\(\int \frac {x^3 (c+d x^2)^2}{a+b x^2} \, dx\) [210]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {(b c-a d)^2 x^2}{2 b^3}+\frac {d (2 b c-a d) x^4}{4 b^2}+\frac {d^2 x^6}{6 b}-\frac {a (b c-a d)^2 \log \left (a+b x^2\right )}{2 b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, 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} \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=-\frac {a (b c-a d)^2 \log \left (a+b x^2\right )}{2 b^4}+\frac {x^2 (b c-a d)^2}{2 b^3}+\frac {d x^4 (2 b c-a d)}{4 b^2}+\frac {d^2 x^6}{6 b} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {b x^2 \left (6 a^2 d^2-3 a b d \left (4 c+d x^2\right )+2 b^2 \left (3 c^2+3 c d x^2+d^2 x^4\right )\right )-6 a (b c-a d)^2 \log \left (a+b x^2\right )}{12 b^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.65 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18

method	result	size
norman	\(\frac {d^{2} x^{6}}{6 b}+\frac {\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) x^{2}}{2 b^{3}}-\frac {d \left (a d -2 b c \right ) x^{4}}{4 b^{2}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(94\)
default	\(\frac {\frac {1}{3} b^{2} d^{2} x^{6}-\frac {1}{2} x^{4} a b \,d^{2}+x^{4} b^{2} c d +a^{2} d^{2} x^{2}-2 a b c d \,x^{2}+b^{2} c^{2} x^{2}}{2 b^{3}}-\frac {a \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \ln \left (b \,x^{2}+a \right )}{2 b^{4}}\)	\(102\)
parallelrisch	\(-\frac {-2 b^{3} d^{2} x^{6}+3 x^{4} a \,b^{2} d^{2}-6 x^{4} b^{3} c d -6 x^{2} a^{2} b \,d^{2}+12 x^{2} a \,b^{2} c d -6 x^{2} b^{3} c^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{3} d^{2}-12 \ln \left (b \,x^{2}+a \right ) a^{2} b c d +6 \ln \left (b \,x^{2}+a \right ) a \,b^{2} c^{2}}{12 b^{4}}\)	\(123\)
risch	\(\frac {d^{2} x^{6}}{6 b}-\frac {x^{4} a \,d^{2}}{4 b^{2}}+\frac {x^{4} c d}{2 b}+\frac {a^{2} d^{2} x^{2}}{2 b^{3}}-\frac {a c d \,x^{2}}{b^{2}}+\frac {c^{2} x^{2}}{2 b}-\frac {a^{3} \ln \left (b \,x^{2}+a \right ) d^{2}}{2 b^{4}}+\frac {a^{2} \ln \left (b \,x^{2}+a \right ) c d}{b^{3}}-\frac {a \ln \left (b \,x^{2}+a \right ) c^{2}}{2 b^{2}}\)	\(124\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {2 \, b^{3} d^{2} x^{6} + 3 \, {\left (2 \, b^{3} c d - a b^{2} d^{2}\right )} x^{4} + 6 \, {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{2} - 6 \, {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, b^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.04 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=- \frac {a \left (a d - b c\right )^{2} \log {\left (a + b x^{2} \right )}}{2 b^{4}} + x^{4} \left (- \frac {a d^{2}}{4 b^{2}} + \frac {c d}{2 b}\right ) + x^{2} \left (\frac {a^{2} d^{2}}{2 b^{3}} - \frac {a c d}{b^{2}} + \frac {c^{2}}{2 b}\right ) + \frac {d^{2} x^{6}}{6 b} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.26 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {2 \, b^{2} d^{2} x^{6} + 3 \, {\left (2 \, b^{2} c d - a b d^{2}\right )} x^{4} + 6 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2}}{12 \, b^{3}} - \frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, b^{4}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=\frac {2 \, b^{2} d^{2} x^{6} + 6 \, b^{2} c d x^{4} - 3 \, a b d^{2} x^{4} + 6 \, b^{2} c^{2} x^{2} - 12 \, a b c d x^{2} + 6 \, a^{2} d^{2} x^{2}}{12 \, b^{3}} - \frac {{\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.32 \[ \int \frac {x^3 \left (c+d x^2\right )^2}{a+b x^2} \, dx=x^2\,\left (\frac {c^2}{2\,b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{2\,b}\right )-x^4\,\left (\frac {a\,d^2}{4\,b^2}-\frac {c\,d}{2\,b}\right )-\frac {\ln \left (b\,x^2+a\right )\,\left (a^3\,d^2-2\,a^2\,b\,c\,d+a\,b^2\,c^2\right )}{2\,b^4}+\frac {d^2\,x^6}{6\,b} \]

[In]

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

[Out]

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