3.1340 \(\int \frac{(A+B x) (d+e x)^3}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=161 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (3 a e^2+c d^2\right )+3 a B e \left (c d^2-a e^2\right )\right )}{2 a^{3/2} c^{5/2}}+\frac{e^2 \log \left (a+c x^2\right ) (A e+3 B d)}{2 c^2}-\frac{e^2 x (A c d-3 a B e)}{2 a c^2}-\frac{(d+e x)^2 (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.208161, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {819, 774, 635, 205, 260} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (3 a e^2+c d^2\right )+3 a B e \left (c d^2-a e^2\right )\right )}{2 a^{3/2} c^{5/2}}+\frac{e^2 \log \left (a+c x^2\right ) (A e+3 B d)}{2 c^2}-\frac{e^2 x (A c d-3 a B e)}{2 a c^2}-\frac{(d+e x)^2 (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^2*(A*c*d - 3*a*B*e)*x)/(2*a*c^2) - ((d + e*x)^2*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(2*a*c*(a + c*x^2)) +
((3*a*B*e*(c*d^2 - a*e^2) + A*c*d*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*c^(5/2)) + (e^2*
(3*B*d + A*e)*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 774

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

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

Mathematica [A]  time = 0.13762, size = 171, normalized size = 1.06 \[ \frac{\frac{\sqrt{c} \left (a^2 e^2 (A e+3 B d+B e x)-a c d (3 A e (d+e x)+B d (d+3 e x))+A c^2 d^3 x\right )}{a \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (3 a e^2+c d^2\right )+3 a B e \left (c d^2-a e^2\right )\right )}{a^{3/2}}+\sqrt{c} e^2 \log \left (a+c x^2\right ) (A e+3 B d)+2 B \sqrt{c} e^3 x}{2 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 296, normalized size = 1.8 \begin{align*}{\frac{B{e}^{3}x}{{c}^{2}}}-{\frac{3\,xAd{e}^{2}}{2\,c \left ( c{x}^{2}+a \right ) }}+{\frac{xA{d}^{3}}{ \left ( 2\,c{x}^{2}+2\,a \right ) a}}+{\frac{aBx{e}^{3}}{2\,{c}^{2} \left ( c{x}^{2}+a \right ) }}-{\frac{3\,Bx{d}^{2}e}{2\,c \left ( c{x}^{2}+a \right ) }}+{\frac{aA{e}^{3}}{2\,{c}^{2} \left ( c{x}^{2}+a \right ) }}-{\frac{3\,A{d}^{2}e}{2\,c \left ( c{x}^{2}+a \right ) }}+{\frac{3\,aBd{e}^{2}}{2\,{c}^{2} \left ( c{x}^{2}+a \right ) }}-{\frac{B{d}^{3}}{2\,c \left ( c{x}^{2}+a \right ) }}+{\frac{\ln \left ( c{x}^{2}+a \right ) A{e}^{3}}{2\,{c}^{2}}}+{\frac{3\,\ln \left ( c{x}^{2}+a \right ) Bd{e}^{2}}{2\,{c}^{2}}}+{\frac{3\,Ad{e}^{2}}{2\,c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{A{d}^{3}}{2\,a}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{3\,B{e}^{3}a}{2\,{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{3\,B{d}^{2}e}{2\,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)^3/(c*x^2+a)^2,x)

[Out]

B*e^3*x/c^2-3/2/c/(c*x^2+a)*x*A*d*e^2+1/2/(c*x^2+a)/a*x*A*d^3+1/2/c^2/(c*x^2+a)*a*x*B*e^3-3/2/c/(c*x^2+a)*x*B*
d^2*e+1/2/c^2/(c*x^2+a)*a*A*e^3-3/2/c/(c*x^2+a)*A*d^2*e+3/2/c^2/(c*x^2+a)*a*B*d*e^2-1/2/c/(c*x^2+a)*B*d^3+1/2/
c^2*ln(c*x^2+a)*A*e^3+3/2/c^2*ln(c*x^2+a)*B*d*e^2+3/2/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d*e^2+1/2/a/(a*c
)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^3-3/2/c^2*a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*e^3+3/2/c/(a*c)^(1/2)*ar
ctan(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14458, size = 1270, normalized size = 7.89 \begin{align*} \left [\frac{4 \, B a^{2} c^{2} e^{3} x^{3} - 2 \, B a^{2} c^{2} d^{3} - 6 \, A a^{2} c^{2} d^{2} e + 6 \, B a^{3} c d e^{2} + 2 \, A a^{3} c e^{3} +{\left (A a c^{2} d^{3} + 3 \, B a^{2} c d^{2} e + 3 \, A a^{2} c d e^{2} - 3 \, B a^{3} e^{3} +{\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - 3 \, B a^{2} 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 (A a c^{3} d^{3} - 3 \, B a^{2} c^{2} d^{2} e - 3 \, A a^{2} c^{2} d e^{2} + 3 \, B a^{3} c e^{3}\right )} x + 2 \,{\left (3 \, B a^{3} c d e^{2} + A a^{3} c e^{3} +{\left (3 \, B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{4 \,{\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac{2 \, B a^{2} c^{2} e^{3} x^{3} - B a^{2} c^{2} d^{3} - 3 \, A a^{2} c^{2} d^{2} e + 3 \, B a^{3} c d e^{2} + A a^{3} c e^{3} +{\left (A a c^{2} d^{3} + 3 \, B a^{2} c d^{2} e + 3 \, A a^{2} c d e^{2} - 3 \, B a^{3} e^{3} +{\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - 3 \, B a^{2} c e^{3}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) +{\left (A a c^{3} d^{3} - 3 \, B a^{2} c^{2} d^{2} e - 3 \, A a^{2} c^{2} d e^{2} + 3 \, B a^{3} c e^{3}\right )} x +{\left (3 \, B a^{3} c d e^{2} + A a^{3} c e^{3} +{\left (3 \, B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{2 \,{\left (a^{2} c^{4} x^{2} + a^{3} 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)^2,x, algorithm="fricas")

