Integrand size = 31, antiderivative size = 245 \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\frac {(8 c f-7 b g) x^2 \sqrt {a+b x+c x^2}}{24 c^2}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}+\frac {\left (192 c^3 d-16 c^2 (9 b e+8 a f)-105 b^3 g+20 b c (6 b f+11 a g)+2 c \left (48 c^2 e-40 b c f+35 b^2 g-36 a c g\right ) x\right ) \sqrt {a+b x+c x^2}}{192 c^4}-\frac {\left (40 b^3 c f+32 b c^2 (2 c d-3 a f)-35 b^4 g-24 b^2 c (2 c e-5 a g)+16 a c^2 (4 c e-3 a g)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{9/2}} \]
-1/128*(40*b^3*c*f+32*b*c^2*(-3*a*f+2*c*d)-35*b^4*g-24*b^2*c*(-5*a*g+2*c*e )+16*a*c^2*(-3*a*g+4*c*e))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/ 2))/c^(9/2)+1/24*(-7*b*g+8*c*f)*x^2*(c*x^2+b*x+a)^(1/2)/c^2+1/4*g*x^3*(c*x ^2+b*x+a)^(1/2)/c+1/192*(192*c^3*d-16*c^2*(8*a*f+9*b*e)-105*b^3*g+20*b*c*( 11*a*g+6*b*f)+2*c*(-36*a*c*g+35*b^2*g-40*b*c*f+48*c^2*e)*x)*(c*x^2+b*x+a)^ (1/2)/c^4
Time = 0.71 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.81 \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (-105 b^3 g+10 b c (12 b f+22 a g+7 b g x)-8 c^2 \left (18 b e+16 a f+10 b f x+9 a g x+7 b g x^2\right )+16 c^3 \left (12 d+x \left (6 e+4 f x+3 g x^2\right )\right )\right )+3 \left (40 b^3 c f+32 b c^2 (2 c d-3 a f)-35 b^4 g-24 b^2 c (2 c e-5 a g)+16 a c^2 (4 c e-3 a g)\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{384 c^{9/2}} \]
(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-105*b^3*g + 10*b*c*(12*b*f + 22*a*g + 7 *b*g*x) - 8*c^2*(18*b*e + 16*a*f + 10*b*f*x + 9*a*g*x + 7*b*g*x^2) + 16*c^ 3*(12*d + x*(6*e + 4*f*x + 3*g*x^2))) + 3*(40*b^3*c*f + 32*b*c^2*(2*c*d - 3*a*f) - 35*b^4*g - 24*b^2*c*(2*c*e - 5*a*g) + 16*a*c^2*(4*c*e - 3*a*g))*L og[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(384*c^(9/2))
Time = 0.57 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {2184, 27, 2184, 27, 1225, 1092, 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.
| \(\displaystyle \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {\int \frac {x \left ((8 c f-7 b g) x^2+2 (4 c e-3 a g) x+8 c d\right )}{2 \sqrt {c x^2+b x+a}}dx}{4 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {x \left ((8 c f-7 b g) x^2+2 (4 c e-3 a g) x+8 c d\right )}{\sqrt {c x^2+b x+a}}dx}{8 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
| \(\Big \downarrow \) 2184 |
| \(\displaystyle \frac {\frac {\int \frac {x \left (4 \left (12 d c^2-8 a f c+7 a b g\right )+\left (35 g b^2-40 c f b+48 c^2 e-36 a c g\right ) x\right )}{2 \sqrt {c x^2+b x+a}}dx}{3 c}+\frac {x^2 \sqrt {a+b x+c x^2} (8 c f-7 b g)}{3 c}}{8 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {x \left (4 \left (12 d c^2-8 a f c+7 a b g\right )+\left (35 g b^2-40 c f b+48 c^2 e-36 a c g\right ) x\right )}{\sqrt {c x^2+b x+a}}dx}{6 c}+\frac {x^2 \sqrt {a+b x+c x^2} (8 c f-7 b g)}{3 c}}{8 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {\frac {\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-36 a c g+35 b^2 g-40 b c f+48 c^2 e\right )-16 c^2 (8 a f+9 b e)+20 b c (11 a g+6 b f)-105 b^3 g+192 c^3 d\right )}{4 c^2}-\frac {3 \left (-24 b^2 c (2 c e-5 a g)+32 b c^2 (2 c d-3 a f)+16 