Integrand size = 23, antiderivative size = 521 \[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=-\frac {5 A b^2-3 a b B-14 a A c}{3 a^2 \left (b^2-4 a c\right ) x^{3/2}}-\frac {a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )}{a^3 \left (b^2-4 a c\right ) \sqrt {x}}+\frac {A b^2-a b B-2 a A c+(A b-2 a B) c x}{a \left (b^2-4 a c\right ) x^{3/2} \left (a+b x+c x^2\right )}-\frac {\sqrt {c} \left (a B \left (3 b^3-16 a b c+3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}\right )-A \left (5 b^4-29 a b^2 c+28 a^2 c^2+5 b^3 \sqrt {b^2-4 a c}-19 a b c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^3 \left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\sqrt {c} \left (a B \left (3 b^3-16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right )-A \left (5 b^4-29 a b^2 c+28 a^2 c^2-5 b^3 \sqrt {b^2-4 a c}+19 a b c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} a^3 \left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}} \]
1/3*(14*A*a*c-5*A*b^2+3*B*a*b)/a^2/(-4*a*c+b^2)/x^(3/2)+(A*b^2-a*b*B-2*A*a *c+(A*b-2*B*a)*c*x)/a/(-4*a*c+b^2)/x^(3/2)/(c*x^2+b*x+a)+(-a*B*(-10*a*c+3* b^2)+A*(-19*a*b*c+5*b^3))/a^3/(-4*a*c+b^2)/x^(1/2)-1/2*arctan(2^(1/2)*c^(1 /2)*x^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(a*B*(3*b^3-16*a*b*c+3*b ^2*(-4*a*c+b^2)^(1/2)-10*a*c*(-4*a*c+b^2)^(1/2))-A*(5*b^4-29*a*b^2*c+28*a^ 2*c^2+5*b^3*(-4*a*c+b^2)^(1/2)-19*a*b*c*(-4*a*c+b^2)^(1/2)))/a^3/(-4*a*c+b ^2)^(3/2)*2^(1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)+1/2*arctan(2^(1/2)*c^(1/2)* x^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*c^(1/2)*(a*B*(3*b^3-16*a*b*c-3*b^2*( -4*a*c+b^2)^(1/2)+10*a*c*(-4*a*c+b^2)^(1/2))-A*(5*b^4-29*a*b^2*c+28*a^2*c^ 2-5*b^3*(-4*a*c+b^2)^(1/2)+19*a*b*c*(-4*a*c+b^2)^(1/2)))/a^3/(-4*a*c+b^2)^ (3/2)*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 3.55 (sec) , antiderivative size = 518, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\frac {-\frac {2 \left (8 a^3 c (A+3 B x)+15 A b^3 x^2 (b+c x)+a^2 \left (-2 A \left (b^2+20 b c x-7 c^2 x^2\right )+3 B x \left (-2 b^2+11 b c x+10 c^2 x^2\right )\right )-a b x \left (9 b B x (b+c x)+A \left (-10 b^2+62 b c x+57 c^2 x^2\right )\right )\right )}{x^{3/2} (a+x (b+c x))}-\frac {3 \sqrt {2} \sqrt {c} \left (a B \left (-3 b^3+16 a b c-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}\right )+A \left (5 b^4-29 a b^2 c+28 a^2 c^2+5 b^3 \sqrt {b^2-4 a c}-19 a b c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {3 \sqrt {2} \sqrt {c} \left (a B \left (-3 b^3+16 a b c+3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}\right )+A \left (5 b^4-29 a b^2 c+28 a^2 c^2-5 b^3 \sqrt {b^2-4 a c}+19 a b c \sqrt {b^2-4 a c}\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {b+\sqrt {b^2-4 a c}}}}{6 a^3 \left (-b^2+4 a c\right )} \]
