3.1348 \(\int \frac{(A+B x) (d+e x)^3}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=125 \[ \frac{3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+A c d)}{8 a^{5/2} c^{5/2}}-\frac{3 (d+e x) (a e-c d x) (a B e+A c d)}{8 a^2 c^2 \left (a+c x^2\right )}-\frac{(d+e x)^3 (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0582483, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {805, 723, 205} \[ \frac{3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+A c d)}{8 a^{5/2} c^{5/2}}-\frac{3 (d+e x) (a e-c d x) (a B e+A c d)}{8 a^2 c^2 \left (a+c x^2\right )}-\frac{(d+e x)^3 (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 805

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

Rule 723

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]

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

Mathematica [A]  time = 0.107883, size = 186, normalized size = 1.49 \[ \frac{-a^2 e^2 (4 A e+12 B d+5 B e x)+3 a c d e x (A e+B d)+3 A c^2 d^3 x}{8 a^2 c^2 \left (a+c x^2\right )}+\frac{a^2 e^2 (A e+3 B d+B e x)-a c d (3 A e (d+e x)+B d (d+3 e x))+A c^2 d^3 x}{4 a c^2 \left (a+c x^2\right )^2}+\frac{3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+A c d)}{8 a^{5/2} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*c^2*d^3*x + 3*a*c*d*e*(B*d + A*e)*x - a^2*e^2*(12*B*d + 4*A*e + 5*B*e*x))/(8*a^2*c^2*(a + c*x^2)) + (A*c^
2*d^3*x + a^2*e^2*(3*B*d + A*e + B*e*x) - a*c*d*(3*A*e*(d + e*x) + B*d*(d + 3*e*x)))/(4*a*c^2*(a + c*x^2)^2) +
(3*(A*c*d + a*B*e)*(c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(5/2))

