Integrand size = 25, antiderivative size = 553 \[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\frac {\left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right ) \sqrt {a+c x^2}}{2 f^3}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\sqrt {c} e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 f^4}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^4 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (e^4-3 d e^2 f+d^2 f^2\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} f^4 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}} \]
1/3*(c*x^2+a)^(3/2)/f-1/2*e*(3*a*f^2+2*c*(-2*d*f+e^2))*arctanh(x*c^(1/2)/( c*x^2+a)^(1/2))*c^(1/2)/f^4+1/2*(2*a*f^2+2*c*(-d*f+e^2)-c*e*f*x)*(c*x^2+a) ^(1/2)/f^3-1/2*arctanh(1/2*(2*a*f-c*x*(e-(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x ^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2))*(2*c*d*e*f *(2*a*f^2+c*(-2*d*f+e^2))-(a^2*f^4+2*a*c*f^2*(-d*f+e^2)+c^2*(d^2*f^2-3*d*e ^2*f+e^4))*(e-(-4*d*f+e^2)^(1/2)))/f^4*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2 +c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)+1/2*arctanh(1/2*(2*a*f-c*x*(e+( -4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d* f+e^2)^(1/2)))^(1/2))*(2*c*d*e*f*(2*a*f^2+c*(-2*d*f+e^2))-(a^2*f^4+2*a*c*f ^2*(-d*f+e^2)+c^2*(d^2*f^2-3*d*e^2*f+e^4))*(e+(-4*d*f+e^2)^(1/2)))/f^4*2^( 1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+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 = 0.68 (sec) , antiderivative size = 755, normalized size of antiderivative = 1.37 \[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\frac {f \sqrt {a+c x^2} \left (8 a f^2+c \left (6 e^2-6 d f-3 e f x+2 f^2 x^2\right )\right )+3 \sqrt {c} e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )-6 \text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a c^2 e^4 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-3 a c^2 d e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a c^2 d^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 a^2 c e^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-2 a^2 c d f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a^3 f^4 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 c^{5/2} d e^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-4 c^{5/2} d^2 e f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+4 a c^{3/2} d e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-c^2 e^4 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+3 c^2 d e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-c^2 d^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-2 a c e^2 f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 a c d f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a^2 f^4 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{6 f^4} \]
(f*Sqrt[a + c*x^2]*(8*a*f^2 + c*(6*e^2 - 6*d*f - 3*e*f*x + 2*f^2*x^2)) + 3 *Sqrt[c]*e*(3*a*f^2 + 2*c*(e^2 - 2*d*f))*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2 ]] - 6*RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2*Sqrt [c]*e*#1^3 + f*#1^4 & , (a*c^2*e^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1 ] - 3*a*c^2*d*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + a*c^2*d^2*f ^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*a^2*c*e^2*f^2*Log[-(Sqrt[c ]*x) + Sqrt[a + c*x^2] - #1] - 2*a^2*c*d*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c *x^2] - #1] + a^3*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] + 2*c^(5/2) *d*e^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - 4*c^(5/2)*d^2*e*f*Log [-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 4*a*c^(3/2)*d*e*f^2*Log[-(Sqrt[ c]*x) + Sqrt[a + c*x^2] - #1]*#1 - c^2*e^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x ^2] - #1]*#1^2 + 3*c^2*d*e^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 ^2 - c^2*d^2*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - 2*a*c*e^2 *f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 2*a*c*d*f^3*Log[-(Sqr t[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - a^2*f^4*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ])/(6*f^4)
Time = 1.35 (sec) , antiderivative size = 569, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {1353, 27, 2027, 2139, 2145, 27, 224, 219, 1367, 488, 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.
