Integrand size = 27, antiderivative size = 496 \[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\frac {a \sqrt {a+c x^2}}{d}+\frac {(c d-a f) \sqrt {a+c x^2}}{d f}-\frac {c^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f^2}-\frac {\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (c^2 d e^2-f (c d-a f)^2\right ) \left (e-\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (c^2 d e^2-f (c d-a f)^2\right ) \left (e+\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}-\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d} \]
-c^(3/2)*e*arctanh(x*c^(1/2)/(c*x^2+a)^(1/2))/f^2-a^(3/2)*arctanh((c*x^2+a )^(1/2)/a^(1/2))/d+a*(c*x^2+a)^(1/2)/d+(-a*f+c*d)*(c*x^2+a)^(1/2)/d/f-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*e*f*(-a^2*f^2+c^2*d^2) -(c^2*d*e^2-f*(-a*f+c*d)^2)*(e-(-4*d*f+e^2)^(1/2)))/d/f^2*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*e*f*(-a^2*f^2+c^2*d^2)-(c^2*d* e^2-f*(-a*f+c*d)^2)*(e+(-4*d*f+e^2)^(1/2)))/d/f^2*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 = 0.55 (sec) , antiderivative size = 552, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\frac {c d f \sqrt {a+c x^2}+2 a^{3/2} f^2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )+c^{3/2} d e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )+\text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {-a c^2 d e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a c^2 d^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-2 a^2 c d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+a^3 f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-2 c^{5/2} d^2 e \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+2 a^2 \sqrt {c} e f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}+c^2 d e^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-c^2 d^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+2 a c d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2-a^2 f^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{d f^2} \]
(c*d*f*Sqrt[a + c*x^2] + 2*a^(3/2)*f^2*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2 ])/Sqrt[a]] + c^(3/2)*d*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]] + 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*d*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]) + a*c^2*d^ 2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - 2*a^2*c*d*f^2*Log[-(Sqrt[c] *x) + Sqrt[a + c*x^2] - #1] + a^3*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1] - 2*c^(5/2)*d^2*e*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + 2*a^2 *Sqrt[c]*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 + c^2*d*e^2*Log [-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 - c^2*d^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + 2*a*c*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2 ] - #1]*#1^2 - a^2*f^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*S qrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ])/(d*f^2)
Time = 1.97 (sec) , antiderivative size = 496, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {7279, 2009}
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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {\left (a+c x^2\right )^{3/2} (-e-f x)}{d \left (d+e x+f x^2\right )}+\frac {\left (a+c x^2\right )^{3/2}}{d x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{d}-\frac {\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (2 e f \left (c^2 d^2-a^2 f^2\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (c^2 d e^2-f (c d-a f)^2\right )\right ) \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {c^{3/2} e \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{f^2}+\frac {\sqrt {a+c x^2} (c d-a f)}{d f}+\frac {a \sqrt {a+c x^2}}{d}\) |
(a*Sqrt[a + c*x^2])/d + ((c*d - a*f)*Sqrt[a + c*x^2])/(d*f) - (c^(3/2)*e*A rcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/f^2 - ((2*e*f*(c^2*d^2 - a^2*f^2) - ( c^2*d*e^2 - f*(c*d - a*f)^2)*(e - Sqrt[e^2 - 4*d*f]))*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]*d*f^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*e*f*(c^2*d^2 - a^2*f ^2) - (c^2*d*e^2 - f*(c*d - a*f)^2)*(e + Sqrt[e^2 - 4*d*f]))*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]*d*f^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])]) - (a^(3/2)*ArcTanh[ Sqrt[a + c*x^2]/Sqrt[a]])/d
3.1.61.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(2378\) vs. \(2(443)=886\).
Time = 0.65 (sec) , antiderivative size = 2379, normalized size of antiderivative = 4.80
-4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))*(1/3*(c*x^2+a)^(3/2)+a *((c*x^2+a)^(1/2)-a^(1/2)*ln((2*a+2*a^(1/2)*(c*x^2+a)^(1/2))/x)))+2*f/(e+( -4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/2)*(1/3*((x+1/2*(e+(-4*d*f+e^2)^(1/2))/ f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*((- 4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(3/2)-1/2*c*(e+(-4*d*f+e^ 2)^(1/2))/f*(1/4*(2*c*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)-c*(e+(-4*d*f+e^2)^( 1/2))/f)/c*((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/ f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+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)+1/8*(2*c*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c* e^2)/f^2-c^2*(e+(-4*d*f+e^2)^(1/2))^2/f^2)/c^(3/2)*ln((-1/2*c*(e+(-4*d*f+e ^2)^(1/2))/f+c*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))/c^(1/2)+((x+1/2*(e+(-4*d* f+e^2)^(1/2))/f)^2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/ 2))/f)+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)))+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+(-4* d*f+e^2)^(1/2))/f)^2*c-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2) ^(1/2))/f)+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)-1/2 *c^(1/2)*(e+(-4*d*f+e^2)^(1/2))/f*ln((-1/2*c*(e+(-4*d*f+e^2)^(1/2))/f+c*(x +1/2*(e+(-4*d*f+e^2)^(1/2))/f))/c^(1/2)+((x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)^ 2*c-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+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))-1/2*((-4*d*f+e^2)^...
Timed out. \[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {\left (a + c x^{2}\right )^{\frac {3}{2}}}{x \left (d + e x + f x^{2}\right )}\, dx \]
\[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int { \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}}}{{\left (f x^{2} + e x + d\right )} x} \,d x } \]
Exception generated. \[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\left (a+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {{\left (c\,x^2+a\right )}^{3/2}}{x\,\left (f\,x^2+e\,x+d\right )} \,d x \]