[Out]

[1/4*(4*B*a^2*c^2*e^3*x^3 - 2*B*a^2*c^2*d^3 - 6*A*a^2*c^2*d^2*e + 6*B*a^3*c*d*e^2 + 2*A*a^3*c*e^3 + (A*a*c^2*d
^3 + 3*B*a^2*c*d^2*e + 3*A*a^2*c*d*e^2 - 3*B*a^3*e^3 + (A*c^3*d^3 + 3*B*a*c^2*d^2*e + 3*A*a*c^2*d*e^2 - 3*B*a^
2*c*e^3)*x^2)*sqrt(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(A*a*c^3*d^3 - 3*B*a^2*c^2*d^2*e -
3*A*a^2*c^2*d*e^2 + 3*B*a^3*c*e^3)*x + 2*(3*B*a^3*c*d*e^2 + A*a^3*c*e^3 + (3*B*a^2*c^2*d*e^2 + A*a^2*c^2*e^3)*
x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3), 1/2*(2*B*a^2*c^2*e^3*x^3 - B*a^2*c^2*d^3 - 3*A*a^2*c^2*d^2*e + 3
*B*a^3*c*d*e^2 + A*a^3*c*e^3 + (A*a*c^2*d^3 + 3*B*a^2*c*d^2*e + 3*A*a^2*c*d*e^2 - 3*B*a^3*e^3 + (A*c^3*d^3 + 3
*B*a*c^2*d^2*e + 3*A*a*c^2*d*e^2 - 3*B*a^2*c*e^3)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (A*a*c^3*d^3 - 3*B*a^
2*c^2*d^2*e - 3*A*a^2*c^2*d*e^2 + 3*B*a^3*c*e^3)*x + (3*B*a^3*c*d*e^2 + A*a^3*c*e^3 + (3*B*a^2*c^2*d*e^2 + A*a
^2*c^2*e^3)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3)]

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Sympy [B]  time = 7.56111, size = 583, normalized size = 3.62 \begin{align*} \frac{B e^{3} x}{c^{2}} + \left (\frac{e^{2} \left (A e + 3 B d\right )}{2 c^{2}} - \frac{\sqrt{- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right ) \log{\left (x + \frac{2 A a^{2} e^{3} + 6 B a^{2} d e^{2} - 4 a^{2} c^{2} \left (\frac{e^{2} \left (A e + 3 B d\right )}{2 c^{2}} - \frac{\sqrt{- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right )}{- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \left (\frac{e^{2} \left (A e + 3 B d\right )}{2 c^{2}} + \frac{\sqrt{- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right ) \log{\left (x + \frac{2 A a^{2} e^{3} + 6 B a^{2} d e^{2} - 4 a^{2} c^{2} \left (\frac{e^{2} \left (A e + 3 B d\right )}{2 c^{2}} + \frac{\sqrt{- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right )}{- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \frac{A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + x \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} 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)**3/(c*x**2+a)**2,x)

[Out]

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

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Giac [A]  time = 1.22674, size = 242, normalized size = 1.5 \begin{align*} \frac{B x e^{3}}{c^{2}} + \frac{{\left (3 \, B d e^{2} + A e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac{{\left (A c^{2} d^{3} + 3 \, B a c d^{2} e + 3 \, A a c d e^{2} - 3 \, B a^{2} e^{3}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{2 \, \sqrt{a c} a c^{2}} - \frac{B a c d^{3} + 3 \, A a c d^{2} e - 3 \, B a^{2} d e^{2} - A a^{2} e^{3} -{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} x}{2 \,{\left (c x^{2} + a\right )} a 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)^2,x, algorithm="giac")

[Out]

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