Integrand size = 23, antiderivative size = 350 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\frac {\left (512 a^2 A c^3+b \left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )\right )+2 c \left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 b^2 B-12 A b c-20 a B c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{512 c^3}-\frac {\left (5 b^3 B-12 A b^2 c-20 a b B c-64 a A c^2+2 c \left (5 b^2 B-12 A b c-20 a B c\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{192 c^2}+\frac {(5 b B+12 A c+10 B c x) \left (a+b x+c x^2\right )^{5/2}}{60 c}-a^{5/2} A \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )+\frac {\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (5 b^4 B-12 A b^3 c-40 a b^2 B c+112 a A b c^2+80 a^2 B c^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{1024 c^{7/2}} \]
-1/192*(5*B*b^3-12*A*b^2*c-20*B*a*b*c-64*A*a*c^2+2*c*(-12*A*b*c-20*B*a*c+5 *B*b^2)*x)*(c*x^2+b*x+a)^(3/2)/c^2+1/60*(10*B*c*x+12*A*c+5*B*b)*(c*x^2+b*x +a)^(5/2)/c-a^(5/2)*A*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))+1 /1024*(512*a^2*A*b*c^3-(-4*a*c+b^2)*(112*A*a*b*c^2-12*A*b^3*c+80*B*a^2*c^2 -40*B*a*b^2*c+5*B*b^4))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2)) /c^(7/2)+1/512*(512*A*a^2*c^3+b*(64*a*A*b*c^2+(-4*a*c+b^2)*(-12*A*b*c-20*B *a*c+5*B*b^2))+2*c*(64*a*A*b*c^2+(-4*a*c+b^2)*(-12*A*b*c-20*B*a*c+5*B*b^2) )*x)*(c*x^2+b*x+a)^(1/2)/c^3
Time = 1.65 (sec) , antiderivative size = 324, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\frac {\sqrt {a+x (b+c x)} \left (75 b^5 B-10 b^4 c (18 A+5 B x)+40 b^3 c (-20 a B+c x (3 A+B x))+48 b^2 c^2 \left (5 a (9 A+2 B x)+c x^2 (62 A+45 B x)\right )+32 c^3 \left (8 c^2 x^4 (6 A+5 B x)+2 a c x^2 (88 A+65 B x)+a^2 (368 A+165 B x)\right )+16 b c^2 \left (165 a^2 B+4 c^2 x^3 (63 A+50 B x)+2 a c x (311 A+195 B x)\right )\right )}{7680 c^3}+2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-\frac {\left (-5 b^6 B+12 A b^5 c+60 a b^4 B c-160 a A b^3 c^2-240 a^2 b^2 B c^2+960 a^2 A b c^3+320 a^3 B c^3\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{1024 c^{7/2}} \]
(Sqrt[a + x*(b + c*x)]*(75*b^5*B - 10*b^4*c*(18*A + 5*B*x) + 40*b^3*c*(-20 *a*B + c*x*(3*A + B*x)) + 48*b^2*c^2*(5*a*(9*A + 2*B*x) + c*x^2*(62*A + 45 *B*x)) + 32*c^3*(8*c^2*x^4*(6*A + 5*B*x) + 2*a*c*x^2*(88*A + 65*B*x) + a^2 *(368*A + 165*B*x)) + 16*b*c^2*(165*a^2*B + 4*c^2*x^3*(63*A + 50*B*x) + 2* a*c*x*(311*A + 195*B*x))))/(7680*c^3) + 2*a^(5/2)*A*ArcTanh[(Sqrt[c]*x - S qrt[a + x*(b + c*x)])/Sqrt[a]] - ((-5*b^6*B + 12*A*b^5*c + 60*a*b^4*B*c - 160*a*A*b^3*c^2 - 240*a^2*b^2*B*c^2 + 960*a^2*A*b*c^3 + 320*a^3*B*c^3)*Log [b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(1024*c^(7/2))
Time = 0.75 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {1231, 27, 1231, 27, 1231, 27, 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.
