Integrand size = 30, antiderivative size = 709 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\frac {c \sqrt {a+b x+c x^2}}{f}-\frac {a^{3/2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{d}-\frac {b \sqrt {c} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 f}-\frac {\sqrt {c} (c e-2 b f) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\left (2 f (c d-a f) (c d e-2 b d f+a e f)+\left (e-\sqrt {e^2-4 d f}\right ) \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}+\frac {\left (2 f (c d-a f) (c d e-2 b d f+a e f)+\left (e+\sqrt {e^2-4 d f}\right ) \left (2 c d f (b e-a f)-f^2 \left (b^2 d-a^2 f\right )-c^2 d \left (e^2-d f\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}} \] Output:
c*(c*x^2+b*x+a)^(1/2)/f-a^(3/2)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a )^(1/2))/d-1/2*b*c^(1/2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2) )/f-c^(1/2)*(-2*b*f+c*e)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2) )/f^2-1/2*(2*f*(-a*f+c*d)*(a*e*f-2*b*d*f+c*d*e)+(e-(-4*d*f+e^2)^(1/2))*(2* c*d*f*(-a*f+b*e)-f^2*(-a^2*f+b^2*d)-c^2*d*(-d*f+e^2)))*arctanh(1/4*(4*a*f- b*(e-(-4*d*f+e^2)^(1/2))+2*(b*f-c*(e-(-4*d*f+e^2)^(1/2)))*x)*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)/(c*x^2+b*x+a) ^(1/2))*2^(1/2)/d/f^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*(2*f*(-a*f+c*d)*(a*e*f-2*b*d*f+c*d*e) +(e+(-4*d*f+e^2)^(1/2))*(2*c*d*f*(-a*f+b*e)-f^2*(-a^2*f+b^2*d)-c^2*d*(-d*f +e^2)))*arctanh(1/4*(4*a*f-b*(e+(-4*d*f+e^2)^(1/2))+2*(b*f-c*(e+(-4*d*f+e^ 2)^(1/2)))*x)*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)/(c*x^2+b*x+a)^(1/2))*2^(1/2)/d/f^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 = 1.72 (sec) , antiderivative size = 1032, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx =\text {Too large to display} \] Input:
Integrate[(a + b*x + c*x^2)^(3/2)/(x*(d + e*x + f*x^2)),x]
Output:
(2*c*d*f*Sqrt[a + x*(b + c*x)] + 4*a^(3/2)*f^2*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]] + 2*c^(3/2)*d*e*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]] - 3*b*Sqrt[c]*d*f*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]] + 2*RootSum[b^2*d - a*b*e + a^2*f - 4*b*Sqrt[c]*d*#1 + 2*a*Sqrt[ c]*e*#1 + 4*c*d*#1^2 + b*e*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (-(b*c^2*d^2*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]) + a*c^2* d*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*b^2*c*d^2*f*Log[- (Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a*c^2*d^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - 2*a*b*c*d*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b *x + c*x^2] - #1] - a*b^2*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*a^2*c*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + a^2* b*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1] - a^3*f^3*Log[-(Sqr t[c]*x) + Sqrt[a + b*x + c*x^2] - #1] + 2*c^(5/2)*d^2*e*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - 4*b*c^(3/2)*d^2*f*Log[-(Sqrt[c]*x) + Sqr t[a + b*x + c*x^2] - #1]*#1 + 4*a*b*Sqrt[c]*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[ a + b*x + c*x^2] - #1]*#1 - 2*a^2*Sqrt[c]*e*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1 - c^2*d*e^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^ 2] - #1]*#1^2 + c^2*d^2*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*# 1^2 + 2*b*c*d*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - b^ 2*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2] - #1]*#1^2 - 2*a*c*d*f...
