3.15.57 \(\int (a+b x) (a c+(b c+a d) x+b d x^2)^2 \, dx\)

Optimal. Leaf size=65 \[ \frac {2 d (a+b x)^5 (b c-a d)}{5 b^3}+\frac {(a+b x)^4 (b c-a d)^2}{4 b^3}+\frac {d^2 (a+b x)^6}{6 b^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {626, 43} \begin {gather*} \frac {2 d (a+b x)^5 (b c-a d)}{5 b^3}+\frac {(a+b x)^4 (b c-a d)^2}{4 b^3}+\frac {d^2 (a+b x)^6}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(a + b*x)^4)/(4*b^3) + (2*d*(b*c - a*d)*(a + b*x)^5)/(5*b^3) + (d^2*(a + b*x)^6)/(6*b^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 626

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

Rubi steps

\begin {align*} \int (a+b x) \left (a c+(b c+a d) x+b d x^2\right )^2 \, dx &=\int (a+b x)^3 (c+d x)^2 \, dx\\ &=\int \left (\frac {(b c-a d)^2 (a+b x)^3}{b^2}+\frac {2 d (b c-a d) (a+b x)^4}{b^2}+\frac {d^2 (a+b x)^5}{b^2}\right ) \, dx\\ &=\frac {(b c-a d)^2 (a+b x)^4}{4 b^3}+\frac {2 d (b c-a d) (a+b x)^5}{5 b^3}+\frac {d^2 (a+b x)^6}{6 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 122, normalized size = 1.88 \begin {gather*} a^3 c^2 x+\frac {1}{4} b x^4 \left (3 a^2 d^2+6 a b c d+b^2 c^2\right )+\frac {1}{3} a x^3 \left (a^2 d^2+6 a b c d+3 b^2 c^2\right )+\frac {1}{2} a^2 c x^2 (2 a d+3 b c)+\frac {1}{5} b^2 d x^5 (3 a d+2 b c)+\frac {1}{6} b^3 d^2 x^6 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*c^2*x + (a^2*c*(3*b*c + 2*a*d)*x^2)/2 + (a*(3*b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^3)/3 + (b*(b^2*c^2 + 6*a*b*
c*d + 3*a^2*d^2)*x^4)/4 + (b^2*d*(2*b*c + 3*a*d)*x^5)/5 + (b^3*d^2*x^6)/6

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.35, size = 130, normalized size = 2.00 \begin {gather*} \frac {1}{6} x^{6} d^{2} b^{3} + \frac {2}{5} x^{5} d c b^{3} + \frac {3}{5} x^{5} d^{2} b^{2} a + \frac {1}{4} x^{4} c^{2} b^{3} + \frac {3}{2} x^{4} d c b^{2} a + \frac {3}{4} x^{4} d^{2} b a^{2} + x^{3} c^{2} b^{2} a + 2 x^{3} d c b a^{2} + \frac {1}{3} x^{3} d^{2} a^{3} + \frac {3}{2} x^{2} c^{2} b a^{2} + x^{2} d c a^{3} + x c^{2} a^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*d^2*b^3 + 2/5*x^5*d*c*b^3 + 3/5*x^5*d^2*b^2*a + 1/4*x^4*c^2*b^3 + 3/2*x^4*d*c*b^2*a + 3/4*x^4*d^2*b*a^
2 + x^3*c^2*b^2*a + 2*x^3*d*c*b*a^2 + 1/3*x^3*d^2*a^3 + 3/2*x^2*c^2*b*a^2 + x^2*d*c*a^3 + x*c^2*a^3

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giac [B]  time = 0.15, size = 130, normalized size = 2.00 \begin {gather*} \frac {1}{6} \, b^{3} d^{2} x^{6} + \frac {2}{5} \, b^{3} c d x^{5} + \frac {3}{5} \, a b^{2} d^{2} x^{5} + \frac {1}{4} \, b^{3} c^{2} x^{4} + \frac {3}{2} \, a b^{2} c d x^{4} + \frac {3}{4} \, a^{2} b d^{2} x^{4} + a b^{2} c^{2} x^{3} + 2 \, a^{2} b c d x^{3} + \frac {1}{3} \, a^{3} d^{2} x^{3} + \frac {3}{2} \, a^{2} b c^{2} x^{2} + a^{3} c d x^{2} + a^{3} c^{2} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*d^2*x^6 + 2/5*b^3*c*d*x^5 + 3/5*a*b^2*d^2*x^5 + 1/4*b^3*c^2*x^4 + 3/2*a*b^2*c*d*x^4 + 3/4*a^2*b*d^2*x^
4 + a*b^2*c^2*x^3 + 2*a^2*b*c*d*x^3 + 1/3*a^3*d^2*x^3 + 3/2*a^2*b*c^2*x^2 + a^3*c*d*x^2 + a^3*c^2*x

