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

Optimal. Leaf size=240 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (B e+3 C d)-c d^2 (3 B e+C d)\right )\right )}{\sqrt {a} c^{5/2}}+\frac {\log \left (a+c x^2\right ) \left (e (A c-a C) \left (3 c d^2-a e^2\right )+B c d \left (c d^2-3 a e^2\right )\right )}{2 c^3}-\frac {e x^2 \left (a C e^2-c \left (e (A e+3 B d)+3 C d^2\right )\right )}{2 c^2}-\frac {x \left (a e^2 (B e+3 C d)-c d \left (3 e (A e+B d)+C d^2\right )\right )}{c^2}+\frac {e^2 x^3 (B e+3 C d)}{3 c}+\frac {C e^3 x^4}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 237, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1629, 635, 205, 260} \[ \frac {e x^2 \left (-a C e^2+c e (A e+3 B d)+3 c C d^2\right )}{2 c^2}+\frac {\log \left (a+c x^2\right ) \left (e (A c-a C) \left (3 c d^2-a e^2\right )+B c d \left (c d^2-3 a e^2\right )\right )}{2 c^3}+\frac {x \left (-a e^2 (B e+3 C d)+3 c d e (A e+B d)+c C d^3\right )}{c^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (B e+3 C d)-c d^2 (3 B e+C d)\right )\right )}{\sqrt {a} c^{5/2}}+\frac {e^2 x^3 (B e+3 C d)}{3 c}+\frac {C e^3 x^4}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 223, normalized size = 0.93 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a \left (a e^2 (B e+3 C d)-c d^2 (3 B e+C d)\right )\right )}{\sqrt {a} c^{5/2}}+\frac {6 \log \left (a+c x^2\right ) \left (e (A c-a C) \left (3 c d^2-a e^2\right )+B c d \left (c d^2-3 a e^2\right )\right )+c x \left (-6 a e^2 (2 B e+6 C d+C e x)+2 c e \left (3 A e (6 d+e x)+B \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 c C \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right )}{12 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*c*d*(c*d^2 - 3*a*e^2) + a*(a*e^2*(3*C*d + B*e) - c*d^2*(C*d + 3*B*e)))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[
a]*c^(5/2)) + (c*x*(-6*a*e^2*(6*C*d + 2*B*e + C*e*x) + 3*c*C*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 2*c
*e*(3*A*e*(6*d + e*x) + B*(18*d^2 + 9*d*e*x + 2*e^2*x^2))) + 6*(B*c*d*(c*d^2 - 3*a*e^2) + (A*c - a*C)*e*(3*c*d
^2 - a*e^2))*Log[a + c*x^2])/(12*c^3)

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fricas [A]  time = 0.90, size = 592, normalized size = 2.47 \[ \left [\frac {3 \, C a c^{2} e^{3} x^{4} + 4 \, {\left (3 \, C a c^{2} d e^{2} + B a c^{2} e^{3}\right )} x^{3} + 6 \, {\left (3 \, C a c^{2} d^{2} e + 3 \, B a c^{2} d e^{2} - {\left (C a^{2} c - A a c^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (3 \, B a c d^{2} e - B a^{2} e^{3} + {\left (C a c - A c^{2}\right )} d^{3} - 3 \, {\left (C a^{2} - A a c\right )} d e^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 12 \, {\left (C a c^{2} d^{3} + 3 \, B a c^{2} d^{2} e - B a^{2} c e^{3} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d e^{2}\right )} x + 6 \, {\left (B a c^{2} d^{3} - 3 \, B a^{2} c d e^{2} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d^{2} e + {\left (C a^{3} - A a^{2} c\right )} e^{3}\right )} \log \left (c x^{2} + a\right )}{12 \, a c^{3}}, \frac {3 \, C a c^{2} e^{3} x^{4} + 4 \, {\left (3 \, C a c^{2} d e^{2} + B a