Integrand size = 25, antiderivative size = 132 \[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=-\frac {\sqrt {b} d \arctan \left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}+\frac {\sqrt {b} d \text {arctanh}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}} \]
-d*arctan((b*sin(f*x+e))^(1/2)/b^(1/2))*b^(1/2)*(b*tan(f*x+e))^(1/2)/f/(d* sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1/2)+d*arctanh((b*sin(f*x+e))^(1/2)/b^(1 /2))*b^(1/2)*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^(1/2)/(b*sin(f*x+e))^(1 /2)
Time = 0.55 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.77 \[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=\frac {\left (-\arctan \left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )+\text {arctanh}\left (\frac {\sqrt {\tan (e+f x)}}{\sqrt [4]{\sec ^2(e+f x)}}\right )\right ) \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}{f \sqrt [4]{\sec ^2(e+f x)} \sqrt {\tan (e+f x)}} \]
((-ArcTan[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4)] + ArcTanh[Sqrt[Tan[e + f*x]]/(Sec[e + f*x]^2)^(1/4)])*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]] )/(f*(Sec[e + f*x]^2)^(1/4)*Sqrt[Tan[e + f*x]])
Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.67, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3096, 3042, 3044, 27, 266, 827, 216, 219}
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.
| \(\displaystyle \int \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}dx\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle \frac {d \sqrt {b \tan (e+f x)} \int \sec (e+f x) \sqrt {b \sin (e+f x)}dx}{\sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {d \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{\cos (e+f x)}dx}{\sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3044 |
| \(\displaystyle \frac {d \sqrt {b \tan (e+f x)} \int \frac {b^2 \sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{b f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b d \sqrt {b \tan (e+f x)} \int \frac {\sqrt {b \sin (e+f x)}}{b^2-b^2 \sin ^2(e+f x)}d(b \sin (e+f x))}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2 b d \sqrt {b \tan (e+f x)} \int \frac {b^2 \sin ^2(e+f x)}{b^2-b^4 \sin ^4(e+f x)}d\sqrt {b \sin (e+f x)}}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {2 b d \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sin ^2(e+f x)+b}d\sqrt {b \sin (e+f x)}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {2 b d \sqrt {b \tan (e+f x)} \left (\frac {1}{2} \int \frac {1}{b-b^2 \sin ^2(e+f x)}d\sqrt {b \sin (e+f x)}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 b d \sqrt {b \tan (e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sin (e+f x)\right )}{2 \sqrt {b}}\right )}{f \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
(2*b*d*(-1/2*ArcTan[Sqrt[b]*Sin[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sin[e + f*x]]/(2*Sqrt[b]))*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e + f*x]]*Sqrt[b* Sin[e + f*x]])
3.3.93.3.1 Defintions of rubi rules used
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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[a^(m + n)*((b*Tan[e + f*x])^n/((a*Sec[e + f*x])^n*(b* Sin[e + f*x])^n)) Int[(b*Sin[e + f*x])^n/Cos[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[n + 1/2] && IntegerQ[m + 1/2]
Time = 17.51 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(\frac {\sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, \left (\operatorname {arctanh}\left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )+\arctan \left (\sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )\right )\right ) \cos \left (f x +e \right )}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(129\) |
1/f*(d*sec(f*x+e))^(1/2)*(b*tan(f*x+e))^(1/2)*(arctanh((sin(f*x+e)/(cos(f* x+e)+1)^2)^(1/2)*(cot(f*x+e)+csc(f*x+e)))+arctan((sin(f*x+e)/(cos(f*x+e)+1 )^2)^(1/2)*(cot(f*x+e)+csc(f*x+e))))*cos(f*x+e)/(cos(f*x+e)+1)/(sin(f*x+e) /(cos(f*x+e)+1)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (108) = 216\).
Time = 0.38 (sec) , antiderivative size = 654, normalized size of antiderivative = 4.95 \[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=\left [-\frac {2 \, \sqrt {-b d} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d - {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) - \sqrt {-b d} \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {-b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d + 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right )}{8 \, f}, -\frac {2 \, \sqrt {b d} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (b d \cos \left (f x + e\right )^{2} - b d + {\left (b d \cos \left (f x + e\right ) + b d\right )} \sin \left (f x + e\right )\right )}}\right ) - \sqrt {b d} \log \left (\frac {b d \cos \left (f x + e\right )^{4} - 72 \, b d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \cos \left (f x + e\right )\right )} \sqrt {b d} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 72 \, b d - 28 \, {\left (b d \cos \left (f x + e\right )^{2} - 2 \, b d\right )} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right )}{8 \, f}\right ] \]
[-1/8*(2*sqrt(-b*d)*arctan(1/4*(cos(f*x + e)^3 - 5*cos(f*x + e)^2 - (cos(f *x + e)^2 + 6*cos(f*x + e) + 4)*sin(f*x + e) - 2*cos(f*x + e) + 4)*sqrt(-b *d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b*d*cos(f*x + e)^2 - b*d - (b*d*cos(f*x + e) + b*d)*sin(f*x + e))) - sqrt(-b*d)*log((b*d *cos(f*x + e)^4 - 72*b*d*cos(f*x + e)^2 - 8*(7*cos(f*x + e)^3 - (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(-b*d)*sqrt(b*s in(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)) + 72*b*d + 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 - 4*(cos( f*x + e)^2 - 2)*sin(f*x + e) + 8)))/f, -1/8*(2*sqrt(b*d)*arctan(1/4*(cos(f *x + e)^3 - 5*cos(f*x + e)^2 + (cos(f*x + e)^2 + 6*cos(f*x + e) + 4)*sin(f *x + e) - 2*cos(f*x + e) + 4)*sqrt(b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))* sqrt(d/cos(f*x + e))/(b*d*cos(f*x + e)^2 - b*d + (b*d*cos(f*x + e) + b*d)* sin(f*x + e))) - sqrt(b*d)*log((b*d*cos(f*x + e)^4 - 72*b*d*cos(f*x + e)^2 - 8*(7*cos(f*x + e)^3 + (cos(f*x + e)^3 - 8*cos(f*x + e))*sin(f*x + e) - 8*cos(f*x + e))*sqrt(b*d)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)) + 72*b*d - 28*(b*d*cos(f*x + e)^2 - 2*b*d)*sin(f*x + e))/(cos(f*x + e)^4 - 8*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - 2)*sin(f*x + e) + 8)))/f]
\[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=\int \sqrt {b \tan {\left (e + f x \right )}} \sqrt {d \sec {\left (e + f x \right )}}\, dx \]
\[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=\int { \sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} \,d x } \]
\[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=\int { \sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} \,d x } \]
Timed out. \[ \int \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)} \, dx=\int \sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}} \,d x \]