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

Optimal. Leaf size=190 \[ -\frac{x (7 b d-3 a e)}{8 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x (b d-a e)}{4 a^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 \left (a+b x^2\right ) (5 b d-a e) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right )}{a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.185439, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {1250, 456, 453, 205} \[ -\frac{x (7 b d-3 a e)}{8 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{x (b d-a e)}{4 a^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 \left (a+b x^2\right ) (5 b d-a e) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b} \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right )}{a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1250

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

Rule 456

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[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 453

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

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}{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}{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^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{-\frac{4 d}{a b}+\frac{3 (b d-a e) x^2}{a^2 b}}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{4 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{(7 b d-3 a e) x}{8 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{\frac{8 d}{a^2 b^2}-\frac{(7 b d-3 a e) x^2}{a^3 b^2}}{x^2 \left (a b+b^2 x^2\right )} \, dx}{8 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{(7 b d-3 a e) x}{8 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right )}{a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (3 (5 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^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=-\frac{(7 b d-3 a e) x}{8 a^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{(b d-a e) x}{4 a^2 \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{d \left (a+b x^2\right )}{a^3 x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{3 (5 b d-a e) \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0683688, size = 124, normalized size = 0.65 \[ \frac{\sqrt{a} \sqrt{b} \left (a^2 \left (5 e x^2-8 d\right )+a b \left (3 e x^4-25 d x^2\right )-15 b^2 d x^4\right )+3 x \left (a+b x^2\right )^2 (a e-5 b d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b} x \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 206, normalized size = 1.1 \begin{align*}{\frac{b{x}^{2}+a}{8\,x{a}^{3}} \left ( 3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{5}a{b}^{2}e-15\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{5}{b}^{3}d+3\,\sqrt{ab}{x}^{4}abe-15\,\sqrt{ab}{x}^{4}{b}^{2}d+6\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{3}{a}^{2}be-30\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{3}a{b}^{2}d+5\,\sqrt{ab}{x}^{2}{a}^{2}e-25\,\sqrt{ab}{x}^{2}abd+3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) x{a}^{3}e-15\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) x{a}^{2}bd-8\,\sqrt{ab}{a}^{2}d \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)/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/8*(3*arctan(b*x/(a*b)^(1/2))*x^5*a*b^2*e-15*arctan(b*x/(a*b)^(1/2))*x^5*b^3*d+3*(a*b)^(1/2)*x^4*a*b*e-15*(a*
b)^(1/2)*x^4*b^2*d+6*arctan(b*x/(a*b)^(1/2))*x^3*a^2*b*e-30*arctan(b*x/(a*b)^(1/2))*x^3*a*b^2*d+5*(a*b)^(1/2)*
x^2*a^2*e-25*(a*b)^(1/2)*x^2*a*b*d+3*arctan(b*x/(a*b)^(1/2))*x*a^3*e-15*arctan(b*x/(a*b)^(1/2))*x*a^2*b*d-8*(a
*b)^(1/2)*a^2*d)*(b*x^2+a)/(a*b)^(1/2)/x/a^3/((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)/x^2/(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 = 1.59615, size = 686, normalized size = 3.61 \begin{align*} \left [-\frac{16 \, a^{3} b d + 6 \,{\left (5 \, a b^{3} d - a^{2} b^{2} e\right )} x^{4} + 10 \,{\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x^{2} - 3 \,{\left ({\left (5 \, b^{3} d - a b^{2} e\right )} x^{5} + 2 \,{\left (5 \, a b^{2} d - a^{2} b e\right )} x^{3} +{\left (5 \, a^{2} b d - a^{3} e\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{16 \,{\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac{8 \, a^{3} b d + 3 \,{\left (5 \, a b^{3} d - a^{2} b^{2} e\right )} x^{4} + 5 \,{\left (5 \, a^{2} b^{2} d - a^{3} b e\right )} x^{2} + 3 \,{\left ({\left (5 \, b^{3} d - a b^{2} e\right )} x^{5} + 2 \,{\left (5 \, a b^{2} d - a^{2} b e\right )} x^{3} +{\left (5 \, a^{2} b d - a^{3} e\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{8 \,{\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x^{2}}{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)/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

[Out]

sage0*x