right )} x^{2} + 2 \, {\left (6 \, a c^{2} + {\left (b^{2} c + 2 \, a c^{2}\right )} m\right )} x\right )} {\left (2 \, c d x + b d\right )}^{m}}{4 \, {\left (c^{2} m^{2} + 4 \, c^{2} m + 3 \, c^{2}\right )}} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.20, size = 209, normalized size = 3.22 \[ \frac {4 \, {\left (2 \, c d x + b d\right )}^{m} c^{3} m x^{3} + 6 \, {\left (2 \, c d x + b d\right )}^{m} b c^{2} m x^{2} + 4 \, {\left (2 \, c d x + b d\right )}^{m} c^{3} x^{3} + 2 \, {\left (2 \, c d x + b d\right )}^{m} b^{2} c m x + 4 \, {\left (2 \, c d x + b d\right )}^{m} a c^{2} m x + 6 \, {\left (2 \, c d x + b d\right )}^{m} b c^{2} x^{2} + 2 \, {\left (2 \, c d x + b d\right )}^{m} a b c m + 12 \, {\left (2 \, c d x + b d\right )}^{m} a c^{2} x - {\left (2 \, c d x + b d\right )}^{m} b^{3} + 6 \, {\left (2 \, c d x + b d\right )}^{m} a b c}{4 \, {\left (c^{2} m^{2} + 4 \, c^{2} m + 3 \, c^{2}\right )}} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 76, normalized size = 1.17 \[ \frac {\left (2 c^{2} m \,x^{2}+2 b c m x +2 c^{2} x^{2}+2 a c m +2 b c x +6 a c -b^{2}\right ) \left (2 c x +b \right ) \left (2 c d x +b d \right )^{m}}{4 \left (m^{2}+4 m +3\right ) c^{2}} \]

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)

________________________________________________________________________________________

maxima [B]  time = 1.48, size = 167, normalized size = 2.57 \[ \frac {{\left (4 \, c^{2} d^{m} {\left (m + 1\right )} x^{2} + 2 \, b c d^{m} m x - b^{2} d^{m}\right )} {\left (2 \, c x + b\right )}^{m} b}{4 \, {\left (m^{2} + 3 \, m + 2\right )} c^{2}} + \frac {{\left (4 \, {\left (m^{2} + 3 \, m + 2\right )} c^{3} d^{m} x^{3} + 2 \, {\left (m^{2} + m\right )} b c^{2} d^{m} x^{2} - 2 \, b^{2} c d^{m} m x + b^{3} d^{m}\right )} {\left (2 \, c x + b\right )}^{m}}{4 \, {\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} c^{2}} + \frac {{\left (2 \, c d x + b d\right )}^{m + 1} a}{2 \, c d {\left (m + 1\right )}} \]

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]

1/4*(4*c^2*d^m*(m + 1)*x^2 + 2*b*c*d^m*m*x - b^2*d^m)*(2*c*x + b)^m*b/((m^2 + 3*m + 2)*c^2) + 1/4*(4*(m^2 + 3*
m + 2)*c^3*d^m*x^3 + 2*(m^2 + m)*b*c^2*d^m*x^2 - 2*b^2*c*d^m*m*x + b^3*d^m)*(2*c*x + b)^m/((m^3 + 6*m^2 + 11*m
+ 6)*c^2) + 1/2*(2*c*d*x + b*d)^(m + 1)*a/(c*d*(m + 1))

________________________________________________________________________________________

mupad [B]  time = 0.58, size = 120, normalized size = 1.85 \[ {\left (b\,d+2\,c\,d\,x\right )}^m\,\left (\frac {b\,\left (-b^2+6\,a\,c+2\,a\,c\,m\right )}{4\,c^2\,\left (m^2+4\,m+3\right )}+\frac {x\,\left (12\,a\,c^2+4\,a\,c^2\,m+2\,b^2\,c\,m\right )}{4\,c^2\,\left (m^2+4\,m+3\right )}+\frac {3\,b\,x^2\,\left (m+1\right )}{2\,\left (m^2+4\,m+3\right )}+\frac {c\,x^3\,\left (m+1\right )}{m^2+4\,m+3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.33, size = 707, normalized size = 10.88 \[ \begin {cases} \left (b d\right )^{m} \left (a x + \frac {b x^{2}}{2}\right ) & \text {for}\: c = 0 \\- \frac {4 a c}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac {2 b^{2} \log {\left (\frac {b}{2 c} + x \right )}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac {b^{2}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac {8 b c x \log {\left (\frac {b}{2 c} + x \right )}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} + \frac {8 c^{2} x^{2} \log {\left (\frac {b}{2 c} + x \right )}}{16 b^{2} c^{2} d^{3} + 64 b c^{3} d^{3} x + 64 c^{4} d^{3} x^{2}} & \text {for}\: m = -3 \\\frac {a \log {\left (\frac {b}{2 c} + x \right )}}{2 c d} - \frac {b^{2} \log {\left (\frac {b}{2 c} + x \right )}}{8 c^{2} d} + \frac {b x}{4 c d} + \frac {x^{2}}{4 d} & \text {for}\: m = -1 \\\frac {2 a b c m \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {6 a b c \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {4 a c^{2} m x \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {12 a c^{2} x \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} - \frac {b^{3} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {2 b^{2} c m x \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {6 b c^{2} m x^{2} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {6 b c^{2} x^{2} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {4 c^{3} m x^{3} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} + \frac {4 c^{3} x^{3} \left (b d + 2 c d x\right )^{m}}{4 c^{2} m^{2} + 16 c^{2} m + 12 c^{2}} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________