((-2*(8*a^3*c*(A + 3*B*x) + 15*A*b^3*x^2*(b + c*x) + a^2*(-2*A*(b^2 + 20*b *c*x - 7*c^2*x^2) + 3*B*x*(-2*b^2 + 11*b*c*x + 10*c^2*x^2)) - a*b*x*(9*b*B *x*(b + c*x) + A*(-10*b^2 + 62*b*c*x + 57*c^2*x^2))))/(x^(3/2)*(a + x*(b + c*x))) - (3*Sqrt[2]*Sqrt[c]*(a*B*(-3*b^3 + 16*a*b*c - 3*b^2*Sqrt[b^2 - 4* a*c] + 10*a*c*Sqrt[b^2 - 4*a*c]) + A*(5*b^4 - 29*a*b^2*c + 28*a^2*c^2 + 5* b^3*Sqrt[b^2 - 4*a*c] - 19*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[b^2 - 4*a*c]*Sqrt[b - Sqrt [b^2 - 4*a*c]]) + (3*Sqrt[2]*Sqrt[c]*(a*B*(-3*b^3 + 16*a*b*c + 3*b^2*Sqrt[ b^2 - 4*a*c] - 10*a*c*Sqrt[b^2 - 4*a*c]) + A*(5*b^4 - 29*a*b^2*c + 28*a^2* c^2 - 5*b^3*Sqrt[b^2 - 4*a*c] + 19*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[b^2 - 4*a*c]*Sqrt[ b + Sqrt[b^2 - 4*a*c]]))/(6*a^3*(-b^2 + 4*a*c))
Time = 0.93 (sec) , antiderivative size = 524, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {1235, 27, 1198, 1198, 25, 1197, 1480, 218}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\int -\frac {5 A b^2-3 a B b-14 a A c+5 (A b-2 a B) c x}{2 x^{5/2} \left (c x^2+b x+a\right )}dx}{a \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {5 A b^2-3 a B b-14 a A c+5 (A b-2 a B) c x}{x^{5/2} \left (c x^2+b x+a\right )}dx}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {\frac {\int \frac {a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )-c \left (5 A b^2-3 a B b-14 a A c\right ) x}{x^{3/2} \left (c x^2+b x+a\right )}dx}{a}-\frac {2 \left (-14 a A c-3 a b B+5 A b^2\right )}{3 a x^{3/2}}}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {\frac {\frac {\int -\frac {a b B \left (3 b^2-13 a c\right )-A \left (5 b^4-24 a c b^2+14 a^2 c^2\right )+c \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right ) x}{\sqrt {x} \left (c x^2+b x+a\right )}dx}{a}-\frac {2 \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right )}{a \sqrt {x}}}{a}-\frac {2 \left (-14 a A c-3 a b B+5 A b^2\right )}{3 a x^{3/2}}}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {\int \frac {a b B \left (3 b^2-13 a c\right )-A \left (5 b^4-24 a c b^2+14 a^2 c^2\right )+c \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right ) x}{\sqrt {x} \left (c x^2+b x+a\right )}dx}{a}-\frac {2 \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right )}{a \sqrt {x}}}{a}-\frac {2 \left (-14 a A c-3 a b B+5 A b^2\right )}{3 a x^{3/2}}}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {\frac {-\frac {2 \int \frac {a b B \left (3 b^2-13 a c\right )-A \left (5 b^4-24 a c b^2+14 a^2 c^2\right )+c \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right ) x}{c x^2+b x+a}d\sqrt {x}}{a}-\frac {2 \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right )}{a \sqrt {x}}}{a}-\frac {2 \left (-14 a A c-3 a b B+5 A b^2\right )}{3 a x^{3/2}}}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {-\frac {2 \left (\frac {c \left (a B \left (3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}-16 a b c+3 b^3\right )-A \left (28 a^2 c^2-29 a b^2 c-19 a b c \sqrt {b^2-4 a c}+5 b^3 \sqrt {b^2-4 a c}+5 b^4\right )\right ) \int \frac {1}{\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )+c x}d\sqrt {x}}{2 \sqrt {b^2-4 a c}}-\frac {c \left (a B \left (-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}-16 a b c+3 b^3\right )-A \left (28 a^2 c^2-29 a b^2 c+19 a b c \sqrt {b^2-4 a c}-5 b^3 \sqrt {b^2-4 a c}+5 b^4\right )\right ) \int \frac {1}{\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )+c