Integrand size = 28, antiderivative size = 360 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=-\frac {\left (64 c^3 d^3-b^3 e^3+4 b c e^2 (12 b d-11 a e)-16 c^2 d e (7 b d-4 a e)-2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{32 c e^4}-\frac {(8 c d-7 b e-6 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^2}+\frac {\left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2} e^5}-\frac {(2 c d-b e) \left (c d^2-b d e+a e^2\right )^{3/2} \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^5} \]
-1/12*(-6*c*e*x-7*b*e+8*c*d)*(c*x^2+b*x+a)^(3/2)/e^2+1/64*(128*c^4*d^4-b^4 *e^4-8*b^2*c*e^3*(-3*a*e+2*b*d)-64*c^3*d^2*e*(-3*a*e+4*b*d)+48*c^2*e^2*(a^ 2*e^2-4*a*b*d*e+3*b^2*d^2))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1 /2))/c^(3/2)/e^5-(-b*e+2*c*d)*(a*e^2-b*d*e+c*d^2)^(3/2)*arctanh(1/2*(b*d-2 *a*e+(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^5-1/ 32*(64*c^3*d^3-b^3*e^3+4*b*c*e^2*(-11*a*e+12*b*d)-16*c^2*d*e*(-4*a*e+7*b*d )-2*c*e*(16*c^2*d^2+b^2*e^2-4*c*e*(-3*a*e+4*b*d))*x)*(c*x^2+b*x+a)^(1/2)/c /e^4
Time = 2.86 (sec) , antiderivative size = 353, normalized size of antiderivative = 0.98 \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\frac {\frac {e \sqrt {a+x (b+c x)} \left (3 b^3 e^3+2 b c e^2 (-72 b d+94 a e+31 b e x)-16 c^3 \left (12 d^3-6 d^2 e x+4 d e^2 x^2-3 e^3 x^3\right )+8 c^2 e \left (a e (-32 d+15 e x)+b \left (42 d^2-20 d e x+13 e^2 x^2\right )\right )\right )}{c}-192 (2 c d-b e) \sqrt {-c d^2+e (b d-a e)} \left (c d^2+e (-b d+a e)\right ) \arctan \left (\frac {\sqrt {-c d^2+e (b d-a e)} x}{\sqrt {a} (d+e x)-d \sqrt {a+x (b+c x)}}\right )+\frac {3 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{c^{3/2}}}{96 e^5} \]
((e*Sqrt[a + x*(b + c*x)]*(3*b^3*e^3 + 2*b*c*e^2*(-72*b*d + 94*a*e + 31*b* e*x) - 16*c^3*(12*d^3 - 6*d^2*e*x + 4*d*e^2*x^2 - 3*e^3*x^3) + 8*c^2*e*(a* e*(-32*d + 15*e*x) + b*(42*d^2 - 20*d*e*x + 13*e^2*x^2))))/c - 192*(2*c*d - b*e)*Sqrt[-(c*d^2) + e*(b*d - a*e)]*(c*d^2 + e*(-(b*d) + a*e))*ArcTan[(S qrt[-(c*d^2) + e*(b*d - a*e)]*x)/(Sqrt[a]*(d + e*x) - d*Sqrt[a + x*(b + c* x)])] + (3*(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d ^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTa nh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/c^(3/2))/(96*e^5)
Time = 0.92 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {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.
| \(\displaystyle \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle -\frac {\int \frac {c \left (7 d e b^2-8 \left (c d^2+a e^2\right ) b+4 a c d e-\left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {c x^2+b x+a}}{d+e x}dx}{8 c e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\left (7 d e b^2-8 \left (c d^2+a e^2\right ) b+4 a c d e-\left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {c x^2+b x+a}}{d+e x}dx}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 1231 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\int \frac {d \left (-e b^2+4 c d b-4 a c e\right ) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+4 c e (b d-2 a e) \left (7 d e b^2-8 \left (c d^2+a e^2\right ) b+4 a c d e\right )+\left (128 c^4 d^4-64 c^3 e (4 b d-3 a e) d^2-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b e d+a^2 e^2\right )\right ) x}{2 (d+e x) \sqrt {c x^2+b x+a}}dx}{4 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\int \frac {d \left (-e b^2+4 c d b-4 a c e\right ) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )+4 c e (b d-2 a e) \left (7 d e b^2-8 \left (c d^2+a e^2\right ) b+4 a c d e\right )+\left (128 c^4 d^4-64 c^3 e (4 b d-3 a e) d^2-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b e d+a^2 e^2\right )\right ) x}{(d+e x) \sqrt {c x^2+b x+a}}dx}{8 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 1269 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\frac {\left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{e}-\frac {64 c (2 c d-b e) \left (a e^2-b d e+c d^2\right )^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{e}}{8 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\frac {2 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{e}-\frac {64 c (2 c d-b e) \left (a e^2-b d e+c d^2\right )^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{e}}{8 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right )}{\sqrt {c} e}-\frac {64 c (2 c d-b e) \left (a e^2-b d e+c d^2\right )^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{e}}{8 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\frac {128 c (2 c d-b e) \left (a e^2-b d e+c d^2\right )^2 \int \frac {1}{4 \left (c d^2-b e d+a e^2\right )-\frac {(b d-2 a e+(2 c d-b e) x)^2}{c x^2+b x+a}}d\left (-\frac {b d-2 a e+(2 c d-b e) x}{\sqrt {c x^2+b x+a}}\right )}{e}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right )}{\sqrt {c} e}}{8 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (7 b d-4 a e)+4 b c e^2 (12 b d-11 a e)-b^3 e^3+64 c^3 d^3\right )}{4 c e^2}-\frac {\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right )}{\sqrt {c} e}-\frac {64 c (2 c d-b e) \left (a e^2-b d e+c d^2\right )^{3/2} \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e}}{8 c e^2}}{8 e^2}-\frac {\left (a+b x+c x^2\right )^{3/2} (-7 b e+8 c d-6 c e x)}{12 e^2}\) |
-1/12*((8*c*d - 7*b*e - 6*c*e*x)*(a + b*x + c*x^2)^(3/2))/e^2 - (((64*c^3* d^3 - b^3*e^3 + 4*b*c*e^2*(12*b*d - 11*a*e) - 16*c^2*d*e*(7*b*d - 4*a*e) - 2*c*e*(16*c^2*d^2 + b^2*e^2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c* x^2])/(4*c*e^2) - (((128*c^4*d^4 - b^4*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^ 2))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(Sqrt[c]*e) - (64*c*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2)^(3/2)*ArcTanh[(b*d - 2*a*e + ( 2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/e) /(8*c*e^2))/(8*e^2)
3.16.61.3.1 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 = 0.73 (sec) , antiderivative size = 538, normalized size of antiderivative = 1.49
| method | result | size |
| risch | \(\frac {\left (48 c^{3} e^{3} x^{3}+104 c^{2} e^{3} b \,x^{2}-64 c^{3} d \,e^{2} x^{2}+120 a \,c^{2} e^{3} x +62 b^{2} c \,e^{3} x -160 b \,c^{2} d \,e^{2} x +96 c^{3} d^{2} e x +188 c \,e^{3} b a -256 a \,c^{2} d \,e^{2}+3 b^{3} e^{3}-144 b^{2} c d \,e^{2}+336 b \,c^{2} d^{2} e -192 c^{3} d^{3}\right ) \sqrt {c \,x^{2}+b x +a}}{96 c \,e^{4}}+\frac {\frac {\left (48 a^{2} c^{2} e^{4}+24 a \,b^{2} c \,e^{4}-192 a b \,c^{2} d \,e^{3}+192 a \,c^{3} d^{2} e^{2}-b^{4} e^{4}-16 b^{3} c d \,e^{3}+144 b^{2} c^{2} d^{2} e^{2}-256 b \,c^{3} d^{3} e +128 c^{4} d^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e \sqrt {c}}-\frac {64 \left (a^{2} b \,e^{5}-2 a^{2} c d \,e^{4}-2 a \,b^{2} d \,e^{4}+6 a b c \,d^{2} e^{3}-4 a \,c^{2} d^{3} e^{2}+b^{3} d^{2} e^{3}-4 b^{2} c \,d^{3} e^{2}+5 b \,c^{2} d^{4} e -2 c^{3} d^{5}\right ) c \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}}{64 e^{4} c}\) | \(538\) |
| default | \(\frac {2 c \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 )}{e}+\frac {\left (b e -2 c d \right ) \left (\frac {\left (\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3}+\frac {\left (b e -2 c d \right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right ) \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a -b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{2 e}+\frac {\left (e^{2} a -b d e +c \,d^{2}\right ) \left (\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (e^{2} a -b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}\right )}{e^{2}}\) | \(743\) |
1/96/c*(48*c^3*e^3*x^3+104*b*c^2*e^3*x^2-64*c^3*d*e^2*x^2+120*a*c^2*e^3*x+ 62*b^2*c*e^3*x-160*b*c^2*d*e^2*x+96*c^3*d^2*e*x+188*a*b*c*e^3-256*a*c^2*d* e^2+3*b^3*e^3-144*b^2*c*d*e^2+336*b*c^2*d^2*e-192*c^3*d^3)*(c*x^2+b*x+a)^( 1/2)/e^4+1/64/e^4/c*((48*a^2*c^2*e^4+24*a*b^2*c*e^4-192*a*b*c^2*d*e^3+192* a*c^3*d^2*e^2-b^4*e^4-16*b^3*c*d*e^3+144*b^2*c^2*d^2*e^2-256*b*c^3*d^3*e+1 28*c^4*d^4)/e*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-64*(a^2* b*e^5-2*a^2*c*d*e^4-2*a*b^2*d*e^4+6*a*b*c*d^2*e^3-4*a*c^2*d^3*e^2+b^3*d^2* e^3-4*b^2*c*d^3*e^2+5*b*c^2*d^4*e-2*c^3*d^5)*c/e^2/((a*e^2-b*d*e+c*d^2)/e^ 2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*((a*e^2-b*d *e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2 )/e^2)^(1/2))/(x+d/e)))
Timed out. \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Timed out} \]
\[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\int \frac {\left (b + 2 c x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{d + e x}\, dx \]
Exception generated. \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Exception generated. \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx=\int \frac {\left (b+2\,c\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{d+e\,x} \,d x \]