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

Optimal. Leaf size=553 \[ -\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} f^4 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} f^4 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )}{2 f^4}+\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f} \]

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Rubi [A]  time = 2.43, antiderivative size = 553, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1020, 1068, 1080, 217, 206, 1034, 725} \begin {gather*} -\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\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^4 \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 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\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^4 \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 {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^3}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )}{2 f^4}+\frac {\left (a+c x^2\right )^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(a*f^2 + c*(e^2 - d*f)) - c*e*f*x)*Sqrt[a + c*x^2])/(2*f^3) + (a + c*x^2)^(3/2)/(3*f) - (Sqrt[c]*e*(3*a*f^
2 + 2*c*(e^2 - 2*d*f))*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*f^4) - ((2*c*d*e*f*(2*a*f^2 + c*(e^2 - 2*d*f))
- (e - Sqrt[e^2 - 4*d*f])*(a^2*f^4 + 2*a*c*f^2*(e^2 - d*f) + c^2*(e^4 - 3*d*e^2*f + d^2*f^2)))*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^4*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) + ((2*c*d*e*f*(2*a*f
^2 + c*(e^2 - 2*d*f)) - (e + Sqrt[e^2 - 4*d*f])*(a^2*f^4 + 2*a*c*f^2*(e^2 - d*f) + c^2*(e^4 - 3*d*e^2*f + d^2*
f^2)))*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^4*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 1020

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(h*(a + c*x^2)^p*(d + e*x + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] + Dist[1/(2*f*(p + q + 1)), Int[(a + c*x^2)
^(p - 1)*(d + e*x + f*x^2)^q*Simp[a*h*e*p - a*(h*e - 2*g*f)*(p + q + 1) - 2*h*p*(c*d - a*f)*x - (h*c*e*p + c*(
h*e - 2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h, q}, x] && NeQ[e^2 - 4*d*f, 0] && GtQ[
p, 0] && NeQ[p + q + 1, 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 1068

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

Rubi steps

\begin {align*} \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx &=\frac {\left (a+c x^2\right )^{3/2}}{3 f}+\frac {\int \frac {\sqrt {a+c x^2} \left (-3 (c d-a f) x-3 c e x^2\right )}{d+e x+f x^2} \, dx}{3 f}\\ &=\frac {\left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right ) \sqrt {a+c x^2}}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\int \frac {-3 a c^2 d e f+3 c \left (a c e^2 f+2 (c d-a f) \left (c e^2-c d f+a f^2\right )\right ) x+3 c^2 e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) x^2}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{6 c f^3}\\ &=\frac {\left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right ) \sqrt {a+c x^2}}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\int \frac {-3 a c^2 d e f^2-3 c^2 d e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )+\left (-3 c^2 e^2 \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )+3 c f \left (a c e^2 f+2 (c d-a f) \left (c e^2-c d f+a f^2\right )\right )\right ) x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{6 c f^4}-\frac {\left (c e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 f^4}\\ &=\frac {\left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right ) \sqrt {a+c x^2}}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\left (c e \left (3 a f^2+2 c \left (e^2-2 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^4}+\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{f^4 \sqrt {e^2-4 d f}}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{f^4 \sqrt {e^2-4 d f}}\\ &=\frac {\left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right ) \sqrt {a+c x^2}}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\sqrt {c} e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 f^4}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\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 )}{f^4 \sqrt {e^2-4 d f}}+\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\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 )}{f^4 \sqrt {e^2-4 d f}}\\ &=\frac {\left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right ) \sqrt {a+c x^2}}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\sqrt {c} e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 f^4}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\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^4 \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 (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\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^4 \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 = 2.01, size = 755, normalized size = 1.37 \begin {gather*} \frac {8 f^3 \left (a+c x^2\right )^{5/2} \sqrt {\frac {c x^2}{a}+1} \left (\sqrt {e^2-4 d f}-e\right )+8 f^3 \left (a+c x^2\right )^{5/2} \sqrt {\frac {c x^2}{a}+1} \left (\sqrt {e^2-4 d f}+e\right )+3 \left (e-\sqrt {e^2-4 d f}\right ) \left (2 \sqrt {c} f^2 \sqrt {a+c x^2} \left (e-\sqrt {e^2-4 d f}\right ) \left (a \sqrt {c} x \left (\frac {c x^2}{a}+1\right )^{3/2}+\sqrt {a} \left (a+c x^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )\right )-a \left (\frac {c x^2}{a}+1\right )^{3/2} \left (4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2\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 )\right )-3 \left (\sqrt {e^2-4 d f}+e\right ) \left (2 \sqrt {c} f^2 \sqrt {a+c x^2} \left (\sqrt {e^2-4 d f}+e\right ) \left (a \sqrt {c} x \left (\frac {c x^2}{a}+1\right )^{3/2}+\sqrt {a} \left (a+c x^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )\right )-a \left (\frac {c x^2}{a}+1\right )^{3/2} \left (4 a f^2+c \left (\sqrt {e^2-4 d f}+e\right )^2\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 )\right )}{48 a f^4 \left (\frac {c x^2}{a}+1\right )^{3/2} \sqrt {e^2-4 d f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*f^3*(-e + Sqrt[e^2 - 4*d*f])*(a + c*x^2)^(5/2)*Sqrt[1 + (c*x^2)/a] + 8*f^3*(e + Sqrt[e^2 - 4*d*f])*(a + c*x
^2)^(5/2)*Sqrt[1 + (c*x^2)/a] + 3*(e - Sqrt[e^2 - 4*d*f])*(2*Sqrt[c]*f^2*(e - Sqrt[e^2 - 4*d*f])*Sqrt[a + c*x^
2]*(a*Sqrt[c]*x*(1 + (c*x^2)/a)^(3/2) + Sqrt[a]*(a + c*x^2)*ArcSinh[(Sqrt[c]*x)/Sqrt[a]]) - a*(4*a*f^2 + c*(e
- Sqrt[e^2 - 4*d*f])^2)*(1 + (c*x^2)/a)^(3/2)*(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)/(Sqrt[4*a*f^2 - 2*c*(-e^2 + 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])) - 3*
(e + Sqrt[e^2 - 4*d*f])*(2*Sqrt[c]*f^2*(e + Sqrt[e^2 - 4*d*f])*Sqrt[a + c*x^2]*(a*Sqrt[c]*x*(1 + (c*x^2)/a)^(3
/2) + Sqrt[a]*(a + c*x^2)*ArcSinh[(Sqrt[c]*x)/Sqrt[a]]) - a*(4*a*f^2 + c*(e + Sqrt[e^2 - 4*d*f])^2)*(1 + (c*x^
2)/a)^(3/2)*(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])])))/(48*a*f^4*Sqrt[e^2 - 4*d*f]*(1 + (c*x^2)/a)^
(3/2))

