3.1019 \(\int \frac{\sqrt{x} (A+B x)}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=276 \[ -\frac{\sqrt{x} (-2 a B-x (b B-2 A c)+A b)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{\left (-\frac{4 a B c-4 A b c+b^2 B}{\sqrt{b^2-4 a c}}-2 A c+b B\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (\frac{4 a B c-4 A b c+b^2 B}{\sqrt{b^2-4 a c}}-2 A c+b B\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}} \]

[Out]

-((Sqrt[x]*(A*b - 2*a*B - (b*B - 2*A*c)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2))) + ((b*B - 2*A*c - (b^2*B - 4*A*
b*c + 4*a*B*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt
[c]*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((b*B - 2*A*c + (b^2*B - 4*A*b*c + 4*a*B*c)/Sqrt[b^2 - 4*a*c]
)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b + Sqrt[
b^2 - 4*a*c]])

________________________________________________________________________________________

Rubi [A]  time = 0.957594, antiderivative size = 276, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {820, 826, 1166, 205} \[ -\frac{\sqrt{x} (-2 a B-x (b B-2 A c)+A b)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{\left (-\frac{4 a B c-4 A b c+b^2 B}{\sqrt{b^2-4 a c}}-2 A c+b B\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (\frac{4 a B c-4 A b c+b^2 B}{\sqrt{b^2-4 a c}}-2 A c+b B\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{\sqrt{b^2-4 a c}+b}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[x]*(A + B*x))/(a + b*x + c*x^2)^2,x]

[Out]

-((Sqrt[x]*(A*b - 2*a*B - (b*B - 2*A*c)*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2))) + ((b*B - 2*A*c - (b^2*B - 4*A*
b*c + 4*a*B*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt
[c]*(b^2 - 4*a*c)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((b*B - 2*A*c + (b^2*B - 4*A*b*c + 4*a*B*c)/Sqrt[b^2 - 4*a*c]
)*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*(b^2 - 4*a*c)*Sqrt[b + Sqrt[
b^2 - 4*a*c]])

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/
((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g*(2*a*e*m + b*d*(2*p + 3)) - f*
(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p]
 || IntegersQ[2*m, 2*p])

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
 q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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{\sqrt{x} (A+B x)}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{\sqrt{x} (A b-2 a B-(b B-2 A c) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{\int \frac{\frac{1}{2} (-A b+2 a B)-\frac{1}{2} (b B-2 A c) x}{\sqrt{x} \left (a+b x+c x^2\right )} \, dx}{-b^2+4 a c}\\ &=-\frac{\sqrt{x} (A b-2 a B-(b B-2 A c) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} (-A b+2 a B)+\frac{1}{2} (-b B+2 A c) x^2}{a+b x^2+c x^4} \, dx,x,\sqrt{x}\right )}{b^2-4 a c}\\ &=-\frac{\sqrt{x} (A b-2 a B-(b B-2 A c) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{\left (b B-2 A c-\frac{b^2 B-4 A b c+4 a B c}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b}{2}-\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx,x,\sqrt{x}\right )}{2 \left (b^2-4 a c\right )}+\frac{\left (b B-2 A c+\frac{b^2 B-4 A b c+4 a B c}{\sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b}{2}+\frac{1}{2} \sqrt{b^2-4 a c}+c x^2} \, dx,x,\sqrt{x}\right )}{2 \left (b^2-4 a c\right )}\\ &=-\frac{\sqrt{x} (A b-2 a B-(b B-2 A c) x)}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac{\left (b B-2 A c-\frac{b^2 B-4 A b c+4 a B c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b-\sqrt{b^2-4 a c}}}+\frac{\left (b B-2 A c+\frac{b^2 B-4 A b c+4 a B c}{\sqrt{b^2-4 a c}}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c} \left (b^2-4 a c\right ) \sqrt{b+\sqrt{b^2-4 a c}}}\\ \end{align*}

