Integrand size = 21, antiderivative size = 77 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b \sqrt {d}}+\frac {2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c d-b e}} \] Output:
-2*arctanh((e*x+d)^(1/2)/d^(1/2))/b/d^(1/2)+2*c^(1/2)*arctanh(c^(1/2)*(e*x +d)^(1/2)/(-b*e+c*d)^(1/2))/b/(-b*e+c*d)^(1/2)
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=-\frac {\frac {2 \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{\sqrt {-c d+b e}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{b} \] Input:
Integrate[1/(Sqrt[d + e*x]*(b*x + c*x^2)),x]
Output:
-(((2*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + b*e]])/Sqrt[-(c *d) + b*e] + (2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/Sqrt[d])/b)
Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1149, 1406, 221}
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.
\(\displaystyle \int \frac {1}{\left (b x+c x^2\right ) \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 1149 |
\(\displaystyle 2 e \int \frac {1}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}\) |
\(\Big \downarrow \) 1406 |
\(\displaystyle 2 e \left (\frac {c \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}-\frac {c \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle 2 e \left (\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b e \sqrt {c d-b e}}-\frac {\text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b \sqrt {d} e}\right )\) |
Input:
Int[1/(Sqrt[d + e*x]*(b*x + c*x^2)),x]
Output:
2*e*(-(ArcTanh[Sqrt[d + e*x]/Sqrt[d]]/(b*Sqrt[d]*e)) + (Sqrt[c]*ArcTanh[(S qrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*e*Sqrt[c*d - b*e]))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Sym bol] :> Simp[2*e Subst[Int[1/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[b^ 2 - 4*a*c, 2]}, Simp[c/q Int[1/(b/2 - q/2 + c*x^2), x], x] - Simp[c/q I nt[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c , 0] && PosQ[b^2 - 4*a*c]
Time = 0.51 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79
method | result | size |
pseudoelliptic | \(\frac {-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 c \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{\sqrt {c \left (b e -c d \right )}}}{b}\) | \(61\) |
derivativedivides | \(2 e \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e \sqrt {d}}-\frac {c \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{b e \sqrt {c \left (b e -c d \right )}}\right )\) | \(71\) |
default | \(2 e \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e \sqrt {d}}-\frac {c \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{b e \sqrt {c \left (b e -c d \right )}}\right )\) | \(71\) |
Input:
int(1/(e*x+d)^(1/2)/(c*x^2+b*x),x,method=_RETURNVERBOSE)
Output:
2/b*(-arctanh((e*x+d)^(1/2)/d^(1/2))/d^(1/2)-c/(c*(b*e-c*d))^(1/2)*arctan( c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2)))
Time = 0.11 (sec) , antiderivative size = 355, normalized size of antiderivative = 4.61 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=\left [\frac {d \sqrt {\frac {c}{c d - b e}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, {\left (c d - b e\right )} \sqrt {e x + d} \sqrt {\frac {c}{c d - b e}}}{c x + b}\right ) + \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right )}{b d}, -\frac {2 \, d \sqrt {-\frac {c}{c d - b e}} \arctan \left (\sqrt {e x + d} \sqrt {-\frac {c}{c d - b e}}\right ) - \sqrt {d} \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right )}{b d}, \frac {d \sqrt {\frac {c}{c d - b e}} \log \left (\frac {c e x + 2 \, c d - b e + 2 \, {\left (c d - b e\right )} \sqrt {e x + d} \sqrt {\frac {c}{c d - b e}}}{c x + b}\right ) + 2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x + d}}\right )}{b d}, -\frac {2 \, {\left (d \sqrt {-\frac {c}{c d - b e}} \arctan \left (\sqrt {e x + d} \sqrt {-\frac {c}{c d - b e}}\right ) - \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x + d}}\right )\right )}}{b d}\right ] \] Input:
integrate(1/(e*x+d)^(1/2)/(c*x^2+b*x),x, algorithm="fricas")
Output:
[(d*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) + sqrt(d)*log((e*x - 2*sqrt(e*x + d)* sqrt(d) + 2*d)/x))/(b*d), -(2*d*sqrt(-c/(c*d - b*e))*arctan(sqrt(e*x + d)* sqrt(-c/(c*d - b*e))) - sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/ x))/(b*d), (d*sqrt(c/(c*d - b*e))*log((c*e*x + 2*c*d - b*e + 2*(c*d - b*e) *sqrt(e*x + d)*sqrt(c/(c*d - b*e)))/(c*x + b)) + 2*sqrt(-d)*arctan(sqrt(-d )/sqrt(e*x + d)))/(b*d), -2*(d*sqrt(-c/(c*d - b*e))*arctan(sqrt(e*x + d)*s qrt(-c/(c*d - b*e))) - sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x + d)))/(b*d)]
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (66) = 132\).
