\(\int \frac {(d+e x)^2}{(a+c x^2)^2} \, dx\) [508]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 17, antiderivative size = 72 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=-\frac {(a e-c d x) (d+e x)}{2 a c \left (a+c x^2\right )}+\frac {\left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {737, 211} \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a e^2+c d^2\right )}{2 a^{3/2} c^{3/2}}-\frac {(d+e x) (a e-c d x)}{2 a c \left (a+c x^2\right )} \]

[In]

Int[(d + e*x)^2/(a + c*x^2)^2,x]

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
 + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[(2*p + 3)*((c*d^2 + a*e^2)/(2*a*c*(p + 1))), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {(a e-c d x) (d+e x)}{2 a c \left (a+c x^2\right )}+\frac {\left (c d^2+a e^2\right ) \int \frac {1}{a+c x^2} \, dx}{2 a c} \\ & = -\frac {(a e-c d x) (d+e x)}{2 a c \left (a+c x^2\right )}+\frac {\left (c d^2+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=\frac {-2 a d e+c d^2 x-a e^2 x}{2 a c \left (a+c x^2\right )}+\frac {\left (c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.03

method result size
default \(\frac {-\frac {\left (e^{2} a -c \,d^{2}\right ) x}{2 a c}-\frac {d e}{c}}{c \,x^{2}+a}+\frac {\left (e^{2} a +c \,d^{2}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 a c \sqrt {a c}}\) \(74\)
risch \(\frac {-\frac {\left (e^{2} a -c \,d^{2}\right ) x}{2 a c}-\frac {d e}{c}}{c \,x^{2}+a}-\frac {\ln \left (c x +\sqrt {-a c}\right ) e^{2}}{4 \sqrt {-a c}\, c}-\frac {\ln \left (c x +\sqrt {-a c}\right ) d^{2}}{4 \sqrt {-a c}\, a}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) e^{2}}{4 \sqrt {-a c}\, c}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) d^{2}}{4 \sqrt {-a c}\, a}\) \(143\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.10 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=\left [-\frac {4 \, a^{2} c d e + {\left (a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (a c^{2} d^{2} - a^{2} c e^{2}\right )} x}{4 \, {\left (a^{2} c^{3} x^{2} + a^{3} c^{2}\right )}}, -\frac {2 \, a^{2} c d e - {\left (a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (a c^{2} d^{2} - a^{2} c e^{2}\right )} x}{2 \, {\left (a^{2} c^{3} x^{2} + a^{3} c^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/4*(4*a^2*c*d*e + (a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)
/(c*x^2 + a)) - 2*(a*c^2*d^2 - a^2*c*e^2)*x)/(a^2*c^3*x^2 + a^3*c^2), -1/2*(2*a^2*c*d*e - (a*c*d^2 + a^2*e^2 +
 (c^2*d^2 + a*c*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (a*c^2*d^2 - a^2*c*e^2)*x)/(a^2*c^3*x^2 + a^3*c^2)
]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (63) = 126\).

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.79 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=- \frac {\sqrt {- \frac {1}{a^{3} c^{3}}} \left (a e^{2} + c d^{2}\right ) \log {\left (- a^{2} c \sqrt {- \frac {1}{a^{3} c^{3}}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{3} c^{3}}} \left (a e^{2} + c d^{2}\right ) \log {\left (a^{2} c \sqrt {- \frac {1}{a^{3} c^{3}}} + x \right )}}{4} + \frac {- 2 a d e + x \left (- a e^{2} + c d^{2}\right )}{2 a^{2} c + 2 a c^{2} x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=-\frac {2 \, a d e - {\left (c d^{2} - a e^{2}\right )} x}{2 \, {\left (a c^{2} x^{2} + a^{2} c\right )}} + \frac {{\left (c d^{2} + a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c} \]

[In]

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

[Out]

-1/2*(2*a*d*e - (c*d^2 - a*e^2)*x)/(a*c^2*x^2 + a^2*c) + 1/2*(c*d^2 + a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*
a*c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=\frac {{\left (c d^{2} + a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c} + \frac {c d^{2} x - a e^{2} x - 2 \, a d e}{2 \, {\left (c x^{2} + a\right )} a c} \]

[In]

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

[Out]

1/2*(c*d^2 + a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c) + 1/2*(c*d^2*x - a*e^2*x - 2*a*d*e)/((c*x^2 + a)*a*c
)

Mupad [B] (verification not implemented)

Time = 9.40 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^2}{\left (a+c x^2\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (c\,d^2+a\,e^2\right )}{2\,a^{3/2}\,c^{3/2}}-\frac {\frac {d\,e}{c}+\frac {x\,\left (a\,e^2-c\,d^2\right )}{2\,a\,c}}{c\,x^2+a} \]

[In]

int((d + e*x)^2/(a + c*x^2)^2,x)

[Out]

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