Integrand size = 14, antiderivative size = 78 \[ \int \sqrt {3+4 \tan (e+f x)} \, dx=-\frac {2 \arctan \left (2-\sqrt {3+4 \tan (e+f x)}\right )}{f}+\frac {2 \arctan \left (2+\sqrt {3+4 \tan (e+f x)}\right )}{f}-\frac {\text {arctanh}\left (\frac {\sqrt {3+4 \tan (e+f x)}}{2+\tan (e+f x)}\right )}{f} \] Output:
2*arctan(-2+(3+4*tan(f*x+e))^(1/2))/f+2*arctan(2+(3+4*tan(f*x+e))^(1/2))/f -arctanh((3+4*tan(f*x+e))^(1/2)/(2+tan(f*x+e)))/f
Result contains complex when optimal does not.
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\frac {(2+i) \arctan \left (\left (\frac {1}{5}+\frac {2 i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )}{f}-\frac {(1+2 i) \text {arctanh}\left (\left (\frac {2}{5}+\frac {i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )}{f} \] Input:
Integrate[Sqrt[3 + 4*Tan[e + f*x]],x]
Output:
((2 + I)*ArcTan[(1/5 + (2*I)/5)*Sqrt[3 + 4*Tan[e + f*x]]])/f - ((1 + 2*I)* ArcTanh[(2/5 + I/5)*Sqrt[3 + 4*Tan[e + f*x]]])/f
Time = 0.35 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 3966, 483, 1447, 1475, 1083, 217, 1478, 27, 1103}
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.
\(\displaystyle \int \sqrt {4 \tan (e+f x)+3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {4 \tan (e+f x)+3}dx\) |
\(\Big \downarrow \) 3966 |
\(\displaystyle \frac {4 \int \frac {\sqrt {4 \tan (e+f x)+3}}{16 \tan ^2(e+f x)+16}d(4 \tan (e+f x))}{f}\) |
\(\Big \downarrow \) 483 |
\(\displaystyle \frac {8 \int \frac {16 \tan ^2(e+f x)}{256 \tan ^4(e+f x)-96 \tan ^2(e+f x)+25}d\sqrt {4 \tan (e+f x)+3}}{f}\) |
\(\Big \downarrow \) 1447 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \int \frac {16 \tan ^2(e+f x)+5}{256 \tan ^4(e+f x)-96 \tan ^2(e+f x)+25}d\sqrt {4 \tan (e+f x)+3}-\frac {1}{2} \int \frac {5-16 \tan ^2(e+f x)}{256 \tan ^4(e+f x)-96 \tan ^2(e+f x)+25}d\sqrt {4 \tan (e+f x)+3}\right )}{f}\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{16 \tan ^2(e+f x)-16 \tan (e+f x)+5}d\sqrt {4 \tan (e+f x)+3}+\frac {1}{2} \int \frac {1}{16 \tan ^2(e+f x)+16 \tan (e+f x)+5}d\sqrt {4 \tan (e+f x)+3}\right )-\frac {1}{2} \int \frac {5-16 \tan ^2(e+f x)}{256 \tan ^4(e+f x)-96 \tan ^2(e+f x)+25}d\sqrt {4 \tan (e+f x)+3}\right )}{f}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \left (-\int \frac {1}{-16 \tan ^2(e+f x)-4}d(8 \tan (e+f x)-4)-\int \frac {1}{-16 \tan ^2(e+f x)-4}d(8 \tan (e+f x)+4)\right )-\frac {1}{2} \int \frac {5-16 \tan ^2(e+f x)}{256 \tan ^4(e+f x)-96 \tan ^2(e+f x)+25}d\sqrt {4 \tan (e+f x)+3}\right )}{f}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \arctan (2 \tan (e+f x))-\frac {1}{2} \int \frac {5-16 \tan ^2(e+f x)}{256 \tan ^4(e+f x)-96 \tan ^2(e+f x)+25}d\sqrt {4 \tan (e+f x)+3}\right )}{f}\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \left (\frac {1}{8} \int -\frac {2 (2-4 \tan (e+f x))}{16 \tan ^2(e+f x)-16 \tan (e+f x)+5}d\sqrt {4 \tan (e+f x)+3}+\frac {1}{8} \int -\frac {2 \left (\sqrt {4 \tan (e+f x)+3}+2\right )}{16 \tan ^2(e+f x)+16 \tan (e+f x)+5}d\sqrt {4 \tan (e+f x)+3}\right )+\frac {1}{2} \arctan (2 \tan (e+f x))\right )}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \left (-\frac {1}{4} \int \frac {2-4 \tan (e+f x)}{16 \tan ^2(e+f x)-16 \tan (e+f x)+5}d\sqrt {4 \tan (e+f x)+3}-\frac {1}{4} \int \frac {\sqrt {4 \tan (e+f x)+3}+2}{16 \tan ^2(e+f x)+16 \tan (e+f x)+5}d\sqrt {4 \tan (e+f x)+3}\right )+\frac {1}{2} \arctan (2 \tan (e+f x))\right )}{f}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {8 \left (\frac {1}{2} \arctan (2 \tan (e+f x))+\frac {1}{2} \left (\frac {1}{8} \log \left (16 \tan ^2(e+f x)-16 \tan (e+f x)+5\right )-\frac {1}{8} \log \left (16 \tan ^2(e+f x)+16 \tan (e+f x)+5\right )\right )\right )}{f}\) |
Input:
Int[Sqrt[3 + 4*Tan[e + f*x]],x]
Output:
(8*(ArcTan[2*Tan[e + f*x]]/2 + (Log[5 - 16*Tan[e + f*x] + 16*Tan[e + f*x]^ 2]/8 - Log[5 + 16*Tan[e + f*x] + 16*Tan[e + f*x]^2]/8)/2))/f
