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

Optimal. Leaf size=99 \[ \frac{c (b c-a d)^2}{4 d^4 \left (c+d x^2\right )^2}-\frac{(3 b c-a d) (b c-a d)}{2 d^4 \left (c+d x^2\right )}-\frac{b (3 b c-2 a d) \log \left (c+d x^2\right )}{2 d^4}+\frac{b^2 x^2}{2 d^3} \]

[Out]

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

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Rubi [A]  time = 0.100078, antiderivative size = 99, normalized size of antiderivative = 1., 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 = {446, 77} \[ \frac{c (b c-a d)^2}{4 d^4 \left (c+d x^2\right )^2}-\frac{(3 b c-a d) (b c-a d)}{2 d^4 \left (c+d x^2\right )}-\frac{b (3 b c-2 a d) \log \left (c+d x^2\right )}{2 d^4}+\frac{b^2 x^2}{2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 446

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]]

Rule 77

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]))))

Rubi steps

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

Mathematica [A]  time = 0.0504032, size = 114, normalized size = 1.15 \[ \frac{-a^2 d^2 \left (c+2 d x^2\right )+2 a b c d \left (3 c+4 d x^2\right )-2 b \left (c+d x^2\right )^2 (3 b c-2 a d) \log \left (c+d x^2\right )+b^2 \left (-4 c^2 d x^2-5 c^3+4 c d^2 x^4+2 d^3 x^6\right )}{4 d^4 \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*d^2*(c + 2*d*x^2)) + 2*a*b*c*d*(3*c + 4*d*x^2) + b^2*(-5*c^3 - 4*c^2*d*x^2 + 4*c*d^2*x^4 + 2*d^3*x^6) -
2*b*(3*b*c - 2*a*d)*(c + d*x^2)^2*Log[c + d*x^2])/(4*d^4*(c + d*x^2)^2)

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Maple [A]  time = 0.011, size = 155, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}{x}^{2}}{2\,{d}^{3}}}+{\frac{b\ln \left ( d{x}^{2}+c \right ) a}{{d}^{3}}}-{\frac{3\,{b}^{2}\ln \left ( d{x}^{2}+c \right ) c}{2\,{d}^{4}}}+{\frac{{a}^{2}c}{4\,{d}^{2} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{ab{c}^{2}}{2\,{d}^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{{b}^{2}{c}^{3}}{4\,{d}^{4} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{{a}^{2}}{2\,{d}^{2} \left ( d{x}^{2}+c \right ) }}+2\,{\frac{abc}{{d}^{3} \left ( d{x}^{2}+c \right ) }}-{\frac{3\,{b}^{2}{c}^{2}}{2\,{d}^{4} \left ( d{x}^{2}+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.982362, size = 162, normalized size = 1.64 \begin{align*} \frac{b^{2} x^{2}}{2 \, d^{3}} - \frac{5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + 2 \,{\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}}{4 \,{\left (d^{6} x^{4} + 2 \, c d^{5} x^{2} + c^{2} d^{4}\right )}} - \frac{{\left (3 \, b^{2} c - 2 \, a b d\right )} \log \left (d x^{2} + c\right )}{2 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.47316, size = 365, normalized size = 3.69 \begin{align*} \frac{2 \, b^{2} d^{3} x^{6} + 4 \, b^{2} c d^{2} x^{4} - 5 \, b^{2} c^{3} + 6 \, a b c^{2} d - a^{2} c d^{2} - 2 \,{\left (2 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2} - 2 \,{\left (3 \, b^{2} c^{3} - 2 \, a b c^{2} d +{\left (3 \, b^{2} c d^{2} - 2 \, a b d^{3}\right )} x^{4} + 2 \,{\left (3 \, b^{2} c^{2} d - 2 \, a b c d^{2}\right )} x^{2}\right )} \log \left (d x^{2} + c\right )}{4 \,{\left (d^{6} x^{4} + 2 \, c d^{5} x^{2} + c^{2} d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.6553, size = 122, normalized size = 1.23 \begin{align*} \frac{b^{2} x^{2}}{2 d^{3}} + \frac{b \left (2 a d - 3 b c\right ) \log{\left (c + d x^{2} \right )}}{2 d^{4}} - \frac{a^{2} c d^{2} - 6 a b c^{2} d + 5 b^{2} c^{3} + x^{2} \left (2 a^{2} d^{3} - 8 a b c d^{2} + 6 b^{2} c^{2} d\right )}{4 c^{2} d^{4} + 8 c d^{5} x^{2} + 4 d^{6} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*x**2/(2*d**3) + b*(2*a*d - 3*b*c)*log(c + d*x**2)/(2*d**4) - (a**2*c*d**2 - 6*a*b*c**2*d + 5*b**2*c**3 +
x**2*(2*a**2*d**3 - 8*a*b*c*d**2 + 6*b**2*c**2*d))/(4*c**2*d**4 + 8*c*d**5*x**2 + 4*d**6*x**4)

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Giac [A]  time = 1.20211, size = 144, normalized size = 1.45 \begin{align*} \frac{b^{2} x^{2}}{2 \, d^{3}} - \frac{{\left (3 \, b^{2} c - 2 \, a b d\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, d^{4}} - \frac{5 \, b^{2} c^{3} - 6 \, a b c^{2} d + a^{2} c d^{2} + 2 \,{\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + a^{2} d^{3}\right )} x^{2}}{4 \,{\left (d x^{2} + c\right )}^{2} d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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