Integrand size = 27, antiderivative size = 679 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=-\frac {(4 c e-5 b f-2 c f x) \sqrt {a+b x+c x^2}}{4 f^2}+\frac {\left (3 b^2 f^2-12 c f (b e-a f)+8 c^2 \left (e^2-d f\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {c} f^3}+\frac {\left ((c e-b f) \left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f^3 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}-\frac {\left ((c e-b f) \left (e+\sqrt {e^2-4 d f}\right ) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )-2 f \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} f^3 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \]
1/8*(3*b^2*f^2-12*c*f*(-a*f+b*e)+8*c^2*(-d*f+e^2))*arctanh(1/2*(2*c*x+b)/c ^(1/2)/(c*x^2+b*x+a)^(1/2))/f^3/c^(1/2)-1/4*(-2*c*f*x-5*b*f+4*c*e)*(c*x^2+ b*x+a)^(1/2)/f^2+1/2*arctanh(1/4*(4*a*f+2*x*(b*f-c*(e-(-4*d*f+e^2)^(1/2))) -b*(e-(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+b*x+a)^(1/2)/(c*e^2-2*c*d*f-b*e* f+2*a*f^2-(-b*f+c*e)*(-4*d*f+e^2)^(1/2))^(1/2))*(-2*f*(2*c*d*f*(-a*f+b*e)- f^2*(-a^2*f+b^2*d)-c^2*d*(-d*f+e^2))+(-b*f+c*e)*(f*(-2*a*f+b*e)-c*(-2*d*f+ e^2))*(e-(-4*d*f+e^2)^(1/2)))/f^3*2^(1/2)/(-4*d*f+e^2)^(1/2)/(c*e^2-2*c*d* f-b*e*f+2*a*f^2-(-b*f+c*e)*(-4*d*f+e^2)^(1/2))^(1/2)-1/2*arctanh(1/4*(4*a* f-b*(e+(-4*d*f+e^2)^(1/2))+2*x*(b*f-c*(e+(-4*d*f+e^2)^(1/2))))*2^(1/2)/(c* x^2+b*x+a)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+e^2)^(1/2 ))^(1/2))*(-2*f*(2*c*d*f*(-a*f+b*e)-f^2*(-a^2*f+b^2*d)-c^2*d*(-d*f+e^2))+( -b*f+c*e)*(f*(-2*a*f+b*e)-c*(-2*d*f+e^2))*(e+(-4*d*f+e^2)^(1/2)))/f^3*2^(1 /2)/(-4*d*f+e^2)^(1/2)/(c*e^2-2*c*d*f-b*e*f+2*a*f^2+(-b*f+c*e)*(-4*d*f+e^2 )^(1/2))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.92 (sec) , antiderivative size = 1472, normalized size of antiderivative = 2.17 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx =\text {Too large to display} \]
(f*(-4*c*e + 5*b*f + 2*c*f*x)*Sqrt[a + x*(b + c*x)] + ((3*b^2*f^2 + 12*c*f *(-(b*e) + a*f) + 8*c^2*(e^2 - d*f))*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[ a + x*(b + c*x)])])/Sqrt[c] - 4*RootSum[c^2*d - b*c*e + b^2*f + 2*Sqrt[a]* c*e*#1 - 4*Sqrt[a]*b*f*#1 - 2*c*d*#1^2 + b*e*#1^2 + 4*a*f*#1^2 - 2*Sqrt[a] *e*#1^3 + d*#1^4 & , (c^3*d*e^2*Log[x] - b*c^2*e^3*Log[x] - c^3*d^2*f*Log[ x] + 2*b^2*c*e^2*f*Log[x] - b^2*c*d*f^2*Log[x] + 2*a*c^2*d*f^2*Log[x] - b^ 3*e*f^2*Log[x] - 2*a*b*c*e*f^2*Log[x] + 2*a*b^2*f^3*Log[x] - a^2*c*f^3*Log [x] - c^3*d*e^2*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + b*c^2*e^3*L og[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + c^3*d^2*f*Log[-Sqrt[a] + Sqr t[a + b*x + c*x^2] - x*#1] - 2*b^2*c*e^2*f*Log[-Sqrt[a] + Sqrt[a + b*x + c *x^2] - x*#1] + b^2*c*d*f^2*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] - 2*a*c^2*d*f^2*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + b^3*e*f^2*Lo g[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + 2*a*b*c*e*f^2*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] - 2*a*b^2*f^3*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + a^2*c*f^3*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1] + 2*Sqrt[a]*c^2*e^3*Log[x]*#1 - 4*Sqrt[a]*c^2*d*e*f*Log[x]*#1 - 4*Sqrt[a]*b* c*e^2*f*Log[x]*#1 + 4*Sqrt[a]*b*c*d*f^2*Log[x]*#1 + 2*Sqrt[a]*b^2*e*f^2*Lo g[x]*#1 + 4*a^(3/2)*c*e*f^2*Log[x]*#1 - 4*a^(3/2)*b*f^3*Log[x]*#1 - 2*Sqrt [a]*c^2*e^3*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1 + 4*Sqrt[a]*c^ 2*d*e*f*Log[-Sqrt[a] + Sqrt[a + b*x + c*x^2] - x*#1]*#1 + 4*Sqrt[a]*b*c...
