\(\int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [367]

Optimal result

Integrand size = 26, antiderivative size = 208 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {9 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{32 \sqrt {2} a^{5/2} d}+\frac {9 i a}{28 d (a+i a \tan (c+d x))^{7/2}}+\frac {9 i}{40 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i}{16 a d (a+i a \tan (c+d x))^{3/2}}+\frac {9 i}{32 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^3}{2 d (a+i a \tan (c+d x))^{7/2} \left (a^2-i a^2 \tan (c+d x)\right )} \] Output:

-9/64*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/a^(5 
/2)/d+9/28*I*a/d/(a+I*a*tan(d*x+c))^(7/2)+9/40*I/d/(a+I*a*tan(d*x+c))^(5/2 
)+3/16*I/a/d/(a+I*a*tan(d*x+c))^(3/2)+9/32*I/a^2/d/(a+I*a*tan(d*x+c))^(1/2 
)-1/2*I*a^3/d/(a+I*a*tan(d*x+c))^(7/2)/(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 = 0.17 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.25 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i a \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},2,-\frac {5}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{14 d (a+i a \tan (c+d x))^{7/2}} \] Input:

Integrate[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((I/14)*a*Hypergeometric2F1[-7/2, 2, -5/2, (1 + I*Tan[c + d*x])/2])/(d*(a 
+ I*a*Tan[c + d*x])^(7/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3968, 52, 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.

\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sec (c+d x)^2 (a+i a \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

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

Input:

Int[Cos[c + d*x]^2/(a + I*a*Tan[c + d*x])^(5/2),x]
 

Output:

((-I)*a^3*(1/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(7/2)) + ( 
9*(-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]])/(Sqrt[2]*a^(3/2)) - 1/(a*Sqrt[a + I*a*Tan[c + d 
*x]]))/(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] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 448 vs. \(2 (164 ) = 328\).

Time = 10.18 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.16

method	result	size
default	\(\frac {i \left (1260 \cos \left (d x +c \right )^{2}+630 \cos \left (d x +c \right )-315\right ) \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\left (1260 \cos \left (d x +c \right )^{3}+630 \cos \left (d x +c \right )^{2}-945 \cos \left (d x +c \right )-315\right ) \operatorname {arctanh}\left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+i}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+i \left (630 \cos \left (d x +c \right )^{2}-315\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-630 \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (720 \cos \left (d x +c \right )^{3}+720 \cos \left (d x +c \right )^{2}-1680 \cos \left (d x +c \right )-420\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \cos \left (d x +c \right ) \left (-400 \cos \left (d x +c \right )^{4}-400 \cos \left (d x +c \right )^{3}+2184 \cos \left (d x +c \right )^{2}+924 \cos \left (d x +c \right )-630\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{2240 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+2\right )+2 \cos \left (d x +c \right )^{3}+2 \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )-1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}\)	\(449\)

Input:

int(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/2240/d*(I*(1260*cos(d*x+c)^2+630*cos(d*x+c)-315)*sin(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))+(1260*cos(d* 
x+c)^3+630*cos(d*x+c)^2-945*cos(d*x+c)-315)*arctanh(1/2/(-cos(d*x+c)/(cos( 
d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+I))+I*(630*cos(d*x+c)^2-315)*2^(1/ 
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-630*sin(d*x+c)*2^(1/2)*(-2*cos(d*x 
+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+sin(d*x+c)*cos(d*x+c)*(720*cos(d*x+c) 
^3+720*cos(d*x+c)^2-1680*cos(d*x+c)-420)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)+I*cos(d*x+c)*(-400*cos(d*x+c)^4-400*cos(d*x+c)^3+2184*cos(d*x+c)^2+924*c 
os(d*x+c)-630)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(-cos(d*x+c)/(cos(d*x+c 
)+1))^(1/2)/(I*sin(d*x+c)*cos(d*x+c)*(2*cos(d*x+c)+2)+2*cos(d*x+c)^3+2*cos 
(d*x+c)^2-cos(d*x+c)-1)/(a*(1+I*tan(d*x+c)))^(1/2)/a^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. 305 vs. \(2 (151) = 302\).

Time = 0.09 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.47 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 315 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (7 i \, d x + 7 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-35 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 353 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 544 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 214 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 68 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 10 i\right )}\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{2240 \, a^{3} d} \] Input:

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/2240*(-315*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(7*I*d*x + 7*I*c)*log(4 
*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x 
 + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 
 315*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(7*I*d*x + 7*I*c)*log(-4*(sqrt( 
2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I* 
c) + 1))*sqrt(1/(a^5*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))*(-35*I*e^(10*I*d*x + 10*I*c) + 353*I*e 
^(8*I*d*x + 8*I*c) + 544*I*e^(6*I*d*x + 6*I*c) + 214*I*e^(4*I*d*x + 4*I*c) 
 + 68*I*e^(2*I*d*x + 2*I*c) + 10*I))*e^(-7*I*d*x - 7*I*c)/(a^3*d)
 

Sympy [F]

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

integrate(cos(d*x+c)**2/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral(cos(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.84 \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} - 420 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a - 168 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{2} - 144 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{3} - 160 \, a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2}} + \frac {315 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}}\right )}}{4480 \, a d} \] Input:

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

1/4480*I*(4*(315*(I*a*tan(d*x + c) + a)^4 - 420*(I*a*tan(d*x + c) + a)^3*a 
 - 168*(I*a*tan(d*x + c) + a)^2*a^2 - 144*(I*a*tan(d*x + c) + a)*a^3 - 160 
*a^4)/((I*a*tan(d*x + c) + a)^(9/2)*a - 2*(I*a*tan(d*x + c) + a)^(7/2)*a^2 
) + 315*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^(3/2))/(a*d)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(5/2),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. \[ \int \frac {\cos ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

int(cos(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(5/2),x)
 

Output:

int(cos(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(5/2), x)
 

Reduce [F]

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

int(cos(d*x+c)^2/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

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