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]

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

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Rubi [A]  time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.10, size = 186, normalized size = 1.49 \[ \frac {3 \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) (a B e+A c d)}{8 a^{5/2} c^{5/2}}+\frac {-a^2 e^2 (4 A e+12 B d+5 B e x)+3 a c d e x (A e+B d)+3 A c^2 d^3 x}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {a^2 e^2 (A e+3 B d+B e x)-a c d (3 A e (d+e x)+B d (d+3 e x))+A c^2 d^3 x}{4 a c^2 \left (a+c x^2\right )^2} \]

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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fricas [B]  time = 0.71, size = 752, normalized size = 6.02 \[ \left [-\frac {4 \, B a^{3} c^{2} d^{3} + 12 \, A a^{3} c^{2} d^{2} e + 12 \, B a^{4} c d e^{2} + 4 \, A a^{4} c e^{3} - 2 \, {\left (3 \, A a c^{4} d^{3} + 3 \, B a^{2} c^{3} d^{2} e + 3 \, A a^{2} c^{3} d e^{2} - 5 \, B a^{3} c^{2} e^{3}\right )} x^{3} + 8 \, {\left (3 \, B a^{3} c^{2} d e^{2} + A a^{3} c^{2} e^{3}\right )} x^{2} + 3 \, {\left (A a^{2} c^{2} d^{3} + B a^{3} c d^{2} e + A a^{3} c d e^{2} + B a^{4} e^{3} + {\left (A c^{4} d^{3} + B a c^{3} d^{2} e + A a c^{3} d e^{2} + B a^{2} c^{2} e^{3}\right )} x^{4} + 2 \, {\left (A a c^{3} d^{3} + B a^{2} c^{2} d^{2} e + A a^{2} c^{2} d e^{2} + B a^{3} c e^{3}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (5 \, A a^{2} c^{3} d^{3} - 3 \, B a^{3} c^{2} d^{2} e - 3 \, A a^{3} c^{2} d e^{2} - 3 \, B a^{4} c e^{3}\right )} x}{16 \, {\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}, -\frac {2 \, B a^{3} c^{2} d^{3} + 6 \, A a^{3} c^{2} d^{2} e + 6 \, B a^{4} c d e^{2} + 2 \, A a^{4} c e^{3} - {\left (3 \, A a c^{4} d^{3} + 3 \, B a^{2} c^{3} d^{2} e + 3 \, A a^{2} c^{3} d e^{2} - 5 \, B a^{3} c^{2} e^{3}\right )} x^{3} + 4 \, {\left (3 \, B a^{3} c^{2} d e^{2} + A a^{3} c^{2} e^{3}\right )} x^{2} - 3 \, {\left (A a^{2} c^{2} d^{3} + B a^{3} c d^{2} e + A a^{3} c d e^{2} + B a^{4} e^{3} + {\left (A c^{4} d^{3} + B a c^{3} d^{2} e + A a c^{3} d e^{2} + B a^{2} c^{2} e^{3}\right )} x^{4} + 2 \, {\left (A a c^{3} d^{3} + B a^{2} c^{2} d^{2} e + A a^{2} c^{2} d e^{2} + B a^{3} c e^{3}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, A a^{2} c^{3} d^{3} - 3 \, B a^{3} c^{2} d^{2} e - 3 \, A a^{3} c^{2} d e^{2} - 3 \, B a^{4} c e^{3}\right )} x}{8 \, {\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}\right ] \]

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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giac [B]  time = 0.19, size = 233, normalized size = 1.86 \[ \frac {3 \, {\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{2}} + \frac {3 \, A c^{3} d^{3} x^{3} + 3 \, B a c^{2} d^{2} x^{3} e + 3 \, A a c^{2} d x^{3} e^{2} + 5 \, A a c^{2} d^{3} x - 5 \, B a^{2} c x^{3} e^{3} - 12 \, B a^{2} c d x^{2} e^{2} - 3 \, B a^{2} c d^{2} x e - 2 \, B a^{2} c d^{3} - 4 \, A a^{2} c x^{2} e^{3} - 3 \, A a^{2} c d x e^{2} - 6 \, A a^{2} c d^{2} e - 3 \, B a^{3} x e^{3} - 6 \, B a^{3} d e^{2} - 2 \, A a^{3} e^{3}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \]

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)