Time = 2.07 (sec) , antiderivative size = 684, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1308, 27, 2143, 27, 1092, 219, 1365, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx\) |
\(\Big \downarrow \) 1308 |
\(\displaystyle -\frac {\int -\frac {-5 d f b^2+4 c d e b+\left (8 \left (e^2-d f\right ) c^2-12 f (b e-a f) c+3 b^2 f^2\right ) x^2-4 a f (c d-2 a f)+\left (8 d e c^2+4 \left (b e^2-a f e-4 b d f\right ) c-b f (5 b e-16 a f)\right ) x}{4 \sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{2 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-5 d f b^2+4 c d e b+\left (8 \left (e^2-d f\right ) c^2-12 f (b e-a f) c+3 b^2 f^2\right ) x^2-4 a f (c d-2 a f)+\left (8 d e c^2-4 a e f c+4 b \left (e^2-4 d f\right ) c-b f (5 b e-16 a f)\right ) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 2143 |
\(\displaystyle \frac {\frac {\int \frac {8 \left (-d \left (e^2-d f\right ) c^2+2 d f (b e-a f) c-f^2 \left (b^2 d-a^2 f\right )+(c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right ) x\right )}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}+\frac {\left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {8 \int \frac {-d \left (e^2-d f\right ) c^2+2 d f (b e-a f) c-f^2 \left (b^2 d-a^2 f\right )-(c e-b f) \left (c e^2-b f e+2 a f^2-2 c d f\right ) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}+\frac {\left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {\frac {8 \int \frac {-d \left (e^2-d f\right ) c^2+2 d f (b e-a f) c-f^2 \left (b^2 d-a^2 f\right )-(c e-b f) \left (c e^2-b f e+2 a f^2-2 c d f\right ) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}+\frac {2 \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {8 \int \frac {-d \left (e^2-d f\right ) c^2+2 d f (b e-a f) c-f^2 \left (b^2 d-a^2 f\right )-(c e-b f) \left (c e^2-b f e+2 a f^2-2 c d f\right ) x}{\sqrt {c x^2+b x+a} \left (f x^2+e x+d\right )}dx}{f}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right )}{\sqrt {c} f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 1365 |
\(\displaystyle \frac {\frac {8 \left (\frac {\left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}-\frac {\left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+b x+a}}dx}{\sqrt {e^2-4 d f}}\right )}{f}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right )}{\sqrt {c} f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {\frac {8 \left (\frac {2 \left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \int \frac {1}{4 \left (4 a f^2-2 b \left (e+\sqrt {e^2-4 d f}\right ) f+c \left (e+\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}-\frac {2 \left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \int \frac {1}{4 \left (4 a f^2-2 b \left (e-\sqrt {e^2-4 d f}\right ) f+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )-\frac {\left (4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x\right )^2}{c x^2+b x+a}}d\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {c x^2+b x+a}}}{\sqrt {e^2-4 d f}}\right )}{f}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right )}{\sqrt {c} f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {8 \left (\frac {\left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\left (-2 f^3 \left (b^2 d-a^2 f\right )-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f) \left (f (b e-2 a f)-c \left (e^2-2 d f\right )\right )+4 c d f^2 (b e-a f)-2 c^2 d f \left (e^2-d f\right )\right ) \text {arctanh}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{f}+\frac {\text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-12 c f (b e-a f)+3 b^2 f^2+8 c^2 \left (e^2-d f\right )\right )}{\sqrt {c} f}}{8 f^2}-\frac {\sqrt {a+b x+c x^2} (-5 b f+4 c e-2 c f x)}{4 f^2}\) |
