\(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [325]

Optimal result
Mathematica [C] (verified)
Rubi [A] (warning: unable to verify)
Maple [B] (warning: unable to verify)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 189 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {5 i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}-\frac {i a^6 \sqrt {a+i a \tan (c+d x)}}{6 d (a-i a \tan (c+d x))^3}-\frac {5 i a^7 \sqrt {a+i a \tan (c+d x)}}{48 d \left (a^2-i a^2 \tan (c+d x)\right )^2}-\frac {5 i a^7 \sqrt {a+i a \tan (c+d x)}}{64 d \left (a^4-i a^4 \tan (c+d x)\right )} \] Output:

-5/128*I*a^(7/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^( 
1/2)/d-1/6*I*a^6*(a+I*a*tan(d*x+c))^(1/2)/d/(a-I*a*tan(d*x+c))^3-5/48*I*a^ 
7*(a+I*a*tan(d*x+c))^(1/2)/d/(a^2-I*a^2*tan(d*x+c))^2-5/64*I*a^7*(a+I*a*ta 
n(d*x+c))^(1/2)/d/(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 = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.28 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i a^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},4,\frac {3}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) \sqrt {a+i a \tan (c+d x)}}{8 d} \] Input:

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((-1/8*I)*a^3*Hypergeometric2F1[1/2, 4, 3/2, (1 + I*Tan[c + d*x])/2]*Sqrt[ 
a + I*a*Tan[c + d*x]])/d
 

Rubi [A] (warning: unable to verify)

Time = 0.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 3968, 52, 52, 52, 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.

\(\displaystyle \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{7/2}}{\sec (c+d x)^6}dx\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {5 \int \frac {1}{(a-i a \tan (c+d x))^3 \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{12 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{6 a (a-i a \tan (c+d x))^3}\right )}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {5 \left (\frac {3 \int \frac {1}{(a-i a \tan (c+d x))^2 \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{12 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{6 a (a-i a \tan (c+d x))^3}\right )}{d}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {i a^7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{4 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{12 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{6 a (a-i a \tan (c+d x))^3}\right )}{d}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {i a^7 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{2 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{12 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{6 a (a-i a \tan (c+d x))^3}\right )}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {i a^7 \left (\frac {5 \left (\frac {3 \left (\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{2 \sqrt {2} a^{3/2}}+\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )}{8 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{4 a (a-i a \tan (c+d x))^2}\right )}{12 a}+\frac {\sqrt {a+i a \tan (c+d x)}}{6 a (a-i a \tan (c+d x))^3}\right )}{d}\)

Input:

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(7/2),x]
 

Output:

((-I)*a^7*(Sqrt[a + I*a*Tan[c + d*x]]/(6*a*(a - I*a*Tan[c + d*x])^3) + (5* 
(Sqrt[a + I*a*Tan[c + d*x]]/(4*a*(a - I*a*Tan[c + d*x])^2) + (3*(((I/2)*Ar 
cTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) + Sqrt[a + I*a*Tan 
[c + d*x]]/(2*a*(a - I*a*Tan[c + d*x]))))/(8*a)))/(12*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 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. 531 vs. \(2 (155 ) = 310\).

Time = 2.20 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.81

\[\frac {\cos \left (d x +c \right )^{3} \left (\left (60 \cos \left (d x +c \right )^{3}+60 \cos \left (d x +c \right )^{2}-15 \cos \left (d x +c \right )-15\right ) \sin \left (d x +c \right ) \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \operatorname {arctanh}\left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ) \sqrt {2}}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (60 \cos \left (d x +c \right )^{4}+60 \cos \left (d x +c \right )^{3}-45 \cos \left (d x +c \right )^{2}-45 \cos \left (d x +c \right )\right )+i \left (60 \cos \left (d x +c \right )^{3}+60 \cos \left (d x +c \right )^{2}-15 \cos \left (d x +c \right )-15\right ) \sin \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right )+\cos \left (d x +c \right ) \left (-60 \cos \left (d x +c \right )^{3}-60 \cos \left (d x +c \right )^{2}+45 \cos \left (d x +c \right )+45\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right )+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (32 \cos \left (d x +c \right )^{2}-50 \cos \left (d x +c \right )-15\right )+\cos \left (d x +c \right ) \left (32 \cos \left (d x +c \right )^{3}+82 \cos \left (d x +c \right )^{2}+35 \cos \left (d x +c \right )-15\right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{3} a^{3}}{192 d \left (1+\cos \left (d x +c \right )+i \sin \left (d x +c \right )\right )}\]