c^{2} e^{3}\right )} x^{3} + 6 \, {\left (3 \, C a c^{2} d^{2} e + 3 \, B a c^{2} d e^{2} - {\left (C a^{2} c - A a c^{2}\right )} e^{3}\right )} x^{2} - 12 \, {\left (3 \, B a c d^{2} e - B a^{2} e^{3} + {\left (C a c - A c^{2}\right )} d^{3} - 3 \, {\left (C a^{2} - A a c\right )} d e^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 12 \, {\left (C a c^{2} d^{3} + 3 \, B a c^{2} d^{2} e - B a^{2} c e^{3} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d e^{2}\right )} x + 6 \, {\left (B a c^{2} d^{3} - 3 \, B a^{2} c d e^{2} - 3 \, {\left (C a^{2} c - A a c^{2}\right )} d^{2} e + {\left (C a^{3} - A a^{2} c\right )} e^{3}\right )} \log \left (c x^{2} + a\right )}{12 \, a c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*C*a*c^2*e^3*x^4 + 4*(3*C*a*c^2*d*e^2 + B*a*c^2*e^3)*x^3 + 6*(3*C*a*c^2*d^2*e + 3*B*a*c^2*d*e^2 - (C*a
^2*c - A*a*c^2)*e^3)*x^2 + 6*(3*B*a*c*d^2*e - B*a^2*e^3 + (C*a*c - A*c^2)*d^3 - 3*(C*a^2 - A*a*c)*d*e^2)*sqrt(
-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 12*(C*a*c^2*d^3 + 3*B*a*c^2*d^2*e - B*a^2*c*e^3 - 3*(C*a
^2*c - A*a*c^2)*d*e^2)*x + 6*(B*a*c^2*d^3 - 3*B*a^2*c*d*e^2 - 3*(C*a^2*c - A*a*c^2)*d^2*e + (C*a^3 - A*a^2*c)*
e^3)*log(c*x^2 + a))/(a*c^3), 1/12*(3*C*a*c^2*e^3*x^4 + 4*(3*C*a*c^2*d*e^2 + B*a*c^2*e^3)*x^3 + 6*(3*C*a*c^2*d
^2*e + 3*B*a*c^2*d*e^2 - (C*a^2*c - A*a*c^2)*e^3)*x^2 - 12*(3*B*a*c*d^2*e - B*a^2*e^3 + (C*a*c - A*c^2)*d^3 -
3*(C*a^2 - A*a*c)*d*e^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 12*(C*a*c^2*d^3 + 3*B*a*c^2*d^2*e - B*a^2*c*e^3 - 3
*(C*a^2*c - A*a*c^2)*d*e^2)*x + 6*(B*a*c^2*d^3 - 3*B*a^2*c*d*e^2 - 3*(C*a^2*c - A*a*c^2)*d^2*e + (C*a^3 - A*a^
2*c)*e^3)*log(c*x^2 + a))/(a*c^3)]

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giac [A]  time = 0.17, size = 279, normalized size = 1.16 \[ -\frac {{\left (C a c d^{3} - A c^{2} d^{3} + 3 \, B a c d^{2} e - 3 \, C a^{2} d e^{2} + 3 \, A a c d e^{2} - B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {{\left (B c^{2} d^{3} - 3 \, C a c d^{2} e + 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + C a^{2} e^{3} - A a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {3 \, C c^{3} x^{4} e^{3} + 12 \, C c^{3} d x^{3} e^{2} + 18 \, C c^{3} d^{2} x^{2} e + 12 \, C c^{3} d^{3} x + 4 \, B c^{3} x^{3} e^{3} + 18 \, B c^{3} d x^{2} e^{2} + 36 \, B c^{3} d^{2} x e - 6 \, C a c^{2} x^{2} e^{3} + 6 \, A c^{3} x^{2} e^{3} - 36 \, C a c^{2} d x e^{2} + 36 \, A c^{3} d x e^{2} - 12 \, B a c^{2} x e^{3}}{12 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(C*a*c*d^3 - A*c^2*d^3 + 3*B*a*c*d^2*e - 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 - B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sq
rt(a*c)*c^2) + 1/2*(B*c^2*d^3 - 3*C*a*c*d^2*e + 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + C*a^2*e^3 - A*a*c*e^3)*log(c*x
^2 + a)/c^3 + 1/12*(3*C*c^3*x^4*e^3 + 12*C*c^3*d*x^3*e^2 + 18*C*c^3*d^2*x^2*e + 12*C*c^3*d^3*x + 4*B*c^3*x^3*e
^3 + 18*B*c^3*d*x^2*e^2 + 36*B*c^3*d^2*x*e - 6*C*a*c^2*x^2*e^3 + 6*A*c^3*x^2*e^3 - 36*C*a*c^2*d*x*e^2 + 36*A*c
^3*d*x*e^2 - 12*B*a*c^2*x*e^3)/c^4

