\(\int \frac {a+c x^2}{(d+e x)^3} \, dx\) [457]

Optimal result

Integrand size = 15, antiderivative size = 53 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {-c d^2-a e^2}{2 e^3 (d+e x)^2}+\frac {2 c d}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {711} \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=-\frac {a e^2+c d^2}{2 e^3 (d+e x)^2}+\frac {2 c d}{e^3 (d+e x)}+\frac {c \log (d+e x)}{e^3} \]

[In]

Int[(a + c*x^2)/(d + e*x)^3,x]

[Out]

-1/2*(c*d^2 + a*e^2)/(e^3*(d + e*x)^2) + (2*c*d)/(e^3*(d + e*x)) + (c*Log[d + e*x])/e^3

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {-a e^2+c d (3 d+4 e x)+2 c (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2} \]

[In]

Integrate[(a + c*x^2)/(d + e*x)^3,x]

[Out]

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

Maple [A] (verified)

Time = 2.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.89

method	result	size
norman	\(\frac {-\frac {e^{2} a -3 c \,d^{2}}{2 e^{3}}+\frac {2 c d x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(47\)
risch	\(\frac {-\frac {e^{2} a -3 c \,d^{2}}{2 e^{3}}+\frac {2 c d x}{e^{2}}}{\left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(47\)
default	\(\frac {2 c d}{e^{3} \left (e x +d \right )}-\frac {e^{2} a +c \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}+\frac {c \ln \left (e x +d \right )}{e^{3}}\)	\(50\)
parallelrisch	\(\frac {2 \ln \left (e x +d \right ) x^{2} c \,e^{2}+4 \ln \left (e x +d \right ) x c d e +2 \ln \left (e x +d \right ) c \,d^{2}+4 x c d e -e^{2} a +3 c \,d^{2}}{2 e^{3} \left (e x +d \right )^{2}}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {4 \, c d e x + 3 \, c d^{2} - a e^{2} + 2 \, {\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

[In]

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

[Out]

1/2*(4*c*d*e*x + 3*c*d^2 - a*e^2 + 2*(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*log(e*x + d))/(e^5*x^2 + 2*d*e^4*x + d^2*
e^3)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {c \log {\left (d + e x \right )}}{e^{3}} + \frac {- a e^{2} + 3 c d^{2} + 4 c d e x}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]

[In]

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

[Out]

c*log(d + e*x)/e**3 + (-a*e**2 + 3*c*d**2 + 4*c*d*e*x)/(2*d**2*e**3 + 4*d*e**4*x + 2*e**5*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {4 \, c d e x + 3 \, c d^{2} - a e^{2}}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} + \frac {c \log \left (e x + d\right )}{e^{3}} \]

[In]

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

[Out]

1/2*(4*c*d*e*x + 3*c*d^2 - a*e^2)/(e^5*x^2 + 2*d*e^4*x + d^2*e^3) + c*log(e*x + d)/e^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {c \log \left ({\left | e x + d \right |}\right )}{e^{3}} + \frac {4 \, c d x + \frac {3 \, c d^{2} - a e^{2}}{e}}{2 \, {\left (e x + d\right )}^{2} e^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \frac {a+c x^2}{(d+e x)^3} \, dx=\frac {c\,\ln \left (d+e\,x\right )}{e^3}-\frac {\frac {a\,e^2-3\,c\,d^2}{2\,e^3}-\frac {2\,c\,d\,x}{e^2}}{d^2+2\,d\,e\,x+e^2\,x^2} \]

[In]

int((a + c*x^2)/(d + e*x)^3,x)

[Out]

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