Integrand size = 36, antiderivative size = 138 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {B \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {B \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {B \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {B \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]
-1/2*B*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/2*B*arctan(1+2^(1/2 )*cot(d*x+c)^(1/2))/d*2^(1/2)+1/4*B*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/ 2))/d*2^(1/2)-1/4*B*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)
Time = 0.06 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.80 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {B \left (2 \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right )}{2 \sqrt {2} d} \]
(B*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[C ot[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))/(2*Sqrt[2]*d)
Time = 0.36 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2011, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle B \int \frac {1}{\sqrt {\cot (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle B \int \frac {1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle -\frac {B \int \frac {1}{\sqrt {\cot (c+d x)} \left (\cot ^2(c+d x)+1\right )}d\cot (c+d x)}{d}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {2 B \int \frac {1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {\cot (c+d x)+1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}\right )}{d}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )\right )}{d}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \left (\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\cot (c+d x)}+1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -\frac {2 B \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}\) |
(-2*B*((-(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqr t[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d *x]] + Cot[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*Sqrt[2]))/2))/d
3.7.9.3.1 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[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d Subst[Int [x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(-\frac {B \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{4 d}\) | \(90\) |
-1/4*B/d*2^(1/2)*(ln((1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c) -2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan( -1+2^(1/2)*cot(d*x+c)^(1/2)))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.80 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {1}{2} \, \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B^{3} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + {\left (d^{3} \cos \left (2 \, d x + 2 \, c\right ) + d^{3}\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) - \frac {1}{2} \, \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B^{3} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - {\left (d^{3} \cos \left (2 \, d x + 2 \, c\right ) + d^{3}\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) + \frac {1}{2} i \, \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B^{3} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - {\left (i \, d^{3} \cos \left (2 \, d x + 2 \, c\right ) + i \, d^{3}\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) - \frac {1}{2} i \, \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {B^{3} \sqrt {\frac {\cos \left (2 \, d x + 2 \, c\right ) + 1}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) - {\left (-i \, d^{3} \cos \left (2 \, d x + 2 \, c\right ) - i \, d^{3}\right )} \left (-\frac {B^{4}}{d^{4}}\right )^{\frac {3}{4}}}{\cos \left (2 \, d x + 2 \, c\right ) + 1}\right ) \]
1/2*(-B^4/d^4)^(1/4)*log((B^3*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c) )*sin(2*d*x + 2*c) + (d^3*cos(2*d*x + 2*c) + d^3)*(-B^4/d^4)^(3/4))/(cos(2 *d*x + 2*c) + 1)) - 1/2*(-B^4/d^4)^(1/4)*log((B^3*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) - (d^3*cos(2*d*x + 2*c) + d^3)*(-B^ 4/d^4)^(3/4))/(cos(2*d*x + 2*c) + 1)) + 1/2*I*(-B^4/d^4)^(1/4)*log((B^3*sq rt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) - (I*d^3*cos( 2*d*x + 2*c) + I*d^3)*(-B^4/d^4)^(3/4))/(cos(2*d*x + 2*c) + 1)) - 1/2*I*(- B^4/d^4)^(1/4)*log((B^3*sqrt((cos(2*d*x + 2*c) + 1)/sin(2*d*x + 2*c))*sin( 2*d*x + 2*c) - (-I*d^3*cos(2*d*x + 2*c) - I*d^3)*(-B^4/d^4)^(3/4))/(cos(2* d*x + 2*c) + 1))
\[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=B \int \frac {1}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \]
Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.84 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=-\frac {2 \, \sqrt {2} B \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} B \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} B \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} B \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{4 \, d} \]
-1/4*(2*sqrt(2)*B*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2 *sqrt(2)*B*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + sqrt(2) *B*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*B*log(-s qrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/d
\[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\int { \frac {B b \tan \left (d x + c\right ) + B a}{{\left (b \tan \left (d x + c\right ) + a\right )} \sqrt {\cot \left (d x + c\right )}} \,d x } \]
Time = 9.41 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.34 \[ \int \frac {a B+b B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,B\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\frac {1}{\mathrm {tan}\left (c+d\,x\right )}}\right )\,1{}\mathrm {i}}{d} \]