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maple [A]  time = 0.01, size = 399, normalized size = 1.66 \[ \frac {C \,e^{3} x^{4}}{4 c}+\frac {B \,e^{3} x^{3}}{3 c}+\frac {C d \,e^{2} x^{3}}{c}-\frac {3 A a d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {A \,e^{3} x^{2}}{2 c}+\frac {A \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}+\frac {B \,a^{2} e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {3 B a \,d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {3 B d \,e^{2} x^{2}}{2 c}+\frac {3 C \,a^{2} d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {C a \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}-\frac {C a \,e^{3} x^{2}}{2 c^{2}}+\frac {3 C \,d^{2} e \,x^{2}}{2 c}-\frac {A a \,e^{3} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {3 A \,d^{2} e \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {3 A d \,e^{2} x}{c}-\frac {3 B a d \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}-\frac {B a \,e^{3} x}{c^{2}}+\frac {B \,d^{3} \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {3 B \,d^{2} e x}{c}+\frac {C \,a^{2} e^{3} \ln \left (c \,x^{2}+a \right )}{2 c^{3}}-\frac {3 C a \,d^{2} e \ln \left (c \,x^{2}+a \right )}{2 c^{2}}-\frac {3 C a d \,e^{2} x}{c^{2}}+\frac {C \,d^{3} x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 244, normalized size = 1.02 \[ -\frac {{\left (3 \, B a c d^{2} e - B a^{2} e^{3} + {\left (C a c - A c^{2}\right )} d^{3} - 3 \, {\left (C a^{2} - A a c\right )} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {3 \, C c e^{3} x^{4} + 4 \, {\left (3 \, C c d e^{2} + B c e^{3}\right )} x^{3} + 6 \, {\left (3 \, C c d^{2} e + 3 \, B c d e^{2} - {\left (C a - A c\right )} e^{3}\right )} x^{2} + 12 \, {\left (C c d^{3} + 3 \, B c d^{2} e - B a e^{3} - 3 \, {\left (C a - A c\right )} d e^{2}\right )} x}{12 \, c^{2}} + \frac {{\left (B c^{2} d^{3} - 3 \, B a c d e^{2} - 3 \, {\left (C a c - A c^{2}\right )} d^{2} e + {\left (C a^{2} - A a c\right )} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.99, size = 277, normalized size = 1.15 \[ x^2\,\left (\frac {3\,C\,d^2\,e+3\,B\,d\,e^2+A\,e^3}{2\,c}-\frac {C\,a\,e^3}{2\,c^2}\right )+x\,\left (\frac {C\,d^3+3\,B\,d^2\,e+3\,A\,d\,e^2}{c}-\frac {a\,\left (B\,e^3+3\,C\,d\,e^2\right )}{c^2}\right )+\frac {x^3\,\left (B\,e^3+3\,C\,d\,e^2\right )}{3\,c}+\frac {C\,e^3\,x^4}{4\,c}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,C\,a^2\,d\,e^2+B\,a^2\,e^3-C\,a\,c\,d^3-3\,B\,a\,c\,d^2\,e-3\,A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}{\sqrt {a}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (4\,C\,a^3\,c^3\,e^3-12\,C\,a^2\,c^4\,d^2\,e-12\,B\,a^2\,c^4\,d\,e^2-4\,A\,a^2\,c^4\,e^3+4\,B\,a\,c^5\,d^3+12\,A\,a\,c^5\,d^2\,e\right )}{8\,a\,c^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 5.46, size = 1008, normalized size = 4.20 \[ \frac {C e^{3} x^{4}}{4 c} + x^{3} \left (\frac {B e^{3}}{3 c} + \frac {C d e^{2}}{c}\right ) + x^{2} \left (\frac {A e^{3}}{2 c} + \frac {3 B d e^{2}}{2 c} - \frac {C a e^{3}}{2 c^{2}} + \frac {3 C d^{2} e}{2 c}\right ) + x \left (\frac {3 A d e^{2}}{c} - \frac {B a e^{3}}{c^{2}} + \frac {3 B d^{2} e}{c} - \frac {3 C a d e^{2}}{c^{2}} + \frac {C d^{3}}{c}\right ) + \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} - \frac {\sqrt {- a c^{7}} \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 + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right ) \log {\left (x + \frac {A a^{2} c e^{3} - 3 A a c^{2} d^{2} e + 3 B a^{2} c d e^{2} - B a c^{2} d^{3} - C a^{3} e^{3} + 3 C a^{2} c d^{2} e + 2 a c^{3} \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} - \frac {\sqrt {- a c^{7}} \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 + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right )}{- 3 A a c^{2} d e^{2} + A c^{3} d^{3} + B a^{2} c e^{3} - 3 B a c^{2} d^{2} e + 3 C a^{2} c d e^{2} - C a c^{2} d^{3}} \right )} + \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} + \frac {\sqrt {- a c^{7}} \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 + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right ) \log {\left (x + \frac {A a^{2} c e^{3} - 3 A a c^{2} d^{2} e + 3 B a^{2} c d e^{2} - B a c^{2} d^{3} - C a^{3} e^{3} + 3 C a^{2} c d^{2} e + 2 a c^{3} \left (\frac {- A a c e^{3} + 3 A c^{2} d^{2} e - 3 B a c d e^{2} + B c^{2} d^{3} + C a^{2} e^{3} - 3 C a c d^{2} e}{2 c^{3}} + \frac {\sqrt {- a c^{7}} \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 + 3 C a^{2} d e^{2} - C a c d^{3}\right )}{2 a c^{6}}\right )}{- 3 A a c^{2} d e^{2} + A c^{3} d^{3} + B a^{2} c e^{3} - 3 B a c^{2} d^{2} e + 3 C a^{2} c d e^{2} - C a c^{2} d^{3}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*e**3*x**4/(4*c) + x**3*(B*e**3/(3*c) + C*d*e**2/c) + x**2*(A*e**3/(2*c) + 3*B*d*e**2/(2*c) - C*a*e**3/(2*c**
2) + 3*C*d**2*e/(2*c)) + x*(3*A*d*e**2/c - B*a*e**3/c**2 + 3*B*d**2*e/c - 3*C*a*d*e**2/c**2 + C*d**3/c) + ((-A
*a*c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*a**2*e**3 - 3*C*a*c*d**2*e)/(2*c**3) - sqrt(-a*
c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e + 3*C*a**2*d*e**2 - C*a*c*d**3)/(2*a*c**6)
)*log(x + (A*a**2*c*e**3 - 3*A*a*c**2*d**2*e + 3*B*a**2*c*d*e**2 - B*a*c**2*d**3 - C*a**3*e**3 + 3*C*a**2*c*d*
*2*e + 2*a*c**3*((-A*a*c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*a**2*e**3 - 3*C*a*c*d**2*e)
/(2*c**3) - sqrt(-a*c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e + 3*C*a**2*d*e**2 - C*
a*c*d**3)/(2*a*c**6)))/(-3*A*a*c**2*d*e**2 + A*c**3*d**3 + B*a**2*c*e**3 - 3*B*a*c**2*d**2*e + 3*C*a**2*c*d*e*
*2 - C*a*c**2*d**3)) + ((-A*a*c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*a**2*e**3 - 3*C*a*c*
d**2*e)/(2*c**3) + sqrt(-a*c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d**2*e + 3*C*a**2*d*e*
*2 - C*a*c*d**3)/(2*a*c**6))*log(x + (A*a**2*c*e**3 - 3*A*a*c**2*d**2*e + 3*B*a**2*c*d*e**2 - B*a*c**2*d**3 -
C*a**3*e**3 + 3*C*a**2*c*d**2*e + 2*a*c**3*((-A*a*c*e**3 + 3*A*c**2*d**2*e - 3*B*a*c*d*e**2 + B*c**2*d**3 + C*
a**2*e**3 - 3*C*a*c*d**2*e)/(2*c**3) + sqrt(-a*c**7)*(-3*A*a*c*d*e**2 + A*c**2*d**3 + B*a**2*e**3 - 3*B*a*c*d*
*2*e + 3*C*a**2*d*e**2 - C*a*c*d**3)/(2*a*c**6)))/(-3*A*a*c**2*d*e**2 + A*c**3*d**3 + B*a**2*c*e**3 - 3*B*a*c*
*2*d**2*e + 3*C*a**2*c*d*e**2 - C*a*c**2*d**3))

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