\(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [305]

Optimal result

Integrand size = 26, antiderivative size = 192 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {7 i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{16 \sqrt {2} d}+\frac {7 i a^2 \cos (c+d x)}{24 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 i a \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{16 d}-\frac {7 i a \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{30 d}-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d} \] Output:

7/32*I*a^(3/2)*arctanh(1/2*a^(1/2)*sec(d*x+c)*2^(1/2)/(a+I*a*tan(d*x+c))^( 
1/2))*2^(1/2)/d+7/24*I*a^2*cos(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)-7/16*I*a* 
cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-7/30*I*a*cos(d*x+c)^3*(a+I*a*tan(d*x 
+c))^(1/2)/d-1/5*I*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2)/d
 

Mathematica [A] (verified)

Time = 1.39 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.83 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {i a e^{-3 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-15+101 e^{2 i (c+d x)}+148 e^{4 i (c+d x)}+38 e^{6 i (c+d x)}+6 e^{8 i (c+d x)}-105 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{240 \sqrt {2} d} \] Input:

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-1/240*I)*a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-15 
 + 101*E^((2*I)*(c + d*x)) + 148*E^((4*I)*(c + d*x)) + 38*E^((6*I)*(c + d* 
x)) + 6*E^((8*I)*(c + d*x)) - 105*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c 
 + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*d*E^((3*I)*(c 
+ d*x)))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 3978, 3042, 3978, 3042, 3983, 3042, 3971, 3042, 3970, 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.

\(\displaystyle \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sec (c+d x)^5}dx\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {7}{10} a \int \cos ^3(c+d x) \sqrt {i \tan (c+d x) a+a}dx-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)^3}dx-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \int \frac {\cos (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \int \frac {1}{\sec (c+d x) \sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3983

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \int \cos (c+d x) \sqrt {i \tan (c+d x) a+a}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sec (c+d x)}dx}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3971

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {1}{2} a \int \frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3970

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i a \int \frac {1}{2-\frac {a \sec ^2(c+d x)}{i \tan (c+d x) a+a}}d\frac {\sec (c+d x)}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {7}{10} a \left (\frac {5}{6} a \left (\frac {3 \left (\frac {i \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} d}-\frac {i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{4 a}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}\right )-\frac {i \cos ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\right )-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\)

Input:

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]
 

Output:

((-1/5*I)*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2))/d + (7*a*(((-1/3*I) 
*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d + (5*a*(((I/2)*Cos[c + d*x]) 
/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (3*((I*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + 
d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) - (I*Cos[c + d*x] 
*Sqrt[a + I*a*Tan[c + d*x]])/d))/(4*a)))/6))/10
 

Defintions of rubi rules used

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 3970

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_S 
ymbol] :> Simp[-2*(a/(b*f))   Subst[Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/ 
Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^2, 0 
]
 

rule 3971

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

rule 3978

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/( 
a*f*m)), x] + Simp[a*((m + n)/(m*d^2))   Int[(d*Sec[e + f*x])^(m + 2)*(a + 
b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b 
^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
 

rule 3983

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[a*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/ 
(b*f*(m + 2*n))), x] + Simp[Simplify[m + n]/(a*(m + 2*n))   Int[(d*Sec[e + 
f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x 
] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2* 
n]
 

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. 794 vs. \(2 (155 ) = 310\).

Time = 4.83 (sec) , antiderivative size = 795, normalized size of antiderivative = 4.14

method	result	size
default	\(\text {Expression too large to display}\)	\(795\)

Input:

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/480/d*cos(d*x+c)*(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))*(-cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)*(210*sin(d*x+c)^2+105*tan(d*x+c)*sin(d*x+c))+I*(210*cos(d*x+c)^2 
+105*cos(d*x+c)-105)*tan(d*x+c)*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))*(-cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)+I*(-210*cos(d*x+c)-105)*sin(d*x+c)*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))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(210*cos(d*x+c)^2+105*cos(d 
*x+c)-105)*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))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2) 
+I*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2))*(-cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)*(-210*sin(d*x+c)^2-105*tan(d*x+c)*sin(d*x+c))+(210*co 
s(d*x+c)^2+105*cos(d*x+c)-105)*tan(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(cos(d 
*x+c)+1))^(1/2)*2^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(-210*cos(d*x+ 
c)-105)*sin(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2) 
)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*(-210*cos(d*x+c)^2-105*cos(d*x+c)+1 
05)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2))*(-cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)+I*sin(d*x+c)^2*(168*cos(d*x+c)^2-210)+sin(d*x+c)*cos 
(d*x+c)*(-72*cos(d*x+c)^2+350)+sin(d*x+c)*cos(d*x+c)*(168*cos(d*x+c)^2-210 
)+I*cos(d*x+c)^2*(72*cos(d*x+c)^2-350))*a*(a*(1+I*tan(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.43 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {{\left (105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, d}\right ) - 105 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{8 \, d}\right ) + \sqrt {2} {\left (-6 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 38 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 148 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 101 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{480 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

1/480*(105*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(2*I*d*x + 2*I*c)*log(7/8*(sqrt(2) 
*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 
 2*I*c) + 1)) + I*a^2)*e^(-I*d*x - I*c)/d) - 105*sqrt(1/2)*sqrt(-a^3/d^2)* 
d*e^(2*I*d*x + 2*I*c)*log(-7/8*(sqrt(2)*sqrt(1/2)*(d*e^(2*I*d*x + 2*I*c) + 
 d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) - I*a^2)*e^(-I*d*x - 
I*c)/d) + sqrt(2)*(-6*I*a*e^(8*I*d*x + 8*I*c) - 38*I*a*e^(6*I*d*x + 6*I*c) 
 - 148*I*a*e^(4*I*d*x + 4*I*c) - 101*I*a*e^(2*I*d*x + 2*I*c) + 15*I*a)*sqr 
t(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^5*(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. \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(3/2),x)
 

Output:

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^(3/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

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

int(cos(d*x+c)^5*(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)**5*tan(c + d*x),x)*i 
+ int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**5,x))