3.15.75 \(\int \frac {(a c+(b c+a d) x+b d x^2)^3}{(a+b x)^3} \, dx\)

Optimal. Leaf size=14 \[ \frac {(c+d x)^4}{4 d} \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {626, 32} \begin {gather*} \frac {(c+d x)^4}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^3,x]

[Out]

(c + d*x)^4/(4*d)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {(c+d x)^4}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^3,x]

[Out]

(c + d*x)^4/(4*d)

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IntegrateAlgebraic [A]  time = 0.06, size = 14, normalized size = 1.00 \begin {gather*} \frac {(c+d x)^4}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*c + (b*c + a*d)*x + b*d*x^2)^3/(a + b*x)^3,x]

[Out]

(c + d*x)^4/(4*d)

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fricas [B]  time = 0.38, size = 31, normalized size = 2.21 \begin {gather*} \frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + c^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^3,x, algorithm="fricas")

[Out]

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + c^3*x

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giac [B]  time = 0.17, size = 31, normalized size = 2.21 \begin {gather*} \frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + c^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^3,x, algorithm="giac")

[Out]

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + c^3*x

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maple [A]  time = 0.04, size = 13, normalized size = 0.93 \begin {gather*} \frac {\left (d x +c \right )^{4}}{4 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^3,x)

[Out]

1/4*(d*x+c)^4/d

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maxima [B]  time = 1.05, size = 31, normalized size = 2.21 \begin {gather*} \frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + c^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x^2)^3/(b*x+a)^3,x, algorithm="maxima")

[Out]

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + c^3*x

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mupad [B]  time = 0.05, size = 31, normalized size = 2.21 \begin {gather*} c^3\,x+\frac {3\,c^2\,d\,x^2}{2}+c\,d^2\,x^3+\frac {d^3\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c + x*(a*d + b*c) + b*d*x^2)^3/(a + b*x)^3,x)

[Out]

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

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sympy [B]  time = 0.16, size = 32, normalized size = 2.29 \begin {gather*} c^{3} x + \frac {3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac {d^{3} x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*c+(a*d+b*c)*x+b*d*x**2)**3/(b*x+a)**3,x)

[Out]

c**3*x + 3*c**2*d*x**2/2 + c*d**2*x**3 + d**3*x**4/4

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