\(\int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx\) [444]

Optimal result

Integrand size = 25, antiderivative size = 26 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b x}{d}-\frac {(b c-a d) \log (c+d x)}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1607, 1598, 45} \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b x}{d}-\frac {(b c-a d) \log (c+d x)}{d^2} \]

[In]

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

[Out]

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

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 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b x}{d}+\frac {(-b c+a d) \log (c+d x)}{d^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {x b}{d}+\frac {\left (d a -b c \right ) \ln \left (d x +c \right )}{d^{2}}\)	\(26\)
norman	\(\frac {x b}{d}+\frac {\left (d a -b c \right ) \ln \left (d x +c \right )}{d^{2}}\)	\(26\)
parallelrisch	\(\frac {\ln \left (d x +c \right ) a d -\ln \left (d x +c \right ) b c +b d x}{d^{2}}\)	\(29\)
risch	\(\frac {x b}{d}+\frac {\ln \left (d x +c \right ) a}{d}-\frac {\ln \left (d x +c \right ) b c}{d^{2}}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b d x - {\left (b c - a d\right )} \log \left (d x + c\right )}{d^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b x}{d} + \frac {\left (a d - b c\right ) \log {\left (c + d x \right )}}{d^{2}} \]

[In]

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

[Out]

b*x/d + (a*d - b*c)*log(c + d*x)/d**2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b x}{d} - \frac {{\left (b c - a d\right )} \log \left (d x + c\right )}{d^{2}} \]

[In]

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

[Out]

b*x/d - (b*c - a*d)*log(d*x + c)/d^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {b x}{d} - \frac {{\left (b c - a d\right )} \log \left ({\left | d x + c \right |}\right )}{d^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {a x^2+b x^3}{c x^2+d x^3} \, dx=\frac {\ln \left (c+d\,x\right )\,\left (a\,d-b\,c\right )}{d^2}+\frac {b\,x}{d} \]

[In]

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

[Out]

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