Integrand size = 25, antiderivative size = 218 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (c \left (b^2-2 a c\right ) d-b^3 e+3 a b c e+c \left (b c d-b^2 e+2 a c e\right ) x\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt {a+b x+c x^2}}-\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{3/2} d}-\frac {e^3 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{d \left (c d^2-b d e+a e^2\right )^{3/2}} \] Output:
2*(c*(-2*a*c+b^2)*d-b^3*e+3*a*b*c*e+c*(2*a*c*e-b^2*e+b*c*d)*x)/a/(-4*a*c+b ^2)/(a*e^2-b*d*e+c*d^2)/(c*x^2+b*x+a)^(1/2)-arctanh(1/2*(b*x+2*a)/a^(1/2)/ (c*x^2+b*x+a)^(1/2))/a^(3/2)/d-e^3*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/ (a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/d/(a*e^2-b*d*e+c*d^2)^(3/2)
Time = 1.87 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 b^3 e-2 b c (3 a e+c d x)-2 b^2 c (d-e x)+4 a c^2 (d-e x)}{a \left (-b^2+4 a c\right ) \left (c d^2+e (-b d+a e)\right ) \sqrt {a+x (b+c x)}}-\frac {2 e^3 \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{d \left (c d^2+e (-b d+a e)\right )^2}+\frac {2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{a^{3/2} d} \] Input:
Integrate[1/(x*(d + e*x)*(a + b*x + c*x^2)^(3/2)),x]
Output:
(2*b^3*e - 2*b*c*(3*a*e + c*d*x) - 2*b^2*c*(d - e*x) + 4*a*c^2*(d - e*x))/ (a*(-b^2 + 4*a*c)*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[a + x*(b + c*x)]) - (2*e ^3*Sqrt[-(c*d^2) + b*d*e - a*e^2]*ArcTan[(Sqrt[c]*(d + e*x) - e*Sqrt[a + x *(b + c*x)])/Sqrt[-(c*d^2) + e*(b*d - a*e)]])/(d*(c*d^2 + e*(-(b*d) + a*e) )^2) + (2*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]])/(a^(3/2)*d )
Time = 0.51 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1289, 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.
\(\displaystyle \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1289 |
\(\displaystyle \int \left (\frac {1}{d x \left (a+b x+c x^2\right )^{3/2}}-\frac {e}{d (d+e x) \left (a+b x+c x^2\right )^{3/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{3/2} d}-\frac {e^3 \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{d \left (a e^2-b d e+c d^2\right )^{3/2}}+\frac {2 e \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \left (a e^2-b d e+c d^2\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a d \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\) |
Input:
Int[1/(x*(d + e*x)*(a + b*x + c*x^2)^(3/2)),x]
Output:
(2*(b^2 - 2*a*c + b*c*x))/(a*(b^2 - 4*a*c)*d*Sqrt[a + b*x + c*x^2]) + (2*e *(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x))/((b^2 - 4*a*c)*d*(c*d^2 - b*d*e + a*e^2)*Sqrt[a + b*x + c*x^2]) - ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqr t[a + b*x + c*x^2])]/(a^(3/2)*d) - (e^3*ArcTanh[(b*d - 2*a*e + (2*c*d - b* e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(d*(c*d^2 - b*d*e + a*e^2)^(3/2))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(492\) vs. \(2(200)=400\).
Time = 1.41 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.26
method | result | size |
default | \(\frac {\frac {1}{a \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{a^{\frac {3}{2}}}}{d}-\frac {\frac {e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (\frac {4 c \left (a \,e^{2}-b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}-\frac {e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{d}\) | \(493\) |
Input:
int(1/x/(e*x+d)/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/d*(1/a/(c*x^2+b*x+a)^(1/2)-b/a*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2) -1/a^(3/2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x))-1/d*(1/(a*e^2-b* d*e+c*d^2)*e^2/(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2) ^(1/2)-(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(2*c*(x+d/e)+(b*e-2*c*d)/e)/(4*c* (a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2)/(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d /e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-1/(a*e^2-b*d*e+c*d^2)*e^2/((a*e^2-b*d*e +c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d/e)+2*( (a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b *d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))
Leaf count of result is larger than twice the leaf count of optimal. 1180 vs. \(2 (201) = 402\).
Time = 2.01 (sec) , antiderivative size = 4819, normalized size of antiderivative = 22.11 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/x/(e*x+d)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x \left (d + e x\right ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x/(e*x+d)/(c*x**2+b*x+a)**(3/2),x)
Output:
Integral(1/(x*(d + e*x)*(a + b*x + c*x**2)**(3/2)), x)
\[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{2}} {\left (e x + d\right )} x} \,d x } \] Input:
integrate(1/x/(e*x+d)/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((c*x^2 + b*x + a)^(3/2)*(e*x + d)*x), x)
Exception generated. \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/x/(e*x+d)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
Timed out. \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{x\,\left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \] Input:
int(1/(x*(d + e*x)*(a + b*x + c*x^2)^(3/2)),x)
Output:
int(1/(x*(d + e*x)*(a + b*x + c*x^2)^(3/2)), x)
Time = 5.60 (sec) , antiderivative size = 3348, normalized size of antiderivative = 15.36 \[ \int \frac {1}{x (d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:
int(1/x/(e*x+d)/(c*x^2+b*x+a)^(3/2),x)
Output:
(4*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e* *2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**4*c*e**3 - sqrt(a *e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d* e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**3*b**2*e**3 + 4*sqrt(a*e** 2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**3*b*c*e**3*x + 4*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d **2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**3*c**2*e**3*x**2 - sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d* *2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**2*b**3*e**3*x - sqrt(a*e**2 - b*d* e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a**2*b**2*c*e**3*x**2 - 4*sqrt(a*e**2 - b *d*e + c*d**2)*log(d + e*x)*a**4*c*e**3 + sqrt(a*e**2 - b*d*e + c*d**2)*lo g(d + e*x)*a**3*b**2*e**3 - 4*sqrt(a*e**2 - b*d*e + c*d**2)*log(d + e*x)*a **3*b*c*e**3*x - 4*sqrt(a*e**2 - b*d*e + c*d**2)*log(d + e*x)*a**3*c**2*e* *3*x**2 + sqrt(a*e**2 - b*d*e + c*d**2)*log(d + e*x)*a**2*b**3*e**3*x + sq rt(a*e**2 - b*d*e + c*d**2)*log(d + e*x)*a**2*b**2*c*e**3*x**2 - 6*sqrt(a + b*x + c*x**2)*a**3*b*c*d*e**3 + 4*sqrt(a + b*x + c*x**2)*a**3*c**2*d**2* e**2 - 4*sqrt(a + b*x + c*x**2)*a**3*c**2*d*e**3*x + 2*sqrt(a + b*x + c*x* *2)*a**2*b**3*d*e**3 + 4*sqrt(a + b*x + c*x**2)*a**2*b**2*c*d**2*e**2 +...