Integrand size = 23, antiderivative size = 284 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=-\frac {5 \left (8 a B \left (b^2+4 a c\right )-A \left (b^3-20 a b c\right )-2 c \left (16 a b B+A \left (b^2+12 a c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{64 a x}-\frac {5 \left (4 a (A b+4 a B)+3 \left (8 a b B+A \left (b^2+4 a c\right )\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{96 a x^3}-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}-\frac {5 \left (8 a b B \left (b^2+12 a c\right )-A \left (b^4-24 a b^2 c-48 a^2 c^2\right )\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{128 a^{3/2}}+\frac {5}{8} \sqrt {c} \left (3 b^2 B+4 A b c+4 a B c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \]
-5/96*(4*a*(A*b+4*B*a)+3*(8*a*b*B+A*(4*a*c+b^2))*x)*(c*x^2+b*x+a)^(3/2)/a/ x^3-1/4*(-2*B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4-5/128*(8*a*b*B*(12*a*c+b^2)-A*( -48*a^2*c^2-24*a*b^2*c+b^4))*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^( 1/2))/a^(3/2)+5/8*(4*A*b*c+4*B*a*c+3*B*b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/ (c*x^2+b*x+a)^(1/2))*c^(1/2)-5/64*(8*a*B*(4*a*c+b^2)-A*(-20*a*b*c+b^3)-2*c *(16*a*b*B+A*(12*a*c+b^2))*x)*(c*x^2+b*x+a)^(1/2)/a/x
Time = 2.23 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=-\frac {\sqrt {a+x (b+c x)} \left (15 A b^3 x^3+16 a^3 (3 A+4 B x)+8 a^2 x \left (17 A b+26 b B x+27 A c x+56 B c x^2\right )+2 a x^2 \left (A \left (59 b^2+278 b c x-96 c^2 x^2\right )-12 B x \left (-11 b^2+18 b c x+4 c^2 x^2\right )\right )\right )}{192 a x^4}-\frac {5 A b^4 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{64 a^{3/2}}-\frac {5 \left (b^3 B+3 A b^2 c+12 a b B c+6 a A c^2\right ) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {5}{8} \sqrt {c} \left (3 b^2 B+4 A b c+4 a B c\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right ) \]
-1/192*(Sqrt[a + x*(b + c*x)]*(15*A*b^3*x^3 + 16*a^3*(3*A + 4*B*x) + 8*a^2 *x*(17*A*b + 26*b*B*x + 27*A*c*x + 56*B*c*x^2) + 2*a*x^2*(A*(59*b^2 + 278* b*c*x - 96*c^2*x^2) - 12*B*x*(-11*b^2 + 18*b*c*x + 4*c^2*x^2))))/(a*x^4) - (5*A*b^4*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]])/(64*a^(3/2 )) - (5*(b^3*B + 3*A*b^2*c + 12*a*b*B*c + 6*a*A*c^2)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + x*(b + c*x)])/Sqrt[a]])/(8*Sqrt[a]) - (5*Sqrt[c]*(3*b^2*B + 4* A*b*c + 4*a*B*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/8
Time = 0.67 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {1230, 27, 1229, 27, 1230, 25, 1269, 1092, 219, 1154, 219}
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.
| \(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx\) |
| \(\Big \downarrow \) 1230 |
| \(\displaystyle -\frac {5}{16} \int -\frac {2 (A b+4 a B+2 (b B+A c) x) \left (c x^2+b x+a\right )^{3/2}}{x^4}dx-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{8} \int \frac {(A b+4 a B+2 (b B+A c) x) \left (c x^2+b x+a\right )^{3/2}}{x^4}dx-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 1229 |
| \(\displaystyle \frac {5}{8} \left (-\frac {\int -\frac {\left (8 a B \left (b^2+4 a c\right )-A \left (b^3-20 a b c\right )+2 c \left (16 a b B+A \left (b^2+12 a c\right )\right ) x\right ) \sqrt {c x^2+b x+a}}{2 x^2}dx}{4 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5}{8} \left (\frac {\int \frac {\left (8 a B \left (b^2+4 a c\right )-A \left (b^3-20 a b c\right )+2 c \left (16 a b B+A \left (b^2+12 a c\right )\right ) x\right ) \sqrt {c x^2+b x+a}}{x^2}dx}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 1230 |
| \(\displaystyle \frac {5}{8} \left (\frac {-\frac {1}{2} \int -\frac {8 a b B \left (b^2+12 a c\right )-A \left (b^4-24 a c b^2-48 a^2 c^2\right )+16 a c \left (3 B b^2+4 A c b+4 a B c\right ) x}{x \sqrt {c x^2+b x+a}}dx-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5}{8} \left (\frac {\frac {1}{2} \int \frac {8 a b B \left (b^2+12 a c\right )-A \left (b^4-24 a c b^2-48 a^2 c^2\right )+16 a c \left (3 B b^2+4 A c b+4 a B c\right ) x}{x \sqrt {c x^2+b x+a}}dx-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {5}{8} \left (\frac {\frac {1}{2} \left (\left (8 a b B \left (12 a c+b^2\right )-A \left (-48 a^2 c^2-24 a b^2 c+b^4\right )\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+16 a c \left (4 a B c+4 A b c+3 b^2 B\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx\right )-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {5}{8} \left (\frac {\frac {1}{2} \left (\left (8 a b B \left (12 a c+b^2\right )-A \left (-48 a^2 c^2-24 a b^2 c+b^4\right )\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+32 a c \left (4 a B c+4 A b c+3 b^2 B\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {5}{8} \left (\frac {\frac {1}{2} \left (\left (8 a b B \left (12 a c+b^2\right )-A \left (-48 a^2 c^2-24 a b^2 c+b^4\right )\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+16 a \sqrt {c} \left (4 a B c+4 A b c+3 b^2 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\right )-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {5}{8} \left (\frac {\frac {1}{2} \left (16 a \sqrt {c} \left (4 a B c+4 A b c+3 b^2 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-2 \left (8 a b B \left (12 a c+b^2\right )-A \left (-48 a^2 c^2-24 a b^2 c+b^4\right )\right ) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {5}{8} \left (\frac {\frac {1}{2} \left (16 a \sqrt {c} \left (4 a B c+4 A b c+3 b^2 B\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )-\frac {\left (8 a b B \left (12 a c+b^2\right )-A \left (-48 a^2 c^2-24 a b^2 c+b^4\right )\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}}\right )-\frac {\sqrt {a+b x+c x^2} \left (-A \left (b^3-20 a b c\right )-2 c x \left (A \left (12 a c+b^2\right )+16 a b B\right )+8 a B \left (4 a c+b^2\right )\right )}{x}}{8 a}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (3 x \left (A \left (4 a c+b^2\right )+8 a b B\right )+4 a (4 a B+A b)\right )}{12 a x^3}\right )-\frac {(A-2 B x) \left (a+b x+c x^2\right )^{5/2}}{4 x^4}\) |
-1/4*((A - 2*B*x)*(a + b*x + c*x^2)^(5/2))/x^4 + (5*(-1/12*((4*a*(A*b + 4* a*B) + 3*(8*a*b*B + A*(b^2 + 4*a*c))*x)*(a + b*x + c*x^2)^(3/2))/(a*x^3) + (-(((8*a*B*(b^2 + 4*a*c) - A*(b^3 - 20*a*b*c) - 2*c*(16*a*b*B + A*(b^2 + 12*a*c))*x)*Sqrt[a + b*x + c*x^2])/x) + (-(((8*a*b*B*(b^2 + 12*a*c) - A*(b ^4 - 24*a*b^2*c - 48*a^2*c^2))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/Sqrt[a]) + 16*a*Sqrt[c]*(3*b^2*B + 4*A*b*c + 4*a*B*c)*ArcTanh [(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/2)/(8*a)))/8
3.10.44.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*((a + b*x + c*x^2 )^p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)))*((d*g - e*f*(m + 2))*(c* d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x), x] - Simp[p/(e^2*(m + 1 )*(m + 2)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2 )^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1) - c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c *(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1) - b*(d*g*( m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g }, x] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] && !ILtQ[m + 2*p + 3, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p + 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 1] || EqQ[p, 1] || (IntegerQ[p] && !RationalQ[m])) && NeQ[m, -1] && !ILtQ [m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.64 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.60
| method | result | size |
| risch | \(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (556 A a b c \,x^{3}+15 A \,b^{3} x^{3}+448 a^{2} B c \,x^{3}+264 B a \,b^{2} x^{3}+216 a^{2} A c \,x^{2}+118 A a \,b^{2} x^{2}+208 B \,a^{2} b \,x^{2}+136 A \,a^{2} b x +64 a^{3} B x +48 A \,a^{3}\right )}{192 x^{4} a}+\frac {5 a B \,c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+\frac {5 A b \,c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+\frac {15 b^{2} B \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8}+\frac {B \,c^{2} x \sqrt {c \,x^{2}+b x +a}}{2}+\frac {9 B c b \sqrt {c \,x^{2}+b x +a}}{4}+c^{2} \sqrt {c \,x^{2}+b x +a}\, A -\frac {15 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A \,c^{2}}{8}-\frac {15 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A \,b^{2} c}{16 \sqrt {a}}+\frac {5 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A \,b^{4}}{128 a^{\frac {3}{2}}}-\frac {15 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b B c}{4}-\frac {5 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) B \,b^{3}}{16 \sqrt {a}}\) | \(453\) |
| default | \(\text {Expression too large to display}\) | \(3359\) |
