3.1.60 \(\int \frac {(a+c x^2)^{3/2}}{d+e x+f x^2} \, dx\) [60]

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

[Out]

1/2*(3*a*f^2+2*c*(-d*f+e^2))*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))*c^(1/2)/f^3-1/2*c*(-f*x+2*e)*(c*x^2+a)^(1/2)/f
^2-1/2*arctanh(1/2*(2*a*f-c*x*(e-(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+
e^2)^(1/2)))^(1/2))*(-2*f*(2*a*c*d*f^2-a^2*f^3+c^2*d*(-d*f+e^2))+c*e*(2*a*f^2+c*(-2*d*f+e^2))*(e-(-4*d*f+e^2)^
(1/2)))/f^3*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)+1/2*arctanh(1/2*(2*a
*f-c*x*(e+(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2)))^(1/2))*(-2
*f*(2*a*c*d*f^2-a^2*f^3+c^2*d*(-d*f+e^2))+c*e*(2*a*f^2+c*(-2*d*f+e^2))*(e+(-4*d*f+e^2)^(1/2)))/f^3*2^(1/2)/(-4
*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/2)))^(1/2)

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Rubi [A]
time = 2.76, 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 = {993, 1094, 223, 212, 1048, 739} \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]

-1/2*(c*(2*e - f*x)*Sqrt[a + c*x^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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 739

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 993

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] && N
eQ[p + q, 0] && NeQ[2*p + 2*q + 1, 0] &&  !IGtQ[p, 0] &&  !IGtQ[q, 0]

Rule 1048

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 1094

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 ) \text {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 ) \text {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 ) \text {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 [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.57, size = 542, normalized size = 1.12 \begin {gather*} \frac {c f (-2 e+f x) \sqrt {a+c x^2}+\sqrt {c} \left (-2 c e^2+2 c d f-3 a f^2\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )+2 \text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a c^2 e^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-2 a c^2 d e f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 a^2 c e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 c^{5/2} d e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 c^{5/2} d^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+4 a c^{3/2} d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-2 a^2 \sqrt {c} f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c^2 e^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 c^2 d e f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-2 a c e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{2 f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2260\) vs. \(2(435)=870\).
time = 0.14, size = 2261, normalized size = 4.67

method	result	size
default	\(\text {Expression too large to display}\)	\(2261\)
risch	\(\text {Expression too large to display}\)	\(5464\)

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,method=_RETURNVERBOSE)

[Out]

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

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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?`
for more det

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Fricas [F(-1)] Timed out
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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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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Giac [F(-1)] Timed out
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="giac")

[Out]

Timed out

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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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