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

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

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Rubi [A]  time = 4.24, antiderivative size = 482, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {979, 1080, 217, 206, 1034, 725} \begin {gather*} \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} \end {gather*}

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 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 1.07, size = 603, normalized size = 1.25 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [C]  time = 0.70, size = 557, normalized size = 1.15 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e+a^2 f\&,\frac {2 \text {$\#$1}^2 c^2 d e f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 \left (-c^2\right ) e^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1}^2 a c e f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 a^2 c e f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} a^2 \sqrt {c} f^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} c^{5/2} d^2 f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} c^{5/2} d e^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+4 \text {$\#$1} a c^{3/2} d f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 a c^2 d e f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+a c^2 e^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 \text {$\#$1}^3 f-3 \text {$\#$1}^2 \sqrt {c} e-2 \text {$\#$1} a f+4 \text {$\#$1} c d+a \sqrt {c} e}\&\right ]}{f^3}+\frac {\log \left (\sqrt {a+c x^2}-\sqrt {c} x\right ) \left (-3 a \sqrt {c} f^2+2 c^{3/2} d f-2 c^{3/2} e^2\right )}{2 f^3}-\frac {c \sqrt {a+c x^2} (2 e-f x)}{2 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(c*(2*e - f*x)*Sqrt[a + c*x^2])/f^2 + ((-2*c^(3/2)*e^2 + 2*c^(3/2)*d*f - 3*a*Sqrt[c]*f^2)*Log[-(Sqrt[c]*x
) + Sqrt[a + c*x^2]])/(2*f^3) + RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3
+ f*#1^4 & , (a*c^2*e^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - 2*a*c^2*d*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + c
*x^2] - #1] + 2*a^2*c*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(5/2)*d*e^2*Log[-(Sqrt[c]*x) + Sqrt
[a + c*x^2] - #1]*#1 - 2*c^(5/2)*d^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 4*a*c^(3/2)*d*f^2*Log[-(S
qrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - 2*a^2*Sqrt[c]*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - c^2*e^3
*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 2*c^2*d*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - 2
*a*c*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^
2 + 2*f*#1^3) & ]/f^3

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{2} \end {gather*}

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

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maple [B]  time = 0.01, size = 8954, normalized size = 18.50 \begin {gather*} \text {output too large to display} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` f
or more details)Is 4*d*f-e^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^2+a\right )}^{3/2}}{f\,x^2+e\,x+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{d + e x + f x^{2}}\, dx \end {gather*}

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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