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maple [B]  time = 0.06, size = 260, normalized size = 2.08 \[ \frac {3 A d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a c}+\frac {3 A \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2}}+\frac {3 B \,d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a c}+\frac {3 B \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, c^{2}}+\frac {-\frac {\left (A e +3 B d \right ) e^{2} x^{2}}{2 c}+\frac {\left (3 A a c d \,e^{2}+3 A \,c^{2} d^{3}-5 B \,a^{2} e^{3}+3 B a c \,d^{2} e \right ) x^{3}}{8 a^{2} c}-\frac {\left (3 A a c d \,e^{2}-5 A \,c^{2} d^{3}+3 B \,a^{2} e^{3}+3 B a c \,d^{2} e \right ) x}{8 a \,c^{2}}-\frac {a A \,e^{3}+3 A c \,d^{2} e +3 a B d \,e^{2}+B c \,d^{3}}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}} \]

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

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maxima [B]  time = 1.51, size = 248, normalized size = 1.98 \[ -\frac {2 \, B a^{2} c d^{3} + 6 \, A a^{2} c d^{2} e + 6 \, B a^{3} d e^{2} + 2 \, A a^{3} e^{3} - {\left (3 \, A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - 5 \, B a^{2} c e^{3}\right )} x^{3} + 4 \, {\left (3 \, B a^{2} c d e^{2} + A a^{2} c e^{3}\right )} x^{2} - {\left (5 \, A a c^{2} d^{3} - 3 \, B a^{2} c d^{2} e - 3 \, A a^{2} c d e^{2} - 3 \, B a^{3} e^{3}\right )} x}{8 \, {\left (a^{2} c^{4} x^{4} + 2 \, a^{3} c^{3} x^{2} + a^{4} c^{2}\right )}} + \frac {3 \, {\left (A c^{2} d^{3} + B a c d^{2} e + A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c^{2}} \]

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]

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

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mupad [B]  time = 0.22, size = 265, normalized size = 2.12 \[ \frac {3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x\,\left (A\,c\,d+B\,a\,e\right )\,\left (c\,d^2+a\,e^2\right )}{\sqrt {a}\,\left (B\,a^2\,e^3+B\,a\,c\,d^2\,e+A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}\right )\,\left (A\,c\,d+B\,a\,e\right )\,\left (c\,d^2+a\,e^2\right )}{8\,a^{5/2}\,c^{5/2}}-\frac {\frac {B\,c\,d^3+3\,A\,c\,d^2\,e+3\,B\,a\,d\,e^2+A\,a\,e^3}{4\,c^2}+\frac {x^2\,\left (A\,e^3+3\,B\,d\,e^2\right )}{2\,c}+\frac {x\,\left (3\,B\,a^2\,e^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2-5\,A\,c^2\,d^3\right )}{8\,a\,c^2}-\frac {x^3\,\left (-5\,B\,a^2\,e^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2+3\,A\,c^2\,d^3\right )}{8\,a^2\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 20.05, size = 468, normalized size = 3.74 \[ - \frac {3 \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right ) \log {\left (- \frac {3 a^{3} c^{2} \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right )}{3 A a c d e^{2} + 3 A c^{2} d^{3} + 3 B a^{2} e^{3} + 3 B a c d^{2} e} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right ) \log {\left (\frac {3 a^{3} c^{2} \sqrt {- \frac {1}{a^{5} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (A c d + B a e\right )}{3 A a c d e^{2} + 3 A c^{2} d^{3} + 3 B a^{2} e^{3} + 3 B a c d^{2} e} + x \right )}}{16} + \frac {- 2 A a^{3} e^{3} - 6 A a^{2} c d^{2} e - 6 B a^{3} d e^{2} - 2 B a^{2} c d^{3} + x^{3} \left (3 A a c^{2} d e^{2} + 3 A c^{3} d^{3} - 5 B a^{2} c e^{3} + 3 B a c^{2} d^{2} e\right ) + x^{2} \left (- 4 A a^{2} c e^{3} - 12 B a^{2} c d e^{2}\right ) + x \left (- 3 A a^{2} c d e^{2} + 5 A a c^{2} d^{3} - 3 B a^{3} e^{3} - 3 B a^{2} c d^{2} e\right )}{8 a^{4} c^{2} + 16 a^{3} c^{3} x^{2} + 8 a^{2} c^{4} x^{4}} \]

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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