3.83 \(\int \frac{d+e x^2}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=156 \[ \frac{x (a e+3 b d)}{8 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x (b d-a e)}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) (a e+3 b d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

((3*b*d + a*e)*x)/(8*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((b*d - a*e)*x)/(4*a*b*(a + b*x^2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4]) + ((3*b*d + a*e)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(3/2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0894809, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1148, 385, 199, 205} \[ \frac{x (a e+3 b d)}{8 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{x (b d-a e)}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a+b x^2\right ) (a e+3 b d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

((3*b*d + a*e)*x)/(8*a^2*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((b*d - a*e)*x)/(4*a*b*(a + b*x^2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4]) + ((3*b*d + a*e)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*b^(3/2)*Sqrt[a^2 + 2*
a*b*x^2 + b^2*x^4])

Rule 1148

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^
4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d + e*x^2)^q*(b/2 + c*x^2)^(2*p), x], x] /;
FreeQ[{a, b, c, d, e, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 385

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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{d+e x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{d+e x^2}{\left (a b+b^2 x^2\right )^3} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(b d-a e) x}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left ((3 b d+a e) \left (a b+b^2 x^2\right )\right ) \int \frac{1}{\left (a b+b^2 x^2\right )^2} \, dx}{4 a \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(3 b d+a e) x}{8 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{(b d-a e) x}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left ((3 b d+a e) \left (a b+b^2 x^2\right )\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{8 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(3 b d+a e) x}{8 a^2 b \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{(b d-a e) x}{4 a b \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{(3 b d+a e) \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0532503, size = 108, normalized size = 0.69 \[ \frac{\sqrt{a} \sqrt{b} x \left (a^2 (-e)+a b \left (5 d+e x^2\right )+3 b^2 d x^2\right )+\left (a+b x^2\right )^2 (a e+3 b d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2} \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(Sqrt[a]*Sqrt[b]*x*(-(a^2*e) + 3*b^2*d*x^2 + a*b*(5*d + e*x^2)) + (3*b*d + a*e)*(a + b*x^2)^2*ArcTan[(Sqrt[b]*
x)/Sqrt[a]])/(8*a^(5/2)*b^(3/2)*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.016, size = 186, normalized size = 1.2 \begin{align*}{\frac{b{x}^{2}+a}{8\,{a}^{2}b} \left ( \arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){x}^{4}a{b}^{2}e+3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{4}{b}^{3}d+\sqrt{ab}{x}^{3}abe+3\,\sqrt{ab}{x}^{3}{b}^{2}d+2\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{2}{a}^{2}be+6\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{2}a{b}^{2}d-\sqrt{ab}x{a}^{2}e+5\,\sqrt{ab}xabd+\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){a}^{3}e+3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){a}^{2}bd \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/8*(arctan(b*x/(a*b)^(1/2))*x^4*a*b^2*e+3*arctan(b*x/(a*b)^(1/2))*x^4*b^3*d+(a*b)^(1/2)*x^3*a*b*e+3*(a*b)^(1/
2)*x^3*b^2*d+2*arctan(b*x/(a*b)^(1/2))*x^2*a^2*b*e+6*arctan(b*x/(a*b)^(1/2))*x^2*a*b^2*d-(a*b)^(1/2)*x*a^2*e+5
*(a*b)^(1/2)*x*a*b*d+arctan(b*x/(a*b)^(1/2))*a^3*e+3*arctan(b*x/(a*b)^(1/2))*a^2*b*d)*(b*x^2+a)/(a*b)^(1/2)/b/
a^2/((b*x^2+a)^2)^(3/2)

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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((e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52063, size = 621, normalized size = 3.98 \begin{align*} \left [\frac{2 \,{\left (3 \, a b^{3} d + a^{2} b^{2} e\right )} x^{3} -{\left ({\left (3 \, b^{3} d + a b^{2} e\right )} x^{4} + 3 \, a^{2} b d + a^{3} e + 2 \,{\left (3 \, a b^{2} d + a^{2} b e\right )} x^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right ) + 2 \,{\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x}{16 \,{\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, \frac{{\left (3 \, a b^{3} d + a^{2} b^{2} e\right )} x^{3} +{\left ({\left (3 \, b^{3} d + a b^{2} e\right )} x^{4} + 3 \, a^{2} b d + a^{3} e + 2 \,{\left (3 \, a b^{2} d + a^{2} b e\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) +{\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x}{8 \,{\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x