a c^2 (4 c e-3 a g)-35 b^4 g+40 b^3 c f\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c^2}}{6 c}+\frac {x^2 \sqrt {a+b x+c x^2} (8 c f-7 b g)}{3 c}}{8 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {\frac {\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-36 a c g+35 b^2 g-40 b c f+48 c^2 e\right )-16 c^2 (8 a f+9 b e)+20 b c (11 a g+6 b f)-105 b^3 g+192 c^3 d\right )}{4 c^2}-\frac {3 \left (-24 b^2 c (2 c e-5 a g)+32 b c^2 (2 c d-3 a f)+16 a c^2 (4 c e-3 a g)-35 b^4 g+40 b^3 c f\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c^2}}{6 c}+\frac {x^2 \sqrt {a+b x+c x^2} (8 c f-7 b g)}{3 c}}{8 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-36 a c g+35 b^2 g-40 b c f+48 c^2 e\right )-16 c^2 (8 a f+9 b e)+20 b c (11 a g+6 b f)-105 b^3 g+192 c^3 d\right )}{4 c^2}-\frac {3 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-24 b^2 c (2 c e-5 a g)+32 b c^2 (2 c d-3 a f)+16 a c^2 (4 c e-3 a g)-35 b^4 g+40 b^3 c f\right )}{8 c^{5/2}}}{6 c}+\frac {x^2 \sqrt {a+b x+c x^2} (8 c f-7 b g)}{3 c}}{8 c}+\frac {g x^3 \sqrt {a+b x+c x^2}}{4 c}\) |
(g*x^3*Sqrt[a + b*x + c*x^2])/(4*c) + (((8*c*f - 7*b*g)*x^2*Sqrt[a + b*x + c*x^2])/(3*c) + (((192*c^3*d - 16*c^2*(9*b*e + 8*a*f) - 105*b^3*g + 20*b* c*(6*b*f + 11*a*g) + 2*c*(48*c^2*e - 40*b*c*f + 35*b^2*g - 36*a*c*g)*x)*Sq rt[a + b*x + c*x^2])/(4*c^2) - (3*(40*b^3*c*f + 32*b*c^2*(2*c*d - 3*a*f) - 35*b^4*g - 24*b^2*c*(2*c*e - 5*a*g) + 16*a*c^2*(4*c*e - 3*a*g))*ArcTanh[( b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(5/2)))/(6*c))/(8*c)
3.3.81.3.1 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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c *d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && !(IGt Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
Time = 0.76 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.86
| method | result | size |
| risch | \(\frac {\left (48 g \,c^{3} x^{3}-56 b \,c^{2} g \,x^{2}+64 c^{3} f \,x^{2}-72 a \,c^{2} g x +70 b^{2} c g x -80 b \,c^{2} f x +96 c^{3} e x +220 a b c g -128 a \,c^{2} f -105 b^{3} g +120 b^{2} c f -144 b \,c^{2} e +192 c^{3} d \right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{4}}+\frac {\left (48 a^{2} c^{2} g -120 a \,b^{2} c g +96 a b \,c^{2} f -64 a \,c^{3} e +35 b^{4} g -40 b^{3} f c +48 b^{2} c^{2} e -64 b d \,c^{3}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {9}{2}}}\) | \(210\) |
| default | \(g \left (\frac {x^{3} \sqrt {c \,x^{2}+b x +a}}{4 c}-\frac {7 b \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )}{8 c}-\frac {3 a \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+f \left (\frac {x^{2} \sqrt {c \,x^{2}+b x +a}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{6 c}-\frac {2 a \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{3 c}\right )+e \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+d \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\) | \(669\) |
1/192*(48*c^3*g*x^3-56*b*c^2*g*x^2+64*c^3*f*x^2-72*a*c^2*g*x+70*b^2*c*g*x- 80*b*c^2*f*x+96*c^3*e*x+220*a*b*c*g-128*a*c^2*f-105*b^3*g+120*b^2*c*f-144* b*c^2*e+192*c^3*d)*(c*x^2+b*x+a)^(1/2)/c^4+1/128*(48*a^2*c^2*g-120*a*b^2*c *g+96*a*b*c^2*f-64*a*c^3*e+35*b^4*g-40*b^3*c*f+48*b^2*c^2*e-64*b*c^3*d)/c^ (9/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))
