\(\int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [383]

Optimal result

Integrand size = 26, antiderivative size = 233 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {11 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {11 i}{64 a^3 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-11/128*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(7/2)/d*2^(1/2)+11/64*I/a^3/d/(a+I*a*tan(d*x
+c))^(1/2)+11/36*I*a/d/(a+I*a*tan(d*x+c))^(9/2)-1/2*I*a^2/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c))^(9/2)+11/56*
I/d/(a+I*a*tan(d*x+c))^(7/2)+11/80*I/a/d/(a+I*a*tan(d*x+c))^(5/2)+11/96*I/a^2/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3568, 44, 53, 65, 212} \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {11 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {11 i}{64 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((-11*I)/64)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(7/2)*d) + (((11*I)/36)*a)/(d*
(a + I*a*Tan[c + d*x])^(9/2)) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + ((11*I)/
56)/(d*(a + I*a*Tan[c + d*x])^(7/2)) + ((11*I)/80)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((11*I)/96)/(a^2*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + ((11*I)/64)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac {\left (11 i a^2\right ) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{11/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d} \\ & = \frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}-\frac {(11 i a) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{9/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d} \\ & = \frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}-\frac {(11 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{16 d} \\ & = \frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}-\frac {(11 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{32 a d} \\ & = \frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {(11 i) \text {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{64 a^2 d} \\ & = \frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {11 i}{64 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {(11 i) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{128 a^3 d} \\ & = \frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {11 i}{64 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {(11 i) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{64 a^3 d} \\ & = -\frac {11 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}-\frac {i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {11 i}{64 a^3 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.22 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},2,-\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{18 d (a+i a \tan (c+d x))^{9/2}} \]

[In]

Integrate[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((I/18)*a*Hypergeometric2F1[-9/2, 2, -7/2, (1 + I*Tan[c + d*x])/2])/(d*(a + I*a*Tan[c + d*x])^(9/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 976 vs. \(2 (183 ) = 366\).

Time = 10.36 (sec) , antiderivative size = 977, normalized size of antiderivative = 4.19

method	result	size
default	\(\text {Expression too large to display}\)	\(977\)

[In]

int(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/40320/d/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/(cos(d*x+c)+1)/(1+I*tan(d*x+c))^3/(a*(1+I*tan(d*x+c)))^(1/2)/a^3
*(-50424*I*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-13860*I*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d
*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-12320*sin(d*x+c)*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+
7840*I*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3465*I*sec(d*x+c)^3*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+
c))/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-12320*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin
(d*x+c)-50424*I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+25410*I*sec(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+425
04*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+27720*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d*x+c)+1)/(
-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+27720*I*sec(d*x+c)*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d
*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+25410*I*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+42504*tan(d
*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+13860*tan(d*x+c)*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d*x+c)+1
)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-27720*I*cos(d*x+c)*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d*x+c)+1)
/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+7840*I*cos(d*x+c)^3*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-6930*tan(d*x+c)*se
c(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-13860*tan(d*x+c)*sec(d*x+c)*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))
/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+10395*I*sec(d*x+c)^2*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c)
)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-6930*tan(d*x+c)*sec(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)-3465*tan(d*x+c)*sec(d*x+c)^2*arctan(1/2*(cos(d*x+c)+1+I*sin(d*x+c))/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-315 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 4303 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 7034 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3754 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1798 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 530 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 70 i\right )}\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{40320 \, a^{4} d} \]

[In]

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

[Out]

1/40320*(-3465*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(9*I*d*x + 9*I*c)*log(4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2*I*
d*x + 2*I*c) + a^4*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c
)) + 3465*I*sqrt(1/2)*a^4*d*sqrt(1/(a^7*d^2))*e^(9*I*d*x + 9*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(a^4*d*e^(2*I*d*x
+ 2*I*c) + a^4*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^7*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) +
sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-315*I*e^(12*I*d*x + 12*I*c) + 4303*I*e^(10*I*d*x + 10*I*c) + 7034
*I*e^(8*I*d*x + 8*I*c) + 3754*I*e^(6*I*d*x + 6*I*c) + 1798*I*e^(4*I*d*x + 4*I*c) + 530*I*e^(2*I*d*x + 2*I*c) +
70*I))*e^(-9*I*d*x - 9*I*c)/(a^4*d)

Sympy [F(-1)]

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

[In]

integrate(cos(d*x+c)**2/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \, {\left (\frac {4 \, {\left (3465 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} - 4620 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 1848 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 1584 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 1760 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 2240 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}} + \frac {3465 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}}\right )}}{80640 \, a d} \]

[In]

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

[Out]

1/80640*I*(4*(3465*(I*a*tan(d*x + c) + a)^5 - 4620*(I*a*tan(d*x + c) + a)^4*a - 1848*(I*a*tan(d*x + c) + a)^3*
a^2 - 1584*(I*a*tan(d*x + c) + a)^2*a^3 - 1760*(I*a*tan(d*x + c) + a)*a^4 - 2240*a^5)/((I*a*tan(d*x + c) + a)^
(11/2)*a^2 - 2*(I*a*tan(d*x + c) + a)^(9/2)*a^3) + 3465*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c)
+ a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2))/(a*d)

Giac [F]

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

[In]

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

[Out]

integrate(cos(d*x + c)^2/(I*a*tan(d*x + c) + a)^(7/2), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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