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

Optimal result

Integrand size = 26, antiderivative size = 237 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {11 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} a^{7/2} d}+\frac {11 i a}{36 d (a+i a \tan (c+d x))^{9/2}}+\frac {11 i}{56 d (a+i a \tan (c+d x))^{7/2}}+\frac {11 i}{80 a d (a+i a \tan (c+d x))^{5/2}}+\frac {11 i}{96 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {11 i}{64 a^3 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^3}{2 d (a+i a \tan (c+d x))^{9/2} \left (a^2-i a^2 \tan (c+d x)\right )} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.42 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.22 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (-\frac {9}{2},2,-\frac {7}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{18 d (a+i a \tan (c+d x))^{9/2}} \] Input:

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

Output:

((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))
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3968, 52, 61, 61, 61, 61, 61, 73, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^2 (a+i a \tan (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i a^3 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{11/2}}d(i a \tan (c+d x))}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^3 \left (\frac {11 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{11/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{9/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{7/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\frac {\frac {\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\frac {\frac {\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {i a^3 \left (\frac {11 \left (\frac {\frac {\frac {\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}}{2 a}-\frac {1}{5 a (a+i a \tan (c+d x))^{5/2}}}{2 a}-\frac {1}{7 a (a+i a \tan (c+d x))^{7/2}}}{2 a}-\frac {1}{9 a (a+i a \tan (c+d x))^{9/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{9/2}}\right )}{d}\)

Input:

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

Output:

((-I)*a^3*(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(9/2)) + ( 
11*(-1/9*1/(a*(a + I*a*Tan[c + d*x])^(9/2)) + (-1/7*1/(a*(a + I*a*Tan[c + 
d*x])^(7/2)) + (-1/5*1/(a*(a + I*a*Tan[c + d*x])^(5/2)) + (-1/3*1/(a*(a + 
I*a*Tan[c + d*x])^(3/2)) + ((I*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[2]])/(Sq 
rt[2]*a^(3/2)) - 1/(a*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a))/(2*a))/(2*a))/(2 
*a)))/(4*a)))/d
 

Defintions of rubi rules used

rule 52

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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

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] - Simp[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] && 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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

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])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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]
 

Maple [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (187 ) = 374\).

Time = 9.44 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.92

method	result	size
default	\(\frac {3465 \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2} \left (1-8 \cos \left (d x +c \right )^{3}-4 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+3465 i \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \left (4+8 \cos \left (d x +c \right )-8 \sec \left (d x +c \right )-3 \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{3}\right )+3465 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right ) \left (\sec \left (d x +c \right )^{2}-4\right )+3465 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (3 \sec \left (d x +c \right )^{2}-4\right )+154 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \left (45+80 \cos \left (d x +c \right )^{4}+80 \cos \left (d x +c \right )^{3}-276 \cos \left (d x +c \right )^{2}-96 \cos \left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-11352+3920 \cos \left (d x +c \right )^{3}+3920 \cos \left (d x +c \right )^{2}-25212 \cos \left (d x +c \right )+12705 \sec \left (d x +c \right )+2310 \sec \left (d x +c \right )^{2}\right )}{40320 d \left (-i+\tan \left (d x +c \right )\right )^{3} \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{3}}\)	\(455\)

Input:

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

Output:

1/40320/d/(-I+tan(d*x+c))^3/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(3465*tan(d*x+c)*sec(d*x+c)^2*(1-8*cos(d*x+ 
c)^3-4*cos(d*x+c)^2+4*cos(d*x+c))*arctanh(1/2/(-cos(d*x+c)/(cos(d*x+c)+1)) 
^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))+3465*I*arctanh(1/2/(-cos(d*x+c)/(cos(d*x 
+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))*(4+8*cos(d*x+c)-8*sec(d*x+c)-3*se 
c(d*x+c)^2+sec(d*x+c)^3)+3465*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/ 
2)*tan(d*x+c)*(sec(d*x+c)^2-4)+3465*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1)) 
^(1/2)*(3*sec(d*x+c)^2-4)+154*I*tan(d*x+c)*sec(d*x+c)*(45+80*cos(d*x+c)^4+ 
80*cos(d*x+c)^3-276*cos(d*x+c)^2-96*cos(d*x+c))*(-cos(d*x+c)/(cos(d*x+c)+1 
))^(1/2)+2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(-11352+3920*cos(d*x+c)^3+39 
20*cos(d*x+c)^2-25212*cos(d*x+c)+12705*sec(d*x+c)+2310*sec(d*x+c)^2))/a^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-3465 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3465 i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (9 i \, d x + 9 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-315 i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 4303 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 7034 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3754 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 1798 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 530 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 70 i\right )}\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{40320 \, a^{4} d} \] Input:

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

Output:

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*(sq 
rt(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)) + sqr 
t(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-315*I*e^(12*I*d*x + 12*I*c) + 430 
3*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. \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \, {\left (\frac {4 \, {\left (3465 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} - 4620 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a - 1848 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 1584 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 1760 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 2240 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} a^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a^{3}} + \frac {3465 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}}\right )}}{80640 \, a d} \] Input:

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

Output:

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)^(1 
1/2)*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(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{2} \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{5} i +3 \tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{3} i +2 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right ) i -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{5} i +3 \tan \left (d x +c \right )^{4}-2 \tan \left (d x +c \right )^{3} i +2 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right )\right )}{a^{4}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**2*tan(c + d*x))/(tan 
(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)* 
*2 - 3*tan(c + d*x)*i - 1),x)*i - int((sqrt(tan(c + d*x)*i + 1)*cos(c + d* 
x)**2)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*ta 
n(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)))/a**4