Integrand size = 35, antiderivative size = 215 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}+\frac {2 b (A b-a B)}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]
(I*A-B)*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*A+B)*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*b-B*a)/a/(a^2+b^2)/d/cot(d*x+c)^(1/2)/(a+b*tan(d *x+c))^(1/2)
Time = 1.30 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {\sqrt [4]{-1} (-i a+b) (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} (a-i b) (-i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a \sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right ) d} \]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-1)^(1/4)*((-I)*a + b)*(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] + ((-1)^(1/4)*(a - I*b)*((-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*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(a*Sqrt[a + b*Tan[c + d*x]])) )/((a^2 + b^2)*d)
Time = 1.18 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4729, 3042, 4092, 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.
\(\displaystyle \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4729 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4092 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \int \frac {a (a A+b B)-a (A b-a B) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a (a A+b B)-a (A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \frac {a (a A+b B)-a (A b-a B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}\right )\) |
\(\Big \downarrow \) 4099 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} a (a-i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} a (a-i b) (A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (a+i b) (A-i B) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 4098 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a+i b) (A-i B) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {a (a-i b) (A+i B) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}}{a \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a-i b) (A+i B) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a+i b) (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}}{a \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a+i b) (A-i B) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {a (a-i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}}{a \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {a (a-i b) (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}+\frac {a (a+i b) (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}}{a \left (a^2+b^2\right )}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((a*(a - I*b)*(A + I*B)*ArcTan[(Sqr t[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d ) + (a*(a + I*b)*(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*(A*b - a* B)*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))
3.7.44.3.1 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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) /(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 )) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(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]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 1.96 (sec) , antiderivative size = 1561166, normalized size of antiderivative = 7261.24
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 18703 vs. \(2 (175) = 350\).
Time = 7.11 (sec) , antiderivative size = 18703, normalized size of antiderivative = 86.99 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\cot {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]