3.1341 \(\int \frac{(A+B x) (d+e x)^2}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=112 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (a e (A e+2 B d)+A c d^2\right )}{2 a^{3/2} c^{3/2}}-\frac{(d+e x) (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )}+\frac{B e^2 \log \left (a+c x^2\right )}{2 c^2} \]

[Out]

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

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Rubi [A]  time = 0.0726357, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {819, 635, 205, 260} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (a e (A e+2 B d)+A c d^2\right )}{2 a^{3/2} c^{3/2}}-\frac{(d+e x) (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )}+\frac{B e^2 \log \left (a+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^2)/(a + c*x^2)^2,x]

[Out]

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

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 635

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

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]

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.109414, size = 119, normalized size = 1.06 \[ \frac{\frac{a^2 B e^2-a c (A e (2 d+e x)+B d (d+2 e x))+A c^2 d^2 x}{a \left (a+c x^2\right )}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (a A e^2+2 a B d e+A c d^2\right )}{a^{3/2}}+B e^2 \log \left (a+c x^2\right )}{2 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 151, normalized size = 1.4 \begin{align*}{\frac{1}{c{x}^{2}+a} \left ( -{\frac{ \left ( Aa{e}^{2}-Ac{d}^{2}+2\,aBde \right ) x}{2\,ac}}-{\frac{2\,Acde-aB{e}^{2}+Bc{d}^{2}}{2\,{c}^{2}}} \right ) }+{\frac{B{e}^{2}\ln \left ( c{x}^{2}+a \right ) }{2\,{c}^{2}}}+{\frac{A{e}^{2}}{2\,c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{A{d}^{2}}{2\,a}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{Bde}{c}\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((B*x+A)*(e*x+d)^2/(c*x^2+a)^2,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95079, size = 814, normalized size = 7.27 \begin{align*} \left [-\frac{2 \, B a^{2} c d^{2} + 4 \, A a^{2} c d e - 2 \, B a^{3} e^{2} +{\left (A a c d^{2} + 2 \, B a^{2} d e + A a^{2} e^{2} +{\left (A c^{2} d^{2} + 2 \, B a c d e + A 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 a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c e^{2}\right )} x - 2 \,{\left (B a^{2} c e^{2} x^{2} + B a^{3} e^{2}\right )} \log \left (c x^{2} + a\right )}{4 \,{\left (a^{2} c^{3} x^{2} + a^{3} c^{2}\right )}}, -\frac{B a^{2} c d^{2} + 2 \, A a^{2} c d e - B a^{3} e^{2} -{\left (A a c d^{2} + 2 \, B a^{2} d e + A a^{2} e^{2} +{\left (A c^{2} d^{2} + 2 \, B a c d e + A a c e^{2}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) -{\left (A a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c e^{2}\right )} x -{\left (B a^{2} c e^{2} x^{2} + B a^{3} e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \,{\left (a^{2} c^{3} x^{2} + a^{3} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 3.65074, size = 382, normalized size = 3.41 \begin{align*} \left (\frac{B e^{2}}{2 c^{2}} - \frac{\sqrt{- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right ) \log{\left (x + \frac{- 2 B a^{2} e^{2} + 4 a^{2} c^{2} \left (\frac{B e^{2}}{2 c^{2}} - \frac{\sqrt{- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right )}{A a c e^{2} + A c^{2} d^{2} + 2 B a c d e} \right )} + \left (\frac{B e^{2}}{2 c^{2}} + \frac{\sqrt{- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right ) \log{\left (x + \frac{- 2 B a^{2} e^{2} + 4 a^{2} c^{2} \left (\frac{B e^{2}}{2 c^{2}} + \frac{\sqrt{- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right )}{A a c e^{2} + A c^{2} d^{2} + 2 B a c d e} \right )} - \frac{2 A a c d e - B a^{2} e^{2} + B a c d^{2} + x \left (A a c e^{2} - A c^{2} d^{2} + 2 B a c d e\right )}{2 a^{2} c^{2} + 2 a c^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21046, size = 171, normalized size = 1.53 \begin{align*} \frac{B e^{2} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac{{\left (A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{2 \, \sqrt{a c} a c} + \frac{{\left (A c d^{2} - 2 \, B a d e - A a e^{2}\right )} x - \frac{B a c d^{2} + 2 \, A a c d e - B a^{2} e^{2}}{c}}{2 \,{\left (c x^{2} + a\right )} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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