-1/4*((4*c*e - 5*b*f - 2*c*f*x)*Sqrt[a + b*x + c*x^2])/f^2 + (((3*b^2*f^2 - 12*c*f*(b*e - a*f) + 8*c^2*(e^2 - d*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*S qrt[a + b*x + c*x^2])])/(Sqrt[c]*f) + (8*(-(((4*c*d*f^2*(b*e - a*f) - 2*f^ 3*(b^2*d - a^2*f) - 2*c^2*d*f*(e^2 - d*f) - (c*e - b*f)*(e - Sqrt[e^2 - 4* d*f])*(f*(b*e - 2*a*f) - c*(e^2 - 2*d*f)))*ArcTanh[(4*a*f - b*(e - Sqrt[e^ 2 - 4*d*f]) + 2*(b*f - c*(e - Sqrt[e^2 - 4*d*f]))*x)/(2*Sqrt[2]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 - (c*e - b*f)*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a *f^2 - (c*e - b*f)*Sqrt[e^2 - 4*d*f]])) + ((4*c*d*f^2*(b*e - a*f) - 2*f^3* (b^2*d - a^2*f) - 2*c^2*d*f*(e^2 - d*f) - (c*e - b*f)*(e + Sqrt[e^2 - 4*d* f])*(f*(b*e - 2*a*f) - c*(e^2 - 2*d*f)))*ArcTanh[(4*a*f - b*(e + Sqrt[e^2 - 4*d*f]) + 2*(b*f - c*(e + Sqrt[e^2 - 4*d*f]))*x)/(2*Sqrt[2]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]]*Sqrt[a + b*x + c*x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*Sqrt[c*e^2 - 2*c*d*f - b*e*f + 2*a*f ^2 + (c*e - b*f)*Sqrt[e^2 - 4*d*f]])))/f)/(8*f^2)
3.2.6.3.1 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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x _)^2)^(q_), x_Symbol] :> Simp[(b*f*(3*p + 2*q) - c*e*(2*p + q) + 2*c*f*(p + q)*x)*(a + b*x + c*x^2)^(p - 1)*((d + e*x + f*x^2)^(q + 1)/(2*f^2*(p + q)* (2*p + 2*q + 1))), x] - Simp[1/(2*f^2*(p + q)*(2*p + 2*q + 1)) Int[(a + b *x + c*x^2)^(p - 2)*(d + e*x + f*x^2)^q*Simp[(b*d - a*e)*(c*e - b*f)*(1 - p )*(2*p + q) - (p + q)*(b^2*d*f*(1 - p) - a*(f*(b*e - 2*a*f)*(2*p + 2*q + 1) + c*(2*d*f - e^2*(2*p + q)))) + (2*(c*d - a*f)*(c*e - b*f)*(1 - p)*(2*p + q) - (p + q)*((b^2 - 4*a*c)*e*f*(1 - p) + b*(c*(e^2 - 4*d*f)*(2*p + q) + f* (2*c*d - b*e + 2*a*f)*(2*p + 2*q + 1))))*x + ((c*e - b*f)^2*(1 - p)*p + c*( p + q)*(f*(b*e - 2*a*f)*(4*p + 2*q - 1) - c*(2*d*f*(1 - 2*p) + e^2*(3*p + q - 1))))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[b^2 - 4* a*c, 0] && NeQ[e^2 - 4*d*f, 0] && GtQ[p, 1] && NeQ[p + q, 0] && NeQ[2*p + 2 *q + 1, 0] && !IGtQ[p, 0] && !IGtQ[q, 0]
Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + ( e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Sim p[(2*c*g - h*(b - q))/q Int[1/((b - q + 2*c*x)*Sqrt[d + e*x + f*x^2]), x] , x] - Simp[(2*c*g - h*(b + q))/q Int[1/((b + q + 2*c*x)*Sqrt[d + e*x + f *x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0 ] && NeQ[e^2 - 4*d*f, 0] && PosQ[b^2 - 4*a*c]
Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_ .)*(x_)^2]), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[Px, x, 2]}, Simp[C/c Int[1/Sqrt[d + e*x + f*x^2], x], x] + Simp[ 1/c Int[(A*c - a*C + (B*c - b*C)*x)/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x ^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && PolyQ[Px, x, 2]
Leaf count of result is larger than twice the leaf count of optimal. \(1327\) vs. \(2(618)=1236\).