x}d\sqrt {x}}{2 \sqrt {b^2-4 a c}}\right )}{a}-\frac {2 \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right )}{a \sqrt {x}}}{a}-\frac {2 \left (-14 a A c-3 a b B+5 A b^2\right )}{3 a x^{3/2}}}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {-\frac {2 \left (\frac {\sqrt {c} \left (a B \left (3 b^2 \sqrt {b^2-4 a c}-10 a c \sqrt {b^2-4 a c}-16 a b c+3 b^3\right )-A \left (28 a^2 c^2-29 a b^2 c-19 a b c \sqrt {b^2-4 a c}+5 b^3 \sqrt {b^2-4 a c}+5 b^4\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (a B \left (-3 b^2 \sqrt {b^2-4 a c}+10 a c \sqrt {b^2-4 a c}-16 a b c+3 b^3\right )-A \left (28 a^2 c^2-29 a b^2 c+19 a b c \sqrt {b^2-4 a c}-5 b^3 \sqrt {b^2-4 a c}+5 b^4\right )\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} \sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )}{a}-\frac {2 \left (a B \left (3 b^2-10 a c\right )-A \left (5 b^3-19 a b c\right )\right )}{a \sqrt {x}}}{a}-\frac {2 \left (-14 a A c-3 a b B+5 A b^2\right )}{3 a x^{3/2}}}{2 a \left (b^2-4 a c\right )}+\frac {c x (A b-2 a B)-2 a A c-a b B+A b^2}{a x^{3/2} \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}\) |
(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x)/(a*(b^2 - 4*a*c)*x^(3/2)*(a + b*x + c*x^2)) + ((-2*(5*A*b^2 - 3*a*b*B - 14*a*A*c))/(3*a*x^(3/2)) + ((- 2*(a*B*(3*b^2 - 10*a*c) - A*(5*b^3 - 19*a*b*c)))/(a*Sqrt[x]) - (2*((Sqrt[c ]*(a*B*(3*b^3 - 16*a*b*c + 3*b^2*Sqrt[b^2 - 4*a*c] - 10*a*c*Sqrt[b^2 - 4*a *c]) - A*(5*b^4 - 29*a*b^2*c + 28*a^2*c^2 + 5*b^3*Sqrt[b^2 - 4*a*c] - 19*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[b^2 - 4*a*c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) - (S qrt[c]*(a*B*(3*b^3 - 16*a*b*c - 3*b^2*Sqrt[b^2 - 4*a*c] + 10*a*c*Sqrt[b^2 - 4*a*c]) - A*(5*b^4 - 29*a*b^2*c + 28*a^2*c^2 - 5*b^3*Sqrt[b^2 - 4*a*c] + 19*a*b*c*Sqrt[b^2 - 4*a*c]))*ArcTan[(Sqrt[2]*Sqrt[c]*Sqrt[x])/Sqrt[b + Sq rt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]) ))/a)/a)/(2*a*(b^2 - 4*a*c))
3.11.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[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] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c *d^2 - b*d*e + a*e^2))), x] + Simp[1/(c*d^2 - b*d*e + a*e^2) Int[(d + e*x )^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] && LtQ[m, -1 ]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(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]
Time = 0.42 (sec) , antiderivative size = 485, normalized size of antiderivative = 0.93
| method | result | size |
| risch | \(-\frac {2 \left (-6 A b x +3 a B x +a A \right )}{3 a^{3} x^{\frac {3}{2}}}+\frac {\frac {\frac {2 c \left (3 A a b c -A \,b^{3}-2 B \,a^{2} c +B a \,b^{2}\right ) x^{\frac {3}{2}}}{8 a c -2 b^{2}}-\frac {\left (2 A \,a^{2} c^{2}-4 A a \,b^{2} c +A \,b^{4}+3 a^{2} b B c -B \,b^{3} a \right ) \sqrt {x}}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {4 c \left (\frac {\left (19 A \sqrt {-4 a c +b^{2}}\, a b c -5 A \sqrt {-4 a c +b^{2}}\, b^{3}+28 A \,a^{2} c^{2}-29 A a \,b^{2} c +5 A \,b^{4}-10 B \sqrt {-4 a c +b^{2}}\, a^{2} c +3 B \sqrt {-4 a c +b^{2}}\, a \,b^{2}+16 a^{2} b B c -3 B \,b^{3} a \right ) \sqrt {2}\, \arctan \left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {\left (19 A \sqrt {-4 a c +b^{2}}\, a b c -5 A \sqrt {-4 a c +b^{2}}\, b^{3}-28 A \,a^{2} c^{2}+29 A a \,b^{2} c -5 A \,b^{4}-10 B \sqrt {-4 a c +b^{2}}\, a^{2} c +3 B \sqrt {-4 a c +b^{2}}\, a \,b^{2}-16 a^{2} b B c +3 B \,b^{3} a \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{a^{3}}\) | \(485\) |
