Integrand size = 38, antiderivative size = 156 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \sqrt [4]{-1} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}} \]
-2*(-1)^(1/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+ c))^(1/2))/d/a^(1/2)-(1/2+1/2*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^ (1/2)/(a+I*a*tan(d*x+c))^(1/2))/d/a^(1/2)+(I*A-B)*tan(d*x+c)^(1/2)/d/(a+I* a*tan(d*x+c))^(1/2)
Time = 2.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {\left (-2 \sqrt [4]{-1} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}+(A+i B) \sqrt {\tan (c+d x)}\right ) \sqrt {a+i a \tan (c+d x)}}{a d (-i+\tan (c+d x))} \]
((-1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[ a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d) + ((-2*(-1)^(1/4)*B*ArcSinh[(-1)^(1/4) *Sqrt[Tan[c + d*x]]]*Sqrt[1 + I*Tan[c + d*x]] + (A + I*B)*Sqrt[Tan[c + d*x ]])*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(-I + Tan[c + d*x]))
Time = 0.88 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 4078, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
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.
| \(\displaystyle \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (i A-B)+2 i a B \tan (c+d x))}{2 \sqrt {\tan (c+d x)}}dx}{a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (i A-B)+2 i a B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} (a (i A-B)+2 i a B \tan (c+d x))}{\sqrt {\tan (c+d x)}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {a (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {-\frac {2 i a^3 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-2 B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}}{2 a^2}\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {4 a^2 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {\frac {(1-i) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {4 \sqrt [4]{-1} a^{3/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}\) |
-1/2*((4*(-1)^(1/4)*a^(3/2)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x] ])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((1 - I)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a^2 + (( I*A - B)*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])
3.2.80.3.1 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[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_)^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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[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] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, 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] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*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] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 893 vs. \(2 (123 ) = 246\).
Time = 0.15 (sec) , antiderivative size = 894, normalized size of antiderivative = 5.73
| method | result | size |
| derivativedivides | \(\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i B \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+2 i A \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )-A \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-i B \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a -8 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+4 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+2 B \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )+4 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )-4 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+A \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a +4 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +4 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\) | \(894\) |
| default | \(\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i B \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )+2 i A \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )-A \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-i B \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a -8 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+4 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+2 B \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \tan \left (d x +c \right )+4 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \left (\tan ^{2}\left (d x +c \right )\right )-4 i A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+A \sqrt {i a}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a +4 A \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +4 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{4 d a \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\) | \(894\) |
| parts | \(\text {Expression too large to display}\) | \(926\) |
1/4/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a*(I*B*(I*a)^(1/2)*2^(1/ 2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3* a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+2*I*A*(I*a)^(1/2)*2^(1/2)*ln( (2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan( d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)-A*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*( -I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan (d*x+c)+I))*a*tan(d*x+c)^2-I*B*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1 /2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+ I))*a-8*I*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/ 2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)+4*I*B*(I*a)^(1/2) *(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+2*B*(I*a)^( 1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+ 3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c)+4*B*ln(1/2*(2*I*a*tan (d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2) )*(-I*a)^(1/2)*a*tan(d*x+c)^2-4*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c) *(1+I*tan(d*x+c)))^(1/2)+A*(I*a)^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d *x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)* a+4*A*(I*a)^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d *x+c)-4*B*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2) *(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a+4*B*(I*a)^(1/2)*(-I*a)^(1/2...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 718 vs. \(2 (116) = 232\).
Time = 0.33 (sec) , antiderivative size = 718, normalized size of antiderivative = 4.60 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {{\left (\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - \sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - a d \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} {\left (B e^{\left (3 i \, d x + 3 i \, c\right )} + B e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + {\left (3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) + a d \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {52 \, {\left (4 \, \sqrt {2} {\left (B e^{\left (3 i \, d x + 3 i \, c\right )} + B e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - {\left (3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {-\frac {4 i \, B^{2}}{a d^{2}}}\right )}}{605 \, {\left (B e^{\left (2 i \, d x + 2 i \, c\right )} + B\right )}}\right ) + 2 \, \sqrt {2} {\left ({\left (-i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]
-1/4*(sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e^(I*d*x + I*c)* log((sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2* I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/( 4*I*A + 4*B)) - sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e^(I*d *x + I*c)*log(-(sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e^(I*d *x + I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2 *I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I* c) + 1)))/(4*I*A + 4*B)) - a*d*sqrt(-4*I*B^2/(a*d^2))*e^(I*d*x + I*c)*log( 52/605*(4*sqrt(2)*(B*e^(3*I*d*x + 3*I*c) + B*e^(I*d*x + I*c))*sqrt(a/(e^(2 *I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I* c) + 1)) + (3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-4*I*B^2/(a*d^2)))/(B*e^ (2*I*d*x + 2*I*c) + B)) + a*d*sqrt(-4*I*B^2/(a*d^2))*e^(I*d*x + I*c)*log(5 2/605*(4*sqrt(2)*(B*e^(3*I*d*x + 3*I*c) + B*e^(I*d*x + I*c))*sqrt(a/(e^(2* I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c ) + 1)) - (3*a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(-4*I*B^2/(a*d^2)))/(B*e^( 2*I*d*x + 2*I*c) + B)) + 2*sqrt(2)*((-I*A + B)*e^(2*I*d*x + 2*I*c) - I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e ^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(a*d)
\[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\tan {\left (c + d x \right )}}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0] was discarded and replaced randomly by 0=[-7]Warning, replacing -7 by -3, a substitutio n variabl
Time = 28.33 (sec) , antiderivative size = 4040, normalized size of antiderivative = 25.90 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \]
-(A*a^(5/2)*tan(c + d*x)^(1/2)*4i + 4*A*a^(5/2)*tan(c + d*x)^(3/2) - 4*B*a ^(5/2)*tan(c + d*x)^(1/2) + B*a^(5/2)*tan(c + d*x)^(3/2)*4i + A*a^(3/2)*ta n(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i - 4*B*a^(3/2)*tan(c + d*x)^(1/ 2)*(a + a*tan(c + d*x)*1i) + (1i/8)^(1/2)*A*(-a)^(5/2)*atanh(((1i/8)^(1/2) *(-a)^(15/2)*a^(17/2)*tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*t an(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2) *tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a) ^(15/2)*a^(17/2)*tan(c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*ta n(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a*tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2)) *8i + 8*(1i/8)^(1/2)*B*(-a)^(5/2)*atanh(((1i/8)^(1/2)*(-a)^(15/2)*a^(17/2) *tan(c + d*x)^(1/2)*4i - 4*(1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) - (1i/8)^(1/2)*(-a)^(31/2)*tan(c + d*x)^(1/2)*( a + a*tan(c + d*x)*1i)^(1/2)*8i + 4*(1i/8)^(1/2)*(-a)^(15/2)*a^(17/2)*tan( c + d*x)^(3/2) + (1i/8)^(1/2)*(-a)^(15/2)*a^(15/2)*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)*4i)/(a^15*(a + a*tan(c + d*x)*1i) - 2*a^(31/2)*(a + a* tan(c + d*x)*1i)^(1/2) + a^16 + a^16*tan(c + d*x)^2)) - A*a^2*tan(c + d*x) ^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*8i - 4*A*a^2*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2) + 8*B*a^2*tan(c + d*x)^(1/2)*(a + a*tan(c + d*x)* 1i)^(1/2) - B*a^2*tan(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2)*4i +...