\(\displaystyle \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx\) |
\(\Big \downarrow \) 1353 |
\(\displaystyle \frac {\int -\frac {3 \sqrt {c x^2+a} \left (c e x^2+(c d-a f) x\right )}{f x^2+e x+d}dx}{3 f}+\frac {\left (a+c x^2\right )^{3/2}}{3 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\int \frac {\sqrt {c x^2+a} \left (c e x^2+(c d-a f) x\right )}{f x^2+e x+d}dx}{f}\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {\int \frac {x (c d-a f+c e x) \sqrt {c x^2+a}}{f x^2+e x+d}dx}{f}\) |
\(\Big \downarrow \) 2139 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\int \frac {-e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) x^2 c^2+a d e f c^2-\left (a c f e^2+2 (c d-a f) \left (a f^2+c \left (e^2-d f\right )\right )\right ) x c}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 2145 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {\int \frac {2 c \left (c d e \left (2 a f^2+c \left (e^2-2 d f\right )\right )+\left (a^2 f^4+2 a c \left (e^2-d f\right ) f^2+c^2 \left (e^4-3 d f e^2+d^2 f^2\right )\right ) x\right )}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}-\frac {c^2 e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \int \frac {1}{\sqrt {c x^2+a}}dx}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {2 c \int \frac {c d e \left (2 a f^2+c \left (e^2-2 d f\right )\right )+\left (a^2 f^4+2 a c \left (e^2-d f\right ) f^2+c^2 \left (e^4-3 d f e^2+d^2 f^2\right )\right ) x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}-\frac {c^2 e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \int \frac {1}{\sqrt {c x^2+a}}dx}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {2 c \int \frac {c d e \left (2 a f^2+c \left (e^2-2 d f\right )\right )+\left (a^2 f^4+2 a c \left (e^2-d f\right ) f^2+c^2 \left (e^4-3 d f e^2+d^2 f^2\right )\right ) x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}-\frac {c^2 e \left (3 a f^2+2 c \left (e^2-2 d f\right )\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+a}}d\frac {x}{\sqrt {c x^2+a}}}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {2 c \int \frac {c d e \left (2 a f^2+c \left (e^2-2 d f\right )\right )+\left (a^2 f^4+2 a c \left (e^2-d f\right ) f^2+c^2 \left (e^4-3 d f e^2+d^2 f^2\right )\right ) x}{\sqrt {c x^2+a} \left (f x^2+e x+d\right )}dx}{f}-\frac {c^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 1367 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {2 c \left (\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \int \frac {1}{\left (e+2 f x-\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+a}}dx}{\sqrt {e^2-4 d f}}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \int \frac {1}{\left (e+2 f x+\sqrt {e^2-4 d f}\right ) \sqrt {c x^2+a}}dx}{\sqrt {e^2-4 d f}}\right )}{f}-\frac {c^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {2 c \left (\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \int \frac {1}{4 a f^2+c \left (e+\sqrt {e^2-4 d f}\right )^2-\frac {\left (2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x\right )^2}{c x^2+a}}d\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {c x^2+a}}}{\sqrt {e^2-4 d f}}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \int \frac {1}{4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2-\frac {\left (2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x\right )^2}{c x^2+a}}d\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {c x^2+a}}}{\sqrt {e^2-4 d f}}\right )}{f}-\frac {c^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (a+c x^2\right )^{3/2}}{3 f}-\frac {-\frac {\frac {2 c \left (\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {\left (2 c d e f \left (2 a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a^2 f^4+2 a c f^2 \left (e^2-d f\right )+c^2 \left (d^2 f^2-3 d e^2 f+e^4\right )\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{f}-\frac {c^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right ) \left (3 a f^2+2 c \left (e^2-2 d f\right )\right )}{f}}{2 c f^2}-\frac {\sqrt {a+c x^2} \left (2 \left (a f^2+c \left (e^2-d f\right )\right )-c e f x\right )}{2 f^2}}{f}\) |
(a + c*x^2)^(3/2)/(3*f) - (-1/2*((2*(a*f^2 + c*(e^2 - d*f)) - c*e*f*x)*Sqr t[a + c*x^2])/f^2 - (-((c^(3/2)*e*(3*a*f^2 + 2*c*(e^2 - 2*d*f))*ArcTanh[(S qrt[c]*x)/Sqrt[a + c*x^2]])/f) + (2*c*(-(((2*c*d*e*f*(2*a*f^2 + c*(e^2 - 2 *d*f)) - (e - Sqrt[e^2 - 4*d*f])*(a^2*f^4 + 2*a*c*f^2*(e^2 - d*f) + c^2*(e ^4 - 3*d*e^2*f + d^2*f^2)))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/ (Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c* x^2])])/(Sqrt[2]*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[ e^2 - 4*d*f])])) + ((2*c*d*e*f*(2*a*f^2 + c*(e^2 - 2*d*f)) - (e + Sqrt[e^2 - 4*d*f])*(a^2*f^4 + 2*a*c*f^2*(e^2 - d*f) + c^2*(e^4 - 3*d*e^2*f + d^2*f ^2)))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*Sqrt[ e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])])))/f)/( 2*c*f^2))/f
