3.59 \(\int \frac{(a+b x^2)^2}{(c+d x^2) \sqrt{e+f x^2}} \, dx\)

Optimal. Leaf size=166 \[ \frac{(b c-a d)^2 \tan ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{c} \sqrt{e+f x^2}}\right )}{\sqrt{c} d^2 \sqrt{d e-c f}}-\frac{b (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{d^2 \sqrt{f}}-\frac{b (b e-2 a f) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{2 d f^{3/2}}+\frac{b^2 x \sqrt{e+f x^2}}{2 d f} \]

[Out]

(b^2*x*Sqrt[e + f*x^2])/(2*d*f) + ((b*c - a*d)^2*ArcTan[(Sqrt[d*e - c*f]*x)/(Sqrt[c]*Sqrt[e + f*x^2])])/(Sqrt[
c]*d^2*Sqrt[d*e - c*f]) - (b*(b*c - a*d)*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(d^2*Sqrt[f]) - (b*(b*e - 2*a*f
)*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(2*d*f^(3/2))

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Rubi [A]  time = 0.105244, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {545, 388, 217, 206, 523, 377, 205} \[ \frac{(b c-a d)^2 \tan ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{c} \sqrt{e+f x^2}}\right )}{\sqrt{c} d^2 \sqrt{d e-c f}}-\frac{b (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{d^2 \sqrt{f}}-\frac{b (b e-2 a f) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{2 d f^{3/2}}+\frac{b^2 x \sqrt{e+f x^2}}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*Sqrt[e + f*x^2])/(2*d*f) + ((b*c - a*d)^2*ArcTan[(Sqrt[d*e - c*f]*x)/(Sqrt[c]*Sqrt[e + f*x^2])])/(Sqrt[
c]*d^2*Sqrt[d*e - c*f]) - (b*(b*c - a*d)*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(d^2*Sqrt[f]) - (b*(b*e - 2*a*f
)*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(2*d*f^(3/2))

Rule 545

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/b, Int[
(c + d*x^2)^(q - 1)*(e + f*x^2)^r, x], x] + Dist[(b*c - a*d)/b, Int[((c + d*x^2)^(q - 1)*(e + f*x^2)^r)/(a + b
*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, r}, x] && GtQ[q, 1]

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])

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 377

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

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

Mathematica [A]  time = 0.305973, size = 150, normalized size = 0.9 \[ \frac{\frac{2 f (b c-a d)^2 \tan ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{c} \sqrt{e+f x^2}}\right )+b^2 \sqrt{c} d x \sqrt{e+f x^2} \sqrt{d e-c f}}{\sqrt{c} f \sqrt{d e-c f}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right ) (-4 a d f+2 b c f+b d e)}{f^{3/2}}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.015, size = 1052, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^(1/2),x)

[Out]

