Integrand size = 25, antiderivative size = 211 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d} \]
arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/2 )*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+2*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/( a+b*tan(d*x+c))^(1/2))*b^(1/2)*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-arctanh ((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a+b)^(1/2)*cot( d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d
Time = 0.88 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\sqrt [4]{-1} \left (\sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )+\frac {2 \sqrt {a} \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((-1)^(1/4)*(Sqrt[-a + I*b]*ArcTan[ ((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + Sqrt[a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) + (2*Sqrt[a]*Sqrt[b]*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d* x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a + b*Tan[c + d*x]]))/d
Time = 0.69 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4729, 3042, 4058, 609, 65, 219, 2035, 2257, 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.
\(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{\tan ^2(c+d x)+1}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 609 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (b \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 b \int \frac {1}{1-\frac {b \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\int \frac {b-a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \int \frac {b-a \tan (c+d x)}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 2257 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \int \left (\frac {i (b-i a)}{2 (i-\tan (c+d x)) \sqrt {a+b \tan (c+d x)}}-\frac {i (-i a-b)}{2 (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-2 \left (\frac {1}{2} \sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {1}{2} \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )\right )}{d}\) |
((2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 2*(-1/2*(Sqrt[I*a - b]*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]) + (Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/2))*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x] ])/d
3.9.44.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 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[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_S ymbol] :> Simp[d*(e/b) Int[(e*x)^(m - 1)*(c + d*x)^(n - 1), x], x] - Simp [e/b Int[(e*x)^(m - 1)*(c + d*x)^(n - 1)*((a*d - b*c*x)/(a + b*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[0, n, 1] && LtQ[0, m, 1] && !Integ erQ[m] && !IntegerQ[n]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol ] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a , c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a *d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(1113\) vs. \(2(171)=342\).
Time = 35.98 (sec) , antiderivative size = 1114, normalized size of antiderivative = 5.28
-1/4/d*(a+b*tan(d*x+c))^(1/2)*(-2^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2 +b^2)^(1/2))^(1/2)*ln(-(cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)+2*sin(d*x+c )*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2*a-2*b *csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/2))^(1 /2)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc(d*x+c)*a- 2*(a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))+2^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*( -b+(a^2+b^2)^(1/2))^(1/2)*ln((cos(d*x+c)*cot(d*x+c)*a-2*a*cot(d*x+c)-2*sin (d*x+c)*(csc(d*x+c)*(cot(d*x+c)^2*a-2*cot(d*x+c)*csc(d*x+c)*a+csc(d*x+c)^2 *a-2*b*csc(d*x+c)+2*cot(d*x+c)*b-a)*(cos(d*x+c)-1))^(1/2)*(b+(a^2+b^2)^(1/ 2))^(1/2)+2*(a^2+b^2)^(1/2)*cos(d*x+c)+2*b*cos(d*x+c)-sin(d*x+c)*a+csc(d*x +c)*a-2*(a^2+b^2)^(1/2)-2*b)/(cos(d*x+c)-1))+2*(a^2+b^2)^(1/2)*2^(1/2)*arc tan(((b+(a^2+b^2)^(1/2))^(1/2)*cos(d*x+c)-(-2*(cos(d*x+c)^2*b-sin(d*x+c)*c os(d*x+c)*a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2 ))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))-2*(a^2+b^2)^(1/2)*2^(1/2)*ar ctan(((b+(a^2+b^2)^(1/2))^(1/2)*cos(d*x+c)+(-2*(cos(d*x+c)^2*b-sin(d*x+c)* cos(d*x+c)*a-b)/(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/ 2))/(cos(d*x+c)-1)/(-b+(a^2+b^2)^(1/2))^(1/2))-2*2^(1/2)*arctan(((b+(a^2+b ^2)^(1/2))^(1/2)*cos(d*x+c)-(-2*(cos(d*x+c)^2*b-sin(d*x+c)*cos(d*x+c)*a-b) /(cos(d*x+c)+1)^2)^(1/2)*sin(d*x+c)-(b+(a^2+b^2)^(1/2))^(1/2))/(cos(d*x+c) -1)/(-b+(a^2+b^2)^(1/2))^(1/2))*b+2*2^(1/2)*arctan(((b+(a^2+b^2)^(1/2))...
Leaf count of result is larger than twice the leaf count of optimal. 2838 vs. \(2 (167) = 334\).
Time = 0.63 (sec) , antiderivative size = 5709, normalized size of antiderivative = 27.06 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\sqrt {a + b \tan {\left (c + d x \right )}}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\int { \frac {\sqrt {b \tan \left (d x + c\right ) + a}}{\sqrt {\cot \left (d x + c\right )}} \,d x } \]
Exception generated. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(con st gen &
Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {\cot (c+d x)}} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]