\(\int \frac {(a+b x^2)^3 (A+B x+C x^2+D x^3)}{x^3} \, dx\) [84]

Optimal result

Integrand size = 28, antiderivative size = 135 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=-\frac {a^3 A}{2 x^2}-\frac {a^3 B}{x}+a^2 (3 b B+a D) x+\frac {3}{2} a b (A b+a C) x^2+a b (b B+a D) x^3+\frac {1}{4} b^2 (A b+3 a C) x^4+\frac {1}{5} b^2 (b B+3 a D) x^5+\frac {1}{6} b^3 C x^6+\frac {1}{7} b^3 D x^7+a^2 (3 A b+a C) \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1816} \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=-\frac {a^3 A}{2 x^2}-\frac {a^3 B}{x}+a^2 \log (x) (a C+3 A b)+a^2 x (a D+3 b B)+\frac {1}{4} b^2 x^4 (3 a C+A b)+\frac {3}{2} a b x^2 (a C+A b)+\frac {1}{5} b^2 x^5 (3 a D+b B)+a b x^3 (a D+b B)+\frac {1}{6} b^3 C x^6+\frac {1}{7} b^3 D x^7 \]

[In]

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

[Out]

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

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 (a^2 (3 b B+a D)+\frac {a^3 A}{x^3}+\frac {a^3 B}{x^2}+\frac {a^2 (3 A b+a C)}{x}+3 a b (A b+a C) x+3 a b (b B+a D) x^2+b^2 (A b+3 a C) x^3+b^2 (b B+3 a D) x^4+b^3 C x^5+b^3 D x^6\right ) \, dx \\ & = -\frac {a^3 A}{2 x^2}-\frac {a^3 B}{x}+a^2 (3 b B+a D) x+\frac {3}{2} a b (A b+a C) x^2+a b (b B+a D) x^3+\frac {1}{4} b^2 (A b+3 a C) x^4+\frac {1}{5} b^2 (b B+3 a D) x^5+\frac {1}{6} b^3 C x^6+\frac {1}{7} b^3 D x^7+a^2 (3 A b+a C) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=-\frac {a^3 \left (A+2 B x-2 D x^3\right )}{2 x^2}+\frac {1}{2} a^2 b x (6 B+x (3 C+2 D x))+\frac {1}{20} a b^2 x^2 (30 A+x (20 B+3 x (5 C+4 D x)))+\frac {1}{420} b^3 x^4 (105 A+2 x (42 B+5 x (7 C+6 D x)))+a^2 (3 A b+a C) \log (x) \]

[In]

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

[Out]

-1/2*(a^3*(A + 2*B*x - 2*D*x^3))/x^2 + (a^2*b*x*(6*B + x*(3*C + 2*D*x)))/2 + (a*b^2*x^2*(30*A + x*(20*B + 3*x*
(5*C + 4*D*x))))/20 + (b^3*x^4*(105*A + 2*x*(42*B + 5*x*(7*C + 6*D*x))))/420 + a^2*(3*A*b + a*C)*Log[x]

Maple [A] (verified)

Time = 3.43 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.05

method	result	size
default	\(\frac {b^{3} D x^{7}}{7}+\frac {b^{3} C \,x^{6}}{6}+\frac {b^{3} B \,x^{5}}{5}+\frac {3 D a \,b^{2} x^{5}}{5}+\frac {A \,b^{3} x^{4}}{4}+\frac {3 C a \,b^{2} x^{4}}{4}+B a \,b^{2} x^{3}+D a^{2} b \,x^{3}+\frac {3 a A \,b^{2} x^{2}}{2}+\frac {3 C \,a^{2} b \,x^{2}}{2}+3 a^{2} b B x +D a^{3} x +a^{2} \left (3 A b +C a \right ) \ln \left (x \right )-\frac {a^{3} B}{x}-\frac {a^{3} A}{2 x^{2}}\)	\(142\)
norman	\(\frac {\left (\frac {1}{5} B \,b^{3}+\frac {3}{5} a \,b^{2} D\right ) x^{7}+\left (\frac {1}{4} b^{3} A +\frac {3}{4} C \,b^{2} a \right ) x^{6}+\left (\frac {3}{2} a \,b^{2} A +\frac {3}{2} C \,a^{2} b \right ) x^{4}+\left (a \,b^{2} B +D a^{2} b \right ) x^{5}+\left (3 a^{2} b B +D a^{3}\right ) x^{3}-\frac {a^{3} A}{2}-a^{3} B x +\frac {b^{3} C \,x^{8}}{6}+\frac {b^{3} D x^{9}}{7}}{x^{2}}+\left (3 a^{2} b A +C \,a^{3}\right ) \ln \left (x \right )\)	\(143\)
parallelrisch	\(\frac {60 b^{3} D x^{9}+70 b^{3} C \,x^{8}+84 b^{3} B \,x^{7}+252 D a \,b^{2} x^{7}+105 x^{6} b^{3} A +315 C a \,b^{2} x^{6}+420 B a \,b^{2} x^{5}+420 D a^{2} b \,x^{5}+630 a A \,b^{2} x^{4}+630 C \,a^{2} b \,x^{4}+1260 A \ln \left (x \right ) x^{2} a^{2} b +1260 B \,a^{2} b \,x^{3}+420 C \ln \left (x \right ) x^{2} a^{3}+420 D a^{3} x^{3}-420 a^{3} B x -210 a^{3} A}{420 x^{2}}\)	\(158\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=\frac {60 \, D b^{3} x^{9} + 70 \, C b^{3} x^{8} + 84 \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{7} + 105 \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{6} + 420 \, {\left (D a^{2} b + B a b^{2}\right )} x^{5} - 420 \, B a^{3} x + 630 \, {\left (C a^{2} b + A a b^{2}\right )} x^{4} - 210 \, A a^{3} + 420 \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{3} + 420 \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2} \log \left (x\right )}{420 \, x^{2}} \]

