Integrand size = 28, antiderivative size = 110 \[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=-\frac {2 \sqrt {\frac {\sqrt {-c} (f+g x)}{\sqrt {-c} f+g}} \operatorname {EllipticPi}\left (\frac {2 e}{\sqrt {-c} d+e},\arcsin \left (\frac {\sqrt {1-\sqrt {-c} x}}{\sqrt {2}}\right ),\frac {2 g}{\sqrt {-c} f+g}\right )}{\left (\sqrt {-c} d+e\right ) \sqrt {f+g x}} \]
-2*EllipticPi(1/2*(1-x*(-c)^(1/2))^(1/2)*2^(1/2),2*e/(e+d*(-c)^(1/2)),2^(1 /2)*(g/(g+f*(-c)^(1/2)))^(1/2))*((g*x+f)*(-c)^(1/2)/(g+f*(-c)^(1/2)))^(1/2 )/(e+d*(-c)^(1/2))/(g*x+f)^(1/2)
Result contains complex when optimal does not.
Time = 22.00 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.37 \[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=-\frac {2 i \sqrt {\frac {g \left (\frac {i}{\sqrt {c}}+x\right )}{f+g x}} \sqrt {-\frac {\frac {i g}{\sqrt {c}}-g x}{f+g x}} (f+g x) \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-f-\frac {i g}{\sqrt {c}}}}{\sqrt {f+g x}}\right ),\frac {\sqrt {c} f-i g}{\sqrt {c} f+i g}\right )-\operatorname {EllipticPi}\left (\frac {\sqrt {c} (e f-d g)}{e \left (\sqrt {c} f+i g\right )},i \text {arcsinh}\left (\frac {\sqrt {-f-\frac {i g}{\sqrt {c}}}}{\sqrt {f+g x}}\right ),\frac {\sqrt {c} f-i g}{\sqrt {c} f+i g}\right )\right )}{\sqrt {-f-\frac {i g}{\sqrt {c}}} (e f-d g) \sqrt {1+c x^2}} \]
((-2*I)*Sqrt[(g*(I/Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*g)/Sqrt[c] - g*x)/( f + g*x))]*(f + g*x)*(EllipticF[I*ArcSinh[Sqrt[-f - (I*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*g)/(Sqrt[c]*f + I*g)] - EllipticPi[(Sqrt[c]*(e*f - d*g))/(e*(Sqrt[c]*f + I*g)), I*ArcSinh[Sqrt[-f - (I*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*g)/(Sqrt[c]*f + I*g)]))/(Sqrt[-f - (I*g)/Sqrt[c]]*(e *f - d*g)*Sqrt[1 + c*x^2])
Leaf count is larger than twice the leaf count of optimal. \(826\) vs. \(2(110)=220\).
Time = 1.64 (sec) , antiderivative size = 826, normalized size of antiderivative = 7.51, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {729, 25, 1540, 1416, 2222}
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.
\(\displaystyle \int \frac {1}{\sqrt {c x^2+1} (d+e x) \sqrt {f+g x}} \, dx\) |
\(\Big \downarrow \) 729 |
\(\displaystyle 2 \int -\frac {1}{(e f-d g-e (f+g x)) \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}d\sqrt {f+g x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {1}{(e f-d g-e (f+g x)) \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}d\sqrt {f+g x}\) |
\(\Big \downarrow \) 1540 |
\(\displaystyle 2 \left (\frac {e \sqrt {c f^2+g^2} \left (\sqrt {c} (e f-d g)-e \sqrt {c f^2+g^2}\right ) \int \frac {\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1}{(e f-d g-e (f+g x)) \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}d\sqrt {f+g x}}{g \left (c d (2 e f-d g)+e^2 g\right )}-\frac {\sqrt {c} \left (-\sqrt {c} \sqrt {c f^2+g^2} (e f-d g)+c e f^2+e g^2\right ) \int \frac {1}{\sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}d\sqrt {f+g x}}{g \sqrt {c f^2+g^2} \left (c d (2 e f-d g)+e^2 g\right )}\right )\) |
\(\Big \downarrow \) 1416 |
\(\displaystyle 2 \left (\frac {e \sqrt {c f^2+g^2} \left (\sqrt {c} (e f-d g)-e \sqrt {c f^2+g^2}\right ) \int \frac {\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1}{(e f-d g-e (f+g x)) \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}d\sqrt {f+g x}}{g \left (c d (2 e f-d g)+e^2 g\right )}-\frac {\sqrt [4]{c} \left (\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1\right ) \sqrt {\frac {\frac {c f^2}{g^2}-\frac {2 c f (f+g x)}{g^2}+\frac {c (f+g x)^2}{g^2}+1}{\left (\frac {c f^2}{g^2}+1\right ) \left (\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1\right )^2}} \left (-\sqrt {c} \sqrt {c f^2+g^2} (e f-d g)+c e f^2+e g^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {f+g x}}{\sqrt [4]{c f^2+g^2}}\right ),\frac {1}{2} \left (\frac {\sqrt {c} f}{\sqrt {c f^2+g^2}}+1\right )\right )}{2 g \sqrt [4]{c f^2+g^2} \sqrt {\frac {c f^2}{g^2}-\frac {2 c f (f+g x)}{g^2}+\frac {c (f+g x)^2}{g^2}+1} \left (c d (2 e f-d g)+e^2 g\right )}\right )\) |
\(\Big \downarrow \) 2222 |
