3.60 \(\int \frac{(a+c x^2)^{3/2}}{d+e x+f x^2} \, dx\)

Optimal. Leaf size=484 \[ -\frac{\left (c e \left (e-\sqrt{e^2-4 d f}\right ) \left (2 a f^2+c \left (e^2-2 d f\right )\right )-2 f \left (-a^2 f^3+2 a c d f^2+c^2 d \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\left (c e \left (\sqrt{e^2-4 d f}+e\right ) \left (2 a f^2+c \left (e^2-2 d f\right )\right )-2 f \left (-a^2 f^3+2 a c d f^2+c^2 d \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-d f\right )\right )}{2 f^3}-\frac{c \sqrt{a+c x^2} (2 e-f x)}{2 f^2} \]

[Out]

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

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Rubi [A]  time = 4.23987, antiderivative size = 482, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {979, 1080, 217, 206, 1034, 725} \[ \frac{\left (-2 a^2 f^4-c e \left (e-\sqrt{e^2-4 d f}\right ) \left (2 a f^2+c \left (e^2-2 d f\right )\right )+4 a c d f^3+2 c^2 d f \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (e-\sqrt{e^2-4 d f}\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}-\frac{\left (-2 a^2 f^4-c e \left (\sqrt{e^2-4 d f}+e\right ) \left (2 a f^2+c \left (e^2-2 d f\right )\right )+4 a c d f^3+2 c^2 d f \left (e^2-d f\right )\right ) \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{2} \sqrt{a+c x^2} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt{2} f^3 \sqrt{e^2-4 d f} \sqrt{2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}+\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-d f\right )\right )}{2 f^3}-\frac{c \sqrt{a+c x^2} (2 e-f x)}{2 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 979

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(c*(e*(2*p + q) -
2*f*(p + q)*x)*(a + c*x^2)^(p - 1)*(d + e*x + f*x^2)^(q + 1))/(2*f^2*(p + q)*(2*p + 2*q + 1)), x] - Dist[1/(2
*f^2*(p + q)*(2*p + 2*q + 1)), Int[(a + c*x^2)^(p - 2)*(d + e*x + f*x^2)^q*Simp[-(a*c*e^2*(1 - p)*(2*p + q)) +
a*(p + q)*(-2*a*f^2*(2*p + 2*q + 1) + c*(2*d*f - e^2*(2*p + q))) + (2*(c*d - a*f)*(c*e)*(1 - p)*(2*p + q) + 4
*a*c*e*f*(1 - p)*(p + q))*x + (p*c^2*e^2*(1 - p) - c*(p + q)*(2*a*f^2*(4*p + 2*q - 1) + c*(2*d*f*(1 - 2*p) + e
^2*(3*p + q - 1))))*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && GtQ[p, 1] && NeQ
[p + q, 0] && NeQ[2*p + 2*q + 1, 0] &&  !IGtQ[p, 0] &&  !IGtQ[q, 0]

Rule 1080

Int[((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (f_.)*(x_)^2]), x_Sym
bol] :> Dist[C/c, Int[1/Sqrt[d + f*x^2], x], x] + Dist[1/c, Int[(A*c - a*C + (B*c - b*C)*x)/((a + b*x + c*x^2)
*Sqrt[d + f*x^2]), x], x] /; FreeQ[{a, b, c, d, f, A, B, C}, x] && NeQ[b^2 - 4*a*c, 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 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rubi steps

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

Mathematica [A]  time = 1.15791, size = 603, normalized size = 1.25 \[ \frac{\frac{2 \left (2 a f^2+c \left (-e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right ) \left (-\sqrt{4 a f^2-2 c e \sqrt{e^2-4 d f}-4 c d f+2 c e^2} \tanh ^{-1}\left (\frac{2 a f+c x \left (\sqrt{e^2-4 d f}-e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2-2 c \left (e \sqrt{e^2-4 d f}+2 d f-e^2\right )}}\right )+\sqrt{c} \left (\sqrt{e^2-4 d f}-e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+2 f \sqrt{a+c x^2}\right )}{f^2}+\frac{2 \left (2 a f^2+c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )\right ) \left (\sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )} \tanh ^{-1}\left (\frac{2 a f-c x \left (\sqrt{e^2-4 d f}+e\right )}{\sqrt{a+c x^2} \sqrt{4 a f^2+2 c \left (e \sqrt{e^2-4 d f}-2 d f+e^2\right )}}\right )+\sqrt{c} \left (\sqrt{e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )-2 f \sqrt{a+c x^2}\right )}{f^2}+\frac{2 \sqrt{c} \sqrt{a+c x^2} \left (\sqrt{e^2-4 d f}-e\right ) \left (\sqrt{c} x \sqrt{\frac{c x^2}{a}+1}+\sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )\right )}{\sqrt{\frac{c x^2}{a}+1}}+\frac{2 \sqrt{c} \sqrt{a+c x^2} \left (\sqrt{e^2-4 d f}+e\right ) \left (\sqrt{c} x \sqrt{\frac{c x^2}{a}+1}+\sqrt{a} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )\right )}{\sqrt{\frac{c x^2}{a}+1}}}{8 f \sqrt{e^2-4 d f}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[c]*(-e + Sqrt[e^2 - 4*d*f])*Sqrt[a + c*x^2]*(Sqrt[c]*x*Sqrt[1 + (c*x^2)/a] + Sqrt[a]*ArcSinh[(Sqrt[c]
*x)/Sqrt[a]]))/Sqrt[1 + (c*x^2)/a] + (2*Sqrt[c]*(e + Sqrt[e^2 - 4*d*f])*Sqrt[a + c*x^2]*(Sqrt[c]*x*Sqrt[1 + (c
*x^2)/a] + Sqrt[a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]]))/Sqrt[1 + (c*x^2)/a] + (2*(2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[
e^2 - 4*d*f]))*(2*f*Sqrt[a + c*x^2] + Sqrt[c]*(-e + Sqrt[e^2 - 4*d*f])*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] -
Sqrt[2*c*e^2 - 4*c*d*f + 4*a*f^2 - 2*c*e*Sqrt[e^2 - 4*d*f]]*ArcTanh[(2*a*f + c*(-e + Sqrt[e^2 - 4*d*f])*x)/(Sq
rt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])]))/f^2 + (2*(2*a*f^2 + c*(e^2 - 2*d*f
+ e*Sqrt[e^2 - 4*d*f]))*(-2*f*Sqrt[a + c*x^2] + Sqrt[c]*(e + Sqrt[e^2 - 4*d*f])*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c
*x^2]] + Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)
/(Sqrt[4*a*f^2 + 2*c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])]))/f^2)/(8*f*Sqrt[e^2 - 4*d*f])

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Maple [B]  time = 0.269, size = 8954, normalized size = 18.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**(3/2)/(f*x**2+e*x+d),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2