\(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [299]

Optimal result

Integrand size = 26, antiderivative size = 247 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {105 i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{256 \sqrt {2} d}+\frac {35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}+\frac {105 i a^2}{256 d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i a^7}{16 d (a+i a \tan (c+d x))^{3/2} \left (a^2-i a^2 \tan (c+d x)\right )^2}-\frac {21 i a^7}{64 d (a+i a \tan (c+d x))^{3/2} \left (a^4-i a^4 \tan (c+d x)\right )} \] Output:

-105/512*I*a^(3/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2 
^(1/2)/d+35/128*I*a^3/d/(a+I*a*tan(d*x+c))^(3/2)-1/6*I*a^6/d/(a-I*a*tan(d* 
x+c))^3/(a+I*a*tan(d*x+c))^(3/2)+105/256*I*a^2/d/(a+I*a*tan(d*x+c))^(1/2)- 
3/16*I*a^7/d/(a+I*a*tan(d*x+c))^(3/2)/(a^2-I*a^2*tan(d*x+c))^2-21/64*I*a^7 
/d/(a+I*a*tan(d*x+c))^(3/2)/(a^4-I*a^4*tan(d*x+c))
 

Mathematica [C] (verified)

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

Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.21 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i a^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},4,-\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right )}{24 d (a+i a \tan (c+d x))^{3/2}} \] Input:

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((I/24)*a^3*Hypergeometric2F1[-3/2, 4, -1/2, (1 + I*Tan[c + d*x])/2])/(d*( 
a + I*a*Tan[c + d*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.36 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.96, 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, 52, 52, 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.

\(\displaystyle \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sec (c+d x)^6}dx\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {3 \int \frac {1}{(a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {3 \left (\frac {7 \int \frac {1}{(a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {3 \left (\frac {7 \left (\frac {5 \int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^7 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {i a^7 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {i a^7 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {i a^7 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a+i a \tan (c+d x)}}}{2 a}-\frac {1}{3 a (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{2 a (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}\right )}{8 a}+\frac {1}{4 a (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}\right )}{4 a}+\frac {1}{6 a (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}\right )}{d}\)

Input:

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-I)*a^7*(1/(6*a*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(3/2)) + 
 (3*(1/(4*a*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(3/2)) + (7*(1 
/(2*a*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + (5*(-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)))/(4*a)))/( 
8*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. 814 vs. \(2 (201 ) = 402\).

Time = 5.06 (sec) , antiderivative size = 815, normalized size of antiderivative = 3.30

method	result	size
default	\(\text {Expression too large to display}\)	\(815\)

Input:

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/1536/d*cos(d*x+c)*((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/(cot(d*x 
+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c) 
)*2^(1/2))*(630*sin(d*x+c)^2+315*tan(d*x+c)*sin(d*x+c))+I*(630*cos(d*x+c)^ 
2+315*cos(d*x+c)-315)*tan(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan 
h(1/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c 
)-cot(d*x+c))*2^(1/2))+I*(-630*cos(d*x+c)-315)*sin(d*x+c)*(-cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*arctanh(1/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x 
+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c))*2^(1/2))+(630*cos(d*x+c)^2+315*cos( 
d*x+c)-315)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/(cot(d*x+c)^2-2*c 
ot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c))*2^(1/2) 
)+I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*2^(1/2))*(630*sin(d*x+c)^2+315*tan(d*x+c)*sin(d*x+c))+(-630*c 
os(d*x+c)^2-315*cos(d*x+c)+315)*tan(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1 
/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2))+(630*cos(d*x+ 
c)+315)*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-2*cos(d 
*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2))+I*(630*cos(d*x+c)^2+315*cos(d*x+c)-31 
5)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c 
)+1))^(1/2)*2^(1/2))+I*sin(d*x+c)^2*cos(d*x+c)*(384*cos(d*x+c)^2+840)+sin( 
d*x+c)*(-128*cos(d*x+c)^4-504*cos(d*x+c)^2+630)+sin(d*x+c)*cos(d*x+c)^2*(3 
84*cos(d*x+c)^2+840)+I*cos(d*x+c)*(128*cos(d*x+c)^4+504*cos(d*x+c)^2-63...
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.26 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {{\left (315 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 315 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {2} {\left (-8 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 58 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 215 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 43 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 224 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{1536 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

-1/1536*(315*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2 
)*sqrt(1/2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I* 
d*x + 2*I*c) + 1)) - a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - 315*sqrt(1 
/2)*sqrt(-a^3/d^2)*d*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(-I*d*e 
^(2*I*d*x + 2*I*c) - I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) 
 - a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - sqrt(2)*(-8*I*a*e^(10*I*d*x 
+ 10*I*c) - 58*I*a*e^(8*I*d*x + 8*I*c) - 215*I*a*e^(6*I*d*x + 6*I*c) + 43* 
I*a*e^(4*I*d*x + 4*I*c) + 224*I*a*e^(2*I*d*x + 2*I*c) + 16*I*a)*sqrt(a/(e^ 
(2*I*d*x + 2*I*c) + 1)))*e^(-3*I*d*x - 3*I*c)/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.86 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i \, {\left (315 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 1680 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 2772 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 1152 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 256 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}}\right )}}{3072 \, a d} \] Input:

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

1/3072*I*(315*sqrt(2)*a^(5/2)*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))) + 4*(315*(I*a*tan( 
d*x + c) + a)^4*a^3 - 1680*(I*a*tan(d*x + c) + a)^3*a^4 + 2772*(I*a*tan(d* 
x + c) + a)^2*a^5 - 1152*(I*a*tan(d*x + c) + a)*a^6 - 256*a^7)/((I*a*tan(d 
*x + c) + a)^(9/2) - 6*(I*a*tan(d*x + c) + a)^(7/2)*a + 12*(I*a*tan(d*x + 
c) + a)^(5/2)*a^2 - 8*(I*a*tan(d*x + c) + a)^(3/2)*a^3))/(a*d)
 

Giac [F(-2)]

Exception generated. \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),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. \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^6\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(3/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6}d x \right ) \] Input:

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x)
 

Output:

sqrt(a)*a*(int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**6*tan(c + d*x),x)*i 
+ int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**6,x))