-1/192*(c*x^2+b*x+a)^(1/2)*(556*A*a*b*c*x^3+15*A*b^3*x^3+448*B*a^2*c*x^3+2 64*B*a*b^2*x^3+216*A*a^2*c*x^2+118*A*a*b^2*x^2+208*B*a^2*b*x^2+136*A*a^2*b *x+64*B*a^3*x+48*A*a^3)/x^4/a+5/2*a*B*c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^ 2+b*x+a)^(1/2))+5/2*A*b*c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2) )+15/8*b^2*B*c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*B*c^2 *x*(c*x^2+b*x+a)^(1/2)+9/4*B*c*b*(c*x^2+b*x+a)^(1/2)+c^2*(c*x^2+b*x+a)^(1/ 2)*A-15/8*a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*A*c^2-15/1 6/a^(1/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*A*b^2*c+5/128/a^(3 /2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*A*b^4-15/4*a^(1/2)*ln((2 *a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*b*B*c-5/16/a^(1/2)*ln((2*a+b*x+2* a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*B*b^3
Time = 3.54 (sec) , antiderivative size = 1305, normalized size of antiderivative = 4.60 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=\text {Too large to display} \]
[1/768*(240*(3*B*a^2*b^2 + 4*(B*a^3 + A*a^2*b)*c)*sqrt(c)*x^4*log(-8*c^2*x ^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 15*(8*B*a*b^3 - A*b^4 + 48*A*a^2*c^2 + 24*(4*B*a^2*b + A*a*b^2)*c)*sqrt( a)*x^4*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 4*(96*B*a^2*c^2*x^5 - 48*A*a^4 + 48*(9*B*a^2* b*c + 4*A*a^2*c^2)*x^4 - (264*B*a^2*b^2 + 15*A*a*b^3 + 4*(112*B*a^3 + 139* A*a^2*b)*c)*x^3 - 2*(104*B*a^3*b + 59*A*a^2*b^2 + 108*A*a^3*c)*x^2 - 8*(8* B*a^4 + 17*A*a^3*b)*x)*sqrt(c*x^2 + b*x + a))/(a^2*x^4), -1/768*(480*(3*B* a^2*b^2 + 4*(B*a^3 + A*a^2*b)*c)*sqrt(-c)*x^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 15*(8*B*a*b^3 - A*b^4 + 48*A*a^2*c^2 + 24*(4*B*a^2*b + A*a*b^2)*c)*sqrt(a)*x^4*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x ^2) - 4*(96*B*a^2*c^2*x^5 - 48*A*a^4 + 48*(9*B*a^2*b*c + 4*A*a^2*c^2)*x^4 - (264*B*a^2*b^2 + 15*A*a*b^3 + 4*(112*B*a^3 + 139*A*a^2*b)*c)*x^3 - 2*(10 4*B*a^3*b + 59*A*a^2*b^2 + 108*A*a^3*c)*x^2 - 8*(8*B*a^4 + 17*A*a^3*b)*x)* sqrt(c*x^2 + b*x + a))/(a^2*x^4), 1/384*(15*(8*B*a*b^3 - A*b^4 + 48*A*a^2* c^2 + 24*(4*B*a^2*b + A*a*b^2)*c)*sqrt(-a)*x^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + 120*(3*B*a^2*b^2 + 4 *(B*a^3 + A*a^2*b)*c)*sqrt(c)*x^4*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt( c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 2*(96*B*a^2*c^2*x^5 - 4...
\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{x^{5}}\, dx \]
Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 1163 vs. \(2 (254) = 508\).
Time = 0.39 (sec) , antiderivative size = 1163, normalized size of antiderivative = 4.10 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=\text {Too large to display} \]
1/4*(2*B*c^2*x + (9*B*b*c^2 + 4*A*c^3)/c)*sqrt(c*x^2 + b*x + a) - 5/8*(3*B *b^2*c + 4*B*a*c^2 + 4*A*b*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c) + 5/64*(8*B*a*b^3 - A*b^4 + 96*B*a^2*b*c + 24*A* a*b^2*c + 48*A*a^2*c^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(- a))/(sqrt(-a)*a) + 1/192*(264*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a*b^ 3 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*b^4 + 864*(sqrt(c)*x - sqrt (c*x^2 + b*x + a))^7*B*a^2*b*c + 792*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7 *A*a*b^2*c + 432*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*c^2 + 1152*(s qrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^2*b^2*sqrt(c) + 384*(sqrt(c)*x - s qrt(c*x^2 + b*x + a))^6*A*a*b^3*sqrt(c) + 1152*(sqrt(c)*x - sqrt(c*x^2 + b *x + a))^6*B*a^3*c^(3/2) + 2304*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A*a^ 2*b*c^(3/2) - 584*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*b^3 + 73*(sq rt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a*b^4 - 1248*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^3*b*c - 600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^2* b^2*c - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^3*c^2 - 2304*(sqrt(c) *x - sqrt(c*x^2 + b*x + a))^4*B*a^3*b^2*sqrt(c) - 2688*(sqrt(c)*x - sqrt(c *x^2 + b*x + a))^4*B*a^4*c^(3/2) - 3456*(sqrt(c)*x - sqrt(c*x^2 + b*x + a) )^4*A*a^3*b*c^(3/2) + 440*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^3*b^3 - 55*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*b^4 + 672*(sqrt(c)*x - sq rt(c*x^2 + b*x + a))^3*B*a^4*b*c + 1320*(sqrt(c)*x - sqrt(c*x^2 + b*x +...
Timed out. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^5} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{x^5} \,d x \]