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

Optimal. Leaf size=49 \[ \frac {(b c-a d)^2 \log (a+b x)}{b^3}+\frac {d x (b c-a d)}{b^2}+\frac {(c+d x)^2}{2 b} \]

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {626, 43} \begin {gather*} \frac {d x (b c-a d)}{b^2}+\frac {(b c-a d)^2 \log (a+b x)}{b^3}+\frac {(c+d x)^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*c - a*d)*x)/b^2 + (c + d*x)^2/(2*b) + ((b*c - a*d)^2*Log[a + b*x])/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 \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^3} \, dx &=\int \frac {(c+d x)^2}{a+b x} \, dx\\ &=\int \left (\frac {d (b c-a d)}{b^2}+\frac {(b c-a d)^2}{b^2 (a+b x)}+\frac {d (c+d x)}{b}\right ) \, dx\\ &=\frac {d (b c-a d) x}{b^2}+\frac {(c+d x)^2}{2 b}+\frac {(b c-a d)^2 \log (a+b x)}{b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.88 \begin {gather*} \frac {b d x (-2 a d+4 b c+b d x)+2 (b c-a d)^2 \log (a+b x)}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*x*(4*b*c - 2*a*d + b*d*x) + 2*(b*c - a*d)^2*Log[a + b*x])/(2*b^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a c+(b c+a d) x+b d x^2\right )^2}{(a+b x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 63, normalized size = 1.29 \begin {gather*} \frac {b^{2} d^{2} x^{2} + 2 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} x + 2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x + a\right )}{2 \, 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)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 60, normalized size = 1.22 \begin {gather*} \frac {b d^{2} x^{2} + 4 \, b c d x - 2 \, a d^{2} x}{2 \, b^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{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)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 74, normalized size = 1.51 \begin {gather*} \frac {d^{2} x^{2}}{2 b}+\frac {a^{2} d^{2} \ln \left (b x +a \right )}{b^{3}}-\frac {2 a c d \ln \left (b x +a \right )}{b^{2}}-\frac {a \,d^{2} x}{b^{2}}+\frac {c^{2} \ln \left (b x +a \right )}{b}+\frac {2 c d x}{b} \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)^3,x)

[Out]

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

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maxima [A]  time = 1.12, size = 61, normalized size = 1.24 \begin {gather*} \frac {b d^{2} x^{2} + 2 \, {\left (2 \, b c d - a d^{2}\right )} x}{2 \, b^{2}} + \frac {{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (b x + a\right )}{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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 62, normalized size = 1.27 \begin {gather*} \frac {\ln \left (a+b\,x\right )\,\left (a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2\right )}{b^3}-x\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )+\frac {d^2\,x^2}{2\,b} \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)^3,x)

[Out]

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

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sympy [A]  time = 0.29, size = 44, normalized size = 0.90 \begin {gather*} x \left (- \frac {a d^{2}}{b^{2}} + \frac {2 c d}{b}\right ) + \frac {d^{2} x^{2}}{2 b} + \frac {\left (a d - b c\right )^{2} \log {\left (a + b x \right )}}{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)**3,x)

[Out]

x*(-a*d**2/b**2 + 2*c*d/b) + d**2*x**2/(2*b) + (a*d - b*c)**2*log(a + b*x)/b**3

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