### 3.515 $$\int \frac{(d+e x)^2}{(a+c x^2)^3} \, dx$$

Optimal. Leaf size=113 $\frac{\left (a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}-\frac{2 a d e-x \left (a e^2+3 c d^2\right )}{8 a^2 c \left (a+c x^2\right )}-\frac{(d+e x) (a e-c d x)}{4 a c \left (a+c x^2\right )^2}$

[Out]

-((a*e - c*d*x)*(d + e*x))/(4*a*c*(a + c*x^2)^2) - (2*a*d*e - (3*c*d^2 + a*e^2)*x)/(8*a^2*c*(a + c*x^2)) + ((3
*c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

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Rubi [A]  time = 0.0493133, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {739, 639, 205} $\frac{\left (a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}-\frac{2 a d e-x \left (a e^2+3 c d^2\right )}{8 a^2 c \left (a+c x^2\right )}-\frac{(d+e x) (a e-c d x)}{4 a c \left (a+c x^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^2/(a + c*x^2)^3,x]

[Out]

-((a*e - c*d*x)*(d + e*x))/(4*a*c*(a + c*x^2)^2) - (2*a*d*e - (3*c*d^2 + a*e^2)*x)/(8*a^2*c*(a + c*x^2)) + ((3
*c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

Rule 739

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[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 639

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{(d+e x)^2}{\left (a+c x^2\right )^3} \, dx &=-\frac{(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}+\frac{\int \frac{3 c d^2+a e^2+2 c d e x}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac{(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}-\frac{2 a d e-\left (3 c d^2+a e^2\right ) x}{8 a^2 c \left (a+c x^2\right )}+\frac{\left (3 c d^2+a e^2\right ) \int \frac{1}{a+c x^2} \, dx}{8 a^2 c}\\ &=-\frac{(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}-\frac{2 a d e-\left (3 c d^2+a e^2\right ) x}{8 a^2 c \left (a+c x^2\right )}+\frac{\left (3 c d^2+a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0703977, size = 101, normalized size = 0.89 $\frac{-a^2 e (4 d+e x)+a c x \left (5 d^2+e^2 x^2\right )+3 c^2 d^2 x^3}{8 a^2 c \left (a+c x^2\right )^2}+\frac{\left (a e^2+3 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} c^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(3*c^2*d^2*x^3 - a^2*e*(4*d + e*x) + a*c*x*(5*d^2 + e^2*x^2))/(8*a^2*c*(a + c*x^2)^2) + ((3*c*d^2 + a*e^2)*Arc
Tan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

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Maple [A]  time = 0.047, size = 108, normalized size = 1. \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( a{e}^{2}+3\,c{d}^{2} \right ){x}^{3}}{8\,{a}^{2}}}-{\frac{ \left ( a{e}^{2}-5\,c{d}^{2} \right ) x}{8\,ac}}-{\frac{de}{2\,c}} \right ) }+{\frac{{e}^{2}}{8\,ac}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,{d}^{2}}{8\,{a}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/8*(a*e^2+3*c*d^2)/a^2*x^3-1/8*(a*e^2-5*c*d^2)/a/c*x-1/2*d*e/c)/(c*x^2+a)^2+1/8/a/c/(a*c)^(1/2)*arctan(x*c/(
a*c)^(1/2))*e^2+3/8/a^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*d^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98568, size = 716, normalized size = 6.34 \begin{align*} \left [-\frac{8 \, a^{3} c d e - 2 \,{\left (3 \, a c^{3} d^{2} + a^{2} c^{2} e^{2}\right )} x^{3} +{\left (3 \, a^{2} c d^{2} + a^{3} e^{2} +{\left (3 \, c^{3} d^{2} + a c^{2} e^{2}\right )} x^{4} + 2 \,{\left (3 \, a c^{2} d^{2} + a^{2} 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 (5 \, a^{2} c^{2} d^{2} - a^{3} c e^{2}\right )} x}{16 \,{\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac{4 \, a^{3} c d e -{\left (3 \, a c^{3} d^{2} + a^{2} c^{2} e^{2}\right )} x^{3} -{\left (3 \, a^{2} c d^{2} + a^{3} e^{2} +{\left (3 \, c^{3} d^{2} + a c^{2} e^{2}\right )} x^{4} + 2 \,{\left (3 \, a c^{2} d^{2} + a^{2} c e^{2}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) -{\left (5 \, a^{2} c^{2} d^{2} - a^{3} c e^{2}\right )} x}{8 \,{\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(8*a^3*c*d*e - 2*(3*a*c^3*d^2 + a^2*c^2*e^2)*x^3 + (3*a^2*c*d^2 + a^3*e^2 + (3*c^3*d^2 + a*c^2*e^2)*x^4
+ 2*(3*a*c^2*d^2 + a^2*c*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(5*a^2*c^2*d^
2 - a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(4*a^3*c*d*e - (3*a*c^3*d^2 + a^2*c^2*e^2)*x^3
- (3*a^2*c*d^2 + a^3*e^2 + (3*c^3*d^2 + a*c^2*e^2)*x^4 + 2*(3*a*c^2*d^2 + a^2*c*e^2)*x^2)*sqrt(a*c)*arctan(sq
rt(a*c)*x/a) - (5*a^2*c^2*d^2 - a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2)]

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Sympy [A]  time = 1.11785, size = 172, normalized size = 1.52 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} c^{3}}} \left (a e^{2} + 3 c d^{2}\right ) \log{\left (- a^{3} c \sqrt{- \frac{1}{a^{5} c^{3}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{5} c^{3}}} \left (a e^{2} + 3 c d^{2}\right ) \log{\left (a^{3} c \sqrt{- \frac{1}{a^{5} c^{3}}} + x \right )}}{16} + \frac{- 4 a^{2} d e + x^{3} \left (a c e^{2} + 3 c^{2} d^{2}\right ) + x \left (- a^{2} e^{2} + 5 a c d^{2}\right )}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**5*c**3))*(a*e**2 + 3*c*d**2)*log(-a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + sqrt(-1/(a**5*c**3))*(a*e
**2 + 3*c*d**2)*log(a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + (-4*a**2*d*e + x**3*(a*c*e**2 + 3*c**2*d**2) + x*(-a
**2*e**2 + 5*a*c*d**2))/(8*a**4*c + 16*a**3*c**2*x**2 + 8*a**2*c**3*x**4)

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Giac [A]  time = 1.25419, size = 128, normalized size = 1.13 \begin{align*} \frac{{\left (3 \, c d^{2} + a e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \, \sqrt{a c} a^{2} c} + \frac{3 \, c^{2} d^{2} x^{3} + a c x^{3} e^{2} + 5 \, a c d^{2} x - a^{2} x e^{2} - 4 \, a^{2} d e}{8 \,{\left (c x^{2} + a\right )}^{2} a^{2} c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*c*d^2 + a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c) + 1/8*(3*c^2*d^2*x^3 + a*c*x^3*e^2 + 5*a*c*d^2*x
- a^2*x*e^2 - 4*a^2*d*e)/((c*x^2 + a)^2*a^2*c)