\(\int \frac {x^3 (a+b x^2)^2}{c+d x^2} \, dx\) [169]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {(b c-a d)^2 x^2}{2 d^3}-\frac {b (b c-2 a d) x^4}{4 d^2}+\frac {b^2 x^6}{6 d}-\frac {c (b c-a d)^2 \log \left (c+d x^2\right )}{2 d^4} \]

[Out]

1/2*(-a*d+b*c)^2*x^2/d^3-1/4*b*(-2*a*d+b*c)*x^4/d^2+1/6*b^2*x^6/d-1/2*c*(-a*d+b*c)^2*ln(d*x^2+c)/d^4

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=-\frac {c (b c-a d)^2 \log \left (c+d x^2\right )}{2 d^4}+\frac {x^2 (b c-a d)^2}{2 d^3}-\frac {b x^4 (b c-2 a d)}{4 d^2}+\frac {b^2 x^6}{6 d} \]

[In]

Int[(x^3*(a + b*x^2)^2)/(c + d*x^2),x]

[Out]

((b*c - a*d)^2*x^2)/(2*d^3) - (b*(b*c - 2*a*d)*x^4)/(4*d^2) + (b^2*x^6)/(6*d) - (c*(b*c - a*d)^2*Log[c + d*x^2
])/(2*d^4)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x (a+b x)^2}{c+d x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {(-b c+a d)^2}{d^3}-\frac {b (b c-2 a d) x}{d^2}+\frac {b^2 x^2}{d}-\frac {c (b c-a d)^2}{d^3 (c+d x)}\right ) \, dx,x,x^2\right ) \\ & = \frac {(b c-a d)^2 x^2}{2 d^3}-\frac {b (b c-2 a d) x^4}{4 d^2}+\frac {b^2 x^6}{6 d}-\frac {c (b c-a d)^2 \log \left (c+d x^2\right )}{2 d^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {d x^2 \left (6 a^2 d^2+6 a b d \left (-2 c+d x^2\right )+b^2 \left (6 c^2-3 c d x^2+2 d^2 x^4\right )\right )-6 c (b c-a d)^2 \log \left (c+d x^2\right )}{12 d^4} \]

[In]

Integrate[(x^3*(a + b*x^2)^2)/(c + d*x^2),x]

[Out]

(d*x^2*(6*a^2*d^2 + 6*a*b*d*(-2*c + d*x^2) + b^2*(6*c^2 - 3*c*d*x^2 + 2*d^2*x^4)) - 6*c*(b*c - a*d)^2*Log[c +
d*x^2])/(12*d^4)

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20

method	result	size
norman	\(\frac {b^{2} x^{6}}{6 d}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 d^{3}}+\frac {b \left (2 a d -b c \right ) x^{4}}{4 d^{2}}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (d \,x^{2}+c \right )}{2 d^{4}}\)	\(95\)
default	\(\frac {\frac {1}{3} b^{2} d^{2} x^{6}+x^{4} a b \,d^{2}-\frac {1}{2} x^{4} b^{2} c d +a^{2} d^{2} x^{2}-2 a b c d \,x^{2}+b^{2} c^{2} x^{2}}{2 d^{3}}-\frac {c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (d \,x^{2}+c \right )}{2 d^{4}}\)	\(102\)
parallelrisch	\(-\frac {-2 b^{2} d^{3} x^{6}-6 x^{4} a b \,d^{3}+3 x^{4} b^{2} c \,d^{2}-6 x^{2} a^{2} d^{3}+12 x^{2} a b c \,d^{2}-6 x^{2} b^{2} c^{2} d +6 \ln \left (d \,x^{2}+c \right ) a^{2} c \,d^{2}-12 \ln \left (d \,x^{2}+c \right ) a b \,c^{2} d +6 \ln \left (d \,x^{2}+c \right ) b^{2} c^{3}}{12 d^{4}}\)	\(123\)
risch	\(\frac {b^{2} x^{6}}{6 d}+\frac {x^{4} a b}{2 d}-\frac {x^{4} b^{2} c}{4 d^{2}}+\frac {a^{2} x^{2}}{2 d}-\frac {a b c \,x^{2}}{d^{2}}+\frac {b^{2} c^{2} x^{2}}{2 d^{3}}-\frac {c \ln \left (d \,x^{2}+c \right ) a^{2}}{2 d^{2}}+\frac {c^{2} \ln \left (d \,x^{2}+c \right ) a b}{d^{3}}-\frac {c^{3} \ln \left (d \,x^{2}+c \right ) b^{2}}{2 d^{4}}\)	\(124\)

