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

Optimal. Leaf size=79 \[ -\frac{b x^4 (b c-2 a d)}{4 d^2}+\frac{x^2 (b c-a d)^2}{2 d^3}-\frac{c (b c-a d)^2 \log \left (c+d x^2\right )}{2 d^4}+\frac{b^2 x^6}{6 d} \]

[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)

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Rubi [A]  time = 0.0842049, antiderivative size = 79, 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} \[ -\frac{b x^4 (b c-2 a d)}{4 d^2}+\frac{x^2 (b c-a d)^2}{2 d^3}-\frac{c (b c-a d)^2 \log \left (c+d x^2\right )}{2 d^4}+\frac{b^2 x^6}{6 d} \]

Antiderivative was successfully verified.

[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 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}{c+d x^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (a+b x)^2}{c+d x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{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]  time = 0.0409976, size = 82, normalized size = 1.04 \[ \frac{d x^2 \left (6 a^2 d^2+6 a b d \left (d x^2-2 c\right )+b^2 \left (6 c^2-3 c d x^2+2 d^2 x^4\right )\right )-6 c (b c-a d)^2 \log \left (c+d x^2\right )}{12 d^4} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.003, size = 124, normalized size = 1.6 \begin{align*}{\frac{{b}^{2}{x}^{6}}{6\,d}}+{\frac{ab{x}^{4}}{2\,d}}-{\frac{{b}^{2}c{x}^{4}}{4\,{d}^{2}}}+{\frac{{a}^{2}{x}^{2}}{2\,d}}-{\frac{abc{x}^{2}}{{d}^{2}}}+{\frac{{x}^{2}{b}^{2}{c}^{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 ) ab}{{d}^{3}}}-{\frac{{c}^{3}\ln \left ( d{x}^{2}+c \right ){b}^{2}}{2\,{d}^{4}}} \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),x)

[Out]

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

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Maxima [A]  time = 0.98049, size = 135, normalized size = 1.71 \begin{align*} \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}} \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),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

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Fricas [A]  time = 1.16406, size = 212, normalized size = 2.68 \begin{align*} \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}} \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),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

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Sympy [A]  time = 0.553723, size = 83, normalized size = 1.05 \begin{align*} \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}} + \frac{x^{4} \left (2 a b d - b^{2} c\right )}{4 d^{2}} + \frac{x^{2} \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{2 d^{3}} \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),x)

[Out]

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

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Giac [A]  time = 1.16863, size = 144, normalized size = 1.82 \begin{align*} \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}} \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),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