Mathematica [A]  time = 0.728504, size = 295, normalized size = 1.07 \[ \frac{\frac{x^{3/2} \left (A \left (-2 a c+b^2+b c x\right )-a B (b+2 c x)\right )}{a+x (b+c x)}+\frac{a \left (\frac{\left (-2 A c \sqrt{b^2-4 a c}+b B \sqrt{b^2-4 a c}+4 a B c-4 A b c+b^2 B\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{\sqrt{b^2-4 a c}+b}}\right )}{\sqrt{b^2-4 a c} \sqrt{\sqrt{b^2-4 a c}+b}}-\frac{\left (\frac{4 a B c-4 A b c+b^2 B}{\sqrt{b^2-4 a c}}+2 A c-b B\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{x}}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{b-\sqrt{b^2-4 a c}}}\right )}{\sqrt{2} \sqrt{c}}+\sqrt{x} (2 a B-A b)}{a \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[x]*(A + B*x))/(a + b*x + c*x^2)^2,x]

[Out]

((-(A*b) + 2*a*B)*Sqrt[x] + (x^(3/2)*(-(a*B*(b + 2*c*x)) + A*(b^2 - 2*a*c + b*c*x)))/(a + x*(b + c*x)) + (a*(-
(((-(b*B) + 2*A*c + (b^2*B - 4*A*b*c + 4*a*B*c)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b - S
qrt[b^2 - 4*a*c]]])/Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((b^2*B - 4*A*b*c + 4*a*B*c + b*B*Sqrt[b^2 - 4*a*c] - 2*A*c
*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[b +
 Sqrt[b^2 - 4*a*c]])))/(Sqrt[2]*Sqrt[c]))/(a*(b^2 - 4*a*c))

________________________________________________________________________________________

Maple [B]  time = 0.033, size = 755, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*x^(1/2)/(c*x^2+b*x+a)^2,x)

[Out]

2*(1/2*(2*A*c-B*b)/(4*a*c-b^2)*x^(3/2)+1/2*(A*b-2*B*a)/(4*a*c-b^2)*x^(1/2))/(c*x^2+b*x+a)-1/(4*a*c-b^2)*c*2^(1
/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(x^(1/2)*c*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*A+2/(4*a*c-
b^2)*c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(x^(1/2)*c*2^(1/2)/((-b+(-4*a*c+b^2
)^(1/2))*c)^(1/2))*A*b+1/2/(4*a*c-b^2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(x^(1/2)*c*2^(1/2)/((-
b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b*B-2/(4*a*c-b^2)*c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/
2)*arctanh(x^(1/2)*c*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*a*B-1/2/(4*a*c-b^2)/(-4*a*c+b^2)^(1/2)*2^(1/2)
/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(x^(1/2)*c*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b^2*B+1/(4*a*c
-b^2)*c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(x^(1/2)*c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*A+
2/(4*a*c-b^2)*c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(x^(1/2)*c*2^(1/2)/((b+(-4*a
*c+b^2)^(1/2))*c)^(1/2))*A*b-1/2/(4*a*c-b^2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(x^(1/2)*c*2^(1/2)
/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b*B-2/(4*a*c-b^2)*c/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(
1/2)*arctan(x^(1/2)*c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*a*B-1/2/(4*a*c-b^2)/(-4*a*c+b^2)^(1/2)*2^(1/2)
/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(x^(1/2)*c*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))*b^2*B

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (2 \, B a c - A b c\right )} x^{\frac{5}{2}} +{\left (B a b -{\left (b^{2} - 2 \, a c\right )} A\right )} x^{\frac{3}{2}}}{a^{2} b^{2} - 4 \, a^{3} c +{\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} x^{2} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x} + \int \frac{{\left (2 \, B a c - A b c\right )} x^{\frac{3}{2}} +{\left (3 \, B a b -{\left (b^{2} + 2 \, a c\right )} A\right )} \sqrt{x}}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c +{\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} x^{2} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