3.1.59.3.1 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_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f _.)*(x_)^2)^(q_), x_Symbol] :> Simp[h*(a + c*x^2)^p*((d + e*x + f*x^2)^(q + 1)/(2*f*(p + q + 1))), x] + Simp[1/(2*f*(p + q + 1)) Int[(a + c*x^2)^(p - 1)*(d + e*x + f*x^2)^q*Simp[a*h*e*p - a*(h*e - 2*g*f)*(p + q + 1) - 2*h*p *(c*d - a*f)*x - (h*c*e*p + c*(h*e - 2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h, q}, x] && NeQ[e^2 - 4*d*f, 0] && GtQ[p, 0] && NeQ[p + q + 1, 0]
Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f _.)*(x_)^2]), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(2*c*g - h*( b - q))/q Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Simp[(2*c*g - h*(b + q))/q Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{ a, b, c, d, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(Px_)*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_ ), x_Symbol] :> With[{A = Coeff[Px, x, 0], B = Coeff[Px, x, 1], C = Coeff[P x, x, 2]}, Simp[(B*c*f*(2*p + 2*q + 3) + C*((-c)*e*(2*p + q + 2)) + 2*c*C*f *(p + q + 1)*x)*(a + c*x^2)^p*((d + e*x + f*x^2)^(q + 1)/(2*c*f^2*(p + q + 1)*(2*p + 2*q + 3))), x] - Simp[1/(2*c*f^2*(p + q + 1)*(2*p + 2*q + 3)) I nt[(a + c*x^2)^(p - 1)*(d + e*x + f*x^2)^q*Simp[p*((-a)*e)*(C*(c*e)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)*(a*c*(C*(2*d*f - e^2*(2*p + q + 2)) + f*(B*e - 2*A*f)*(2*p + 2*q + 3))) + (2*p*(c*d - a*f)*(C*(c*e)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)*(C*e*f*p*(-4*a*c)))*x + (p*(c*e)*(C*(c*e)*(q + 1) - c*(C*e - B*f)*(2*p + 2*q + 3)) + (p + q + 1)* (C*f^2*p*(-4*a*c) - c^2*(C*(e^2 - 4*d*f)*(2*p + q + 2) + f*(2*C*d - B*e + 2 *A*f)*(2*p + 2*q + 3))))*x^2, x], x], x]] /; FreeQ[{a, c, d, e, f, q}, x] & & PolyQ[Px, x, 2] && GtQ[p, 0] && NeQ[p + q + 1, 0] && NeQ[2*p + 2*q + 3, 0 ] && !IGtQ[p, 0] && !IGtQ[q, 0]
Int[(Px_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (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 + 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 + f*x^2]), x], x]] /; FreeQ[{a , b, c, d, f}, x] && PolyQ[Px, x, 2]
Leaf count of result is larger than twice the leaf count of optimal. \(1038\) vs. \(2(499)=998\).
Time = 0.76 (sec) , antiderivative size = 1039, normalized size of antiderivative = 1.88
method | result | size |
risch | \(\text {Expression too large to display}\) | \(1039\) |
default | \(\text {Expression too large to display}\) | \(2294\) |
1/6*(2*c*f^2*x^2-3*c*e*f*x+8*a*f^2-6*c*d*f+6*c*e^2)*(c*x^2+a)^(1/2)/f^3-1/ 2/f^3*(1/f*c^(1/2)*e*(3*a*f^2-4*c*d*f+2*c*e^2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2 ))-1/2*(-2*a^2*f^4*(-4*d*f+e^2)^(1/2)+4*a*c*d*f^3*(-4*d*f+e^2)^(1/2)-4*a*c *e^2*f^2*(-4*d*f+e^2)^(1/2)-2*c^2*d^2*f^2*(-4*d*f+e^2)^(1/2)+6*c^2*d*e^2*f *(-4*d*f+e^2)^(1/2)-2*c^2*e^4*(-4*d*f+e^2)^(1/2)+2*e*f^4*a^2-12*f^3*a*c*d* e+4*a*c*e^3*f^2+10*c^2*d^2*e*f^2-10*c^2*d*e^3*f+2*c^2*e^5)/f^2/(-4*d*f+e^2 )^(1/2)*2^(1/2)/((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2 )*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2-c*(e-(-4*d*f+e^2 )^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*2^(1/2)*((-(-4*d*f+e^2)^( 1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x-1/2/f*(-e+(-4*d*f+e^2)^(1 /2)))^2*c-4*c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+2 *(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2))/(x-1/2/f*(-e+ (-4*d*f+e^2)^(1/2))))-1/2*(-2*a^2*f^4*(-4*d*f+e^2)^(1/2)+4*a*c*d*f^3*(-4*d *f+e^2)^(1/2)-4*a*c*e^2*f^2*(-4*d*f+e^2)^(1/2)-2*c^2*d^2*f^2*(-4*d*f+e^2)^ (1/2)+6*c^2*d*e^2*f*(-4*d*f+e^2)^(1/2)-2*c^2*e^4*(-4*d*f+e^2)^(1/2)-2*e*f^ 4*a^2+12*f^3*a*c*d*e-4*a*c*e^3*f^2-10*c^2*d^2*e*f^2+10*c^2*d*e^3*f-2*c^2*e ^5)/f^2/(-4*d*f+e^2)^(1/2)*2^(1/2)/(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d* f+c*e^2)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2 -c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^(1/2)*( ((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x+1/2*(e+...
Timed out. \[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Timed out} \]
\[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\int \frac {x \left (a + c x^{2}\right )^{\frac {3}{2}}}{d + e x + f x^{2}}\, dx \]
Exception generated. \[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \]
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. \[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {x \left (a+c x^2\right )^{3/2}}{d+e x+f x^2} \, dx=\int \frac {x\,{\left (c\,x^2+a\right )}^{3/2}}{f\,x^2+e\,x+d} \,d x \]