Input:

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x)
 

Output:

1/192/d*cos(d*x+c)^3*((60*cos(d*x+c)^3+60*cos(d*x+c)^2-15*cos(d*x+c)-15)*s 
in(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*arctanh(1/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)*(c 
sc(d*x+c)-cot(d*x+c))*2^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(60*cos( 
d*x+c)^4+60*cos(d*x+c)^3-45*cos(d*x+c)^2-45*cos(d*x+c))+I*(60*cos(d*x+c)^3 
+60*cos(d*x+c)^2-15*cos(d*x+c)-15)*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))+cos(d*x+c) 
*(-60*cos(d*x+c)^3-60*cos(d*x+c)^2+45*cos(d*x+c)+45)*(-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*s 
in(d*x+c)*cos(d*x+c)*(32*cos(d*x+c)^2-50*cos(d*x+c)-15)+cos(d*x+c)*(32*cos 
(d*x+c)^3+82*cos(d*x+c)^2+35*cos(d*x+c)-15))*(a*(1+I*tan(d*x+c)))^(1/2)*(- 
tan(d*x+c)+I)^3*a^3/(1+cos(d*x+c)+I*sin(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.47 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - 15 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - \sqrt {2} {\left (-8 i \, a^{3} e^{\left (7 i \, d x + 7 i \, c\right )} - 34 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 59 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 33 i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{384 \, d} \] Input:

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-1/384*(15*sqrt(1/2)*sqrt(-a^7/d^2)*d*log(4*(a^4*e^(I*d*x + I*c) - sqrt(2) 
*sqrt(1/2)*sqrt(-a^7/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d 
*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) - 15*sqrt(1/2)*sqrt(-a^7/d^2)*d*l 
og(4*(a^4*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^7/d^2)*(-I*d*e^(2*I* 
d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^ 
3) - sqrt(2)*(-8*I*a^3*e^(7*I*d*x + 7*I*c) - 34*I*a^3*e^(5*I*d*x + 5*I*c) 
- 59*I*a^3*e^(3*I*d*x + 3*I*c) - 33*I*a^3*e^(I*d*x + I*c))*sqrt(a/(e^(2*I* 
d*x + 2*I*c) + 1)))/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {i \, {\left (15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{5} - 80 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{6} + 132 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}}\right )}}{768 \, a d} \] Input:

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

1/768*I*(15*sqrt(2)*a^(9/2)*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*(15*(I*a*tan(d*x 
 + c) + a)^(5/2)*a^5 - 80*(I*a*tan(d*x + c) + a)^(3/2)*a^6 + 132*sqrt(I*a* 
tan(d*x + c) + a)*a^7)/((I*a*tan(d*x + c) + a)^3 - 6*(I*a*tan(d*x + c) + a 
)^2*a + 12*(I*a*tan(d*x + c) + a)*a^2 - 8*a^3))/(a*d)
 

Giac [F(-2)]

Exception generated. \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cos(d*x+c)^6*(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. \[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^6\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \] Input:

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(7/2),x)
 

Output:

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(7/2), x)
 

Reduce [F]

\[ \int \cos ^6(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6} \tan \left (d x +c \right )^{3}d x \right ) i -3 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6} \tan \left (d x +c \right )^{2}d x \right )+3 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{6}d x \right ) \] Input:

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(7/2),x)
                                                                                    
                                                                                    
 

Output:

sqrt(a)*a**3*( - int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**6*tan(c + d*x) 
**3,x)*i - 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**6*tan(c + d*x)**2, 
x) + 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**6*tan(c + d*x),x)*i + in 
t(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**6,x))