\(\int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [342]

Optimal result

Integrand size = 26, antiderivative size = 73 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

8/35*I*a^2*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(5/2)+2/7*I*a*sec(d*x+c)^5/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

Int[Sec[c + d*x]^5/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((8*I)/35)*a^2*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((2*I)/7)*a*Sec[c + d*x]^5)/(d*(a + I*a*Ta
n[c + d*x])^(3/2))

Rule 3574

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

Rule 3575

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 - 1)/(f*(m + n - 1))), x] + Dist[a*((m + 2*n - 2)/(m + n - 1)), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{7} (4 a) \int \frac {\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx \\ & = \frac {8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \sec ^3(c+d x) (\cos (2 (c+d x))-i \sin (2 (c+d x))) (-9 i+5 \tan (c+d x))}{35 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^5/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(-2*Sec[c + d*x]^3*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(-9*I + 5*Tan[c + d*x]))/(35*d*Sqrt[a + I*a*Tan[c +
d*x]])

Maple [A] (verified)

Time = 7.52 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\frac {16 i \sec \left (d x +c \right )}{35}+\frac {16 \sec \left (d x +c \right ) \tan \left (d x +c \right )}{35}+\frac {2 i \left (\sec ^{3}\left (d x +c \right )\right )}{35}+\frac {2 \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{7}}{d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(72\)

[In]

int(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/35/d/(a*(1+I*tan(d*x+c)))^(1/2)*(8*I*sec(d*x+c)+8*sec(d*x+c)*tan(d*x+c)+I*sec(d*x+c)^3+5*tan(d*x+c)*sec(d*x+
c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )}}{35 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-16/35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-7*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a*d*e^(6*I*d*x + 6*I*c) + 3
*a*d*e^(4*I*d*x + 4*I*c) + 3*a*d*e^(2*I*d*x + 2*I*c) + a*d)

Sympy [F]

\[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

[In]

integrate(sec(d*x+c)**5/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**5/sqrt(I*a*(tan(c + d*x) - I)), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (57) = 114\).

Time = 0.35 (sec) , antiderivative size = 340, normalized size of antiderivative = 4.66 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \, {\left (-9 i \, \sqrt {a} - \frac {26 \, \sqrt {a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 i \, \sqrt {a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {14 \, \sqrt {a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, \sqrt {a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 i \, \sqrt {a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {26 \, \sqrt {a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {9 i \, \sqrt {a} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1}}{35 \, {\left (a - \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d \sqrt {-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/35*(-9*I*sqrt(a) - 26*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 14*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 - 14*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*I
*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 26*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9*I*sqrt(a)*si
n(d*x + c)^8/(cos(d*x + c) + 1)^8)*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*sqrt(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/((a - 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d*sqrt(-2*I*sin(d*x + c)/(cos(d*x + c) +
1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1))

Giac [F]

\[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )^{5}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/sqrt(I*a*tan(d*x + c) + a), x)

Mupad [B] (verification not implemented)

Time = 10.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^5(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {16\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,7{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}}{35\,a\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

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

[Out]

(16*exp(- c*1i - d*x*1i)*(exp(c*2i + d*x*2i)*7i + 2i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x
*2i) + 1))^(1/2))/(35*a*d*(exp(c*2i + d*x*2i) + 1)^3)