-((2*B*a*c - A*b*c)*x^(5/2) + (B*a*b - (b^2 - 2*a*c)*A)*x^(3/2))/(a^2*b^2 - 4*a^3*c + (a*b^2*c - 4*a^2*c^2)*x^
2 + (a*b^3 - 4*a^2*b*c)*x) + integrate(1/2*((2*B*a*c - A*b*c)*x^(3/2) + (3*B*a*b - (b^2 + 2*a*c)*A)*sqrt(x))/(
a^2*b^2 - 4*a^3*c + (a*b^2*c - 4*a^2*c^2)*x^2 + (a*b^3 - 4*a^2*b*c)*x), x)

________________________________________________________________________________________

Fricas [B]  time = 4.75215, size = 7082, normalized size = 25.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-1/2*(sqrt(1/2)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2
- 3*A^2*a*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b^2 + A^2*b^3)*c + (a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64
*a^4*c^4)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5
)))/(a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4))*log(sqrt(1/2)*(2*B^3*a^2*b^4 - A*B^2*a*b^5 - 16*
(2*A^2*B*a^3 - A^3*a^2*b)*c^3 + 8*(4*B^3*a^4 - 2*A*B^2*a^3*b + 2*A^2*B*a^2*b^2 - A^3*a*b^3)*c^2 - (16*B^3*a^3*
b^2 - 8*A*B^2*a^2*b^3 + 2*A^2*B*a*b^4 - A^3*b^5)*c + (192*B*a^4*b^3*c^3 + 256*A*a^5*c^5 - 128*(2*B*a^5*b + A*a
^4*b^2)*c^4 - 8*(6*B*a^3*b^5 - A*a^2*b^6)*c^2 + (4*B*a^2*b^7 - A*a*b^8)*c)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4
*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2 - 3*A^2*a
*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b^2 + A^2*b^3)*c + (a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)
*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))/(a*b^
6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)) - 2*(3*B^4*a^2*b^2 - A*B^3*a*b^3 - 4*A^4*a*c^3 + 3*(4*A^3
*B*a*b - A^4*b^2)*c^2 + (4*B^4*a^3 - 12*A*B^3*a^2*b + A^3*B*b^3)*c)*sqrt(x)) - sqrt(1/2)*(a*b^2 - 4*a^2*c + (b
^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2 - 3*A^2*a*b)*c^2 + (12*B^2*a^2*b - 12
*A*B*a*b^2 + A^2*b^3)*c + (a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)*sqrt((B^4*a^2 - 2*A^2*B^2*a
*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))/(a*b^6*c - 12*a^2*b^4*c^2 + 48*a^
3*b^2*c^3 - 64*a^4*c^4))*log(-sqrt(1/2)*(2*B^3*a^2*b^4 - A*B^2*a*b^5 - 16*(2*A^2*B*a^3 - A^3*a^2*b)*c^3 + 8*(4
*B^3*a^4 - 2*A*B^2*a^3*b + 2*A^2*B*a^2*b^2 - A^3*a*b^3)*c^2 - (16*B^3*a^3*b^2 - 8*A*B^2*a^2*b^3 + 2*A^2*B*a*b^
4 - A^3*b^5)*c + (192*B*a^4*b^3*c^3 + 256*A*a^5*c^5 - 128*(2*B*a^5*b + A*a^4*b^2)*c^4 - 8*(6*B*a^3*b^5 - A*a^2
*b^6)*c^2 + (4*B*a^2*b^7 - A*a*b^8)*c)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3
+ 48*a^4*b^2*c^4 - 64*a^5*c^5)))*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2 - 3*A^2*a*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b
^2 + A^2*b^3)*c + (a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4
*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))/(a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^
3 - 64*a^4*c^4)) - 2*(3*B^4*a^2*b^2 - A*B^3*a*b^3 - 4*A^4*a*c^3 + 3*(4*A^3*B*a*b - A^4*b^2)*c^2 + (4*B^4*a^3 -