| \(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}-\frac {\int -\frac {\left (24 a A c+\left (12 A b c-5 B \left (b^2-4 a c\right )\right ) x\right ) \left (c x^2+b x+a\right )^{3/2}}{2 x}dx}{12 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\left (24 a A c-\left (5 B b^2-12 A c b-20 a B c\right ) x\right ) \left (c x^2+b x+a\right )^{3/2}}{x}dx}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {-\frac {\int -\frac {3 \left (128 a^2 A c^2+\left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 B b^2-12 A c b-20 a B c\right )\right ) x\right ) \sqrt {c x^2+b x+a}}{2 x}dx}{8 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \frac {\left (128 a^2 A c^2+\left (64 a A b c^2+\left (b^2-4 a c\right ) \left (5 B b^2-12 A c b-20 a B c\right )\right ) x\right ) \sqrt {c x^2+b x+a}}{x}dx}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}-\frac {\int -\frac {1024 a^3 A c^3+\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (5 B b^4-12 A c b^3-40 a B c b^2+112 a A c^2 b+80 a^2 B c^2\right )\right ) x}{2 x \sqrt {c x^2+b x+a}}dx}{4 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1024 a^3 A c^3+\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (5 B b^4-12 A c b^3-40 a B c b^2+112 a A c^2 b+80 a^2 B c^2\right )\right ) x}{x \sqrt {c x^2+b x+a}}dx}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle \frac {\frac {3 \left (\frac {1024 a^3 A c^3 \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\frac {3 \left (\frac {1024 a^3 A c^3 \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+2 \left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3 \left (\frac {1024 a^3 A c^3 \int \frac {1}{x \sqrt {c x^2+b x+a}}dx+\frac {\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-2048 a^3 A c^3 \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}}{8 c}+\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\sqrt {a+b x+c x^2} \left (512 a^2 A c^3+2 c x \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )+b \left (\left (b^2-4 a c\right ) \left (-20 a B c-12 A b c+5 b^2 B\right )+64 a A b c^2\right )\right )}{4 c}+\frac {\frac {\left (512 a^2 A b c^3-\left (b^2-4 a c\right ) \left (80 a^2 B c^2+112 a A b c^2-40 a b^2 B c-12 A b^3 c+5 b^4 B\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}-1024 a^{5/2} A c^3 \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 c}\right )}{16 c}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (2 c x \left (-20 a B c-12 A b c+5 b^2 B\right )-64 a A c^2-20 a b B c-12 A b^2 c+5 b^3 B\right )}{8 c}}{24 c}+\frac {\left (a+b x+c x^2\right )^{5/2} (12 A c+5 b B+10 B c x)}{60 c}\) |
((5*b*B + 12*A*c + 10*B*c*x)*(a + b*x + c*x^2)^(5/2))/(60*c) + (-1/8*((5*b ^3*B - 12*A*b^2*c - 20*a*b*B*c - 64*a*A*c^2 + 2*c*(5*b^2*B - 12*A*b*c - 20 *a*B*c)*x)*(a + b*x + c*x^2)^(3/2))/c + (3*(((512*a^2*A*c^3 + b*(64*a*A*b* c^2 + (b^2 - 4*a*c)*(5*b^2*B - 12*A*b*c - 20*a*B*c)) + 2*c*(64*a*A*b*c^2 + (b^2 - 4*a*c)*(5*b^2*B - 12*A*b*c - 20*a*B*c))*x)*Sqrt[a + b*x + c*x^2])/ (4*c) + (-1024*a^(5/2)*A*c^3*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])] + ((512*a^2*A*b*c^3 - (b^2 - 4*a*c)*(5*b^4*B - 12*A*b^3*c - 40*a *b^2*B*c + 112*a*A*b*c^2 + 80*a^2*B*c^2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S qrt[a + b*x + c*x^2])])/Sqrt[c])/(8*c)))/(16*c))/(24*c)