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IntegrateAlgebraic [C]  time = 0.89, size = 771, normalized size = 1.39 \begin {gather*} -\frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e+a^2 f\&,\frac {-\text {$\#$1}^2 a^2 f^4 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 c^2 d^2 f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+3 \text {$\#$1}^2 c^2 d e^2 f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 \left (-c^2\right ) e^4 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1}^2 a c d f^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1}^2 a c e^2 f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+a^3 f^4 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-2 a^2 c d f^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 a^2 c e^2 f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-4 \text {$\#$1} c^{5/2} d^2 e f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} c^{5/2} d e^3 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+4 \text {$\#$1} a c^{3/2} d e f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+a c^2 d^2 f^2 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )-3 a c^2 d e^2 f \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )+a c^2 e^4 \log \left (-\text {$\#$1}+\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 \text {$\#$1}^3 f-3 \text {$\#$1}^2 \sqrt {c} e-2 \text {$\#$1} a f+4 \text {$\#$1} c d+a \sqrt {c} e}\&\right ]}{f^4}+\frac {\log \left (\sqrt {a+c x^2}-\sqrt {c} x\right ) \left (3 a \sqrt {c} e f^2-4 c^{3/2} d e f+2 c^{3/2} e^3\right )}{2 f^4}+\frac {\sqrt {a+c x^2} \left (8 a f^2-6 c d f+6 c e^2-3 c e f x+2 c f^2 x^2\right )}{6 f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(6*c*e^2 - 6*c*d*f + 8*a*f^2 - 3*c*e*f*x + 2*c*f^2*x^2))/(6*f^3) + ((2*c^(3/2)*e^3 - 4*c^(3/2
)*d*e*f + 3*a*Sqrt[c]*e*f^2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(2*f^4) - 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^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] -
3*a*c^2*d*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + a*c^2*d^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] -
#1] + 2*a^2*c*e^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - 2*a^2*c*d*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*
x^2] - #1] + a^3*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(5/2)*d*e^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*
x^2] - #1]*#1 - 4*c^(5/2)*d^2*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 4*a*c^(3/2)*d*e*f^2*Log[-(Sqrt
[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - c^2*e^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 3*c^2*d*e^2*f*Log[
-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - c^2*d^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - 2*a*c*
e^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 2*a*c*d*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#
1^2 - a^2*f^4*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^4

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

[Out]

sage2

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maple [B]  time = 0.02, size = 14709, normalized size = 26.60 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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