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]

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

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Rubi [A]  time = 0.21, antiderivative size = 161, normalized size of antiderivative = 1.00, 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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.12, size = 171, normalized size = 1.06 \[ \frac {\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (3 a e^2+c d^2\right )+3 a B e \left (c d^2-a e^2\right )\right )}{a^{3/2}}+\frac {\sqrt {c} \left (a^2 e^2 (A e+3 B d+B e x)-a c d (3 A e (d+e x)+B d (d+3 e x))+A c^2 d^3 x\right )}{a \left (a+c x^2\right )}+\sqrt {c} e^2 \log \left (a+c x^2\right ) (A e+3 B d)+2 B \sqrt {c} e^3 x}{2 c^{5/2}} \]

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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fricas [B]  time = 0.52, size = 609, normalized size = 3.78 \[ \left [\frac {4 \, B a^{2} c^{2} e^{3} x^{3} - 2 \, B a^{2} c^{2} d^{3} - 6 \, A a^{2} c^{2} d^{2} e + 6 \, B a^{3} c d e^{2} + 2 \, A a^{3} c e^{3} + {\left (A a c^{2} d^{3} + 3 \, B a^{2} c d^{2} e + 3 \, A a^{2} c d e^{2} - 3 \, B a^{3} e^{3} + {\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - 3 \, B a^{2} c e^{3}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (A a c^{3} d^{3} - 3 \, B a^{2} c^{2} d^{2} e - 3 \, A a^{2} c^{2} d e^{2} + 3 \, B a^{3} c e^{3}\right )} x + 2 \, {\left (3 \, B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (3 \, B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac {2 \, B a^{2} c^{2} e^{3} x^{3} - B a^{2} c^{2} d^{3} - 3 \, A a^{2} c^{2} d^{2} e + 3 \, B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (A a c^{2} d^{3} + 3 \, B a^{2} c d^{2} e + 3 \, A a^{2} c d e^{2} - 3 \, B a^{3} e^{3} + {\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - 3 \, B a^{2} c e^{3}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (A a c^{3} d^{3} - 3 \, B a^{2} c^{2} d^{2} e - 3 \, A a^{2} c^{2} d e^{2} + 3 \, B a^{3} c e^{3}\right )} x + {\left (3 \, B a^{3} c d e^{2} + A a^{3} c e^{3} + {\left (3 \, B a^{2} c^{2} d e^{2} + A a^{2} c^{2} e^{3}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}\right ] \]

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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giac [A]  time = 0.19, size = 179, normalized size = 1.11 \[ \frac {B x e^{3}}{c^{2}} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c^{2} d^{3} + 3 \, B a c d^{2} e + 3 \, A a c d e^{2} - 3 \, B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} - \frac {B a c d^{3} + 3 \, A a c d^{2} e - 3 \, B a^{2} d e^{2} - A a^{2} e^{3} - {\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} x}{2 \, {\left (c x^{2} + a\right )} a c^{2}} \]

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)

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maple [B]  time = 0.05, size = 296, normalized size = 1.84 \[ \frac {A \,d^{3} x}{2 \left (c \,x^{2}+a \right ) a}+\frac {A \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}-\frac {3 A d \,e^{2} x}{2 \left (c \,x^{2}+a \right ) c}+\frac {3 A d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}+\frac {B a \,e^{3} x}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {3 B a \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}-\frac {3 B \,d^{2} e x}{2 \left (c \,x^{2}+a \right ) c}+\frac {3 B \,d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}+\frac {A a \,e^{3}}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {3 A \,d^{2} e}{2 \left (c \,x^{2}+a \right ) c}+\frac {A \,e^{3} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {3 B a d \,e^{2}}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {B \,d^{3}}{2 \left (c \,x^{2}+a \right ) c}+\frac {3 B d \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {B \,e^{3} x}{c^{2}} \]

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

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maxima [A]  time = 1.15, size = 188, normalized size = 1.17 \[ \frac {B e^{3} x}{c^{2}} - \frac {B a c d^{3} + 3 \, A a c d^{2} e - 3 \, B a^{2} d e^{2} - A a^{2} e^{3} - {\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} x}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c^{2} d^{3} + 3 \, B a c d^{2} e + 3 \, A a c d e^{2} - 3 \, B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} \]

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]

B*e^3*x/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)/(a*c^3*x^2 + a^2*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)

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mupad [B]  time = 0.19, size = 193, normalized size = 1.20 \[ \frac {\frac {x\,\left (B\,a^2\,e^3-3\,B\,a\,c\,d^2\,e-3\,A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}{2\,a}+\frac {A\,a\,e^3}{2}-\frac {B\,c\,d^3}{2}+\frac {3\,B\,a\,d\,e^2}{2}-\frac {3\,A\,c\,d^2\,e}{2}}{c^3\,x^2+a\,c^2}+\frac {\ln \left (c\,x^2+a\right )\,\left (16\,A\,a^3\,c^3\,e^3+48\,B\,d\,a^3\,c^3\,e^2\right )}{32\,a^3\,c^5}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-3\,B\,a^2\,e^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}{2\,a^{3/2}\,c^{5/2}}+\frac {B\,e^3\,x}{c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((x*(A*c^2*d^3 + B*a^2*e^3 - 3*A*a*c*d*e^2 - 3*B*a*c*d^2*e))/(2*a) + (A*a*e^3)/2 - (B*c*d^3)/2 + (3*B*a*d*e^2)
/2 - (3*A*c*d^2*e)/2)/(a*c^2 + c^3*x^2) + (log(a + c*x^2)*(16*A*a^3*c^3*e^3 + 48*B*a^3*c^3*d*e^2))/(32*a^3*c^5
) + (atan((c^(1/2)*x)/a^(1/2))*(A*c^2*d^3 - 3*B*a^2*e^3 + 3*A*a*c*d*e^2 + 3*B*a*c*d^2*e))/(2*a^(3/2)*c^(5/2))
+ (B*e^3*x)/c^2

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sympy [B]  time = 6.04, size = 583, normalized size = 3.62 \[ \frac {B e^{3} x}{c^{2}} + \left (\frac {e^{2} \left (A e + 3 B d\right )}{2 c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right ) \log {\left (x + \frac {2 A a^{2} e^{3} + 6 B a^{2} d e^{2} - 4 a^{2} c^{2} \left (\frac {e^{2} \left (A e + 3 B d\right )}{2 c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right )}{- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \left (\frac {e^{2} \left (A e + 3 B d\right )}{2 c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right ) \log {\left (x + \frac {2 A a^{2} e^{3} + 6 B a^{2} d e^{2} - 4 a^{2} c^{2} \left (\frac {e^{2} \left (A e + 3 B d\right )}{2 c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e\right )}{4 a^{3} c^{5}}\right )}{- 3 A a c d e^{2} - A c^{2} d^{3} + 3 B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \frac {A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + x \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a^{2} c^{2} + 2 a c^{3} x^{2}} \]

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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