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[Sqrt[(c_) + (d_.)*(x_)]/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[2*d Subst[Int[x^2/(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4), x], x, Sqrt[c + d*x]], x ] /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a/c, 2]}, Simp[1/2 Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b ^2 - 4*a*c, 0] && PosQ[a*c]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^ 2, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ [c*d^2 - a*e^2, 0] && !GtQ[b^2 - 4*a*c, 0]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Su bst[Int[(a + x)^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && NeQ[a^2 + b^2, 0]
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+2 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )-\frac {\ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+2 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{f}\) | \(94\) |
default | \(\frac {\frac {\ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+2 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )-\frac {\ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+2 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{f}\) | \(94\) |
Input:
int((3+4*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/f*(1/2*ln(8+4*tan(f*x+e)-4*(3+4*tan(f*x+e))^(1/2))+2*arctan(-2+(3+4*tan( f*x+e))^(1/2))-1/2*ln(8+4*tan(f*x+e)+4*(3+4*tan(f*x+e))^(1/2))+2*arctan(2+ (3+4*tan(f*x+e))^(1/2)))
Time = 0.10 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\frac {4 \, \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 4 \, \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) - \log \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) + \log \left (-\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right )}{2 \, f} \] Input:
integrate((3+4*tan(f*x+e))^(1/2),x, algorithm="fricas")
Output:
1/2*(4*arctan(sqrt(4*tan(f*x + e) + 3) + 2) + 4*arctan(sqrt(4*tan(f*x + e) + 3) - 2) - log(sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) + log(-sqrt( 4*tan(f*x + e) + 3) + tan(f*x + e) + 2))/f
\[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\int \sqrt {4 \tan {\left (e + f x \right )} + 3}\, dx \] Input:
integrate((3+4*tan(f*x+e))**(1/2),x)
Output:
Integral(sqrt(4*tan(e + f*x) + 3), x)
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.18 \[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\frac {4 \, \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 4 \, \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) - \log \left (4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right ) + \log \left (-4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right )}{2 \, f} \] Input:
integrate((3+4*tan(f*x+e))^(1/2),x, algorithm="maxima")
Output:
1/2*(4*arctan(sqrt(4*tan(f*x + e) + 3) + 2) + 4*arctan(sqrt(4*tan(f*x + e) + 3) - 2) - log(4*sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8) + log(-4 *sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8))/f
Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.29 \[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\frac {2 \, \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right )}{f} + \frac {2 \, \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right )}{f} - \frac {\log \left (4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right )}{2 \, f} + \frac {\log \left (-4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right )}{2 \, f} \] Input:
integrate((3+4*tan(f*x+e))^(1/2),x, algorithm="giac")
Output:
2*arctan(sqrt(4*tan(f*x + e) + 3) + 2)/f + 2*arctan(sqrt(4*tan(f*x + e) + 3) - 2)/f - 1/2*log(4*sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8)/f + 1 /2*log(-4*sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8)/f
Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.63 \[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\frac {\mathrm {atan}\left (\sqrt {4\,\mathrm {tan}\left (e+f\,x\right )+3}\,\left (\frac {1}{5}-\frac {2}{5}{}\mathrm {i}\right )\right )\,\left (2-\mathrm {i}\right )}{f}+\frac {\mathrm {atan}\left (\sqrt {4\,\mathrm {tan}\left (e+f\,x\right )+3}\,\left (\frac {1}{5}+\frac {2}{5}{}\mathrm {i}\right )\right )\,\left (2+1{}\mathrm {i}\right )}{f} \] Input:
int((4*tan(e + f*x) + 3)^(1/2),x)
Output:
(atan((4*tan(e + f*x) + 3)^(1/2)*(1/5 - 2i/5))*(2 - 1i))/f + (atan((4*tan( e + f*x) + 3)^(1/2)*(1/5 + 2i/5))*(2 + 1i))/f
\[ \int \sqrt {3+4 \tan (e+f x)} \, dx=\int \sqrt {4 \tan \left (f x +e \right )+3}d x \] Input:
int((3+4*tan(f*x+e))^(1/2),x)
Output:
int(sqrt(4*tan(e + f*x) + 3),x)