Time = 2.73 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b \sqrt {\frac {b e - c d}{c}}} + \frac {e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}}{\sqrt {d}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(e*x+d)**(1/2)/(c*x**2+b*x),x)
Output:
Piecewise((2*(-e*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*sqrt((b*e - c* d)/c)) + e*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)))/e, Ne(e, 0)), ((-2*c *Piecewise(((b/(2*c) + x)/b, Eq(c, 0)), (-log(b - 2*c*(b/(2*c) + x))/(2*c) , True))/b - 2*c*Piecewise(((b/(2*c) + x)/b, Eq(c, 0)), (log(b + 2*c*(b/(2 *c) + x))/(2*c), True))/b)/sqrt(d), True))
Exception generated. \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x+d)^(1/2)/(c*x^2+b*x),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m ore detail
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=-\frac {2 \, c \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b} + \frac {2 \, \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} \] Input:
integrate(1/(e*x+d)^(1/2)/(c*x^2+b*x),x, algorithm="giac")
Output:
-2*c*arctan(sqrt(e*x + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b) + 2*arctan(sqrt(e*x + d)/sqrt(-d))/(b*sqrt(-d))
Time = 5.27 (sec) , antiderivative size = 625, normalized size of antiderivative = 8.12 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {d+e\,x}}{\sqrt {d}}\right )}{b\,\sqrt {d}}+\frac {\mathrm {atan}\left (\frac {\frac {\left (16\,c^3\,e^2\,\sqrt {d+e\,x}+\frac {\sqrt {c^2\,d-b\,c\,e}\,\left (8\,b^2\,c^2\,e^3+\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\sqrt {c^2\,d-b\,c\,e}\,\sqrt {d+e\,x}}{b^2\,e-b\,c\,d}\right )}{b^2\,e-b\,c\,d}\right )\,\sqrt {c^2\,d-b\,c\,e}\,1{}\mathrm {i}}{b^2\,e-b\,c\,d}+\frac {\left (16\,c^3\,e^2\,\sqrt {d+e\,x}-\frac {\sqrt {c^2\,d-b\,c\,e}\,\left (8\,b^2\,c^2\,e^3-\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\sqrt {c^2\,d-b\,c\,e}\,\sqrt {d+e\,x}}{b^2\,e-b\,c\,d}\right )}{b^2\,e-b\,c\,d}\right )\,\sqrt {c^2\,d-b\,c\,e}\,1{}\mathrm {i}}{b^2\,e-b\,c\,d}}{\frac {\left (16\,c^3\,e^2\,\sqrt {d+e\,x}+\frac {\sqrt {c^2\,d-b\,c\,e}\,\left (8\,b^2\,c^2\,e^3+\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\sqrt {c^2\,d-b\,c\,e}\,\sqrt {d+e\,x}}{b^2\,e-b\,c\,d}\right )}{b^2\,e-b\,c\,d}\right )\,\sqrt {c^2\,d-b\,c\,e}}{b^2\,e-b\,c\,d}-\frac {\left (16\,c^3\,e^2\,\sqrt {d+e\,x}-\frac {\sqrt {c^2\,d-b\,c\,e}\,\left (8\,b^2\,c^2\,e^3-\frac {\left (8\,b^3\,c^2\,e^3-16\,b^2\,c^3\,d\,e^2\right )\,\sqrt {c^2\,d-b\,c\,e}\,\sqrt {d+e\,x}}{b^2\,e-b\,c\,d}\right )}{b^2\,e-b\,c\,d}\right )\,\sqrt {c^2\,d-b\,c\,e}}{b^2\,e-b\,c\,d}}\right )\,\sqrt {c^2\,d-b\,c\,e}\,2{}\mathrm {i}}{b^2\,e-b\,c\,d} \] Input:
int(1/((b*x + c*x^2)*(d + e*x)^(1/2)),x)
Output:
(atan((((16*c^3*e^2*(d + e*x)^(1/2) + ((c^2*d - b*c*e)^(1/2)*(8*b^2*c^2*e^ 3 + ((8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2)*(c^2*d - b*c*e)^(1/2)*(d + e*x)^(1 /2))/(b^2*e - b*c*d)))/(b^2*e - b*c*d))*(c^2*d - b*c*e)^(1/2)*1i)/(b^2*e - b*c*d) + ((16*c^3*e^2*(d + e*x)^(1/2) - ((c^2*d - b*c*e)^(1/2)*(8*b^2*c^2 *e^3 - ((8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2)*(c^2*d - b*c*e)^(1/2)*(d + e*x) ^(1/2))/(b^2*e - b*c*d)))/(b^2*e - b*c*d))*(c^2*d - b*c*e)^(1/2)*1i)/(b^2* e - b*c*d))/(((16*c^3*e^2*(d + e*x)^(1/2) + ((c^2*d - b*c*e)^(1/2)*(8*b^2* c^2*e^3 + ((8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2)*(c^2*d - b*c*e)^(1/2)*(d + e *x)^(1/2))/(b^2*e - b*c*d)))/(b^2*e - b*c*d))*(c^2*d - b*c*e)^(1/2))/(b^2* e - b*c*d) - ((16*c^3*e^2*(d + e*x)^(1/2) - ((c^2*d - b*c*e)^(1/2)*(8*b^2* c^2*e^3 - ((8*b^3*c^2*e^3 - 16*b^2*c^3*d*e^2)*(c^2*d - b*c*e)^(1/2)*(d + e *x)^(1/2))/(b^2*e - b*c*d)))/(b^2*e - b*c*d))*(c^2*d - b*c*e)^(1/2))/(b^2* e - b*c*d)))*(c^2*d - b*c*e)^(1/2)*2i)/(b^2*e - b*c*d) - (2*atanh((d + e*x )^(1/2)/d^(1/2)))/(b*d^(1/2))
Time = 0.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sqrt {d+e x} \left (b x+c x^2\right )} \, dx=\frac {-2 \sqrt {c}\, \sqrt {b e -c d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c}{\sqrt {c}\, \sqrt {b e -c d}}\right ) d +\sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) b e -\sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}-\sqrt {d}\right ) c d -\sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) b e +\sqrt {d}\, \mathrm {log}\left (\sqrt {e x +d}+\sqrt {d}\right ) c d}{b d \left (b e -c d \right )} \] Input:
int(1/(e*x+d)^(1/2)/(c*x^2+b*x),x)
Output:
( - 2*sqrt(c)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c *d)))*d + sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*b*e - sqrt(d)*log(sqrt(d + e*x) - sqrt(d))*c*d - sqrt(d)*log(sqrt(d + e*x) + sqrt(d))*b*e + sqrt(d)*l og(sqrt(d + e*x) + sqrt(d))*c*d)/(b*d*(b*e - c*d))