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

Optimal. Leaf size=127 \[ -\frac{\left (-a^2 d^2-6 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{3/2} d^{7/2}}+\frac{x^3 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}+\frac{x (b c-a d) (a d+7 b c)}{8 c d^3 \left (c+d x^2\right )}+\frac{b^2 x}{d^3} \]

[Out]

(b^2*x)/d^3 + ((b*c - a*d)^2*x^3)/(4*c*d^2*(c + d*x^2)^2) + ((b*c - a*d)*(7*b*c + a*d)*x)/(8*c*d^3*(c + d*x^2)
) - ((15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(3/2)*d^(7/2))

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Rubi [A]  time = 0.124717, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {463, 455, 388, 205} \[ -\frac{\left (-a^2 d^2-6 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{3/2} d^{7/2}}+\frac{x^3 (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2}+\frac{x (b c-a d) (a d+7 b c)}{8 c d^3 \left (c+d x^2\right )}+\frac{b^2 x}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x)/d^3 + ((b*c - a*d)^2*x^3)/(4*c*d^2*(c + d*x^2)^2) + ((b*c - a*d)*(7*b*c + a*d)*x)/(8*c*d^3*(c + d*x^2)
) - ((15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(3/2)*d^(7/2))

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 388

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^3} \, dx &=\frac{(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}-\frac{\int \frac{x^2 \left (-4 a^2 d^2+3 (b c-a d)^2-4 b^2 c d x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c d^2}\\ &=\frac{(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac{(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}+\frac{\int \frac{-d (b c-a d) (7 b c+a d)+8 b^2 c d^2 x^2}{c+d x^2} \, dx}{8 c d^4}\\ &=\frac{b^2 x}{d^3}+\frac{(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac{(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \int \frac{1}{c+d x^2} \, dx}{8 c d^3}\\ &=\frac{b^2 x}{d^3}+\frac{(b c-a d)^2 x^3}{4 c d^2 \left (c+d x^2\right )^2}+\frac{(b c-a d) (7 b c+a d) x}{8 c d^3 \left (c+d x^2\right )}-\frac{\left (15 b^2 c^2-6 a b c d-a^2 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{3/2} d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0995385, size = 130, normalized size = 1.02 \[ \frac{x \left (a^2 d^2 \left (d x^2-c\right )-2 a b c d \left (3 c+5 d x^2\right )+b^2 c \left (15 c^2+25 c d x^2+8 d^2 x^4\right )\right )}{8 c d^3 \left (c+d x^2\right )^2}-\frac{\left (-a^2 d^2-6 a b c d+15 b^2 c^2\right ) \tan ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{8 c^{3/2} d^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a^2*d^2*(-c + d*x^2) - 2*a*b*c*d*(3*c + 5*d*x^2) + b^2*c*(15*c^2 + 25*c*d*x^2 + 8*d^2*x^4)))/(8*c*d^3*(c +
d*x^2)^2) - ((15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*ArcTan[(Sqrt[d]*x)/Sqrt[c]])/(8*c^(3/2)*d^(7/2))

