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

Optimal. Leaf size=14 \[ \frac {(c+d x)^3}{3 d} \]

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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)^3}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} \frac {(c+d x)^3}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^3/(3*d)

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IntegrateAlgebraic [A]  time = 0.04, size = 14, normalized size = 1.00 \begin {gather*} \frac {(c+d x)^3}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^3/(3*d)

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fricas [A]  time = 0.39, size = 20, normalized size = 1.43 \begin {gather*} \frac {1}{3} \, d^{2} x^{3} + c d x^{2} + c^{2} 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)^2/(b*x+a)^2,x, algorithm="fricas")

[Out]

1/3*d^2*x^3 + c*d*x^2 + c^2*x

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giac [B]  time = 0.17, size = 84, normalized size = 6.00 \begin {gather*} \frac {{\left (\frac {3 \, b^{2} c^{2}}{{\left (b x + a\right )}^{2}} + \frac {3 \, b c d}{b x + a} - \frac {6 \, a b c d}{{\left (b x + a\right )}^{2}} - \frac {3 \, a d^{2}}{b x + a} + \frac {3 \, a^{2} d^{2}}{{\left (b x + a\right )}^{2}} + d^{2}\right )} {\left (b x + a\right )}^{3}}{3 \, b^{3}} \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)^2/(b*x+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 13, normalized size = 0.93 \begin {gather*} \frac {\left (d x +c \right )^{3}}{3 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)^2/(b*x+a)^2,x)

[Out]

1/3*(d*x+c)^3/d

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maxima [A]  time = 1.09, size = 20, normalized size = 1.43 \begin {gather*} \frac {1}{3} \, d^{2} x^{3} + c d x^{2} + c^{2} 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)^2/(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3*d^2*x^3 + c*d*x^2 + c^2*x

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mupad [B]  time = 0.03, size = 20, normalized size = 1.43 \begin {gather*} c^2\,x+c\,d\,x^2+\frac {d^2\,x^3}{3} \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)^2/(a + b*x)^2,x)

[Out]

c^2*x + (d^2*x^3)/3 + c*d*x^2

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sympy [B]  time = 0.13, size = 19, normalized size = 1.36 \begin {gather*} c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3} \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)**2/(b*x+a)**2,x)

[Out]

c**2*x + c*d*x**2 + d**2*x**3/3

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