| derivativedivides | \(-\frac {2 \left (\frac {-\frac {c \left (3 A a b c -A \,b^{3}-2 B \,a^{2} c +B a \,b^{2}\right ) x^{\frac {3}{2}}}{2 \left (4 a c -b^{2}\right )}+\frac {\left (2 A \,a^{2} c^{2}-4 A a \,b^{2} c +A \,b^{4}+3 a^{2} b B c -B \,b^{3} a \right ) \sqrt {x}}{8 a c -2 b^{2}}}{c \,x^{2}+b x +a}+\frac {2 c \left (-\frac {\left (-19 A \sqrt {-4 a c +b^{2}}\, a b c +5 A \sqrt {-4 a c +b^{2}}\, b^{3}+28 A \,a^{2} c^{2}-29 A a \,b^{2} c +5 A \,b^{4}+10 B \sqrt {-4 a c +b^{2}}\, a^{2} c -3 B \sqrt {-4 a c +b^{2}}\, a \,b^{2}+16 a^{2} b B c -3 B \,b^{3} a \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-19 A \sqrt {-4 a c +b^{2}}\, a b c +5 A \sqrt {-4 a c +b^{2}}\, b^{3}-28 A \,a^{2} c^{2}+29 A a \,b^{2} c -5 A \,b^{4}+10 B \sqrt {-4 a c +b^{2}}\, a^{2} c -3 B \sqrt {-4 a c +b^{2}}\, a \,b^{2}-16 a^{2} b B c +3 B \,b^{3} a \right ) \sqrt {2}\, \arctan \left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}\right )}{a^{3}}-\frac {2 A}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 \left (-2 A b +B a \right )}{a^{3} \sqrt {x}}\) | \(488\) |
| default | \(-\frac {2 \left (\frac {-\frac {c \left (3 A a b c -A \,b^{3}-2 B \,a^{2} c +B a \,b^{2}\right ) x^{\frac {3}{2}}}{2 \left (4 a c -b^{2}\right )}+\frac {\left (2 A \,a^{2} c^{2}-4 A a \,b^{2} c +A \,b^{4}+3 a^{2} b B c -B \,b^{3} a \right ) \sqrt {x}}{8 a c -2 b^{2}}}{c \,x^{2}+b x +a}+\frac {2 c \left (-\frac {\left (-19 A \sqrt {-4 a c +b^{2}}\, a b c +5 A \sqrt {-4 a c +b^{2}}\, b^{3}+28 A \,a^{2} c^{2}-29 A a \,b^{2} c +5 A \,b^{4}+10 B \sqrt {-4 a c +b^{2}}\, a^{2} c -3 B \sqrt {-4 a c +b^{2}}\, a \,b^{2}+16 a^{2} b B c -3 B \,b^{3} a \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-19 A \sqrt {-4 a c +b^{2}}\, a b c +5 A \sqrt {-4 a c +b^{2}}\, b^{3}-28 A \,a^{2} c^{2}+29 A a \,b^{2} c -5 A \,b^{4}+10 B \sqrt {-4 a c +b^{2}}\, a^{2} c -3 B \sqrt {-4 a c +b^{2}}\, a \,b^{2}-16 a^{2} b B c +3 B \,b^{3} a \right ) \sqrt {2}\, \arctan \left (\frac {c \sqrt {x}\, \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}\right )}{a^{3}}-\frac {2 A}{3 a^{2} x^{\frac {3}{2}}}-\frac {2 \left (-2 A b +B a \right )}{a^{3} \sqrt {x}}\) | \(488\) |
-2/3*(-6*A*b*x+3*B*a*x+A*a)/a^3/x^(3/2)+1/a^3*(2*(1/2*c*(3*A*a*b*c-A*b^3-2 *B*a^2*c+B*a*b^2)/(4*a*c-b^2)*x^(3/2)-1/2*(2*A*a^2*c^2-4*A*a*b^2*c+A*b^4+3 *B*a^2*b*c-B*a*b^3)/(4*a*c-b^2)*x^(1/2))/(c*x^2+b*x+a)+4/(4*a*c-b^2)*c*(1/ 8*(19*A*(-4*a*c+b^2)^(1/2)*a*b*c-5*A*(-4*a*c+b^2)^(1/2)*b^3+28*A*a^2*c^2-2 9*A*a*b^2*c+5*A*b^4-10*B*(-4*a*c+b^2)^(1/2)*a^2*c+3*B*(-4*a*c+b^2)^(1/2)*a *b^2+16*a^2*b*B*c-3*B*b^3*a)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^( 1/2))*c)^(1/2)*arctan(c*x^(1/2)*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))*c)^(1/2))- 1/8*(19*A*(-4*a*c+b^2)^(1/2)*a*b*c-5*A*(-4*a*c+b^2)^(1/2)*b^3-28*A*a^2*c^2 +29*A*a*b^2*c-5*A*b^4-10*B*(-4*a*c+b^2)^(1/2)*a^2*c+3*B*(-4*a*c+b^2)^(1/2) *a*b^2-16*a^2*b*B*c+3*B*b^3*a)/(-4*a*c+b^2)^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2 )^(1/2))*c)^(1/2)*arctanh(c*x^(1/2)*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))*c)^(1 /2))))
