\(\int (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx\) [682]

Optimal result

Integrand size = 30, antiderivative size = 179 \[ \int (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {12 i a (e \cos (c+d x))^{7/2} \sec ^2(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {32 i a (e \cos (c+d x))^{7/2} \sec ^4(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {16 i (e \cos (c+d x))^{7/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d} \] Output:

12/35*I*a*(e*cos(d*x+c))^(7/2)*sec(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)+32/ 
35*I*a*(e*cos(d*x+c))^(7/2)*sec(d*x+c)^4/d/(a+I*a*tan(d*x+c))^(1/2)-2/7*I* 
(e*cos(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^(1/2)/d-16/35*I*(e*cos(d*x+c))^(7/ 
2)*sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 1.66 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {e^3 \sqrt {e \cos (c+d x)} (35 i \cos (c+d x)+i \cos (3 (c+d x))+70 \sin (c+d x)+6 \sin (3 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{70 d} \] Input:

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

Output:

(e^3*Sqrt[e*Cos[c + d*x]]*((35*I)*Cos[c + d*x] + I*Cos[3*(c + d*x)] + 70*S 
in[c + d*x] + 6*Sin[3*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]])/(70*d)
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3998, 3042, 3978, 3042, 3983, 3042, 3978, 3042, 3969}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3998

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3978

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3983

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3978

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3969

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3969

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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ 
[Simplify[m + 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]
 

rule 3998

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m   Int[( 
a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m 
, n}, x] &&  !IntegerQ[m]
 

Maple [A] (verified)

Time = 10.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.44

method	result	size
default	\(-\frac {2 e^{3} \sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-6 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}-i \cos \left (d x +c \right )^{3}-16 \sin \left (d x +c \right )-8 i \cos \left (d x +c \right )\right )}{35 d}\)	\(79\)

Input:

int((e*cos(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/35/d*e^3*(e*cos(d*x+c))^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(-6*sin(d*x+c) 
*cos(d*x+c)^2-I*cos(d*x+c)^3-16*sin(d*x+c)-8*I*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.56 \[ \int (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-5 i \, e^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 35 i \, e^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 105 i \, e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i \, e^{3}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{140 \, d} \] Input:

integrate((e*cos(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fric 
as")
 

Output:

1/140*sqrt(2)*sqrt(1/2)*(-5*I*e^3*e^(6*I*d*x + 6*I*c) - 35*I*e^3*e^(4*I*d* 
x + 4*I*c) + 105*I*e^3*e^(2*I*d*x + 2*I*c) + 7*I*e^3)*sqrt(e*e^(2*I*d*x + 
2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-5/2*I*d*x - 5/2*I*c)/d
 

Sympy [F(-1)]

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

integrate((e*cos(d*x+c))**(7/2)*(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.13 \[ \int (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (7 i \, e^{3} \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, e^{3} \cos \left (\frac {7}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) - 35 i \, e^{3} \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 105 i \, e^{3} \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 7 \, e^{3} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, e^{3} \sin \left (\frac {7}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 35 \, e^{3} \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 105 \, e^{3} \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} \sqrt {a} \sqrt {e}}{140 \, d} \] Input:

integrate((e*cos(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxi 
ma")
 

Output:

1/140*(7*I*e^3*cos(5/2*d*x + 5/2*c) - 5*I*e^3*cos(7/5*arctan2(sin(5/2*d*x 
+ 5/2*c), cos(5/2*d*x + 5/2*c))) - 35*I*e^3*cos(3/5*arctan2(sin(5/2*d*x + 
5/2*c), cos(5/2*d*x + 5/2*c))) + 105*I*e^3*cos(1/5*arctan2(sin(5/2*d*x + 5 
/2*c), cos(5/2*d*x + 5/2*c))) + 7*e^3*sin(5/2*d*x + 5/2*c) + 5*e^3*sin(7/5 
*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 35*e^3*sin(3/5*arc 
tan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 105*e^3*sin(1/5*arctan 
2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))))*sqrt(a)*sqrt(e)/d
 

Giac [F(-2)]

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

integrate((e*cos(d*x+c))^(7/2)*(a+I*a*tan(d*x+c))^(1/2),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 [B] (verification not implemented)

Time = 2.56 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.54 \[ \int (e \cos (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {e^3\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\sin \left (c+d\,x\right )+\frac {3\,\sin \left (3\,c+3\,d\,x\right )}{35}+\frac {\cos \left (c+d\,x\right )\,1{}\mathrm {i}}{2}+\frac {\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{70}\right )}{d} \] Input:

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

Output:

(e^3*(e*cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 
1))/(cos(2*c + 2*d*x) + 1))^(1/2)*((cos(c + d*x)*1i)/2 + sin(c + d*x) + (c 
os(3*c + 3*d*x)*1i)/70 + (3*sin(3*c + 3*d*x))/35))/d
 

Reduce [F]

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

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

Output:

sqrt(e)*sqrt(a)*int(sqrt(tan(c + d*x)*i + 1)*sqrt(cos(c + d*x))*cos(c + d* 
x)**3,x)*e**3