 12*A*B^3*a^2*b + A^3*B*b^3)*c)*sqrt(x)) + sqrt(1/2)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c
)*x)*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2 - 3*A^2*a*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b^2 + A^2*b^3)*c - (a*b^6*c -
 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*
b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))/(a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4))*log(sqrt(1/
2)*(2*B^3*a^2*b^4 - A*B^2*a*b^5 - 16*(2*A^2*B*a^3 - A^3*a^2*b)*c^3 + 8*(4*B^3*a^4 - 2*A*B^2*a^3*b + 2*A^2*B*a^
2*b^2 - A^3*a*b^3)*c^2 - (16*B^3*a^3*b^2 - 8*A*B^2*a^2*b^3 + 2*A^2*B*a*b^4 - A^3*b^5)*c - (192*B*a^4*b^3*c^3 +
 256*A*a^5*c^5 - 128*(2*B*a^5*b + A*a^4*b^2)*c^4 - 8*(6*B*a^3*b^5 - A*a^2*b^6)*c^2 + (4*B*a^2*b^7 - A*a*b^8)*c
)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))*sqrt
(-(B^2*a*b^3 - 4*(4*A*B*a^2 - 3*A^2*a*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b^2 + A^2*b^3)*c - (a*b^6*c - 12*a^2*b
^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 +
 48*a^4*b^2*c^4 - 64*a^5*c^5)))/(a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)) - 2*(3*B^4*a^2*b^2 -
 A*B^3*a*b^3 - 4*A^4*a*c^3 + 3*(4*A^3*B*a*b - A^4*b^2)*c^2 + (4*B^4*a^3 - 12*A*B^3*a^2*b + A^3*B*b^3)*c)*sqrt(
x)) - sqrt(1/2)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2
- 3*A^2*a*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b^2 + A^2*b^3)*c - (a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64
*a^4*c^4)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5
)))/(a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4))*log(-sqrt(1/2)*(2*B^3*a^2*b^4 - A*B^2*a*b^5 - 16
*(2*A^2*B*a^3 - A^3*a^2*b)*c^3 + 8*(4*B^3*a^4 - 2*A*B^2*a^3*b + 2*A^2*B*a^2*b^2 - A^3*a*b^3)*c^2 - (16*B^3*a^3
*b^2 - 8*A*B^2*a^2*b^3 + 2*A^2*B*a*b^4 - A^3*b^5)*c - (192*B*a^4*b^3*c^3 + 256*A*a^5*c^5 - 128*(2*B*a^5*b + A*
a^4*b^2)*c^4 - 8*(6*B*a^3*b^5 - A*a^2*b^6)*c^2 + (4*B*a^2*b^7 - A*a*b^8)*c)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^
4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))*sqrt(-(B^2*a*b^3 - 4*(4*A*B*a^2 - 3*A^2*
a*b)*c^2 + (12*B^2*a^2*b - 12*A*B*a*b^2 + A^2*b^3)*c - (a*b^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4
)*sqrt((B^4*a^2 - 2*A^2*B^2*a*c + A^4*c^2)/(a^2*b^6*c^2 - 12*a^3*b^4*c^3 + 48*a^4*b^2*c^4 - 64*a^5*c^5)))/(a*b
^6*c - 12*a^2*b^4*c^2 + 48*a^3*b^2*c^3 - 64*a^4*c^4)) - 2*(3*B^4*a^2*b^2 - A*B^3*a*b^3 - 4*A^4*a*c^3 + 3*(4*A^
3*B*a*b - A^4*b^2)*c^2 + (4*B^4*a^3 - 12*A*B^3*a^2*b + A^3*B*b^3)*c)*sqrt(x)) - 2*(2*B*a - A*b + (B*b - 2*A*c)
*x)*sqrt(x))/(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)

________________________________________________________________________________________

Sympy [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((B*x+A)*x**(1/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [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((B*x+A)*x^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

Timed out