\(\int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [536]

Optimal result

Integrand size = 38, antiderivative size = 198 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 (13 A-5 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 (i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d} \]

[Out]

(-1-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))*a^(1/2)*cot(d*x+c)^(1/2)*tan(d
*x+c)^(1/2)/d-2/15*(I*A+5*B)*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/5*A*cot(d*x+c)^(5/2)*(a+I*a*tan(d*x
+c))^(1/2)/d+2/15*(13*A-5*I*B)*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {4326, 3679, 12, 3625, 211} \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {(1+i) \sqrt {a} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 (5 B+i A) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}+\frac {2 (13 A-5 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d} \]

[In]

Int[Cot[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

((-1 - I)*Sqrt[a]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[
c + d*x]]*Sqrt[Tan[c + d*x]])/d + (2*(13*A - (5*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) -
(2*(I*A + 5*B)*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (2*A*Cot[c + d*x]^(5/2)*Sqrt[a + I*a*Ta
n[c + d*x]])/(5*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-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] && NeQ[c^2 + d^2, 0]

Rule 3679

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac {7}{2}}(c+d x)} \, dx \\ & = -\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (i A+5 B)-2 a A \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{5 a} \\ & = -\frac {2 (i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (13 A-5 i B)-\frac {1}{2} a^2 (i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^2} \\ & = \frac {2 (13 A-5 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 (i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {15 a^3 (i A+B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{15 a^3} \\ & = \frac {2 (13 A-5 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 (i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\left ((i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 (13 A-5 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 (i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (2 i a^2 (i A+B) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \\ & = -\frac {(1-i) \sqrt {a} (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)}}{d}+\frac {2 (13 A-5 i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 (i A+5 B) \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d}-\frac {2 A \cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.88 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.78 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {i \left (\frac {15 \sqrt {2} a (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {i a \tan (c+d x)}}+2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-3 i A+(A-5 i B) \tan (c+d x)+(13 i A+5 B) \tan ^2(c+d x)\right )\right )}{15 d \sqrt {\cot (c+d x)}} \]

[In]

Integrate[Cot[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

((-1/15*I)*((15*Sqrt[2]*a*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/Sqrt
[I*a*Tan[c + d*x]] + 2*Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]]*((-3*I)*A + (A - (5*I)*B)*Tan[c + d*x] + ((13
*I)*A + 5*B)*Tan[c + d*x]^2)))/(d*Sqrt[Cot[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 637 vs. \(2 (161 ) = 322\).

Time = 0.68 (sec) , antiderivative size = 638, normalized size of antiderivative = 3.22

method	result	size
derivativedivides	\(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {7}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i B \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 \tan \left (d x +c \right )^{4}+15 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 \tan \left (d x +c \right )^{3}-15 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 \tan \left (d x +c \right )^{4}-20 i B \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 )^{3}+15 B \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 \tan \left (d x +c \right )^{3}-56 i A \sqrt {-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 )}+52 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 )^{3}+20 i B \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-40 B \sqrt {-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 )}+12 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-16 A \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{30 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(638\)
default	\(-\frac {\left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {7}{2}} \tan \left (d x +c \right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (15 i B \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 \tan \left (d x +c \right )^{4}+15 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 \tan \left (d x +c \right )^{3}-15 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 \tan \left (d x +c \right )^{4}-20 i B \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 )^{3}+15 B \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 \tan \left (d x +c \right )^{3}-56 i A \sqrt {-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 )}+52 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 )^{3}+20 i B \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-40 B \sqrt {-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 )}+12 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-16 A \sqrt {-i a}\, \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{30 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(638\)

[In]

int(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/30/d*(1/tan(d*x+c))^(7/2)*tan(d*x+c)*(a*(1+I*tan(d*x+c)))^(1/2)*(15*I*B*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)^4+15*I*A*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
)^3-15*A*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)^4-20*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+15*B*2^(1/2)*l
n(-(-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)^3-56*I*A*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+52*A*(-I*a)^(1/2)*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+20*I*B*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-40*B
*(-I*a)^(1/2)*tan(d*x+c)^2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+12*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*
x+c)))^(1/2)-16*A*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)
))^(1/2)/(-tan(d*x+c)+I)/(-I*a)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (151) = 302\).

Time = 0.26 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.43 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {15 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} \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 )}}{i \, A + B}\right ) - 15 \, \sqrt {2} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} \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 )}}{i \, A + B}\right ) - 4 \, \sqrt {2} {\left ({\left (17 \, A - 10 i \, B\right )} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 \, {\left (2 \, A - i \, B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, A 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}}}{30 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/30*(15*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a/d^2)*
log(-4*((A - I*B)*a*e^(I*d*x + I*c) - (I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a/d^2)*sq
rt(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)/
(I*A + B)) - 15*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a
/d^2)*log(-4*((A - I*B)*a*e^(I*d*x + I*c) - (-I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a/
d^2)*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)/(I*A + B)) - 4*sqrt(2)*((17*A - 10*I*B)*e^(5*I*d*x + 5*I*c) - 10*(2*A - I*B)*e^(3*I*d*x + 3*I*c) + 15*A
*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)
))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**(7/2)*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1409 vs. \(2 (151) = 302\).