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maple [B]  time = 0.04, size = 147, normalized size = 2.26 \begin {gather*} \frac {b^{3} d^{2} x^{6}}{6}+a^{3} c^{2} x +\frac {\left (a \,b^{2} d^{2}+2 \left (a d +b c \right ) b^{2} d \right ) x^{5}}{5}+\frac {\left (2 \left (a d +b c \right ) a b d +\left (2 a b c d +\left (a d +b c \right )^{2}\right ) b \right ) x^{4}}{4}+\frac {\left (2 \left (a d +b c \right ) a b c +\left (2 a b c d +\left (a d +b c \right )^{2}\right ) a \right ) x^{3}}{3}+\frac {\left (a^{2} b \,c^{2}+2 \left (a d +b c \right ) a^{2} c \right ) x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*d^2*x^6+1/5*(a*b^2*d^2+2*b^2*(a*d+b*c)*d)*x^5+1/4*(2*a*(a*d+b*c)*b*d+b*(2*a*b*c*d+(a*d+b*c)^2))*x^4+1/
3*(a*(2*a*b*c*d+(a*d+b*c)^2)+2*b*a*c*(a*d+b*c))*x^3+1/2*(2*a^2*c*(a*d+b*c)+a^2*b*c^2)*x^2+a^3*c^2*x

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maxima [B]  time = 1.05, size = 124, normalized size = 1.91 \begin {gather*} \frac {1}{6} \, b^{3} d^{2} x^{6} + a^{3} c^{2} x + \frac {1}{5} \, {\left (2 \, b^{3} c d + 3 \, a b^{2} d^{2}\right )} x^{5} + \frac {1}{4} \, {\left (b^{3} c^{2} + 6 \, a b^{2} c d + 3 \, a^{2} b d^{2}\right )} x^{4} + \frac {1}{3} \, {\left (3 \, a b^{2} c^{2} + 6 \, a^{2} b c d + a^{3} d^{2}\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*d^2*x^6 + a^3*c^2*x + 1/5*(2*b^3*c*d + 3*a*b^2*d^2)*x^5 + 1/4*(b^3*c^2 + 6*a*b^2*c*d + 3*a^2*b*d^2)*x^
4 + 1/3*(3*a*b^2*c^2 + 6*a^2*b*c*d + a^3*d^2)*x^3 + 1/2*(3*a^2*b*c^2 + 2*a^3*c*d)*x^2

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mupad [B]  time = 0.58, size = 115, normalized size = 1.77 \begin {gather*} x^3\,\left (\frac {a^3\,d^2}{3}+2\,a^2\,b\,c\,d+a\,b^2\,c^2\right )+x^4\,\left (\frac {3\,a^2\,b\,d^2}{4}+\frac {3\,a\,b^2\,c\,d}{2}+\frac {b^3\,c^2}{4}\right )+a^3\,c^2\,x+\frac {b^3\,d^2\,x^6}{6}+\frac {a^2\,c\,x^2\,\left (2\,a\,d+3\,b\,c\right )}{2}+\frac {b^2\,d\,x^5\,\left (3\,a\,d+2\,b\,c\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*((a^3*d^2)/3 + a*b^2*c^2 + 2*a^2*b*c*d) + x^4*((b^3*c^2)/4 + (3*a^2*b*d^2)/4 + (3*a*b^2*c*d)/2) + a^3*c^2*
x + (b^3*d^2*x^6)/6 + (a^2*c*x^2*(2*a*d + 3*b*c))/2 + (b^2*d*x^5*(3*a*d + 2*b*c))/5

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sympy [B]  time = 0.11, size = 133, normalized size = 2.05 \begin {gather*} a^{3} c^{2} x + \frac {b^{3} d^{2} x^{6}}{6} + x^{5} \left (\frac {3 a b^{2} d^{2}}{5} + \frac {2 b^{3} c d}{5}\right ) + x^{4} \left (\frac {3 a^{2} b d^{2}}{4} + \frac {3 a b^{2} c d}{2} + \frac {b^{3} c^{2}}{4}\right ) + x^{3} \left (\frac {a^{3} d^{2}}{3} + 2 a^{2} b c d + a b^{2} c^{2}\right ) + x^{2} \left (a^{3} c d + \frac {3 a^{2} b c^{2}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*c**2*x + b**3*d**2*x**6/6 + x**5*(3*a*b**2*d**2/5 + 2*b**3*c*d/5) + x**4*(3*a**2*b*d**2/4 + 3*a*b**2*c*d/
2 + b**3*c**2/4) + x**3*(a**3*d**2/3 + 2*a**2*b*c*d + a*b**2*c**2) + x**2*(a**3*c*d + 3*a**2*b*c**2/2)

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