3.10.40.3.1 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)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ (c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c ^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x ] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] || !R ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) && !ILtQ[m + 2*p, 0] && (Integer Q[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 = 0.38 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.24
| method | result | size |
| default | \(B \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{12 c}+\frac {5 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{24 c}\right )+A \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}}}{5}+\frac {b \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2}+a \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2}+a \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )\right )\right )\) | \(433\) |
B*(1/12*(2*c*x+b)/c*(c*x^2+b*x+a)^(5/2)+5/24*(4*a*c-b^2)/c*(1/8*(2*c*x+b)/ c*(c*x^2+b*x+a)^(3/2)+3/16*(4*a*c-b^2)/c*(1/4*(2*c*x+b)/c*(c*x^2+b*x+a)^(1 /2)+1/8*(4*a*c-b^2)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))) +A*(1/5*(c*x^2+b*x+a)^(5/2)+1/2*b*(1/8*(2*c*x+b)/c*(c*x^2+b*x+a)^(3/2)+3/1 6*(4*a*c-b^2)/c*(1/4*(2*c*x+b)/c*(c*x^2+b*x+a)^(1/2)+1/8*(4*a*c-b^2)/c^(3/ 2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))))+a*(1/3*(c*x^2+b*x+a)^(3/2 )+1/2*b*(1/4*(2*c*x+b)/c*(c*x^2+b*x+a)^(1/2)+1/8*(4*a*c-b^2)/c^(3/2)*ln((1 /2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))+a*((c*x^2+b*x+a)^(1/2)+1/2*b*ln((1 /2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-a^(1/2)*ln((2*a+b*x+2*a^(1/ 2)*(c*x^2+b*x+a)^(1/2))/x))))
Time = 9.60 (sec) , antiderivative size = 1575, normalized size of antiderivative = 4.50 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\text {Too large to display} \]
[1/30720*(15360*A*a^(5/2)*c^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) - 15*(5*B*b^6 - 320*(B*a ^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b^2 + 2*A*a*b^3)*c^2 - 12*(5*B*a*b^4 + A *b^5)*c)*sqrt(c)*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) + 4*(1280*B*c^6*x^5 + 75*B*b^5*c + 11776*A*a^ 2*c^4 + 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 240*(11*B*a^2*b + 9*A*a*b^2)*c^3 + 16*(135*B*b^2*c^4 + 4*(65*B*a + 63*A*b)*c^5)*x^3 - 20*(40*B*a*b^3 + 9*A *b^4)*c^2 + 8*(5*B*b^3*c^3 + 704*A*a*c^5 + 12*(65*B*a*b + 31*A*b^2)*c^4)*x ^2 - 2*(25*B*b^4*c^2 - 16*(165*B*a^2 + 311*A*a*b)*c^4 - 60*(4*B*a*b^2 + A* b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/15360*(7680*A*a^(5/2)*c^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) + 15*(5*B*b^6 - 320*(B*a^3 + 3*A*a^2*b)*c^3 + 80*(3*B*a^2*b ^2 + 2*A*a*b^3)*c^2 - 12*(5*B*a*b^4 + A*b^5)*c)*sqrt(-c)*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)) + 2*(1280*B* c^6*x^5 + 75*B*b^5*c + 11776*A*a^2*c^4 + 128*(25*B*b*c^5 + 12*A*c^6)*x^4 + 240*(11*B*a^2*b + 9*A*a*b^2)*c^3 + 16*(135*B*b^2*c^4 + 4*(65*B*a + 63*A*b )*c^5)*x^3 - 20*(40*B*a*b^3 + 9*A*b^4)*c^2 + 8*(5*B*b^3*c^3 + 704*A*a*c^5 + 12*(65*B*a*b + 31*A*b^2)*c^4)*x^2 - 2*(25*B*b^4*c^2 - 16*(165*B*a^2 + 31 1*A*a*b)*c^4 - 60*(4*B*a*b^2 + A*b^3)*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/30720*(30720*A*sqrt(-a)*a^2*c^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x...
\[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\int \frac {\left (A + B x\right ) \left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{x}\, dx \]
Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\text {Exception raised: ValueError} \]
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
Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{5/2}}{x} \,d x \]