\(\int \frac {x^4 (A+B x+C x^2+D x^3)}{a+b x^2} \, dx\) [86]

Optimal result

Integrand size = 28, antiderivative size = 151 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=-\frac {a (A b-a C) x}{b^3}-\frac {a (b B-a D) x^2}{2 b^3}+\frac {(A b-a C) x^3}{3 b^2}+\frac {(b B-a D) x^4}{4 b^2}+\frac {C x^5}{5 b}+\frac {D x^6}{6 b}+\frac {a^{3/2} (A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 (b B-a D) \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

-a*(A*b-C*a)*x/b^3-1/2*a*(B*b-D*a)*x^2/b^3+1/3*(A*b-C*a)*x^3/b^2+1/4*(B*b-D*a)*x^4/b^2+1/5*C*x^5/b+1/6*D*x^6/b
+a^(3/2)*(A*b-C*a)*arctan(x*b^(1/2)/a^(1/2))/b^(7/2)+1/2*a^2*(B*b-D*a)*ln(b*x^2+a)/b^4

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1816, 649, 211, 266} \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=\frac {a^{3/2} (A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 (b B-a D) \log \left (a+b x^2\right )}{2 b^4}-\frac {a x (A b-a C)}{b^3}+\frac {x^3 (A b-a C)}{3 b^2}-\frac {a x^2 (b B-a D)}{2 b^3}+\frac {x^4 (b B-a D)}{4 b^2}+\frac {C x^5}{5 b}+\frac {D x^6}{6 b} \]

[In]

Int[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2),x]

[Out]

-((a*(A*b - a*C)*x)/b^3) - (a*(b*B - a*D)*x^2)/(2*b^3) + ((A*b - a*C)*x^3)/(3*b^2) + ((b*B - a*D)*x^4)/(4*b^2)
+ (C*x^5)/(5*b) + (D*x^6)/(6*b) + (a^(3/2)*(A*b - a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(7/2) + (a^2*(b*B - a*D
)*Log[a + b*x^2])/(2*b^4)

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 1816

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a (A b-a C)}{b^3}-\frac {a (b B-a D) x}{b^3}+\frac {(A b-a C) x^2}{b^2}+\frac {(b B-a D) x^3}{b^2}+\frac {C x^4}{b}+\frac {D x^5}{b}+\frac {a^2 (A b-a C)+a^2 (b B-a D) x}{b^3 \left (a+b x^2\right )}\right ) \, dx \\ & = -\frac {a (A b-a C) x}{b^3}-\frac {a (b B-a D) x^2}{2 b^3}+\frac {(A b-a C) x^3}{3 b^2}+\frac {(b B-a D) x^4}{4 b^2}+\frac {C x^5}{5 b}+\frac {D x^6}{6 b}+\frac {\int \frac {a^2 (A b-a C)+a^2 (b B-a D) x}{a+b x^2} \, dx}{b^3} \\ & = -\frac {a (A b-a C) x}{b^3}-\frac {a (b B-a D) x^2}{2 b^3}+\frac {(A b-a C) x^3}{3 b^2}+\frac {(b B-a D) x^4}{4 b^2}+\frac {C x^5}{5 b}+\frac {D x^6}{6 b}+\frac {\left (a^2 (A b-a C)\right ) \int \frac {1}{a+b x^2} \, dx}{b^3}+\frac {\left (a^2 (b B-a D)\right ) \int \frac {x}{a+b x^2} \, dx}{b^3} \\ & = -\frac {a (A b-a C) x}{b^3}-\frac {a (b B-a D) x^2}{2 b^3}+\frac {(A b-a C) x^3}{3 b^2}+\frac {(b B-a D) x^4}{4 b^2}+\frac {C x^5}{5 b}+\frac {D x^6}{6 b}+\frac {a^{3/2} (A b-a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{7/2}}+\frac {a^2 (b B-a D) \log \left (a+b x^2\right )}{2 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=\frac {b x \left (30 a^2 (2 C+D x)-5 a b (12 A+x (6 B+x (4 C+3 D x)))+b^2 x^2 (20 A+x (15 B+2 x (6 C+5 D x)))\right )-60 a^{3/2} \sqrt {b} (-A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-30 a^2 (-b B+a D) \log \left (a+b x^2\right )}{60 b^4} \]

[In]

Integrate[(x^4*(A + B*x + C*x^2 + D*x^3))/(a + b*x^2),x]

[Out]