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Maple [B]  time = 0.008, size = 260, normalized size = 2.1 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{2}} \left ({\frac{ \left ( 3\,Aacd{e}^{2}+3\,A{d}^{3}{c}^{2}-5\,{a}^{2}B{e}^{3}+3\,Bac{d}^{2}e \right ){x}^{3}}{8\,{a}^{2}c}}-{\frac{{e}^{2} \left ( Ae+3\,Bd \right ){x}^{2}}{2\,c}}-{\frac{ \left ( 3\,Aacd{e}^{2}-5\,A{d}^{3}{c}^{2}+3\,{a}^{2}B{e}^{3}+3\,Bac{d}^{2}e \right ) x}{8\,a{c}^{2}}}-{\frac{aA{e}^{3}+3\,Ac{d}^{2}e+3\,aBd{e}^{2}+Bc{d}^{3}}{4\,{c}^{2}}} \right ) }+{\frac{3\,Ad{e}^{2}}{8\,ac}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,A{d}^{3}}{8\,{a}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,B{e}^{3}}{8\,{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,B{d}^{2}e}{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)^3/(c*x^2+a)^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.9381, size = 1544, normalized size = 12.35 \begin{align*} \left [-\frac{4 \, B a^{3} c^{2} d^{3} + 12 \, A a^{3} c^{2} d^{2} e + 12 \, B a^{4} c d e^{2} + 4 \, A a^{4} c e^{3} - 2 \,{\left (3 \, A a c^{4} d^{3} + 3 \, B a^{2} c^{3} d^{2} e + 3 \, A a^{2} c^{3} d e^{2} - 5 \, B a^{3} c^{2} e^{3}\right )} x^{3} + 8 \,{\left (3 \, B a^{3} c^{2} d e^{2} + A a^{3} c^{2} e^{3}\right )} x^{2} + 3 \,{\left (A a^{2} c^{2} d^{3} + B a^{3} c d^{2} e + A a^{3} c d e^{2} + B a^{4} e^{3} +{\left (A c^{4} d^{3} + B a c^{3} d^{2} e + A a c^{3} d e^{2} + B a^{2} c^{2} e^{3}\right )} x^{4} + 2 \,{\left (A a c^{3} d^{3} + B a^{2} c^{2} d^{2} e + A a^{2} c^{2} d e^{2} + B a^{3} c e^{3}\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^{3} d^{3} - 3 \, B a^{3} c^{2} d^{2} e - 3 \, A a^{3} c^{2} d e^{2} - 3 \, B a^{4} c e^{3}\right )} x}{16 \,{\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}, -\frac{2 \, B a^{3} c^{2} d^{3} + 6 \, A a^{3} c^{2} d^{2} e + 6 \, B a^{4} c d e^{2} + 2 \, A a^{4} c e^{3} -{\left (3 \, A a c^{4} d^{3} + 3 \, B a^{2} c^{3} d^{2} e + 3 \, A a^{2} c^{3} d e^{2} - 5 \, B a^{3} c^{2} e^{3}\right )} x^{3} + 4 \,{\left (3 \, B a^{3} c^{2} d e^{2} + A a^{3} c^{2} e^{3}\right )} x^{2} - 3 \,{\left (A a^{2} c^{2} d^{3} + B a^{3} c d^{2} e + A a^{3} c d e^{2} + B a^{4} e^{3} +{\left (A c^{4} d^{3} + B a c^{3} d^{2} e + A a c^{3} d e^{2} + B a^{2} c^{2} e^{3}\right )} x^{4} + 2 \,{\left (A a c^{3} d^{3} + B a^{2} c^{2} d^{2} e + A a^{2} c^{2} d e^{2} + B a^{3} c e^{3}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) -{\left (5 \, A a^{2} c^{3} d^{3} - 3 \, B a^{3} c^{2} d^{2} e - 3 \, A a^{3} c^{2} d e^{2} - 3 \, B a^{4} c e^{3}\right )} x}{8 \,{\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(4*B*a^3*c^2*d^3 + 12*A*a^3*c^2*d^2*e + 12*B*a^4*c*d*e^2 + 4*A*a^4*c*e^3 - 2*(3*A*a*c^4*d^3 + 3*B*a^2*c
^3*d^2*e + 3*A*a^2*c^3*d*e^2 - 5*B*a^3*c^2*e^3)*x^3 + 8*(3*B*a^3*c^2*d*e^2 + A*a^3*c^2*e^3)*x^2 + 3*(A*a^2*c^2
*d^3 + B*a^3*c*d^2*e + A*a^3*c*d*e^2 + B*a^4*e^3 + (A*c^4*d^3 + B*a*c^3*d^2*e + A*a*c^3*d*e^2 + B*a^2*c^2*e^3)
*x^4 + 2*(A*a*c^3*d^3 + B*a^2*c^2*d^2*e + A*a^2*c^2*d*e^2 + B*a^3*c*e^3)*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^3*d^3 - 3*B*a^3*c^2*d^2*e - 3*A*a^3*c^2*d*e^2 - 3*B*a^4*c*e^3)*x)/(a^3
*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3), -1/8*(2*B*a^3*c^2*d^3 + 6*A*a^3*c^2*d^2*e + 6*B*a^4*c*d*e^2 + 2*A*a^4*c*e
^3 - (3*A*a*c^4*d^3 + 3*B*a^2*c^3*d^2*e + 3*A*a^2*c^3*d*e^2 - 5*B*a^3*c^2*e^3)*x^3 + 4*(3*B*a^3*c^2*d*e^2 + A*
a^3*c^2*e^3)*x^2 - 3*(A*a^2*c^2*d^3 + B*a^3*c*d^2*e + A*a^3*c*d*e^2 + B*a^4*e^3 + (A*c^4*d^3 + B*a*c^3*d^2*e +
A*a*c^3*d*e^2 + B*a^2*c^2*e^3)*x^4 + 2*(A*a*c^3*d^3 + B*a^2*c^2*d^2*e + A*a^2*c^2*d*e^2 + B*a^3*c*e^3)*x^2)*s
qrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*A*a^2*c^3*d^3 - 3*B*a^3*c^2*d^2*e - 3*A*a^3*c^2*d*e^2 - 3*B*a^4*c*e^3)*x)/
(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3)]

