Integrand size = 25, antiderivative size = 199 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}+\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]
I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c) ^(1/2)*tan(d*x+c)^(1/2)/(I*a-b)^(3/2)/d+I*arctanh((I*a+b)^(1/2)*tan(d*x+c) ^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/(I*a+b)^( 3/2)/d+2*b^2/a/(a^2+b^2)/d/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)
Time = 1.67 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {(-1)^{3/4} (a+i b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} (i a+b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 b^2 \sqrt {\tan (c+d x)}}{a \sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right ) d} \]
-((Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-1)^(3/4)*(a + I*b)*ArcTan[((- 1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqr t[-a + I*b] + ((-1)^(1/4)*(I*a + 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] - (2*b^2*Sqrt[Tan[ c + d*x]])/(a*Sqrt[a + b*Tan[c + d*x]])))/((a^2 + b^2)*d))
Time = 0.99 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4729, 3042, 4052, 27, 3042, 4099, 3042, 4098, 104, 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.
| \(\displaystyle \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4052 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \int \frac {a^2-a b \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a^2-a b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a^2-a b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} a (a-i b) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} a (a-i b) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a+i b) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a (a-i b) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}}{a \left (a^2+b^2\right )}\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a-i b) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}}{a \left (a^2+b^2\right )}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a+i b) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a-i b) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}}{a \left (a^2+b^2\right )}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b^2 \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a-i b) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a (a+i b) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}}{a \left (a^2+b^2\right )}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((a*(a - I*b)*ArcTan[(Sqrt[I*a - b] *Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) + (a*(a + I*b)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] ])/(Sqrt[I*a + b]*d))/(a*(a^2 + b^2)) + (2*b^2*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))
3.9.69.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / ; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^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. \(2630\) vs. \(2(165)=330\).
Time = 40.51 (sec) , antiderivative size = 2631, normalized size of antiderivative = 13.22
1/4/d*(-1/(1-cos(d*x+c))*(csc(d*x+c)*(1-cos(d*x+c))^2-sin(d*x+c)))^(1/2)*( (csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)/(csc(d*x+c )^2*(1-cos(d*x+c))^2-1))^(1/2)*(ln(-1/(1-cos(d*x+c))*(-csc(d*x+c)*a*(1-cos (d*x+c))^2+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))-2*sin(d*x+c)*(-csc(d*x+c)*(csc (d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)) )^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*a^2*(- csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a) *(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2 )-ln(-1/(1-cos(d*x+c))*(-csc(d*x+c)*a*(1-cos(d*x+c))^2+2*(a^2+b^2)^(1/2)*( 1-cos(d*x+c))-2*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2 *b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1 /2)+2*b*(1-cos(d*x+c))+sin(d*x+c)*a))*b^2*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1- cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2 +b^2)^(1/2))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)+(a^2+b^2)^(1/2)*ln(-1/(1-cos (d*x+c))*(-csc(d*x+c)*a*(1-cos(d*x+c))^2+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))- 2*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c) -cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)+2*b*(1-cos (d*x+c))+sin(d*x+c)*a))*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b* (csc(d*x+c)-cot(d*x+c))-a)*(1-cos(d*x+c)))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2) *(-b+(a^2+b^2)^(1/2))^(1/2)*b-ln(1/(1-cos(d*x+c))*(-csc(d*x+c)*a*(1-cos...
Leaf count of result is larger than twice the leaf count of optimal. 7703 vs. \(2 (159) = 318\).
Time = 1.43 (sec) , antiderivative size = 7703, normalized size of antiderivative = 38.71 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\cot {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {\sqrt {\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]