\(\displaystyle 2 \left (\frac {e \sqrt {c f^2+g^2} \left (\sqrt {c} (e f-d g)-e \sqrt {c f^2+g^2}\right ) \left (\frac {\left (e+\frac {\sqrt {c} (e f-d g)}{\sqrt {c f^2+g^2}}\right ) \text {arctanh}\left (\frac {\sqrt {c d^2+e^2} \sqrt {f+g x}}{\sqrt {e} \sqrt {e f-d g} \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}\right )}{2 \sqrt {e} \sqrt {c d^2+e^2} \sqrt {e f-d g}}-\frac {\left (\frac {\sqrt {c}}{e}-\frac {\sqrt {c f^2+g^2}}{e f-d g}\right ) \left (\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1\right ) \sqrt {\frac {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}{\left (\frac {c f^2}{g^2}+1\right ) \left (\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1\right )^2}} \operatorname {EllipticPi}\left (\frac {\left (\sqrt {c f^2+g^2} e+\sqrt {c} (e f-d g)\right )^2}{4 \sqrt {c} e (e f-d g) \sqrt {c f^2+g^2}},2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {f+g x}}{\sqrt [4]{c f^2+g^2}}\right ),\frac {1}{2} \left (\frac {\sqrt {c} f}{\sqrt {c f^2+g^2}}+1\right )\right )}{4 \sqrt [4]{c} \sqrt [4]{c f^2+g^2} \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}\right )}{g \left (g e^2+c d (2 e f-d g)\right )}-\frac {\sqrt [4]{c} \left (c e f^2+e g^2-\sqrt {c} (e f-d g) \sqrt {c f^2+g^2}\right ) \left (\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1\right ) \sqrt {\frac {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}{\left (\frac {c f^2}{g^2}+1\right ) \left (\frac {\sqrt {c} (f+g x)}{\sqrt {c f^2+g^2}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {f+g x}}{\sqrt [4]{c f^2+g^2}}\right ),\frac {1}{2} \left (\frac {\sqrt {c} f}{\sqrt {c f^2+g^2}}+1\right )\right )}{2 g \sqrt [4]{c f^2+g^2} \left (g e^2+c d (2 e f-d g)\right ) \sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+1}}\right )\) |
2*(-1/2*(c^(1/4)*(c*e*f^2 + e*g^2 - Sqrt[c]*(e*f - d*g)*Sqrt[c*f^2 + g^2]) *(1 + (Sqrt[c]*(f + g*x))/Sqrt[c*f^2 + g^2])*Sqrt[(1 + (c*f^2)/g^2 - (2*c* f*(f + g*x))/g^2 + (c*(f + g*x)^2)/g^2)/((1 + (c*f^2)/g^2)*(1 + (Sqrt[c]*( f + g*x))/Sqrt[c*f^2 + g^2])^2)]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[f + g*x] )/(c*f^2 + g^2)^(1/4)], (1 + (Sqrt[c]*f)/Sqrt[c*f^2 + g^2])/2])/(g*(c*f^2 + g^2)^(1/4)*(e^2*g + c*d*(2*e*f - d*g))*Sqrt[1 + (c*f^2)/g^2 - (2*c*f*(f + g*x))/g^2 + (c*(f + g*x)^2)/g^2]) + (e*Sqrt[c*f^2 + g^2]*(Sqrt[c]*(e*f - d*g) - e*Sqrt[c*f^2 + g^2])*(((e + (Sqrt[c]*(e*f - d*g))/Sqrt[c*f^2 + g^2 ])*ArcTanh[(Sqrt[c*d^2 + e^2]*Sqrt[f + g*x])/(Sqrt[e]*Sqrt[e*f - d*g]*Sqrt [1 + (c*f^2)/g^2 - (2*c*f*(f + g*x))/g^2 + (c*(f + g*x)^2)/g^2])])/(2*Sqrt [e]*Sqrt[c*d^2 + e^2]*Sqrt[e*f - d*g]) - ((Sqrt[c]/e - Sqrt[c*f^2 + g^2]/( e*f - d*g))*(1 + (Sqrt[c]*(f + g*x))/Sqrt[c*f^2 + g^2])*Sqrt[(1 + (c*f^2)/ g^2 - (2*c*f*(f + g*x))/g^2 + (c*(f + g*x)^2)/g^2)/((1 + (c*f^2)/g^2)*(1 + (Sqrt[c]*(f + g*x))/Sqrt[c*f^2 + g^2])^2)]*EllipticPi[(Sqrt[c]*(e*f - d*g ) + e*Sqrt[c*f^2 + g^2])^2/(4*Sqrt[c]*e*(e*f - d*g)*Sqrt[c*f^2 + g^2]), 2* ArcTan[(c^(1/4)*Sqrt[f + g*x])/(c*f^2 + g^2)^(1/4)], (1 + (Sqrt[c]*f)/Sqrt [c*f^2 + g^2])/2])/(4*c^(1/4)*(c*f^2 + g^2)^(1/4)*Sqrt[1 + (c*f^2)/g^2 - ( 2*c*f*(f + g*x))/g^2 + (c*(f + g*x)^2)/g^2])))/(g*(e^2*g + c*d*(2*e*f - d* g))))
3.7.54.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))*Sqrt[(a_) + (b_.)*(x_) ^2]), x_Symbol] :> Simp[2 Subst[Int[1/((d*e - c*f + f*x^2)*Sqrt[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^2/d^2) + b*(x^4/d^2)]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c /a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ (2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) ], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_S ymbol] :> With[{q = Rt[c/a, 2]}, Simp[(c*d + a*e*q)/(c*d^2 - a*e^2) Int[1 /Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[(a*e*(e + d*q))/(c*d^2 - a*e^2) I nt[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A rcTanh[Rt[b - c*(d/e) - a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ b - c*(d/e) - a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*Ell ipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && NegQ[-b + c*(d/e) + a*(e/d)]
Leaf count of result is larger than twice the leaf count of optimal. \(214\) vs. \(2(95)=190\).
