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

Optimal. Leaf size=106 \[ \frac{b x \sqrt{c+d x^2} (3 b c-2 a d)}{2 c d^2}-\frac{b (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 d^{5/2}}-\frac{x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt{c+d x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0539976, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {413, 388, 217, 206} \[ \frac{b x \sqrt{c+d x^2} (3 b c-2 a d)}{2 c d^2}-\frac{b (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 d^{5/2}}-\frac{x \left (a+b x^2\right ) (b c-a d)}{c d \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 413

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

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 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{\left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=-\frac{(b c-a d) x \left (a+b x^2\right )}{c d \sqrt{c+d x^2}}+\frac{\int \frac{a b c+b (3 b c-2 a d) x^2}{\sqrt{c+d x^2}} \, dx}{c d}\\ &=-\frac{(b c-a d) x \left (a+b x^2\right )}{c d \sqrt{c+d x^2}}+\frac{b (3 b c-2 a d) x \sqrt{c+d x^2}}{2 c d^2}-\frac{(b (3 b c-4 a d)) \int \frac{1}{\sqrt{c+d x^2}} \, dx}{2 d^2}\\ &=-\frac{(b c-a d) x \left (a+b x^2\right )}{c d \sqrt{c+d x^2}}+\frac{b (3 b c-2 a d) x \sqrt{c+d x^2}}{2 c d^2}-\frac{(b (3 b c-4 a d)) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+d x^2}}\right )}{2 d^2}\\ &=-\frac{(b c-a d) x \left (a+b x^2\right )}{c d \sqrt{c+d x^2}}+\frac{b (3 b c-2 a d) x \sqrt{c+d x^2}}{2 c d^2}-\frac{b (3 b c-4 a d) \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{2 d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0998388, size = 93, normalized size = 0.88 \[ \sqrt{c+d x^2} \left (\frac{x (b c-a d)^2}{c d^2 \left (c+d x^2\right )}+\frac{b^2 x}{2 d^2}\right )-\frac{b (3 b c-4 a d) \log \left (\sqrt{d} \sqrt{c+d x^2}+d x\right )}{2 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 123, normalized size = 1.2 \begin{align*}{\frac{{b}^{2}{x}^{3}}{2\,d}{\frac{1}{\sqrt{d{x}^{2}+c}}}}+{\frac{3\,{b}^{2}cx}{2\,{d}^{2}}{\frac{1}{\sqrt{d{x}^{2}+c}}}}-{\frac{3\,{b}^{2}c}{2}\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){d}^{-{\frac{5}{2}}}}-2\,{\frac{abx}{d\sqrt{d{x}^{2}+c}}}+2\,{\frac{ab\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) }{{d}^{3/2}}}+{\frac{{a}^{2}x}{c}{\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^3/d/(d*x^2+c)^(1/2)+3/2*b^2*c/d^2*x/(d*x^2+c)^(1/2)-3/2*b^2*c/d^(5/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))-
2*a*b*x/d/(d*x^2+c)^(1/2)+2*a*b/d^(3/2)*ln(x*d^(1/2)+(d*x^2+c)^(1/2))+a^2*x/c/(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((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 = 1.36837, size = 595, normalized size = 5.61 \begin{align*} \left [-\frac{{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d +{\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\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 (b^{2} c d^{2} x^{3} +{\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{4 \,{\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}, \frac{{\left (3 \, b^{2} c^{3} - 4 \, a b c^{2} d +{\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2}\right )} x^{2}\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) +{\left (b^{2} c d^{2} x^{3} +{\left (3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}\right )} x\right )} \sqrt{d x^{2} + c}}{2 \,{\left (c d^{4} x^{2} + c^{2} d^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((3*b^2*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*c*d^2)*x^2)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sq
rt(d)*x - c) - 2*(b^2*c*d^2*x^3 + (3*b^2*c^2*d - 4*a*b*c*d^2 + 2*a^2*d^3)*x)*sqrt(d*x^2 + c))/(c*d^4*x^2 + c^2
*d^3), 1/2*((3*b^2*c^3 - 4*a*b*c^2*d + (3*b^2*c^2*d - 4*a*b*c*d^2)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2
+ c)) + (b^2*c*d^2*x^3 + (3*b^2*c^2*d - 4*a*b*c*d^2 + 2*a^2*d^3)*x)*sqrt(d*x^2 + c))/(c*d^4*x^2 + c^2*d^3)]

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

[Out]

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

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Giac [A]  time = 1.13102, size = 124, normalized size = 1.17 \begin{align*} \frac{{\left (\frac{b^{2} x^{2}}{d} + \frac{3 \, b^{2} c^{2} d - 4 \, a b c d^{2} + 2 \, a^{2} d^{3}}{c d^{3}}\right )} x}{2 \, \sqrt{d x^{2} + c}} + \frac{{\left (3 \, b^{2} c - 4 \, a b d\right )} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right )}{2 \, d^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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