Time = 5.79 (sec) , antiderivative size = 956, normalized size of antiderivative = 1.35, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, 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.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {(-e-f x) \left (a+b x+c x^2\right )^{3/2}}{d \left (d+e x+f x^2\right )}+\frac {\left (a+b x+c x^2\right )^{3/2}}{d x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {c x^2+b x+a}}\right ) a^{3/2}}{d}-\frac {\left (-f^2 b^3+12 a c f^2 b-24 c^2 d f b+16 c^3 d e\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right )}{16 c^{3/2} d f^2}-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {c x^2+b x+a}}\right )}{16 c^{3/2} d}+\frac {\left (d \left (e^3-\sqrt {e^2-4 d f} e^2-3 d f e+d f \sqrt {e^2-4 d f}\right ) c^2+2 d f \left (a f \left (e-\sqrt {e^2-4 d f}\right )-b \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )\right ) c-f^2 \left (-f \left (e+\sqrt {e^2-4 d f}\right ) a^2+4 b d f a-b^2 d \left (e-\sqrt {e^2-4 d f}\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-b f e+2 a f^2-2 c d f-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {c x^2+b x+a}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {c \left (e^2-\sqrt {e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e-\sqrt {e^2-4 d f}\right )\right )}}-\frac {\left (d \left (e^3+\sqrt {e^2-4 d f} e^2-3 d f e-d f \sqrt {e^2-4 d f}\right ) c^2+2 d f \left (a f \left (e+\sqrt {e^2-4 d f}\right )-b \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )\right ) c-f^2 \left (-f \left (e-\sqrt {e^2-4 d f}\right ) a^2+4 b d f a-b^2 d \left (e+\sqrt {e^2-4 d f}\right )\right )\right ) \text {arctanh}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-b f e+2 a f^2-2 c d f+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {c x^2+b x+a}}\right )}{\sqrt {2} d f^2 \sqrt {e^2-4 d f} \sqrt {c \left (e^2+\sqrt {e^2-4 d f} e-2 d f\right )+f \left (2 a f-b \left (e+\sqrt {e^2-4 d f}\right )\right )}}+\frac {\left (b^2+2 c x b+8 a c\right ) \sqrt {c x^2+b x+a}}{8 c d}+\frac {\left (-f b^2-2 c f x b+8 c^2 d-8 a c f\right ) \sqrt {c x^2+b x+a}}{8 c d f}\) |
Input:
Int[(a + b*x + c*x^2)^(3/2)/(x*(d + e*x + f*x^2)),x]
Output:
((b^2 + 8*a*c + 2*b*c*x)*Sqrt[a + b*x + c*x^2])/(8*c*d) + ((8*c^2*d - b^2* f - 8*a*c*f - 2*b*c*f*x)*Sqrt[a + b*x + c*x^2])/(8*c*d*f) - (a^(3/2)*ArcTa nh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/d - (b*(b^2 - 12*a*c)*A rcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2)*d) - (( 16*c^3*d*e - 24*b*c^2*d*f - b^3*f^2 + 12*a*b*c*f^2)*ArcTanh[(b + 2*c*x)/(2 *Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*c^(3/2)*d*f^2) + ((c^2*d*(e^3 - 3*d* e*f - e^2*Sqrt[e^2 - 4*d*f] + d*f*Sqrt[e^2 - 4*d*f]) - f^2*(4*a*b*d*f - b^ 2*d*(e - Sqrt[e^2 - 4*d*f]) - a^2*f*(e + Sqrt[e^2 - 4*d*f])) + 2*c*d*f*(a* f*(e - Sqrt[e^2 - 4*d*f]) - b*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])))*ArcTan h[(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]*d*f^2*Sqrt[e^2 - 4*d*f]*Sqr t[c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f]) + f*(2*a*f - b*(e - Sqrt[e^2 - 4*d *f]))]) - ((c^2*d*(e^3 - 3*d*e*f + e^2*Sqrt[e^2 - 4*d*f] - d*f*Sqrt[e^2 - 4*d*f]) - f^2*(4*a*b*d*f - a^2*f*(e - Sqrt[e^2 - 4*d*f]) - b^2*d*(e + Sqrt [e^2 - 4*d*f])) + 2*c*d*f*(a*f*(e + Sqrt[e^2 - 4*d*f]) - b*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*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 ]*d*f^2*Sqrt[e^2 - 4*d*f]*Sqrt[c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f]) + ...