[In]

int(x^3*(b*x^2+a)^2/(d*x^2+c),x,method=_RETURNVERBOSE)

[Out]

1/6*b^2*x^6/d+1/2*(a^2*d^2-2*a*b*c*d+b^2*c^2)/d^3*x^2+1/4*b*(2*a*d-b*c)/d^2*x^4-1/2*c*(a^2*d^2-2*a*b*c*d+b^2*c
^2)/d^4*ln(d*x^2+c)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {2 \, b^{2} d^{3} x^{6} - 3 \, {\left (b^{2} c d^{2} - 2 \, a b d^{3}\right )} x^{4} + 6 \, {\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2} - 6 \, {\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \log \left (d x^{2} + c\right )}{12 \, d^{4}} \]

[In]

integrate(x^3*(b*x^2+a)^2/(d*x^2+c),x, algorithm="fricas")

[Out]

1/12*(2*b^2*d^3*x^6 - 3*(b^2*c*d^2 - 2*a*b*d^3)*x^4 + 6*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x^2 - 6*(b^2*c^3 -
2*a*b*c^2*d + a^2*c*d^2)*log(d*x^2 + c))/d^4

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.05 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {b^{2} x^{6}}{6 d} - \frac {c \left (a d - b c\right )^{2} \log {\left (c + d x^{2} \right )}}{2 d^{4}} + x^{4} \left (\frac {a b}{2 d} - \frac {b^{2} c}{4 d^{2}}\right ) + x^{2} \left (\frac {a^{2}}{2 d} - \frac {a b c}{d^{2}} + \frac {b^{2} c^{2}}{2 d^{3}}\right ) \]

[In]

integrate(x**3*(b*x**2+a)**2/(d*x**2+c),x)

[Out]

b**2*x**6/(6*d) - c*(a*d - b*c)**2*log(c + d*x**2)/(2*d**4) + x**4*(a*b/(2*d) - b**2*c/(4*d**2)) + x**2*(a**2/
(2*d) - a*b*c/d**2 + b**2*c**2/(2*d**3))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {2 \, b^{2} d^{2} x^{6} - 3 \, {\left (b^{2} c d - 2 \, a b d^{2}\right )} x^{4} + 6 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2}}{12 \, d^{3}} - \frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, d^{4}} \]

[In]

integrate(x^3*(b*x^2+a)^2/(d*x^2+c),x, algorithm="maxima")

[Out]

1/12*(2*b^2*d^2*x^6 - 3*(b^2*c*d - 2*a*b*d^2)*x^4 + 6*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2)/d^3 - 1/2*(b^2*c^3
- 2*a*b*c^2*d + a^2*c*d^2)*log(d*x^2 + c)/d^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.35 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=\frac {2 \, b^{2} d^{2} x^{6} - 3 \, b^{2} c d x^{4} + 6 \, a b d^{2} x^{4} + 6 \, b^{2} c^{2} x^{2} - 12 \, a b c d x^{2} + 6 \, a^{2} d^{2} x^{2}}{12 \, d^{3}} - \frac {{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, d^{4}} \]

[In]

integrate(x^3*(b*x^2+a)^2/(d*x^2+c),x, algorithm="giac")

[Out]

1/12*(2*b^2*d^2*x^6 - 3*b^2*c*d*x^4 + 6*a*b*d^2*x^4 + 6*b^2*c^2*x^2 - 12*a*b*c*d*x^2 + 6*a^2*d^2*x^2)/d^3 - 1/
2*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)*log(abs(d*x^2 + c))/d^4

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 \left (a+b x^2\right )^2}{c+d x^2} \, dx=x^2\,\left (\frac {a^2}{2\,d}+\frac {c\,\left (\frac {b^2\,c}{d^2}-\frac {2\,a\,b}{d}\right )}{2\,d}\right )-x^4\,\left (\frac {b^2\,c}{4\,d^2}-\frac {a\,b}{2\,d}\right )+\frac {b^2\,x^6}{6\,d}-\frac {\ln \left (d\,x^2+c\right )\,\left (a^2\,c\,d^2-2\,a\,b\,c^2\,d+b^2\,c^3\right )}{2\,d^4} \]

[In]

int((x^3*(a + b*x^2)^2)/(c + d*x^2),x)

[Out]

x^2*(a^2/(2*d) + (c*((b^2*c)/d^2 - (2*a*b)/d))/(2*d)) - x^4*((b^2*c)/(4*d^2) - (a*b)/(2*d)) + (b^2*x^6)/(6*d)
- (log(c + d*x^2)*(b^2*c^3 + a^2*c*d^2 - 2*a*b*c^2*d))/(2*d^4)