Integrand size = 25, antiderivative size = 189 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}-\frac {2 b}{\left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]
-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+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/(a^2+b^2)/d/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)
Time = 2.05 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}+\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{3/2}}-\frac {2 b \sqrt {\tan (c+d x)}}{\left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )}{d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqr t[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a + I*b)^(3/2 ) + ((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[ a + b*Tan[c + d*x]]])/(a + I*b)^(3/2) - (2*b*Sqrt[Tan[c + d*x]])/((a^2 + b ^2)*Sqrt[a + b*Tan[c + d*x]])))/d
Time = 0.94 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.04, 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, 4051, 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.
| \(\displaystyle \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\sqrt {\tan (c+d x)}}{(a+b \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4051 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 \int -\frac {b+a \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {b+a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {b+a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (-b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (-b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {(-b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}}{a^2+b^2}\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(b+i a) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {(-b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}}{a^2+b^2}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(b+i a) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(-b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}}{a^2+b^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {2 b \sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(b+i a) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(-b+i a) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}}{a^2+b^2}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((((I*a + b)*ArcTan[(Sqrt[I*a - b]*S qrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) - ((I*a - b)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/( Sqrt[I*a + b]*d))/(a^2 + b^2) - (2*b*Sqrt[Tan[c + d*x]])/((a^2 + b^2)*d*Sq rt[a + b*Tan[c + d*x]]))
3.9.70.3.1 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*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(a^2 + b^2 )) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c *(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && GtQ[n, 0] && Int egerQ[2*m]
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. \(2531\) vs. \(2(157)=314\).
Time = 40.45 (sec) , antiderivative size = 2532, normalized size of antiderivative = 13.40
-1/4/d*((csc(d*x+c)^2*a*(1-cos(d*x+c))^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)/(c sc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(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)-(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*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*(a^2+b^2)^(1/2)*(1-cos(d*x+c))-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 )-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*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-c os(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)+2*ln(-1/(1-cos(d*x+c))*(csc(d *x+c)*a*(1-cos(d*x+c))^2+2*sin(d*x+c)*(-csc(d*x+c)*(csc(d*x+c)^2*a*(1-c...
Leaf count of result is larger than twice the leaf count of optimal. 7585 vs. \(2 (153) = 306\).
Time = 1.41 (sec) , antiderivative size = 7585, normalized size of antiderivative = 40.13 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cot {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\cot \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]