Integrand size = 35, antiderivative size = 168 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}-\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d} \]
(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I *a-b)^(1/2)+2*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d /b^(1/2)-(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^( 1/2))/d/(I*a+b)^(1/2)
Time = 1.40 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\sqrt [4]{-1} \left (-\frac {(A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {(A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}\right )+\frac {2 \sqrt {a} B \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{d} \]
((-1)^(1/4)*(-(((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]) + ((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]) + (2*Sqrt[a]*B*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt [1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]))/d
Time = 1.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4097, 3042, 4099, 3042, 4098, 104, 216, 219, 4117, 65, 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.
| \(\displaystyle \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4097 |
| \(\displaystyle \int \frac {A \tan (c+d x)-B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+B \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \tan (c+d x)-B}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle \frac {1}{2} (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} (-B+i A) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} (B+i A) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {(B+i A) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+\frac {(-B+i A) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}+B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {(-B+i A) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {(B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {(B+i A) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle B \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx+\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}+\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {2 B \int \frac {1}{1-\frac {b \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}+\frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(-B+i A) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {-b+i a}}-\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d \sqrt {b+i a}}+\frac {2 B \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b} d}\) |
((I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d* x]]])/(Sqrt[I*a - b]*d) + (2*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[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)
3.5.53.3.1 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_))^(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[(Sqrt[(a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> In t[Simp[a*A - b*B + (A*b + a*B)*Tan[e + f*x], x]/(Sqrt[a + b*Tan[e + f*x]]*S qrt[c + d*Tan[e + f*x]]), x] + Simp[b*B Int[(1 + Tan[e + f*x]^2)/(Sqrt[a + b*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e , f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^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^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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.95 (sec) , antiderivative size = 1885950, normalized size of antiderivative = 11225.89
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 10116 vs. \(2 (132) = 264\).
Time = 4.15 (sec) , antiderivative size = 20234, normalized size of antiderivative = 120.44 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
\[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\tan \left (d x + c\right )}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Time = 103.96 (sec) , antiderivative size = 30600, normalized size of antiderivative = 182.14 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
atan(((((B^2 - A^2 + A*B*2i)/(4*(a*d^2*1i + b*d^2)))^(1/2)*((((B^2 - A^2 + A*B*2i)/(4*(a*d^2*1i + b*d^2)))^(1/2)*((((B^2 - A^2 + A*B*2i)/(4*(a*d^2*1 i + b*d^2)))^(1/2)*((((274877906944*(1600*a^12*b^34*d^8 - 16640*a^14*b^32* d^8 + 22784*a^16*b^30*d^8 + 106496*a^18*b^28*d^8 + 65536*a^20*b^26*d^8))/d ^8 - (274877906944*tan(c + d*x)*(1600*a^12*b^35*d^8 - 48000*a^14*b^33*d^8 + 155136*a^16*b^31*d^8 + 466944*a^18*b^29*d^8 + 262144*a^20*b^27*d^8))/(d^ 8*((a + b*tan(c + d*x))^(1/2) - a^(1/2))^2))*((B^2 - A^2 + A*B*2i)/(4*(a*d ^2*1i + b*d^2)))^(1/2) - (2199023255552*tan(c + d*x)^(1/2)*(240*A*a^13*b^3 3*d^6 + 3064*A*a^15*b^31*d^6 + 8960*A*a^17*b^29*d^6 + 6144*A*a^19*b^27*d^6 + 6920*B*a^14*b^32*d^6 + 9472*B*a^16*b^30*d^6 - 5632*B*a^18*b^28*d^6 - 81 92*B*a^20*b^26*d^6))/(d^7*((a + b*tan(c + d*x))^(1/2) - a^(1/2))))*((B^2 - A^2 + A*B*2i)/(4*(a*d^2*1i + b*d^2)))^(1/2) - (274877906944*(19216*A^2*a^ 14*b^31*d^6 - 1440*A^2*a^12*b^33*d^6 - 22016*A^2*a^16*b^29*d^6 - 45056*A^2 *a^18*b^27*d^6 + 1200*B^2*a^12*b^33*d^6 - 16640*B^2*a^14*b^31*d^6 + 279040 *B^2*a^16*b^29*d^6 + 561152*B^2*a^18*b^27*d^6 + 262144*B^2*a^20*b^25*d^6 + 16480*A*B*a^13*b^32*d^6 - 25792*A*B*a^15*b^30*d^6 + 34816*A*B*a^17*b^28*d ^6 + 81920*A*B*a^19*b^26*d^6))/d^8 + (274877906944*tan(c + d*x)*(46704*A^2 *a^14*b^32*d^6 - 1440*A^2*a^12*b^34*d^6 - 137216*A^2*a^16*b^30*d^6 - 12288 0*A^2*a^18*b^28*d^6 + 65536*A^2*a^20*b^26*d^6 + 1200*B^2*a^12*b^34*d^6 - 5 2320*B^2*a^14*b^32*d^6 + 1200640*B^2*a^16*b^30*d^6 + 2306048*B^2*a^18*b...