\(\int \frac {(d+e x)^2 (A+B x+C x^2)}{a+c x^2} \, dx\) [44]

Optimal result

Integrand size = 27, antiderivative size = 168 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=-\frac {\left (a C e^2-c \left (C d^2+e (2 B d+A e)\right )\right ) x}{c^2}+\frac {e (2 C d+B e) x^2}{2 c}+\frac {C e^2 x^3}{3 c}+\frac {\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d+2 B e)\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {\left (B c d^2+2 A c d e-2 a C d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1643, 649, 211, 266} \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (2 B e+C d)\right )\right )}{\sqrt {a} c^{5/2}}+\frac {\log \left (a+c x^2\right ) \left (-a B e^2-2 a C d e+2 A c d e+B c d^2\right )}{2 c^2}+\frac {x \left (-a C e^2+c e (A e+2 B d)+c C d^2\right )}{c^2}+\frac {e x^2 (B e+2 C d)}{2 c}+\frac {C e^2 x^3}{3 c} \]

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 266

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 649

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 1643

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d+2 B e)\right )\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {x \left (-6 a C e^2+3 c e (4 B d+2 A e+B e x)+2 c C \left (3 d^2+3 d e x+e^2 x^2\right )\right )+3 \left (B c d^2+2 A c d e-2 a C d e-a B e^2\right ) \log \left (a+c x^2\right )}{6 c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\frac {1}{3} c C \,x^{3} e^{2}+\frac {1}{2} B c \,e^{2} x^{2}+C c d e \,x^{2}+A c \,e^{2} x +2 B c d e x -a C \,e^{2} x +C c \,d^{2} x}{c^{2}}+\frac {\frac {\left (2 A \,c^{2} d e -B \,e^{2} a c +B \,c^{2} d^{2}-2 a c d e C \right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (-A a c \,e^{2}+A \,c^{2} d^{2}-2 B a c d e +a^{2} C \,e^{2}-C a c \,d^{2}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{c^{2}}\)	\(169\)
risch	\(\text {Expression too large to display}\)	\(1241\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.40 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\left [\frac {2 \, C a c^{2} e^{2} x^{3} + 3 \, {\left (2 \, C a c^{2} d e + B a c^{2} e^{2}\right )} x^{2} - 3 \, {\left (2 \, B a c d e + {\left (C a c - A c^{2}\right )} d^{2} - {\left (C a^{2} - A a c\right )} e^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 6 \, {\left (C a c^{2} d^{2} + 2 \, B a c^{2} d e - {\left (C a^{2} c - A a c^{2}\right )} e^{2}\right )} x + 3 \, {\left (B a c^{2} d^{2} - B a^{2} c e^{2} - 2 \, {\left (C a^{2} c - A a c^{2}\right )} d e\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}, \frac {2 \, C a c^{2} e^{2} x^{3} + 3 \, {\left (2 \, C a c^{2} d e + B a c^{2} e^{2}\right )} x^{2} - 6 \, {\left (2 \, B a c d e + {\left (C a c - A c^{2}\right )} d^{2} - {\left (C a^{2} - A a c\right )} e^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 6 \, {\left (C a c^{2} d^{2} + 2 \, B a c^{2} d e - {\left (C a^{2} c - A a c^{2}\right )} e^{2}\right )} x + 3 \, {\left (B a c^{2} d^{2} - B a^{2} c e^{2} - 2 \, {\left (C a^{2} c - A a c^{2}\right )} d e\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 638 vs. \(2 (156) = 312\).

Time = 1.39 (sec) , antiderivative size = 638, normalized size of antiderivative = 3.80 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {C e^{2} x^{3}}{3 c} + x^{2} \left (\frac {B e^{2}}{2 c} + \frac {C d e}{c}\right ) + x \left (\frac {A e^{2}}{c} + \frac {2 B d e}{c} - \frac {C a e^{2}}{c^{2}} + \frac {C d^{2}}{c}\right ) + \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2} + 2 C a d e}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}\right )}{2 a c^{5}}\right ) \log {\left (x + \frac {- 2 A a c d e + B a^{2} e^{2} - B a c d^{2} + 2 C a^{2} d e + 2 a c^{2} \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2} + 2 C a d e}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}\right )}{2 a c^{5}}\right )}{- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}} \right )} + \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2} + 2 C a d e}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}\right )}{2 a c^{5}}\right ) \log {\left (x + \frac {- 2 A a c d e + B a^{2} e^{2} - B a c d^{2} + 2 C a^{2} d e + 2 a c^{2} \left (- \frac {- 2 A c d e + B a e^{2} - B c d^{2} + 2 C a d e}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}\right )}{2 a c^{5}}\right )}{- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e + C a^{2} e^{2} - C a c d^{2}} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {{\left (B c d^{2} - B a e^{2} - 2 \, {\left (C a - A c\right )} d e\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} - \frac {{\left (2 \, B a c d e + {\left (C a c - A c^{2}\right )} d^{2} - {\left (C a^{2} - A a c\right )} e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {2 \, C c e^{2} x^{3} + 3 \, {\left (2 \, C c d e + B c e^{2}\right )} x^{2} + 6 \, {\left (C c d^{2} + 2 \, B c d e - {\left (C a - A c\right )} e^{2}\right )} x}{6 \, c^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=\frac {{\left (B c d^{2} - 2 \, C a d e + 2 \, A c d e - B a e^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} - \frac {{\left (C a c d^{2} - A c^{2} d^{2} + 2 \, B a c d e - C a^{2} e^{2} + A a c e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {2 \, C c^{2} e^{2} x^{3} + 6 \, C c^{2} d e x^{2} + 3 \, B c^{2} e^{2} x^{2} + 6 \, C c^{2} d^{2} x + 12 \, B c^{2} d e x - 6 \, C a c e^{2} x + 6 \, A c^{2} e^{2} x}{6 \, c^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.92 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^2 \left (A+B x+C x^2\right )}{a+c x^2} \, dx=x\,\left (\frac {C\,d^2+2\,B\,d\,e+A\,e^2}{c}-\frac {C\,a\,e^2}{c^2}\right )+\frac {x^2\,\left (B\,e^2+2\,C\,d\,e\right )}{2\,c}+\frac {C\,e^2\,x^3}{3\,c}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-C\,a^2\,e^2+C\,a\,c\,d^2+2\,B\,a\,c\,d\,e+A\,a\,c\,e^2-A\,c^2\,d^2\right )}{\sqrt {a}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (-8\,C\,a^2\,c^3\,d\,e-4\,B\,a^2\,c^3\,e^2+4\,B\,a\,c^4\,d^2+8\,A\,a\,c^4\,d\,e\right )}{8\,a\,c^5} \]

[In]

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

[Out]

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