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Sympy [B]  time = 50.6199, size = 466, normalized size = 3.73 \begin{align*} - \frac{3 \sqrt{- \frac{1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right ) \log{\left (- \frac{3 a^{3} c^{2} \sqrt{- \frac{1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right )}{3 A a c d e^{2} + 3 A c^{2} d^{3} + 3 B a^{2} e^{3} + 3 B a c d^{2} e} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right ) \log{\left (\frac{3 a^{3} c^{2} \sqrt{- \frac{1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right )}{3 A a c d e^{2} + 3 A c^{2} d^{3} + 3 B a^{2} e^{3} + 3 B a c d^{2} e} + x \right )}}{16} - \frac{2 A a^{3} e^{3} + 6 A a^{2} c d^{2} e + 6 B a^{3} d e^{2} + 2 B a^{2} c d^{3} + x^{3} \left (- 3 A a c^{2} d e^{2} - 3 A c^{3} d^{3} + 5 B a^{2} c e^{3} - 3 B a c^{2} d^{2} e\right ) + x^{2} \left (4 A a^{2} c e^{3} + 12 B a^{2} c d e^{2}\right ) + x \left (3 A a^{2} c d e^{2} - 5 A a c^{2} d^{3} + 3 B a^{3} e^{3} + 3 B a^{2} c d^{2} e\right )}{8 a^{4} c^{2} + 16 a^{3} c^{3} x^{2} + 8 a^{2} c^{4} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(a**5*c**5))*(a*e**2 + c*d**2)*(A*c*d + B*a*e)*log(-3*a**3*c**2*sqrt(-1/(a**5*c**5))*(a*e**2 + c*d*
*2)*(A*c*d + B*a*e)/(3*A*a*c*d*e**2 + 3*A*c**2*d**3 + 3*B*a**2*e**3 + 3*B*a*c*d**2*e) + x)/16 + 3*sqrt(-1/(a**
5*c**5))*(a*e**2 + c*d**2)*(A*c*d + B*a*e)*log(3*a**3*c**2*sqrt(-1/(a**5*c**5))*(a*e**2 + c*d**2)*(A*c*d + B*a
*e)/(3*A*a*c*d*e**2 + 3*A*c**2*d**3 + 3*B*a**2*e**3 + 3*B*a*c*d**2*e) + x)/16 - (2*A*a**3*e**3 + 6*A*a**2*c*d*
*2*e + 6*B*a**3*d*e**2 + 2*B*a**2*c*d**3 + x**3*(-3*A*a*c**2*d*e**2 - 3*A*c**3*d**3 + 5*B*a**2*c*e**3 - 3*B*a*
c**2*d**2*e) + x**2*(4*A*a**2*c*e**3 + 12*B*a**2*c*d*e**2) + x*(3*A*a**2*c*d*e**2 - 5*A*a*c**2*d**3 + 3*B*a**3
*e**3 + 3*B*a**2*c*d**2*e))/(8*a**4*c**2 + 16*a**3*c**3*x**2 + 8*a**2*c**4*x**4)

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Giac [B]  time = 1.17474, size = 315, normalized size = 2.52 \begin{align*} \frac{3 \,{\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \, \sqrt{a c} a^{2} c^{2}} + \frac{3 \, A c^{3} d^{3} x^{3} + 3 \, B a c^{2} d^{2} x^{3} e + 3 \, A a c^{2} d x^{3} e^{2} + 5 \, A a c^{2} d^{3} x - 5 \, B a^{2} c x^{3} e^{3} - 12 \, B a^{2} c d x^{2} e^{2} - 3 \, B a^{2} c d^{2} x e - 2 \, B a^{2} c d^{3} - 4 \, A a^{2} c x^{2} e^{3} - 3 \, A a^{2} c d x e^{2} - 6 \, A a^{2} c d^{2} e - 3 \, B a^{3} x e^{3} - 6 \, B a^{3} d e^{2} - 2 \, A a^{3} e^{3}}{8 \,{\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(A*c^2*d^3 + B*a*c*d^2*e + A*a*c*d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c^2) + 1/8*(3*A*c
^3*d^3*x^3 + 3*B*a*c^2*d^2*x^3*e + 3*A*a*c^2*d*x^3*e^2 + 5*A*a*c^2*d^3*x - 5*B*a^2*c*x^3*e^3 - 12*B*a^2*c*d*x^
2*e^2 - 3*B*a^2*c*d^2*x*e - 2*B*a^2*c*d^3 - 4*A*a^2*c*x^2*e^3 - 3*A*a^2*c*d*x*e^2 - 6*A*a^2*c*d^2*e - 3*B*a^3*
x*e^3 - 6*B*a^3*d*e^2 - 2*A*a^3*e^3)/((c*x^2 + a)^2*a^2*c^2)