Time = 0.74 (sec) , antiderivative size = 1409, normalized size of antiderivative = 7.12 \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/30*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((30*((I + 1)*A - (I - 1)*B)*cos(
3*d*x + 3*c) + (-(39*I + 39)*A + (25*I - 25)*B)*cos(d*x + c) + 30*((I - 1)*A + (I + 1)*B)*sin(3*d*x + 3*c) + (
-(39*I - 39)*A - (25*I + 25)*B)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (30*(
-(I - 1)*A - (I + 1)*B)*cos(3*d*x + 3*c) + ((39*I - 39)*A + (25*I + 25)*B)*cos(d*x + c) + 30*((I + 1)*A - (I -
1)*B)*sin(3*d*x + 3*c) + (-(39*I + 39)*A + (25*I - 25)*B)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c) - 1)))*sqrt(a) + 15*(2*((-(I - 1)*A - (I + 1)*B)*cos(2*d*x + 2*c)^2 + (-(I - 1)*A - (I + 1)*B)*s
in(2*d*x + 2*c)^2 + 2*((I - 1)*A + (I + 1)*B)*cos(2*d*x + 2*c) - (I - 1)*A - (I + 1)*B)*arctan2(2*(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + ((-(I + 1)*A + (I - 1)*B)*cos(2*d*x + 2*c)
^2 + (-(I + 1)*A + (I - 1)*B)*sin(2*d*x + 2*c)^2 + 2*((I + 1)*A - (I - 1)*B)*cos(2*d*x + 2*c) - (I + 1)*A + (I
- 1)*B)*log(4*cos(d*x + c)^2 + 4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*
x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) - 1))^2) + 8*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*
x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) - 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a)
+ ((30*((I + 1)*A - (I - 1)*B)*cos(5*d*x + 5*c) + 5*((I + 1)*A + (5*I - 5)*B)*cos(3*d*x + 3*c) + ((13*I + 13)*
A + (5*I - 5)*B)*cos(d*x + c) + 30*((I - 1)*A + (I + 1)*B)*sin(5*d*x + 5*c) + 5*((I - 1)*A - (5*I + 5)*B)*sin(
3*d*x + 3*c) + ((13*I - 13)*A - (5*I + 5)*B)*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
- 1)) + 4*((((2*I + 2)*A + (5*I - 5)*B)*cos(d*x + c) + ((2*I - 2)*A - (5*I + 5)*B)*sin(d*x + c))*cos(2*d*x + 2
*c)^2 + (((2*I + 2)*A + (5*I - 5)*B)*cos(d*x + c) + ((2*I - 2)*A - (5*I + 5)*B)*sin(d*x + c))*sin(2*d*x + 2*c)
^2 + 2*((-(2*I + 2)*A - (5*I - 5)*B)*cos(d*x + c) + (-(2*I - 2)*A + (5*I + 5)*B)*sin(d*x + c))*cos(2*d*x + 2*c
) + ((2*I + 2)*A + (5*I - 5)*B)*cos(d*x + c) + ((2*I - 2)*A - (5*I + 5)*B)*sin(d*x + c))*cos(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (30*(-(I - 1)*A - (I + 1)*B)*cos(5*d*x + 5*c) + 5*(-(I - 1)*A + (5*I + 5
)*B)*cos(3*d*x + 3*c) + (-(13*I - 13)*A + (5*I + 5)*B)*cos(d*x + c) + 30*((I + 1)*A - (I - 1)*B)*sin(5*d*x + 5
*c) + 5*((I + 1)*A + (5*I - 5)*B)*sin(3*d*x + 3*c) + ((13*I + 13)*A + (5*I - 5)*B)*sin(d*x + c))*sin(5/2*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 4*(((-(2*I - 2)*A + (5*I + 5)*B)*cos(d*x + c) + ((2*I + 2)*A + (
5*I - 5)*B)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((-(2*I - 2)*A + (5*I + 5)*B)*cos(d*x + c) + ((2*I + 2)*A + (5*
I - 5)*B)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 2*(((2*I - 2)*A - (5*I + 5)*B)*cos(d*x + c) + (-(2*I + 2)*A - (5*
I - 5)*B)*sin(d*x + c))*cos(2*d*x + 2*c) + (-(2*I - 2)*A + (5*I + 5)*B)*cos(d*x + c) + ((2*I + 2)*A + (5*I - 5
)*B)*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + s
in(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

Giac [F]

\[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {7}{2}} \,d x } \]

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \cot ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

[In]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int(cot(c + d*x)^(7/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(1/2), x)