Time = 0.30 (sec) , antiderivative size = 499, normalized size of antiderivative = 2.04 \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\left [-\frac {3 \, {\left (64 \, b c^{3} d - 16 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} e + 8 \, {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} f - {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} g\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (48 \, c^{4} g x^{3} + 192 \, c^{4} d - 144 \, b c^{3} e + 8 \, {\left (8 \, c^{4} f - 7 \, b c^{3} g\right )} x^{2} + 8 \, {\left (15 \, b^{2} c^{2} - 16 \, a c^{3}\right )} f - 5 \, {\left (21 \, b^{3} c - 44 \, a b c^{2}\right )} g + 2 \, {\left (48 \, c^{4} e - 40 \, b c^{3} f + {\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} g\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{5}}, \frac {3 \, {\left (64 \, b c^{3} d - 16 \, {\left (3 \, b^{2} c^{2} - 4 \, a c^{3}\right )} e + 8 \, {\left (5 \, b^{3} c - 12 \, a b c^{2}\right )} f - {\left (35 \, b^{4} - 120 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} g\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} g x^{3} + 192 \, c^{4} d - 144 \, b c^{3} e + 8 \, {\left (8 \, c^{4} f - 7 \, b c^{3} g\right )} x^{2} + 8 \, {\left (15 \, b^{2} c^{2} - 16 \, a c^{3}\right )} f - 5 \, {\left (21 \, b^{3} c - 44 \, a b c^{2}\right )} g + 2 \, {\left (48 \, c^{4} e - 40 \, b c^{3} f + {\left (35 \, b^{2} c^{2} - 36 \, a c^{3}\right )} g\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{5}}\right ] \]
[-1/768*(3*(64*b*c^3*d - 16*(3*b^2*c^2 - 4*a*c^3)*e + 8*(5*b^3*c - 12*a*b* c^2)*f - (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*g)*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*(4 8*c^4*g*x^3 + 192*c^4*d - 144*b*c^3*e + 8*(8*c^4*f - 7*b*c^3*g)*x^2 + 8*(1 5*b^2*c^2 - 16*a*c^3)*f - 5*(21*b^3*c - 44*a*b*c^2)*g + 2*(48*c^4*e - 40*b *c^3*f + (35*b^2*c^2 - 36*a*c^3)*g)*x)*sqrt(c*x^2 + b*x + a))/c^5, 1/384*( 3*(64*b*c^3*d - 16*(3*b^2*c^2 - 4*a*c^3)*e + 8*(5*b^3*c - 12*a*b*c^2)*f - (35*b^4 - 120*a*b^2*c + 48*a^2*c^2)*g)*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*(48*c^4*g*x^3 + 1 92*c^4*d - 144*b*c^3*e + 8*(8*c^4*f - 7*b*c^3*g)*x^2 + 8*(15*b^2*c^2 - 16* a*c^3)*f - 5*(21*b^3*c - 44*a*b*c^2)*g + 2*(48*c^4*e - 40*b*c^3*f + (35*b^ 2*c^2 - 36*a*c^3)*g)*x)*sqrt(c*x^2 + b*x + a))/c^5]
Time = 0.84 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.96 \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\begin {cases} \left (- \frac {a \left (- \frac {3 a g}{4 c} - \frac {5 b \left (- \frac {7 b g}{8 c} + f\right )}{6 c} + e\right )}{2 c} - \frac {b \left (- \frac {2 a \left (- \frac {7 b g}{8 c} + f\right )}{3 c} - \frac {3 b \left (- \frac {3 a g}{4 c} - \frac {5 b \left (- \frac {7 b g}{8 c} + f\right )}{6 c} + e\right )}{4 c} + d\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {a + b x + c x^{2}} \left (\frac {g x^{3}}{4 c} + \frac {x^{2} \left (- \frac {7 b g}{8 c} + f\right )}{3 c} + \frac {x \left (- \frac {3 a g}{4 c} - \frac {5 b \left (- \frac {7 b g}{8 