Time = 1.81 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.95
method | result | size |
default | \(\frac {2 \left (g +f \sqrt {-c}\right ) \Pi \left (\sqrt {\frac {\left (g x +f \right ) \sqrt {-c}}{g +f \sqrt {-c}}}, -\frac {\left (g +f \sqrt {-c}\right ) e}{\sqrt {-c}\, \left (d g -e f \right )}, \sqrt {\frac {g +f \sqrt {-c}}{f \sqrt {-c}-g}}\right ) \sqrt {-\frac {\left (x \sqrt {-c}-1\right ) g}{g +f \sqrt {-c}}}\, \sqrt {-\frac {\left (x \sqrt {-c}+1\right ) g}{f \sqrt {-c}-g}}\, \sqrt {\frac {\left (g x +f \right ) \sqrt {-c}}{g +f \sqrt {-c}}}\, \sqrt {c \,x^{2}+1}\, \sqrt {g x +f}}{\sqrt {-c}\, \left (d g -e f \right ) \left (c g \,x^{3}+c f \,x^{2}+g x +f \right )}\) | \(215\) |
elliptic | \(\frac {2 \sqrt {\left (g x +f \right ) \left (c \,x^{2}+1\right )}\, \left (\frac {f}{g}+\frac {1}{\sqrt {-c}}\right ) \sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}+\frac {1}{\sqrt {-c}}}}\, \sqrt {\frac {x +\frac {1}{\sqrt {-c}}}{-\frac {f}{g}+\frac {1}{\sqrt {-c}}}}\, \sqrt {\frac {x -\frac {1}{\sqrt {-c}}}{-\frac {f}{g}-\frac {1}{\sqrt {-c}}}}\, \Pi \left (\sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}+\frac {1}{\sqrt {-c}}}}, \frac {-\frac {f}{g}-\frac {1}{\sqrt {-c}}}{-\frac {f}{g}+\frac {d}{e}}, \sqrt {\frac {-\frac {f}{g}-\frac {1}{\sqrt {-c}}}{-\frac {f}{g}+\frac {1}{\sqrt {-c}}}}\right )}{\sqrt {g x +f}\, \sqrt {c \,x^{2}+1}\, e \sqrt {c g \,x^{3}+c f \,x^{2}+g x +f}\, \left (-\frac {f}{g}+\frac {d}{e}\right )}\) | \(240\) |
2*(g+f*(-c)^(1/2))/(-c)^(1/2)*EllipticPi(((g*x+f)*(-c)^(1/2)/(g+f*(-c)^(1/ 2)))^(1/2),-(g+f*(-c)^(1/2))*e/(-c)^(1/2)/(d*g-e*f),((g+f*(-c)^(1/2))/(f*( -c)^(1/2)-g))^(1/2))*(-(x*(-c)^(1/2)-1)*g/(g+f*(-c)^(1/2)))^(1/2)*(-(x*(-c )^(1/2)+1)*g/(f*(-c)^(1/2)-g))^(1/2)*((g*x+f)*(-c)^(1/2)/(g+f*(-c)^(1/2))) ^(1/2)*(c*x^2+1)^(1/2)*(g*x+f)^(1/2)/(d*g-e*f)/(c*g*x^3+c*f*x^2+g*x+f)
Timed out. \[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=\int \frac {1}{\left (d + e x\right ) \sqrt {f + g x} \sqrt {c x^{2} + 1}}\, dx \]
\[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + 1} {\left (e x + d\right )} \sqrt {g x + f}} \,d x } \]
\[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + 1} {\left (e x + d\right )} \sqrt {g x + f}} \,d x } \]
Timed out. \[ \int \frac {1}{(d+e x) \sqrt {f+g x} \sqrt {1+c x^2}} \, dx=\int \frac {1}{\sqrt {f+g\,x}\,\sqrt {c\,x^2+1}\,\left (d+e\,x\right )} \,d x \]