(b*x*(30*a^2*(2*C + D*x) - 5*a*b*(12*A + x*(6*B + x*(4*C + 3*D*x))) + b^2*x^2*(20*A + x*(15*B + 2*x*(6*C + 5*D
*x)))) - 60*a^(3/2)*Sqrt[b]*(-(A*b) + a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]] - 30*a^2*(-(b*B) + a*D)*Log[a + b*x^2])
/(60*b^4)

Maple [A] (verified)

Time = 3.44 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {-\frac {1}{6} b^{2} D x^{6}-\frac {1}{5} b^{2} C \,x^{5}-\frac {1}{4} b^{2} B \,x^{4}+\frac {1}{4} D a b \,x^{4}-\frac {1}{3} A \,b^{2} x^{3}+\frac {1}{3} C a b \,x^{3}+\frac {1}{2} B a b \,x^{2}-\frac {1}{2} D a^{2} x^{2}+a A b x -C \,a^{2} x}{b^{3}}+\frac {a^{2} \left (\frac {\left (B b -D a \right ) \ln \left (b \,x^{2}+a \right )}{2 b}+\frac {\left (A b -C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{b^{3}}\)	\(141\)

[In]

int(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

-1/b^3*(-1/6*b^2*D*x^6-1/5*b^2*C*x^5-1/4*b^2*B*x^4+1/4*D*a*b*x^4-1/3*A*b^2*x^3+1/3*C*a*b*x^3+1/2*B*a*b*x^2-1/2
*D*a^2*x^2+a*A*b*x-C*a^2*x)+a^2/b^3*(1/2*(B*b-D*a)/b*ln(b*x^2+a)+(A*b-C*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))
)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.20 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=\left [\frac {10 \, D b^{3} x^{6} + 12 \, C b^{3} x^{5} - 15 \, {\left (D a b^{2} - B b^{3}\right )} x^{4} - 20 \, {\left (C a b^{2} - A b^{3}\right )} x^{3} + 30 \, {\left (D a^{2} b - B a b^{2}\right )} x^{2} - 30 \, {\left (C a^{2} b - A a b^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right ) + 60 \, {\left (C a^{2} b - A a b^{2}\right )} x - 30 \, {\left (D a^{3} - B a^{2} b\right )} \log \left (b x^{2} + a\right )}{60 \, b^{4}}, \frac {10 \, D b^{3} x^{6} + 12 \, C b^{3} x^{5} - 15 \, {\left (D a b^{2} - B b^{3}\right )} x^{4} - 20 \, {\left (C a b^{2} - A b^{3}\right )} x^{3} + 30 \, {\left (D a^{2} b - B a b^{2}\right )} x^{2} - 60 \, {\left (C a^{2} b - A a b^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right ) + 60 \, {\left (C a^{2} b - A a b^{2}\right )} x - 30 \, {\left (D a^{3} - B a^{2} b\right )} \log \left (b x^{2} + a\right )}{60 \, b^{4}}\right ] \]

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a),x, algorithm="fricas")

[Out]

[1/60*(10*D*b^3*x^6 + 12*C*b^3*x^5 - 15*(D*a*b^2 - B*b^3)*x^4 - 20*(C*a*b^2 - A*b^3)*x^3 + 30*(D*a^2*b - B*a*b
^2)*x^2 - 30*(C*a^2*b - A*a*b^2)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 60*(C*a^2*b - A*
a*b^2)*x - 30*(D*a^3 - B*a^2*b)*log(b*x^2 + a))/b^4, 1/60*(10*D*b^3*x^6 + 12*C*b^3*x^5 - 15*(D*a*b^2 - B*b^3)*
x^4 - 20*(C*a*b^2 - A*b^3)*x^3 + 30*(D*a^2*b - B*a*b^2)*x^2 - 60*(C*a^2*b - A*a*b^2)*sqrt(a/b)*arctan(b*x*sqrt
(a/b)/a) + 60*(C*a^2*b - A*a*b^2)*x - 30*(D*a^3 - B*a^2*b)*log(b*x^2 + a))/b^4]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (134) = 268\).