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Maple [A]  time = 0.008, size = 196, normalized size = 1.5 \begin{align*}{\frac{{b}^{2}x}{{d}^{3}}}+{\frac{{x}^{3}{a}^{2}}{8\, \left ( d{x}^{2}+c \right ) ^{2}c}}-{\frac{5\,{x}^{3}ab}{4\,d \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{9\,{x}^{3}{b}^{2}c}{8\,{d}^{2} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{{a}^{2}x}{8\,d \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{3\,abcx}{4\,{d}^{2} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{7\,{b}^{2}{c}^{2}x}{8\,{d}^{3} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{{a}^{2}}{8\,cd}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}+{\frac{3\,ab}{4\,{d}^{2}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}}-{\frac{15\,{b}^{2}c}{8\,{d}^{3}}\arctan \left ({dx{\frac{1}{\sqrt{cd}}}} \right ){\frac{1}{\sqrt{cd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x/d^3+1/8/(d*x^2+c)^2/c*x^3*a^2-5/4/d/(d*x^2+c)^2*x^3*a*b+9/8/d^2/(d*x^2+c)^2*x^3*b^2*c-1/8/d/(d*x^2+c)^2*
a^2*x-3/4/d^2/(d*x^2+c)^2*c*a*b*x+7/8/d^3/(d*x^2+c)^2*b^2*c^2*x+1/8/d/c/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a^
2+3/4/d^2/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*a*b-15/8/d^3*c/(c*d)^(1/2)*arctan(x*d/(c*d)^(1/2))*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52891, size = 963, normalized size = 7.58 \begin{align*} \left [\frac{16 \, b^{2} c^{2} d^{3} x^{5} + 2 \,{\left (25 \, b^{2} c^{3} d^{2} - 10 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{3} +{\left (15 \, b^{2} c^{4} - 6 \, a b c^{3} d - a^{2} c^{2} d^{2} +{\left (15 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} - a^{2} d^{4}\right )} x^{4} + 2 \,{\left (15 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{-c d} \log \left (\frac{d x^{2} - 2 \, \sqrt{-c d} x - c}{d x^{2} + c}\right ) + 2 \,{\left (15 \, b^{2} c^{4} d - 6 \, a b c^{3} d^{2} - a^{2} c^{2} d^{3}\right )} x}{16 \,{\left (c^{2} d^{6} x^{4} + 2 \, c^{3} d^{5} x^{2} + c^{4} d^{4}\right )}}, \frac{8 \, b^{2} c^{2} d^{3} x^{5} +{\left (25 \, b^{2} c^{3} d^{2} - 10 \, a b c^{2} d^{3} + a^{2} c d^{4}\right )} x^{3} -{\left (15 \, b^{2} c^{4} - 6 \, a b c^{3} d - a^{2} c^{2} d^{2} +{\left (15 \, b^{2} c^{2} d^{2} - 6 \, a b c d^{3} - a^{2} d^{4}\right )} x^{4} + 2 \,{\left (15 \, b^{2} c^{3} d - 6 \, a b c^{2} d^{2} - a^{2} c d^{3}\right )} x^{2}\right )} \sqrt{c d} \arctan \left (\frac{\sqrt{c d} x}{c}\right ) +{\left (15 \, b^{2} c^{4} d - 6 \, a b c^{3} d^{2} - a^{2} c^{2} d^{3}\right )} x}{8 \,{\left (c^{2} d^{6} x^{4} + 2 \, c^{3} d^{5} x^{2} + c^{4} d^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(16*b^2*c^2*d^3*x^5 + 2*(25*b^2*c^3*d^2 - 10*a*b*c^2*d^3 + a^2*c*d^4)*x^3 + (15*b^2*c^4 - 6*a*b*c^3*d -
a^2*c^2*d^2 + (15*b^2*c^2*d^2 - 6*a*b*c*d^3 - a^2*d^4)*x^4 + 2*(15*b^2*c^3*d - 6*a*b*c^2*d^2 - a^2*c*d^3)*x^2)
*sqrt(-c*d)*log((d*x^2 - 2*sqrt(-c*d)*x - c)/(d*x^2 + c)) + 2*(15*b^2*c^4*d - 6*a*b*c^3*d^2 - a^2*c^2*d^3)*x)/
(c^2*d^6*x^4 + 2*c^3*d^5*x^2 + c^4*d^4), 1/8*(8*b^2*c^2*d^3*x^5 + (25*b^2*c^3*d^2 - 10*a*b*c^2*d^3 + a^2*c*d^4
)*x^3 - (15*b^2*c^4 - 6*a*b*c^3*d - a^2*c^2*d^2 + (15*b^2*c^2*d^2 - 6*a*b*c*d^3 - a^2*d^4)*x^4 + 2*(15*b^2*c^3
*d - 6*a*b*c^2*d^2 - a^2*c*d^3)*x^2)*sqrt(c*d)*arctan(sqrt(c*d)*x/c) + (15*b^2*c^4*d - 6*a*b*c^3*d^2 - a^2*c^2
*d^3)*x)/(c^2*d^6*x^4 + 2*c^3*d^5*x^2 + c^4*d^4)]

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Sympy [A]  time = 1.76807, size = 223, normalized size = 1.76 \begin{align*} \frac{b^{2} x}{d^{3}} - \frac{\sqrt{- \frac{1}{c^{3} d^{7}}} \left (a^{2} d^{2} + 6 a b c d - 15 b^{2} c^{2}\right ) \log{\left (- c^{2} d^{3} \sqrt{- \frac{1}{c^{3} d^{7}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{c^{3} d^{7}}} \left (a^{2} d^{2} + 6 a b c d - 15 b^{2} c^{2}\right ) \log{\left (c^{2} d^{3} \sqrt{- \frac{1}{c^{3} d^{7}}} + x \right )}}{16} + \frac{x^{3} \left (a^{2} d^{3} - 10 a b c d^{2} + 9 b^{2} c^{2} d\right ) + x \left (- a^{2} c d^{2} - 6 a b c^{2} d + 7 b^{2} c^{3}\right )}{8 c^{3} d^{3} + 16 c^{2} d^{4} x^{2} + 8 c d^{5} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*x/d**3 - sqrt(-1/(c**3*d**7))*(a**2*d**2 + 6*a*b*c*d - 15*b**2*c**2)*log(-c**2*d**3*sqrt(-1/(c**3*d**7))
+ x)/16 + sqrt(-1/(c**3*d**7))*(a**2*d**2 + 6*a*b*c*d - 15*b**2*c**2)*log(c**2*d**3*sqrt(-1/(c**3*d**7)) + x)/
16 + (x**3*(a**2*d**3 - 10*a*b*c*d**2 + 9*b**2*c**2*d) + x*(-a**2*c*d**2 - 6*a*b*c**2*d + 7*b**2*c**3))/(8*c**
3*d**3 + 16*c**2*d**4*x**2 + 8*c*d**5*x**4)

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Giac [A]  time = 1.19722, size = 180, normalized size = 1.42 \begin{align*} \frac{b^{2} x}{d^{3}} - \frac{{\left (15 \, b^{2} c^{2} - 6 \, a b c d - a^{2} d^{2}\right )} \arctan \left (\frac{d x}{\sqrt{c d}}\right )}{8 \, \sqrt{c d} c d^{3}} + \frac{9 \, b^{2} c^{2} d x^{3} - 10 \, a b c d^{2} x^{3} + a^{2} d^{3} x^{3} + 7 \, b^{2} c^{3} x - 6 \, a b c^{2} d x - a^{2} c d^{2} x}{8 \,{\left (d x^{2} + c\right )}^{2} c d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x/d^3 - 1/8*(15*b^2*c^2 - 6*a*b*c*d - a^2*d^2)*arctan(d*x/sqrt(c*d))/(sqrt(c*d)*c*d^3) + 1/8*(9*b^2*c^2*d*
x^3 - 10*a*b*c*d^2*x^3 + a^2*d^3*x^3 + 7*b^2*c^3*x - 6*a*b*c^2*d*x - a^2*c*d^2*x)/((d*x^2 + c)^2*c*d^3)