Leaf count of result is larger than twice the leaf count of optimal. 10203 vs. \(2 (460) = 920\).
Time = 65.88 (sec) , antiderivative size = 10203, normalized size of antiderivative = 19.58 \[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x + a\right )}^{2} x^{\frac {5}{2}}} \,d x } \]
1/3*(3*((5*b^4*c - 24*a*b^2*c^2 + 14*a^2*c^3)*A - (3*a*b^3*c - 13*a^2*b*c^ 2)*B)*x^(5/2) + 3*((5*b^5 - 19*a*b^3*c - 5*a^2*b*c^2)*A - (3*a*b^4 - 10*a^ 2*b^2*c - 10*a^3*c^2)*B)*x^(3/2) + 2*((15*a*b^4 - 67*a^2*b^2*c + 28*a^3*c^ 2)*A - 9*(a^2*b^3 - 4*a^3*b*c)*B)*sqrt(x) - 2*(a^3*b^2 - 4*a^4*c)*A/x^(3/2 ) + 2*(5*(a^2*b^3 - 4*a^3*b*c)*A - 3*(a^3*b^2 - 4*a^4*c)*B)/sqrt(x))/(a^5* b^2 - 4*a^6*c + (a^4*b^2*c - 4*a^5*c^2)*x^2 + (a^4*b^3 - 4*a^5*b*c)*x) + i ntegrate(-1/2*(((5*b^4*c - 24*a*b^2*c^2 + 14*a^2*c^3)*A - (3*a*b^3*c - 13* a^2*b*c^2)*B)*x^(3/2) + ((5*b^5 - 29*a*b^3*c + 33*a^2*b*c^2)*A - (3*a*b^4 - 16*a^2*b^2*c + 10*a^3*c^2)*B)*sqrt(x))/(a^5*b^2 - 4*a^6*c + (a^4*b^2*c - 4*a^5*c^2)*x^2 + (a^4*b^3 - 4*a^5*b*c)*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 6335 vs. \(2 (460) = 920\).
Time = 1.53 (sec) , antiderivative size = 6335, normalized size of antiderivative = 12.16 \[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]
-(B*a*b^2*c*x^(3/2) - A*b^3*c*x^(3/2) - 2*B*a^2*c^2*x^(3/2) + 3*A*a*b*c^2* x^(3/2) + B*a*b^3*sqrt(x) - A*b^4*sqrt(x) - 3*B*a^2*b*c*sqrt(x) + 4*A*a*b^ 2*c*sqrt(x) - 2*A*a^2*c^2*sqrt(x))/((a^3*b^2 - 4*a^4*c)*(c*x^2 + b*x + a)) - 1/8*((10*b^5*c^2 - 78*a*b^3*c^3 + 152*a^2*b*c^4 - 5*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5 + 39*sqrt(2)*sqrt(b^2 - 4*a*c)* sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c + 10*sqrt(2)*sqrt(b^2 - 4*a*c)*sqr t(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c - 76*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 - 38*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 - 5*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 + 19*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqr t(b^2 - 4*a*c)*c)*a*b*c^3 - 10*(b^2 - 4*a*c)*b^3*c^2 + 38*(b^2 - 4*a*c)*a* b*c^3)*(a^3*b^2 - 4*a^4*c)^2*A - (6*a*b^4*c^2 - 44*a^2*b^2*c^3 + 80*a^3*c^ 4 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4 + 22 *sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c + 6*s qrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c - 40*sqrt (2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*c^2 - 20*sqrt(2) *sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^2 - 3*sqrt(2)*s qrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 + 10*sqrt(2)*sq rt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^3 - 6*(b^2 - 4*a*c)* a*b^2*c^2 + 20*(b^2 - 4*a*c)*a^2*c^3)*(a^3*b^2 - 4*a^4*c)^2*B - 2*(5*sq...
