3.58 \(\int \frac{(a+b x^2)^3}{(c+d x^2) \sqrt{e+f x^2}} \, dx\)

Optimal. Leaf size=304 \[ \frac{b \left (8 a^2 f^2-8 a b e f+3 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{8 d f^{5/2}}-\frac{b^2 x \sqrt{e+f x^2} (b c-a d)}{2 d^2 f}-\frac{3 b^2 x \sqrt{e+f x^2} (b e-2 a f)}{8 d f^2}+\frac{b^2 x \left (a+b x^2\right ) \sqrt{e+f x^2}}{4 d f}+\frac{b (b c-a d) (b e-2 a f) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{2 d^2 f^{3/2}}-\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{c} \sqrt{e+f x^2}}\right )}{\sqrt{c} d^3 \sqrt{d e-c f}}+\frac{b (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{d^3 \sqrt{f}} \]

[Out]

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

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Rubi [A]  time = 0.299454, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {545, 416, 388, 217, 206, 523, 377, 205} \[ \frac{b \left (8 a^2 f^2-8 a b e f+3 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{8 d f^{5/2}}-\frac{b^2 x \sqrt{e+f x^2} (b c-a d)}{2 d^2 f}-\frac{3 b^2 x \sqrt{e+f x^2} (b e-2 a f)}{8 d f^2}+\frac{b^2 x \left (a+b x^2\right ) \sqrt{e+f x^2}}{4 d f}+\frac{b (b c-a d) (b e-2 a f) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{2 d^2 f^{3/2}}-\frac{(b c-a d)^3 \tan ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{c} \sqrt{e+f x^2}}\right )}{\sqrt{c} d^3 \sqrt{d e-c f}}+\frac{b (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{d^3 \sqrt{f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 416

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

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

Mathematica [A]  time = 0.342896, size = 265, normalized size = 0.87 \[ \frac{\frac{8 b \left (3 a^2 d^2-3 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{\sqrt{f}}+\frac{4 b^2 d e (b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )}{f^{3/2}}-\frac{4 b^2 d x \sqrt{e+f x^2} (b c-3 a d)}{f}+\frac{8 (a d-b c)^3 \tan ^{-1}\left (\frac{x \sqrt{d e-c f}}{\sqrt{c} \sqrt{e+f x^2}}\right )}{\sqrt{c} \sqrt{d e-c f}}+\frac{3 b^3 d^2 e \left (e \tanh ^{-1}\left (\frac{\sqrt{f} x}{\sqrt{e+f x^2}}\right )-\sqrt{f} x \sqrt{e+f x^2}\right )}{f^{5/2}}+\frac{2 b^3 d^2 x^3 \sqrt{e+f x^2}}{f}}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.039, size = 1541, normalized size = 5.1 \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)^3/(d*x^2+c)/(f*x^2+e)^(1/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 43.4153, size = 3518, normalized size = 11.57 \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)^3/(d*x^2+c)/(f*x^2+e)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-c*d*e + c^2*f)*f^3*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)) + (3*b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f +
4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3)*e*f^2 - 8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqr
t(f)*log(-2*f*x^2 - 2*sqrt(f*x^2 + e)*sqrt(f)*x - e) + 2*(2*(b^3*c*d^3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c
*d^3*e^2*f + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(
c*d^4*e*f^3 - c^2*d^3*f^4), -1/16*(8*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c*d*e - c^2*f)*f
^3*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)) - (3*b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f + 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 +
6*a^2*b*c*d^3)*e*f^2 - 8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqrt(f)*log(-2*f*x^2 - 2*sqrt(f*x^2
+ e)*sqrt(f)*x - e) - 2*(2*(b^3*c*d^3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c*d^3*e^2*f + (b^3*c^2*d^2 - 12*a*
b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(c*d^4*e*f^3 - c^2*d^3*f^4), 1/8*(
2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-c*d*e + c^2*f)*f^3*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)*sqr
t(f*x^2 + e))/(d^2*x^4 + 2*c*d*x^2 + c^2)) - (3*b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f + 4*(b^3*
c^3*d - 3*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3)*e*f^2 - 8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqrt(-f)*a
rctan(sqrt(-f)*x/sqrt(f*x^2 + e)) + (2*(b^3*c*d^3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c*d^3*e^2*f + (b^3*c^2
*d^2 - 12*a*b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(c*d^4*e*f^3 - c^2*d^3
*f^4), -1/8*(4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c*d*e - c^2*f)*f^3*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)) + (3*
b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f + 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3)*e*f^2 -
8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqrt(-f)*arctan(sqrt(-f)*x/sqrt(f*x^2 + e)) - (2*(b^3*c*d^
3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c*d^3*e^2*f + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*
a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(c*d^4*e*f^3 - c^2*d^3*f^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{3}}{\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)**3/(d*x**2+c)/(f*x**2+e)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.7549, size = 392, normalized size = 1.29 \begin{align*} \frac{1}{8} \,{\left (\frac{2 \, b^{3} x^{2}}{d f} - \frac{4 \, b^{3} c d^{4} f^{2} - 12 \, a b^{2} d^{5} f^{2} + 3 \, b^{3} d^{5} f e}{d^{6} f^{3}}\right )} \sqrt{f x^{2} + e} x + \frac{{\left (b^{3} c^{3} \sqrt{f} - 3 \, a b^{2} c^{2} d \sqrt{f} + 3 \, a^{2} b c d^{2} \sqrt{f} - a^{3} d^{3} \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^{3}} - \frac{{\left (8 \, b^{3} c^{2} f^{2} - 24 \, a b^{2} c d f^{2} + 24 \, a^{2} b d^{2} f^{2} + 4 \, b^{3} c d f e - 12 \, a b^{2} d^{2} f e + 3 \, b^{3} d^{2} e^{2}\right )} \log \left ({\left (\sqrt{f} x - \sqrt{f x^{2} + e}\right )}^{2}\right )}{16 \, d^{3} f^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*b^3*x^2/(d*f) - (4*b^3*c*d^4*f^2 - 12*a*b^2*d^5*f^2 + 3*b^3*d^5*f*e)/(d^6*f^3))*sqrt(f*x^2 + e)*x + (b^
3*c^3*sqrt(f) - 3*a*b^2*c^2*d*sqrt(f) + 3*a^2*b*c*d^2*sqrt(f) - a^3*d^3*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^3) - 1/16*(8*b^3*c^2*f^2
- 24*a*b^2*c*d*f^2 + 24*a^2*b*d^2*f^2 + 4*b^3*c*d*f*e - 12*a*b^2*d^2*f*e + 3*b^3*d^2*e^2)*log((sqrt(f)*x - sq
rt(f*x^2 + e))^2)/(d^3*f^(5/2))