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

Optimal. Leaf size=38 \[ \frac {(a+b x)^4 (b c-a d)}{4 b^2}+\frac {d (a+b x)^5}{5 b^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 38, 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 {(a+b x)^4 (b c-a d)}{4 b^2}+\frac {d (a+b x)^5}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 67, normalized size = 1.76 \begin {gather*} a^3 c x+\frac {1}{2} a^2 x^2 (a d+3 b c)+\frac {1}{4} b^2 x^4 (3 a d+b c)+a b x^3 (a d+b c)+\frac {1}{5} b^3 d x^5 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.34, size = 72, normalized size = 1.89 \begin {gather*} \frac {1}{5} x^{5} d b^{3} + \frac {1}{4} x^{4} c b^{3} + \frac {3}{4} x^{4} d b^{2} a + x^{3} c b^{2} a + x^{3} d b a^{2} + \frac {3}{2} x^{2} c b a^{2} + \frac {1}{2} x^{2} d a^{3} + x c a^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.15, size = 72, normalized size = 1.89 \begin {gather*} \frac {1}{5} \, b^{3} d x^{5} + \frac {1}{4} \, b^{3} c x^{4} + \frac {3}{4} \, a b^{2} d x^{4} + a b^{2} c x^{3} + a^{2} b d x^{3} + \frac {3}{2} \, a^{2} b c x^{2} + \frac {1}{2} \, a^{3} d x^{2} + a^{3} c x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 94, normalized size = 2.47 \begin {gather*} \frac {b^{3} d \,x^{5}}{5}+a^{3} c x +\frac {\left (2 a \,b^{2} d +\left (a d +b c \right ) b^{2}\right ) x^{4}}{4}+\frac {\left (a^{2} b d +a \,b^{2} c +2 \left (a d +b c \right ) a b \right ) x^{3}}{3}+\frac {\left (2 a^{2} b c +\left (a d +b c \right ) a^{2}\right ) x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.06, size = 69, normalized size = 1.82 \begin {gather*} \frac {1}{5} \, b^{3} d x^{5} + a^{3} c x + \frac {1}{4} \, {\left (b^{3} c + 3 \, a b^{2} d\right )} x^{4} + {\left (a b^{2} c + a^{2} b d\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b c + a^{3} d\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.55, size = 65, normalized size = 1.71 \begin {gather*} x^4\,\left (\frac {c\,b^3}{4}+\frac {3\,a\,d\,b^2}{4}\right )+x^2\,\left (\frac {d\,a^3}{2}+\frac {3\,b\,c\,a^2}{2}\right )+\frac {b^3\,d\,x^5}{5}+a^3\,c\,x+a\,b\,x^3\,\left (a\,d+b\,c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.09, size = 73, normalized size = 1.92 \begin {gather*} a^{3} c x + \frac {b^{3} d x^{5}}{5} + x^{4} \left (\frac {3 a b^{2} d}{4} + \frac {b^{3} c}{4}\right ) + x^{3} \left (a^{2} b d + a b^{2} c\right ) + x^{2} \left (\frac {a^{3} d}{2} + \frac {3 a^{2} b c}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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