3.2.81 \(\int \frac {x^3 (a+b x^2)^2}{(c+d x^2)^2} \, dx\) [181]

Optimal. Leaf size=90 \[ -\frac {b (b c-a d) x^2}{d^3}+\frac {b^2 x^4}{4 d^2}+\frac {c (b c-a d)^2}{2 d^4 \left (c+d x^2\right )}+\frac {(b c-a d) (3 b c-a d) \log \left (c+d x^2\right )}{2 d^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {457, 78} \begin {gather*} \frac {c (b c-a d)^2}{2 d^4 \left (c+d x^2\right )}+\frac {(b c-a d) (3 b c-a d) \log \left (c+d x^2\right )}{2 d^4}-\frac {b x^2 (b c-a d)}{d^3}+\frac {b^2 x^4}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]
time = 0.05, size = 87, normalized size = 0.97 \begin {gather*} \frac {4 b d (-b c+a d) x^2+b^2 d^2 x^4+\frac {2 c (b c-a d)^2}{c+d x^2}+2 \left (3 b^2 c^2-4 a b c d+a^2 d^2\right ) \log \left (c+d x^2\right )}{4 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 80, normalized size = 0.89

method	result	size
default	\(\frac {\left (b d \,x^{2}+2 a d -2 b c \right )^{2}}{4 d^{4}}+\frac {\left (a d -b c \right ) \left (\frac {c \left (a d -b c \right )}{d \left (d \,x^{2}+c \right )}+\frac {\left (a d -3 b c \right ) \ln \left (d \,x^{2}+c \right )}{d}\right )}{2 d^{3}}\)	\(80\)
norman	\(\frac {\frac {b^{2} x^{6}}{4 d}+\frac {b \left (4 a d -3 b c \right ) x^{4}}{4 d^{2}}-\frac {\left (a^{2} c \,d^{2}-4 a b \,c^{2} d +3 b^{2} c^{3}\right ) x^{2}}{2 d^{3} c}}{d \,x^{2}+c}+\frac {\left (a^{2} d^{2}-4 a b c d +3 b^{2} c^{2}\right ) \ln \left (d \,x^{2}+c \right )}{2 d^{4}}\)	\(113\)
risch	\(\frac {b^{2} x^{4}}{4 d^{2}}+\frac {a b \,x^{2}}{d^{2}}-\frac {b^{2} c \,x^{2}}{d^{3}}+\frac {a^{2}}{d^{2}}-\frac {2 a b c}{d^{3}}+\frac {b^{2} c^{2}}{d^{4}}+\frac {c \,a^{2}}{2 d^{2} \left (d \,x^{2}+c \right )}-\frac {c^{2} a b}{d^{3} \left (d \,x^{2}+c \right )}+\frac {c^{3} b^{2}}{2 d^{4} \left (d \,x^{2}+c \right )}+\frac {\ln \left (d \,x^{2}+c \right ) a^{2}}{2 d^{2}}-\frac {2 \ln \left (d \,x^{2}+c \right ) a b c}{d^{3}}+\frac {3 \ln \left (d \,x^{2}+c \right ) b^{2} c^{2}}{2 d^{4}}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 107, normalized size = 1.19 \begin {gather*} \frac {b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}}{2 \, {\left (d^{5} x^{2} + c d^{4}\right )}} + \frac {b^{2} d x^{4} - 4 \, {\left (b^{2} c - a b d\right )} x^{2}}{4 \, d^{3}} + \frac {{\left (3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2}\right )} \log \left (d x^{2} + c\right )}{2 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)/(d^5*x^2 + c*d^4) + 1/4*(b^2*d*x^4 - 4*(b^2*c - a*b*d)*x^2)/d^3 + 1/2*
(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*log(d*x^2 + c)/d^4

________________________________________________________________________________________

Fricas [A]
time = 1.41, size = 161, normalized size = 1.79 \begin {gather*} \frac {b^{2} d^{3} x^{6} + 2 \, b^{2} c^{3} - 4 \, a b c^{2} d + 2 \, a^{2} c d^{2} - {\left (3 \, b^{2} c d^{2} - 4 \, a b d^{3}\right )} x^{4} - 4 \, {\left (b^{2} c^{2} d - a b c d^{2}\right )} x^{2} + 2 \, {\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d + a^{2} c d^{2} + {\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}\right )} \log \left (d x^{2} + c\right )}{4 \, {\left (d^{5} x^{2} + c d^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*d^3*x^6 + 2*b^2*c^3 - 4*a*b*c^2*d + 2*a^2*c*d^2 - (3*b^2*c*d^2 - 4*a*b*d^3)*x^4 - 4*(b^2*c^2*d - a*b*
c*d^2)*x^2 + 2*(3*b^2*c^3 - 4*a*b*c^2*d + a^2*c*d^2 + (3*b^2*c^2*d - 4*a*b*c*d^2 + a^2*d^3)*x^2)*log(d*x^2 + c
))/(d^5*x^2 + c*d^4)

________________________________________________________________________________________

Sympy [A]
time = 0.52, size = 99, normalized size = 1.10 \begin {gather*} \frac {b^{2} x^{4}}{4 d^{2}} + x^{2} \left (\frac {a b}{d^{2}} - \frac {b^{2} c}{d^{3}}\right ) + \frac {a^{2} c d^{2} - 2 a b c^{2} d + b^{2} c^{3}}{2 c d^{4} + 2 d^{5} x^{2}} + \frac {\left (a d - 3 b c\right ) \left (a d - b c\right ) \log {\left (c + d x^{2} \right )}}{2 d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.51, size = 163, normalized size = 1.81 \begin {gather*} \frac {\frac {{\left (d x^{2} + c\right )}^{2} {\left (b^{2} - \frac {2 \, {\left (3 \, b^{2} c d - 2 \, a b d^{2}\right )}}{{\left (d x^{2} + c\right )} d}\right )}}{d^{3}} - \frac {2 \, {\left (3 \, b^{2} c^{2} - 4 \, a b c d + a^{2} d^{2}\right )} \log \left (\frac {{\left | d x^{2} + c \right |}}{{\left (d x^{2} + c\right )}^{2} {\left | d \right |}}\right )}{d^{3}} + \frac {2 \, {\left (\frac {b^{2} c^{3} d^{2}}{d x^{2} + c} - \frac {2 \, a b c^{2} d^{3}}{d x^{2} + c} + \frac {a^{2} c d^{4}}{d x^{2} + c}\right )}}{d^{5}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((d*x^2 + c)^2*(b^2 - 2*(3*b^2*c*d - 2*a*b*d^2)/((d*x^2 + c)*d))/d^3 - 2*(3*b^2*c^2 - 4*a*b*c*d + a^2*d^2)
*log(abs(d*x^2 + c)/((d*x^2 + c)^2*abs(d)))/d^3 + 2*(b^2*c^3*d^2/(d*x^2 + c) - 2*a*b*c^2*d^3/(d*x^2 + c) + a^2
*c*d^4/(d*x^2 + c))/d^5)/d

________________________________________________________________________________________

Mupad [B]
time = 0.07, size = 112, normalized size = 1.24 \begin {gather*} \frac {a^2\,c\,d^2-2\,a\,b\,c^2\,d+b^2\,c^3}{2\,d\,\left (d^4\,x^2+c\,d^3\right )}-x^2\,\left (\frac {b^2\,c}{d^3}-\frac {a\,b}{d^2}\right )+\frac {b^2\,x^4}{4\,d^2}+\frac {\ln \left (d\,x^2+c\right )\,\left (a^2\,d^2-4\,a\,b\,c\,d+3\,b^2\,c^2\right )}{2\,d^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________