1/2*b^2*x*(f*x^2+e)^(1/2)/d/f-1/2*b^2/d*e/f^(3/2)*ln(x*f^(1/2)+(f*x^2+e)^(1/2))+2*b/d*a*ln(x*f^(1/2)+(f*x^2+e)
^(1/2))/f^(1/2)-b^2/d^2*c*ln(x*f^(1/2)+(f*x^2+e)^(1/2))/f^(1/2)-1/2/(-c*d)^(1/2)/(-(c*f-d*e)/d)^(1/2)*ln((-2*(
c*f-d*e)/d+2*f*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*(-(c*f-d*e)/d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*f+2*f*(-c*d)^(1/
2)/d*(x-(-c*d)^(1/2)/d)-(c*f-d*e)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*a^2+1/d/(-c*d)^(1/2)/(-(c*f-d*e)/d)^(1/2)*ln((
-2*(c*f-d*e)/d+2*f*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*(-(c*f-d*e)/d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*f+2*f*(-c*d)
^(1/2)/d*(x-(-c*d)^(1/2)/d)-(c*f-d*e)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*c*a*b-1/2/d^2/(-c*d)^(1/2)/(-(c*f-d*e)/d)^
(1/2)*ln((-2*(c*f-d*e)/d+2*f*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)+2*(-(c*f-d*e)/d)^(1/2)*((x-(-c*d)^(1/2)/d)^2*f+
2*f*(-c*d)^(1/2)/d*(x-(-c*d)^(1/2)/d)-(c*f-d*e)/d)^(1/2))/(x-(-c*d)^(1/2)/d))*b^2*c^2+1/2/(-c*d)^(1/2)/(-(c*f-
d*e)/d)^(1/2)*ln((-2*(c*f-d*e)/d-2*f*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+2*(-(c*f-d*e)/d)^(1/2)*((x+(-c*d)^(1/2)
/d)^2*f-2*f*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)-(c*f-d*e)/d)^(1/2))/(x+(-c*d)^(1/2)/d))*a^2-1/d/(-c*d)^(1/2)/(-(
c*f-d*e)/d)^(1/2)*ln((-2*(c*f-d*e)/d-2*f*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+2*(-(c*f-d*e)/d)^(1/2)*((x+(-c*d)^(
1/2)/d)^2*f-2*f*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)-(c*f-d*e)/d)^(1/2))/(x+(-c*d)^(1/2)/d))*c*a*b+1/2/d^2/(-c*d)
^(1/2)/(-(c*f-d*e)/d)^(1/2)*ln((-2*(c*f-d*e)/d-2*f*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)+2*(-(c*f-d*e)/d)^(1/2)*((
x+(-c*d)^(1/2)/d)^2*f-2*f*(-c*d)^(1/2)/d*(x+(-c*d)^(1/2)/d)-(c*f-d*e)/d)^(1/2))/(x+(-c*d)^(1/2)/d))*b^2*c^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)/(f*x^2+e)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 11.0741, size = 2346, normalized size = 14.13 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-c*d*e + c^2*f)*f^2*log(((d^2*e^2 - 8*c*d*e*f + 8*c^2*f^2)*x^4 + c
^2*e^2 - 2*(3*c*d*e^2 - 4*c^2*e*f)*x^2 - 4*((d*e - 2*c*f)*x^3 - c*e*x)*sqrt(-c*d*e + c^2*f)*sqrt(f*x^2 + e))/(
d^2*x^4 + 2*c*d*x^2 + c^2)) - 2*(b^2*c*d^2*e*f - b^2*c^2*d*f^2)*sqrt(f*x^2 + e)*x + (b^2*c*d^2*e^2 + (b^2*c^2*
d - 4*a*b*c*d^2)*e*f - 2*(b^2*c^3 - 2*a*b*c^2*d)*f^2)*sqrt(f)*log(-2*f*x^2 - 2*sqrt(f*x^2 + e)*sqrt(f)*x - e))
/(c*d^3*e*f^2 - c^2*d^2*f^3), 1/4*(2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(c*d*e - c^2*f)*f^2*arctan(1/2*sqrt(c
*d*e - c^2*f)*((d*e - 2*c*f)*x^2 - c*e)*sqrt(f*x^2 + e)/((c*d*e*f - c^2*f^2)*x^3 + (c*d*e^2 - c^2*e*f)*x)) + 2
*(b^2*c*d^2*e*f - b^2*c^2*d*f^2)*sqrt(f*x^2 + e)*x - (b^2*c*d^2*e^2 + (b^2*c^2*d - 4*a*b*c*d^2)*e*f - 2*(b^2*c
^3 - 2*a*b*c^2*d)*f^2)*sqrt(f)*log(-2*f*x^2 - 2*sqrt(f*x^2 + e)*sqrt(f)*x - e))/(c*d^3*e*f^2 - c^2*d^2*f^3), -
1/4*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(-c*d*e + c^2*f)*f^2*log(((d^2*e^2 - 8*c*d*e*f + 8*c^2*f^2)*x^4 + c^2
*e^2 - 2*(3*c*d*e^2 - 4*c^2*e*f)*x^2 - 4*((d*e - 2*c*f)*x^3 - c*e*x)*sqrt(-c*d*e + c^2*f)*sqrt(f*x^2 + e))/(d^
2*x^4 + 2*c*d*x^2 + c^2)) - 2*(b^2*c*d^2*e*f - b^2*c^2*d*f^2)*sqrt(f*x^2 + e)*x - 2*(b^2*c*d^2*e^2 + (b^2*c^2*
d - 4*a*b*c*d^2)*e*f - 2*(b^2*c^3 - 2*a*b*c^2*d)*f^2)*sqrt(-f)*arctan(sqrt(-f)*x/sqrt(f*x^2 + e)))/(c*d^3*e*f^
2 - c^2*d^2*f^3), 1/2*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(c*d*e - c^2*f)*f^2*arctan(1/2*sqrt(c*d*e - c^2*f)*
((d*e - 2*c*f)*x^2 - c*e)*sqrt(f*x^2 + e)/((c*d*e*f - c^2*f^2)*x^3 + (c*d*e^2 - c^2*e*f)*x)) + (b^2*c*d^2*e*f
- b^2*c^2*d*f^2)*sqrt(f*x^2 + e)*x + (b^2*c*d^2*e^2 + (b^2*c^2*d - 4*a*b*c*d^2)*e*f - 2*(b^2*c^3 - 2*a*b*c^2*d
)*f^2)*sqrt(-f)*arctan(sqrt(-f)*x/sqrt(f*x^2 + e)))/(c*d^3*e*f^2 - c^2*d^2*f^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 ) \sqrt{e + f x^{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)/(f*x**2+e)**(1/2),x)

[Out]

Integral((a + b*x**2)**2/((c + d*x**2)*sqrt(e + f*x**2)), x)

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Giac [A]  time = 1.65003, size = 248, normalized size = 1.49 \begin{align*} \frac{\sqrt{f x^{2} + e} b^{2} x}{2 \, d f} - \frac{{\left (b^{2} c^{2} \sqrt{f} - 2 \, a b c d \sqrt{f} + a^{2} d^{2} \sqrt{f}\right )} \arctan \left (\frac{{\left (\sqrt{f} x - \sqrt{f x^{2} + e}\right )}^{2} d + 2 \, c f - d e}{2 \, \sqrt{-c^{2} f^{2} + c d f e}}\right )}{\sqrt{-c^{2} f^{2} + c d f e} d^{2}} + \frac{{\left (2 \, b^{2} c f^{\frac{3}{2}} - 4 \, a b d f^{\frac{3}{2}} + b^{2} d \sqrt{f} e\right )} \log \left ({\left (\sqrt{f} x - \sqrt{f x^{2} + e}\right )}^{2}\right )}{4 \, d^{2} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2/(d*x^2+c)/(f*x^2+e)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(f*x^2 + e)*b^2*x/(d*f) - (b^2*c^2*sqrt(f) - 2*a*b*c*d*sqrt(f) + a^2*d^2*sqrt(f))*arctan(1/2*((sqrt(f)
*x - sqrt(f*x^2 + e))^2*d + 2*c*f - d*e)/sqrt(-c^2*f^2 + c*d*f*e))/(sqrt(-c^2*f^2 + c*d*f*e)*d^2) + 1/4*(2*b^2
*c*f^(3/2) - 4*a*b*d*f^(3/2) + b^2*d*sqrt(f)*e)*log((sqrt(f)*x - sqrt(f*x^2 + e))^2)/(d^2*f^2)