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. \(3087\) vs. \(2(634)=1268\).
Time = 2.80 (sec) , antiderivative size = 3088, normalized size of antiderivative = 4.36
Input:
int((c*x^2+b*x+a)^(3/2)/x/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
Output:
-4*f/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))*(1/3*(c*x^2+b*x+a)^(3/ 2)+1/2*b*(1/4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c+1/8*(4*a*c-b^2)/c^(3/2)*ln(( 1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))+a*((c*x^2+b*x+a)^(1/2)+1/2*b*ln(( 1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))/c^(1/2)-a^(1/2)*ln((2*a+b*x+2*a^(1 /2)*(c*x^2+b*x+a)^(1/2))/x)))+2*f/(-e+(-4*d*f+e^2)^(1/2))/(-4*d*f+e^2)^(1/ 2)*(1/3*(c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2+(c*(-4*d*f+e^2)^(1/2)+f*b-c *e)/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*(f*b*(-4*d*f+e^2)^(1/2)-(-4*d* f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(3/2)+1/2*(c*(-4*d*f+e^ 2)^(1/2)+f*b-c*e)/f*(1/4*(2*c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+(c*(-4*d*f +e^2)^(1/2)+f*b-c*e)/f)/c*(c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2+(c*(-4*d* f+e^2)^(1/2)+f*b-c*e)/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*(f*b*(-4*d*f +e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2) +1/8*(2*c*(f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d *f*c+c*e^2)/f^2-(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)^2/f^2)/c^(3/2)*ln((1/2*(c*( -4*d*f+e^2)^(1/2)+f*b-c*e)/f+c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2))))/c^(1/2)+ (c*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2+(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/f*(x -1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+1/2*(f*b*(-4*d*f+e^2)^(1/2)-(-4*d*f+e^2)^( 1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2)^(1/2)))+1/2*(f*b*(-4*d*f+e^2)^( 1/2)-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-b*e*f-2*d*f*c+c*e^2)/f^2*(1/2*(4*c*(x- 1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2+4*(c*(-4*d*f+e^2)^(1/2)+f*b-c*e)/f*(x-...
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \] Input:
integrate((c*x^2+b*x+a)^(3/2)/x/(f*x^2+e*x+d),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{x \left (d + e x + f x^{2}\right )}\, dx \] Input:
integrate((c*x**2+b*x+a)**(3/2)/x/(f*x**2+e*x+d),x)
Output:
Integral((a + b*x + c*x**2)**(3/2)/(x*(d + e*x + f*x**2)), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}}{{\left (f x^{2} + e x + d\right )} x} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(3/2)/x/(f*x^2+e*x+d),x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x + a)^(3/2)/((f*x^2 + e*x + d)*x), x)
Exception generated. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c*x^2+b*x+a)^(3/2)/x/(f*x^2+e*x+d),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio n over extensionNot implemented, e.g. for multivariate mod/approx polynomi alsError:
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x\,\left (f\,x^2+e\,x+d\right )} \,d x \] Input:
int((a + b*x + c*x^2)^(3/2)/(x*(d + e*x + f*x^2)),x)
Output:
int((a + b*x + c*x^2)^(3/2)/(x*(d + e*x + f*x^2)), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{3/2}}{x \left (d+e x+f x^2\right )} \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{x \left (f \,x^{2}+e x +d \right )}d x \] Input:
int((c*x^2+b*x+a)^(3/2)/x/(f*x^2+e*x+d),x)
Output:
int((c*x^2+b*x+a)^(3/2)/x/(f*x^2+e*x+d),x)