Integrand size = 38, antiderivative size = 243 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 (-1)^{3/4} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{3/2} d}+\frac {i A-B}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {A+3 i B}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]
2*(-1)^(3/4)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c ))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/a^(3/2)/d+(1/4+1/4*I)*(I*A+B)* arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c )^(1/2)*tan(d*x+c)^(1/2)/a^(3/2)/d+1/2*(A+3*I*B)/a/d/cot(d*x+c)^(1/2)/(a+I *a*tan(d*x+c))^(1/2)+1/3*(I*A-B)/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/ 2)
Time = 5.87 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{12}+\frac {i}{12}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left ((12+12 i) \sqrt [4]{-1} \sqrt {a} B \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) (1+i \tan (c+d x))^{3/2}+3 (i A+B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}+(1+i) \sqrt {a} \sqrt {\tan (c+d x)} (-3 (A+3 i B)+(-5 i A+11 B) \tan (c+d x))\right )}{a^{3/2} d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]
((1/12 + I/12)*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((12 + 12*I)*(-1)^(1/ 4)*Sqrt[a]*B*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*(1 + I*Tan[c + d*x])^( 3/2) + 3*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I *a*Tan[c + d*x]]]*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]] + (1 + I) *Sqrt[a]*Sqrt[Tan[c + d*x]]*(-3*(A + (3*I)*B) + ((-5*I)*A + 11*B)*Tan[c + d*x])))/(a^(3/2)*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.48 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.95, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 4729, 3042, 4078, 27, 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.
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\cot (c+d x)^{3/2} (a+i a \tan (c+d x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{3/2} (A+B \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {3 \sqrt {\tan (c+d x)} (a (i A-B)+2 i a B \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (a (i A-B)+2 i a B \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\sqrt {\tan (c+d x)} (a (i A-B)+2 i a B \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {-\frac {\int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{2 \sqrt {\tan (c+d x)}}dx}{a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((A+3 i B) a^2+4 B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 i a^4 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {4 i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {4 i a^3 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {8 i a^3 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {(-B+i A) \tan ^{\frac {3}{2}}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\frac {\frac {(1-i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {8 (-1)^{3/4} a^{5/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{2 a^2}-\frac {a (A+3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{2 a^2}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((I*A - B)*Tan[c + d*x]^(3/2))/(3*d *(a + I*a*Tan[c + d*x])^(3/2)) - (((-8*(-1)^(3/4)*a^(5/2)*B*ArcTan[((-1)^( 3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((1 - I) *a^(5/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I *a*Tan[c + d*x]]])/d)/(2*a^2) - (a*(A + (3*I)*B)*Sqrt[Tan[c + d*x]])/(d*Sq rt[a + I*a*Tan[c + d*x]]))/(2*a^2))
3.6.61.3.1 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]
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1232 vs. \(2 (192 ) = 384\).
Time = 0.54 (sec) , antiderivative size = 1233, normalized size of antiderivative = 5.07
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(1233\) |
| default | \(\text {Expression too large to display}\) | \(1233\) |
1/24*I/d/(1/tan(d*x+c))^(3/2)/tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(2 4*I*B*(-I*a)^(1/2)*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))*a+20*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan( d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+44*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)^2-3*A*2^(1/2)*(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))*a*tan(d*x+c)^3-3*I*A*2^(1/2)*(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))*a-9*I*B*2^(1/2)*(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))/(t an(d*x+c)+I))*a*tan(d*x+c)-32*I*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)+9*B*2^(1/2)*(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))/(ta n(d*x+c)+I))*a*tan(d*x+c)^2+24*B*(-I*a)^(1/2)*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))*a*tan(d*x +c)^3-72*I*B*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*ta n(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)^2+3*I*B*2^(1/2)* (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))*a*tan(d*x+c)^3+9*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*...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 782 vs. \(2 (181) = 362\).
Time = 0.28 (sec) , antiderivative size = 782, normalized size of antiderivative = 3.22 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\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}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\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}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} - {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, a^{2} d \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} {\left (i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a^{3} d 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}} \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) + 3 \, a^{2} d \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} {\left (-i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{3} d 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}} \sqrt {\frac {4 i \, B^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{B}\right ) + \sqrt {2} {\left (2 \, {\left (2 i \, A - 5 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (5 i \, A - 11 \, 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 (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]
-1/12*(3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d *x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)* 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))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2)) + (A - I*B)*a *e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 3*sqrt(1/2)*a^2*d*sqrt((-I *A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt(2)*sqrt(1 /2)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*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))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2)) - (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c) /(I*A + B)) - 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-16* (3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 + sqrt(2)*(I*a^3*d*e^(3*I*d*x + 3*I*c ) - I*a^3*d*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))*sqrt(4*I*B^2/(a^3*d^2)))* e^(-2*I*d*x - 2*I*c)/B) + 3*a^2*d*sqrt(4*I*B^2/(a^3*d^2))*e^(3*I*d*x + 3*I *c)*log(-16*(3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 + sqrt(2)*(-I*a^3*d*e^(3* I*d*x + 3*I*c) + I*a^3*d*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))*sqrt(4*I*B^2 /(a^3*d^2)))*e^(-2*I*d*x - 2*I*c)/B) + sqrt(2)*(2*(2*I*A - 5*B)*e^(4*I*d*x + 4*I*c) - (5*I*A - 11*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)...
\[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]