Time = 1.03 (sec) , antiderivative size = 1328, normalized size of antiderivative = 1.96
method | result | size |
risch | \(\text {Expression too large to display}\) | \(1328\) |
default | \(\text {Expression too large to display}\) | \(2860\) |
1/4*(2*c*f*x+5*b*f-4*c*e)*(c*x^2+b*x+a)^(1/2)/f^2+1/8/f^2*(1/f*(12*a*c*f^2 +3*b^2*f^2-12*b*c*e*f-8*c^2*d*f+8*c^2*e^2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b *x+a)^(1/2))/c^(1/2)-1/2*(16*a*b*f^3*(-4*d*f+e^2)^(1/2)-16*a*c*e*f^2*(-4*d *f+e^2)^(1/2)-8*b^2*e*f^2*(-4*d*f+e^2)^(1/2)-16*b*c*d*f^2*(-4*d*f+e^2)^(1/ 2)+16*b*c*e^2*f*(-4*d*f+e^2)^(1/2)+16*c^2*d*e*f*(-4*d*f+e^2)^(1/2)-8*c^2*e ^3*(-4*d*f+e^2)^(1/2)-16*a^2*f^4+16*a*b*f^3*e+32*a*c*d*f^3-16*a*c*e^2*f^2+ 16*b^2*d*f^3-8*b^2*e^2*f^2-48*b*c*d*e*f^2+16*b*c*e^3*f-16*c^2*d^2*f^2+32*c ^2*d*e^2*f-8*c^2*e^4)/f^2/(-4*d*f+e^2)^(1/2)*2^(1/2)/((-b*f*(-4*d*f+e^2)^( 1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((- b*f*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2) /f^2+1/f*(-c*(-4*d*f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+ 1/2*2^(1/2)*((-b*f*(-4*d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f -2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c+4/f*(-c *(-4*d*f+e^2)^(1/2)+b*f-c*e)*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+2*(-b*f*(-4* d*f+e^2)^(1/2)+(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*c*d*f+c*e^2)/f^2)^(1 /2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))-1/2*(16*a*b*f^3*(-4*d*f+e^2)^(1/2)- 16*a*c*e*f^2*(-4*d*f+e^2)^(1/2)-8*b^2*e*f^2*(-4*d*f+e^2)^(1/2)-16*b*c*d*f^ 2*(-4*d*f+e^2)^(1/2)+16*b*c*e^2*f*(-4*d*f+e^2)^(1/2)+16*c^2*d*e*f*(-4*d*f+ e^2)^(1/2)-8*c^2*e^3*(-4*d*f+e^2)^(1/2)+16*a^2*f^4-16*a*b*f^3*e-32*a*c*d*f ^3+16*a*c*e^2*f^2-16*b^2*d*f^3+8*b^2*e^2*f^2+48*b*c*d*e*f^2-16*b*c*e^3*...
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \]
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*d*f-e^2>0)', see `assume?` for more deta
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{f\,x^2+e\,x+d} \,d x \]