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

Optimal. Leaf size=152 \[ -\frac{x \sqrt{c+d x^2} \left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right )}{8 c d^3}+\frac{\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{7/2}}+\frac{x^3 (b c-a d)^2}{c d^2 \sqrt{c+d x^2}}+\frac{b^2 x^3 \sqrt{c+d x^2}}{4 d^2} \]

[Out]

((b*c - a*d)^2*x^3)/(c*d^2*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(8*c*d
^3) + (b^2*x^3*Sqrt[c + d*x^2])/(4*d^2) + ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(8*d^(7/2))

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Rubi [A]  time = 0.117052, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {463, 459, 321, 217, 206} \[ -\frac{x \sqrt{c+d x^2} \left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right )}{8 c d^3}+\frac{\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{7/2}}+\frac{x^3 (b c-a d)^2}{c d^2 \sqrt{c+d x^2}}+\frac{b^2 x^3 \sqrt{c+d x^2}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*x^3)/(c*d^2*Sqrt[c + d*x^2]) - ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*x*Sqrt[c + d*x^2])/(8*c*d
^3) + (b^2*x^3*Sqrt[c + d*x^2])/(4*d^2) + ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/Sqrt[c +
d*x^2]])/(8*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 459

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

Rule 321

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac{(b c-a d)^2 x^3}{c d^2 \sqrt{c+d x^2}}-\frac{\int \frac{x^2 \left (-a^2 d^2+3 (b c-a d)^2-b^2 c d x^2\right )}{\sqrt{c+d x^2}} \, dx}{c d^2}\\ &=\frac{(b c-a d)^2 x^3}{c d^2 \sqrt{c+d x^2}}+\frac{b^2 x^3 \sqrt{c+d x^2}}{4 d^2}-\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \int \frac{x^2}{\sqrt{c+d x^2}} \, dx}{4 c d^2}\\ &=\frac{(b c-a d)^2 x^3}{c d^2 \sqrt{c+d x^2}}-\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{8 c d^3}+\frac{b^2 x^3 \sqrt{c+d x^2}}{4 d^2}+\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{8 d^3}\\ &=\frac{(b c-a d)^2 x^3}{c d^2 \sqrt{c+d x^2}}-\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{8 c d^3}+\frac{b^2 x^3 \sqrt{c+d x^2}}{4 d^2}+\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{8 d^3}\\ &=\frac{(b c-a d)^2 x^3}{c d^2 \sqrt{c+d x^2}}-\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) x \sqrt{c+d x^2}}{8 c d^3}+\frac{b^2 x^3 \sqrt{c+d x^2}}{4 d^2}+\frac{\left (15 b^2 c^2-24 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{8 d^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.111202, size = 124, normalized size = 0.82 \[ \frac{\left (8 a^2 d^2-24 a b c d+15 b^2 c^2\right ) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{8 d^{7/2}}+\sqrt{c+d x^2} \left (-\frac{x (a d-b c)^2}{d^3 \left (c+d x^2\right )}-\frac{b x (7 b c-8 a d)}{8 d^3}+\frac{b^2 x^3}{4 d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^2]*(-(b*(7*b*c - 8*a*d)*x)/(8*d^3) + (b^2*x^3)/(4*d^2) - ((-(b*c) + a*d)^2*x)/(d^3*(c + d*x^2)))
+ ((15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/(8*d^(7/2))

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Maple [A]  time = 0.008, size = 192, normalized size = 1.3 \begin{align*}{\frac{{b}^{2}{x}^{5}}{4\,d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{5\,{b}^{2}c{x}^{3}}{8\,{d}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{15\,{b}^{2}{c}^{2}x}{8\,{d}^{3}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{15\,{b}^{2}{c}^{2}}{8}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{7}{2}}}}+{\frac{ab{x}^{3}}{d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+3\,{\frac{abcx}{{d}^{2}\sqrt{d{x}^{2}+c}}}-3\,{\frac{abc\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) }{{d}^{5/2}}}-{\frac{{a}^{2}x}{d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{{a}^{2}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

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

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50125, size = 759, normalized size = 4.99 \begin{align*} \left [\frac{{\left (15 \, b^{2} c^{3} - 24 \, a b c^{2} d + 8 \, a^{2} c d^{2} +{\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{d} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \,{\left (2 \, b^{2} d^{3} x^{5} -{\left (5 \, b^{2} c d^{2} - 8 \, a b d^{3}\right )} x^{3} -{\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{16 \,{\left (d^{5} x^{2} + c d^{4}\right )}}, -\frac{{\left (15 \, b^{2} c^{3} - 24 \, a b c^{2} d + 8 \, a^{2} c d^{2} +{\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) -{\left (2 \, b^{2} d^{3} x^{5} -{\left (5 \, b^{2} c d^{2} - 8 \, a b d^{3}\right )} x^{3} -{\left (15 \, b^{2} c^{2} d - 24 \, a b c d^{2} + 8 \, a^{2} d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{8 \,{\left (d^{5} x^{2} + c 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)^(3/2),x, algorithm="fricas")

[Out]

[1/16*((15*b^2*c^3 - 24*a*b*c^2*d + 8*a^2*c*d^2 + (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x^2)*sqrt(d)*log(-
2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(2*b^2*d^3*x^5 - (5*b^2*c*d^2 - 8*a*b*d^3)*x^3 - (15*b^2*c^2*d
- 24*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/(d^5*x^2 + c*d^4), -1/8*((15*b^2*c^3 - 24*a*b*c^2*d + 8*a^2*c*
d^2 + (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b^2*d^3*
x^5 - (5*b^2*c*d^2 - 8*a*b*d^3)*x^3 - (15*b^2*c^2*d - 24*a*b*c*d^2 + 8*a^2*d^3)*x)*sqrt(d*x^2 + c))/(d^5*x^2 +
c*d^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.12833, size = 177, normalized size = 1.16 \begin{align*} \frac{{\left ({\left (\frac{2 \, b^{2} x^{2}}{d} - \frac{5 \, b^{2} c d^{3} - 8 \, a b d^{4}}{d^{5}}\right )} x^{2} - \frac{15 \, b^{2} c^{2} d^{2} - 24 \, a b c d^{3} + 8 \, a^{2} d^{4}}{d^{5}}\right )} x}{8 \, \sqrt{d x^{2} + c}} - \frac{{\left (15 \, b^{2} c^{2} - 24 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{8 \, d^{\frac{7}{2}}} \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)^(3/2),x, algorithm="giac")

[Out]

1/8*((2*b^2*x^2/d - (5*b^2*c*d^3 - 8*a*b*d^4)/d^5)*x^2 - (15*b^2*c^2*d^2 - 24*a*b*c*d^3 + 8*a^2*d^4)/d^5)*x/sq
rt(d*x^2 + c) - 1/8*(15*b^2*c^2 - 24*a*b*c*d + 8*a^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)