c} + f\right )}{6 c} + e\right )}{2 c} + \frac {- \frac {2 a \left (- \frac {7 b g}{8 c} + f\right )}{3 c} - \frac {3 b \left (- \frac {3 a g}{4 c} - \frac {5 b \left (- \frac {7 b g}{8 c} + f\right )}{6 c} + e\right )}{4 c} + d}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {g \left (a + b x\right )^{\frac {9}{2}}}{9 b^{3}} + \frac {\left (a + b x\right )^{\frac {7}{2}} \left (- 4 a g + b f\right )}{7 b^{3}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \cdot \left (6 a^{2} g - 3 a b f + b^{2} e\right )}{5 b^{3}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (- 4 a^{3} g + 3 a^{2} b f - 2 a b^{2} e + b^{3} d\right )}{3 b^{3}} + \frac {\sqrt {a + b x} \left (a^{4} g - a^{3} b f + a^{2} b^{2} e - a b^{3} d\right )}{b^{3}}\right )}{b^{2}} & \text {for}\: b \neq 0 \\\frac {\frac {d x^{2}}{2} + \frac {e x^{3}}{3} + \frac {f x^{4}}{4} + \frac {g x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
Piecewise(((-a*(-3*a*g/(4*c) - 5*b*(-7*b*g/(8*c) + f)/(6*c) + e)/(2*c) - b *(-2*a*(-7*b*g/(8*c) + f)/(3*c) - 3*b*(-3*a*g/(4*c) - 5*b*(-7*b*g/(8*c) + f)/(6*c) + e)/(4*c) + d)/(2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/( 2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(a + b*x + c*x**2)*(g*x** 3/(4*c) + x**2*(-7*b*g/(8*c) + f)/(3*c) + x*(-3*a*g/(4*c) - 5*b*(-7*b*g/(8 *c) + f)/(6*c) + e)/(2*c) + (-2*a*(-7*b*g/(8*c) + f)/(3*c) - 3*b*(-3*a*g/( 4*c) - 5*b*(-7*b*g/(8*c) + f)/(6*c) + e)/(4*c) + d)/c), Ne(c, 0)), (2*(g*( a + b*x)**(9/2)/(9*b**3) + (a + b*x)**(7/2)*(-4*a*g + b*f)/(7*b**3) + (a + b*x)**(5/2)*(6*a**2*g - 3*a*b*f + b**2*e)/(5*b**3) + (a + b*x)**(3/2)*(-4 *a**3*g + 3*a**2*b*f - 2*a*b**2*e + b**3*d)/(3*b**3) + sqrt(a + b*x)*(a**4 *g - a**3*b*f + a**2*b**2*e - a*b**3*d)/b**3)/b**2, Ne(b, 0)), ((d*x**2/2 + e*x**3/3 + f*x**4/4 + g*x**5/5)/sqrt(a), True))
Exception generated. \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.33 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.90 \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (\frac {6 \, g x}{c} + \frac {8 \, c^{3} f - 7 \, b c^{2} g}{c^{4}}\right )} x + \frac {48 \, c^{3} e - 40 \, b c^{2} f + 35 \, b^{2} c g - 36 \, a c^{2} g}{c^{4}}\right )} x + \frac {192 \, c^{3} d - 144 \, b c^{2} e + 120 \, b^{2} c f - 128 \, a c^{2} f - 105 \, b^{3} g + 220 \, a b c g}{c^{4}}\right )} + \frac {{\left (64 \, b c^{3} d - 48 \, b^{2} c^{2} e + 64 \, a c^{3} e + 40 \, b^{3} c f - 96 \, a b c^{2} f - 35 \, b^{4} g + 120 \, a b^{2} c g - 48 \, a^{2} c^{2} g\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {9}{2}}} \]
1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*g*x/c + (8*c^3*f - 7*b*c^2*g)/c^4)*x + (48*c^3*e - 40*b*c^2*f + 35*b^2*c*g - 36*a*c^2*g)/c^4)*x + (192*c^3*d - 144*b*c^2*e + 120*b^2*c*f - 128*a*c^2*f - 105*b^3*g + 220*a*b*c*g)/c^4) + 1/128*(64*b*c^3*d - 48*b^2*c^2*e + 64*a*c^3*e + 40*b^3*c*f - 96*a*b*c^2*f - 35*b^4*g + 120*a*b^2*c*g - 48*a^2*c^2*g)*log(abs(2*(sqrt(c)*x - sqrt(c*x ^2 + b*x + a))*sqrt(c) + b))/c^(9/2)
Timed out. \[ \int \frac {x \left (d+e x+f x^2+g x^3\right )}{\sqrt {a+b x+c x^2}} \, dx=\int \frac {x\,\left (g\,x^3+f\,x^2+e\,x+d\right )}{\sqrt {c\,x^2+b\,x+a}} \,d x \]