3.1350 \(\int \frac{(A+B x) (d+e x)}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=110 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+3 A c d)}{8 a^{5/2} c^{3/2}}+\frac{x (a B e+3 A c d)}{8 a^2 c \left (a+c x^2\right )}-\frac{a (A e+B d)-x (A c d-a B e)}{4 a c \left (a+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0442802, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {778, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+3 A c d)}{8 a^{5/2} c^{3/2}}+\frac{x (a B e+3 A c d)}{8 a^2 c \left (a+c x^2\right )}-\frac{a (A e+B d)-x (A c d-a B e)}{4 a c \left (a+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 778

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.0899762, size = 101, normalized size = 0.92 \[ \frac{-a^2 (2 A e+2 B d+B e x)+a c x \left (5 A d+B e x^2\right )+3 A c^2 d x^3}{8 a^2 c \left (a+c x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+3 A c d)}{8 a^{5/2} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 108, normalized size = 1. \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( 3\,Acd+aBe \right ){x}^{3}}{8\,{a}^{2}}}+{\frac{ \left ( 5\,Acd-aBe \right ) x}{8\,ac}}-{\frac{Ae+Bd}{4\,c}} \right ) }+{\frac{3\,Ad}{8\,{a}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{Be}{8\,ac}\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)/(c*x^2+a)^3,x)

[Out]

(1/8*(3*A*c*d+B*a*e)/a^2*x^3+1/8*(5*A*c*d-B*a*e)/a/c*x-1/4*(A*e+B*d)/c)/(c*x^2+a)^2+3/8/a^2/(a*c)^(1/2)*arctan
(x*c/(a*c)^(1/2))*A*d+1/8/a/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*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)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.83696, size = 753, normalized size = 6.85 \begin{align*} \left [-\frac{4 \, B a^{3} c d + 4 \, A a^{3} c e - 2 \,{\left (3 \, A a c^{3} d + B a^{2} c^{2} e\right )} x^{3} +{\left (3 \, A a^{2} c d + B a^{3} e +{\left (3 \, A c^{3} d + B a c^{2} e\right )} x^{4} + 2 \,{\left (3 \, A a c^{2} d + B a^{2} c e\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 a^{2} c^{2} d - B a^{3} c e\right )} x}{16 \,{\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac{2 \, B a^{3} c d + 2 \, A a^{3} c e -{\left (3 \, A a c^{3} d + B a^{2} c^{2} e\right )} x^{3} -{\left (3 \, A a^{2} c d + B a^{3} e +{\left (3 \, A c^{3} d + B a c^{2} e\right )} x^{4} + 2 \,{\left (3 \, A a c^{2} d + B a^{2} c e\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) -{\left (5 \, A a^{2} c^{2} d - B a^{3} c e\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((B*x+A)*(e*x+d)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/16*(4*B*a^3*c*d + 4*A*a^3*c*e - 2*(3*A*a*c^3*d + B*a^2*c^2*e)*x^3 + (3*A*a^2*c*d + B*a^3*e + (3*A*c^3*d +
B*a*c^2*e)*x^4 + 2*(3*A*a*c^2*d + B*a^2*c*e)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2
*(5*A*a^2*c^2*d - B*a^3*c*e)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(2*B*a^3*c*d + 2*A*a^3*c*e - (3*
A*a*c^3*d + B*a^2*c^2*e)*x^3 - (3*A*a^2*c*d + B*a^3*e + (3*A*c^3*d + B*a*c^2*e)*x^4 + 2*(3*A*a*c^2*d + B*a^2*c
*e)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*A*a^2*c^2*d - B*a^3*c*e)*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 = 3.96053, size = 180, normalized size = 1.64 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} c^{3}}} \left (3 A c d + B a e\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 (3 A c d + B a e\right ) \log{\left (a^{3} c \sqrt{- \frac{1}{a^{5} c^{3}}} + x \right )}}{16} + \frac{- 2 A a^{2} e - 2 B a^{2} d + x^{3} \left (3 A c^{2} d + B a c e\right ) + x \left (5 A a c d - B a^{2} e\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((B*x+A)*(e*x+d)/(c*x**2+a)**3,x)

[Out]

-sqrt(-1/(a**5*c**3))*(3*A*c*d + B*a*e)*log(-a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + sqrt(-1/(a**5*c**3))*(3*A*c
*d + B*a*e)*log(a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + (-2*A*a**2*e - 2*B*a**2*d + x**3*(3*A*c**2*d + B*a*c*e)
+ x*(5*A*a*c*d - B*a**2*e))/(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 = 1.21529, size = 138, normalized size = 1.25 \begin{align*} \frac{{\left (3 \, A c d + B a e\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \, \sqrt{a c} a^{2} c} + \frac{3 \, A c^{2} d x^{3} + B a c x^{3} e + 5 \, A a c d x - B a^{2} x e - 2 \, B a^{2} d - 2 \, A a^{2} 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((B*x+A)*(e*x+d)/(c*x^2+a)^3,x, algorithm="giac")

[Out]

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