Time = 15.93 (sec) , antiderivative size = 21585, normalized size of antiderivative = 41.43 \[ \int \frac {A+B x}{x^{5/2} \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \]
- ((2*A)/(3*a) - (2*x*(5*A*b - 3*B*a))/(3*a^2) + (x^2*(15*A*b^4 + 14*A*a^2 *c^2 - 9*B*a*b^3 - 62*A*a*b^2*c + 33*B*a^2*b*c))/(3*a^3*(4*a*c - b^2)) + ( c*x^3*(5*A*b^3 - 3*B*a*b^2 + 10*B*a^2*c - 19*A*a*b*c))/(a^3*(4*a*c - b^2)) )/(a*x^(3/2) + b*x^(5/2) + c*x^(7/2)) - atan((((-(25*A^2*b^15 + 9*B^2*a^2* b^13 - 25*A^2*b^6*(-(4*a*c - b^2)^9)^(1/2) - 30*A*B*a*b^14 + 6366*A^2*a^2* b^11*c^2 - 35767*A^2*a^3*b^9*c^3 + 116928*A^2*a^4*b^7*c^4 - 219744*A^2*a^5 *b^5*c^5 + 215040*A^2*a^6*b^3*c^6 + 49*A^2*a^3*c^3*(-(4*a*c - b^2)^9)^(1/2 ) - 9*B^2*a^2*b^4*(-(4*a*c - b^2)^9)^(1/2) + 2077*B^2*a^4*b^9*c^2 - 10656* B^2*a^5*b^7*c^3 + 30240*B^2*a^6*b^5*c^4 - 44800*B^2*a^7*b^3*c^5 - 25*B^2*a ^4*c^2*(-(4*a*c - b^2)^9)^(1/2) + 35840*A*B*a^8*c^7 - 615*A^2*a*b^13*c - 8 0640*A^2*a^7*b*c^7 - 213*B^2*a^3*b^11*c + 26880*B^2*a^8*b*c^6 - 246*A^2*a^ 2*b^2*c^2*(-(4*a*c - b^2)^9)^(1/2) - 7278*A*B*a^3*b^10*c^2 + 39132*A*B*a^4 *b^8*c^3 - 119616*A*B*a^5*b^6*c^4 + 201600*A*B*a^6*b^4*c^5 - 161280*A*B*a^ 7*b^2*c^6 + 165*A^2*a*b^4*c*(-(4*a*c - b^2)^9)^(1/2) + 51*B^2*a^3*b^2*c*(- (4*a*c - b^2)^9)^(1/2) + 30*A*B*a*b^5*(-(4*a*c - b^2)^9)^(1/2) + 724*A*B*a ^2*b^12*c - 184*A*B*a^2*b^3*c*(-(4*a*c - b^2)^9)^(1/2) + 186*A*B*a^3*b*c^2 *(-(4*a*c - b^2)^9)^(1/2))/(8*(a^7*b^12 + 4096*a^13*c^6 - 24*a^8*b^10*c + 240*a^9*b^8*c^2 - 1280*a^10*b^6*c^3 + 3840*a^11*b^4*c^4 - 6144*a^12*b^2*c^ 5)))^(1/2)*(x^(1/2)*(-(25*A^2*b^15 + 9*B^2*a^2*b^13 - 25*A^2*b^6*(-(4*a*c - b^2)^9)^(1/2) - 30*A*B*a*b^14 + 6366*A^2*a^2*b^11*c^2 - 35767*A^2*a^3...