Time = 0.55 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.09 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=\frac {C x^{5}}{5 b} + \frac {D x^{6}}{6 b} + x^{4} \left (\frac {B}{4 b} - \frac {D a}{4 b^{2}}\right ) + x^{3} \left (\frac {A}{3 b} - \frac {C a}{3 b^{2}}\right ) + x^{2} \left (- \frac {B a}{2 b^{2}} + \frac {D a^{2}}{2 b^{3}}\right ) + x \left (- \frac {A a}{b^{2}} + \frac {C a^{2}}{b^{3}}\right ) + \left (- \frac {a^{2} \left (- B b + D a\right )}{2 b^{4}} - \frac {\sqrt {- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right ) \log {\left (x + \frac {B a^{2} b - D a^{3} - 2 b^{4} \left (- \frac {a^{2} \left (- B b + D a\right )}{2 b^{4}} - \frac {\sqrt {- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right )}{- A a b^{2} + C a^{2} b} \right )} + \left (- \frac {a^{2} \left (- B b + D a\right )}{2 b^{4}} + \frac {\sqrt {- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right ) \log {\left (x + \frac {B a^{2} b - D a^{3} - 2 b^{4} \left (- \frac {a^{2} \left (- B b + D a\right )}{2 b^{4}} + \frac {\sqrt {- a^{3} b^{9}} \left (- A b + C a\right )}{2 b^{8}}\right )}{- A a b^{2} + C a^{2} b} \right )} \]

[In]

integrate(x**4*(D*x**3+C*x**2+B*x+A)/(b*x**2+a),x)

[Out]

C*x**5/(5*b) + D*x**6/(6*b) + x**4*(B/(4*b) - D*a/(4*b**2)) + x**3*(A/(3*b) - C*a/(3*b**2)) + x**2*(-B*a/(2*b*
*2) + D*a**2/(2*b**3)) + x*(-A*a/b**2 + C*a**2/b**3) + (-a**2*(-B*b + D*a)/(2*b**4) - sqrt(-a**3*b**9)*(-A*b +
C*a)/(2*b**8))*log(x + (B*a**2*b - D*a**3 - 2*b**4*(-a**2*(-B*b + D*a)/(2*b**4) - sqrt(-a**3*b**9)*(-A*b + C*
a)/(2*b**8)))/(-A*a*b**2 + C*a**2*b)) + (-a**2*(-B*b + D*a)/(2*b**4) + sqrt(-a**3*b**9)*(-A*b + C*a)/(2*b**8))
*log(x + (B*a**2*b - D*a**3 - 2*b**4*(-a**2*(-B*b + D*a)/(2*b**4) + sqrt(-a**3*b**9)*(-A*b + C*a)/(2*b**8)))/(
-A*a*b**2 + C*a**2*b))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=-\frac {{\left (C a^{3} - A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {10 \, D b^{2} x^{6} + 12 \, C b^{2} x^{5} - 15 \, {\left (D a b - B b^{2}\right )} x^{4} - 20 \, {\left (C a b - A b^{2}\right )} x^{3} + 30 \, {\left (D a^{2} - B a b\right )} x^{2} + 60 \, {\left (C a^{2} - A a b\right )} x}{60 \, b^{3}} - \frac {{\left (D a^{3} - B a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a),x, algorithm="maxima")

[Out]

-(C*a^3 - A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/60*(10*D*b^2*x^6 + 12*C*b^2*x^5 - 15*(D*a*b - B*b
^2)*x^4 - 20*(C*a*b - A*b^2)*x^3 + 30*(D*a^2 - B*a*b)*x^2 + 60*(C*a^2 - A*a*b)*x)/b^3 - 1/2*(D*a^3 - B*a^2*b)*
log(b*x^2 + a)/b^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.07 \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=-\frac {{\left (C a^{3} - A a^{2} b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {{\left (D a^{3} - B a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} + \frac {10 \, D b^{5} x^{6} + 12 \, C b^{5} x^{5} - 15 \, D a b^{4} x^{4} + 15 \, B b^{5} x^{4} - 20 \, C a b^{4} x^{3} + 20 \, A b^{5} x^{3} + 30 \, D a^{2} b^{3} x^{2} - 30 \, B a b^{4} x^{2} + 60 \, C a^{2} b^{3} x - 60 \, A a b^{4} x}{60 \, b^{6}} \]

[In]

integrate(x^4*(D*x^3+C*x^2+B*x+A)/(b*x^2+a),x, algorithm="giac")

[Out]

-(C*a^3 - A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/2*(D*a^3 - B*a^2*b)*log(b*x^2 + a)/b^4 + 1/60*(10
*D*b^5*x^6 + 12*C*b^5*x^5 - 15*D*a*b^4*x^4 + 15*B*b^5*x^4 - 20*C*a*b^4*x^3 + 20*A*b^5*x^3 + 30*D*a^2*b^3*x^2 -
30*B*a*b^4*x^2 + 60*C*a^2*b^3*x - 60*A*a*b^4*x)/b^6

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (A+B x+C x^2+D x^3\right )}{a+b x^2} \, dx=\int \frac {x^4\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{b\,x^2+a} \,d x \]

[In]

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2),x)

[Out]

int((x^4*(A + B*x + C*x^2 + x^3*D))/(a + b*x^2), x)