3.18.75 \(\int \frac {(a c+(b c+a d) x+b d x^2)^2}{(a+b x)^3} \, dx\) [1775]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {640, 45} \begin {gather*} \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} \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 45

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 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^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 = 0.01, size = 43, normalized size = 0.88 \begin {gather*} \frac {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} \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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Maple [A]
time = 0.67, size = 56, normalized size = 1.14

method	result	size
default	\(-\frac {d \left (-\frac {1}{2} b d \,x^{2}+a d x -2 b c x \right )}{b^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{3}}\)	\(56\)
risch	\(\frac {d^{2} x^{2}}{2 b}-\frac {d^{2} a x}{b^{2}}+\frac {2 d c x}{b}+\frac {\ln \left (b x +a \right ) a^{2} d^{2}}{b^{3}}-\frac {2 \ln \left (b x +a \right ) a c d}{b^{2}}+\frac {\ln \left (b x +a \right ) c^{2}}{b}\)	\(74\)
norman	\(\frac {\frac {a^{2} \left (3 a^{2} d^{2}-8 a b c d \right )}{2 b^{3}}+\frac {b \,d^{2} x^{4}}{2}+\frac {2 a \left (a^{2} d^{2}-3 a b c d \right ) x}{b^{2}}+2 b c d \,x^{3}}{\left (b x +a \right )^{2}}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b x +a \right )}{b^{3}}\)	\(103\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.32, 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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Fricas [A]
time = 3.81, 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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Sympy [A]
time = 0.13, 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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Giac [A]
time = 0.54, 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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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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