\(\int \frac {(a+b x)^2}{(\frac {a d}{b}+d x)^3} \, dx\) [1002]

Optimal result

Integrand size = 20, antiderivative size = 13 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^2 \log (a+b x)}{d^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 31} \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^2 \log (a+b x)}{d^3} \]

[In]

Int[(a + b*x)^2/((a*d)/b + d*x)^3,x]

[Out]

(b^2*Log[a + b*x])/d^3

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {b^3 \int \frac {1}{a+b x} \, dx}{d^3} \\ & = \frac {b^2 \log (a+b x)}{d^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^2 \log (a+b x)}{d^3} \]

[In]

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

[Out]

(b^2*Log[a + b*x])/d^3

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {b^{2} \ln \left (b x +a \right )}{d^{3}}\)	\(14\)
norman	\(\frac {b^{2} \ln \left (b x +a \right )}{d^{3}}\)	\(14\)
risch	\(\frac {b^{2} \ln \left (b x +a \right )}{d^{3}}\)	\(14\)
parallelrisch	\(\frac {b^{2} \ln \left (b x +a \right )}{d^{3}}\)	\(14\)

[In]

int((b*x+a)^2/(a*d/b+d*x)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^{2} \log \left (b x + a\right )}{d^{3}} \]

[In]

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

[Out]

b^2*log(b*x + a)/d^3

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^{2} \log {\left (a d^{3} + b d^{3} x \right )}}{d^{3}} \]

[In]

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

[Out]

b**2*log(a*d**3 + b*d**3*x)/d**3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^{2} \log \left (b x + a\right )}{d^{3}} \]

[In]

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

[Out]

b^2*log(b*x + a)/d^3

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{d^{3}} \]

[In]

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

[Out]

b^2*log(abs(b*x + a))/d^3

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^2}{\left (\frac {a d}{b}+d x\right )^3} \, dx=\frac {b^2\,\ln \left (a+b\,x\right )}{d^3} \]

[In]

int((a + b*x)^2/(d*x + (a*d)/b)^3,x)

[Out]

(b^2*log(a + b*x))/d^3