[In]

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

[Out]

1/420*(60*D*b^3*x^9 + 70*C*b^3*x^8 + 84*(3*D*a*b^2 + B*b^3)*x^7 + 105*(3*C*a*b^2 + A*b^3)*x^6 + 420*(D*a^2*b +
B*a*b^2)*x^5 - 420*B*a^3*x + 630*(C*a^2*b + A*a*b^2)*x^4 - 210*A*a^3 + 420*(D*a^3 + 3*B*a^2*b)*x^3 + 420*(C*a
^3 + 3*A*a^2*b)*x^2*log(x))/x^2

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=\frac {C b^{3} x^{6}}{6} + \frac {D b^{3} x^{7}}{7} + a^{2} \cdot \left (3 A b + C a\right ) \log {\left (x \right )} + x^{5} \left (\frac {B b^{3}}{5} + \frac {3 D a b^{2}}{5}\right ) + x^{4} \left (\frac {A b^{3}}{4} + \frac {3 C a b^{2}}{4}\right ) + x^{3} \left (B a b^{2} + D a^{2} b\right ) + x^{2} \cdot \left (\frac {3 A a b^{2}}{2} + \frac {3 C a^{2} b}{2}\right ) + x \left (3 B a^{2} b + D a^{3}\right ) + \frac {- A a^{3} - 2 B a^{3} x}{2 x^{2}} \]

[In]

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

[Out]

C*b**3*x**6/6 + D*b**3*x**7/7 + a**2*(3*A*b + C*a)*log(x) + x**5*(B*b**3/5 + 3*D*a*b**2/5) + x**4*(A*b**3/4 +
3*C*a*b**2/4) + x**3*(B*a*b**2 + D*a**2*b) + x**2*(3*A*a*b**2/2 + 3*C*a**2*b/2) + x*(3*B*a**2*b + D*a**3) + (-
A*a**3 - 2*B*a**3*x)/(2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=\frac {1}{7} \, D b^{3} x^{7} + \frac {1}{6} \, C b^{3} x^{6} + \frac {1}{5} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{5} + \frac {1}{4} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{4} + {\left (D a^{2} b + B a b^{2}\right )} x^{3} + \frac {3}{2} \, {\left (C a^{2} b + A a b^{2}\right )} x^{2} + {\left (D a^{3} + 3 \, B a^{2} b\right )} x + {\left (C a^{3} + 3 \, A a^{2} b\right )} \log \left (x\right ) - \frac {2 \, B a^{3} x + A a^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=\frac {1}{7} \, D b^{3} x^{7} + \frac {1}{6} \, C b^{3} x^{6} + \frac {3}{5} \, D a b^{2} x^{5} + \frac {1}{5} \, B b^{3} x^{5} + \frac {3}{4} \, C a b^{2} x^{4} + \frac {1}{4} \, A b^{3} x^{4} + D a^{2} b x^{3} + B a b^{2} x^{3} + \frac {3}{2} \, C a^{2} b x^{2} + \frac {3}{2} \, A a b^{2} x^{2} + D a^{3} x + 3 \, B a^{2} b x + {\left (C a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, B a^{3} x + A a^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^3} \, dx=\frac {A\,b^3\,x^4}{4}-\frac {B\,a^3}{x}-\frac {A\,a^3}{2\,x^2}+\frac {B\,b^3\,x^5}{5}+\frac {C\,b^3\,x^6}{6}+C\,a^3\,\ln \left (x\right )+a^3\,x\,D+\frac {b^3\,x^7\,D}{7}+a^2\,b\,x^3\,D+\frac {3\,a\,b^2\,x^5\,D}{5}+3\,B\,a^2\,b\,x+\frac {3\,A\,a\,b^2\,x^2}{2}+B\,a\,b^2\,x^3+\frac {3\,C\,a^2\,b\,x^2}{2}+\frac {3\,C\,a\,b^2\,x^4}{4}+3\,A\,a^2\,b\,\ln \left (x\right ) \]

[In]

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

[Out]

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