\(\int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx\) [422]

Optimal result

Integrand size = 30, antiderivative size = 206 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{315 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}} \]

[Out]

2/9*I/d/(e*sec(d*x+c))^(7/2)/(a+I*a*tan(d*x+c))^(1/2)+32/105*I/d/e^2/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(
1/2)+256/315*I*(e*sec(d*x+c))^(1/2)/d/e^4/(a+I*a*tan(d*x+c))^(1/2)-16/63*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/(e*sec
(d*x+c))^(7/2)-128/315*I*(a+I*a*tan(d*x+c))^(1/2)/a/d/e^2/(e*sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3583, 3578, 3569} \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {256 i \sqrt {e \sec (c+d x)}}{315 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}}+\frac {32 i}{105 d e^2 \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac {2 i}{9 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{7/2}} \]

[In]

Int[1/((e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((2*I)/9)/(d*(e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]) + ((32*I)/105)/(d*e^2*(e*Sec[c + d*x])^(3/2)*S
qrt[a + I*a*Tan[c + d*x]]) + (((256*I)/315)*Sqrt[e*Sec[c + d*x]])/(d*e^4*Sqrt[a + I*a*Tan[c + d*x]]) - (((16*I
)/63)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d*(e*Sec[c + d*x])^(7/2)) - (((128*I)/315)*Sqrt[a + I*a*Tan[c + d*x]])/(a
*d*e^2*(e*Sec[c + d*x])^(3/2))

Rule 3569

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 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[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 3583

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {8 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{7/2}} \, dx}{9 a} \\ & = \frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac {16 \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx}{21 e^2} \\ & = \frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}+\frac {64 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{105 a e^2} \\ & = \frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}}+\frac {128 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{315 e^4} \\ & = \frac {2 i}{9 d (e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{105 d e^2 (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{315 d e^4 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i \sqrt {a+i a \tan (c+d x)}}{63 a d (e \sec (c+d x))^{7/2}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{315 a d e^2 (e \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.42 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e \sec (c+d x)} (945 i-84 i \cos (2 (c+d x))-5 i \cos (4 (c+d x))+336 \sin (2 (c+d x))+40 \sin (4 (c+d x)))}{1260 d e^4 \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[1/((e*Sec[c + d*x])^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(Sqrt[e*Sec[c + d*x]]*(945*I - (84*I)*Cos[2*(c + d*x)] - (5*I)*Cos[4*(c + d*x)] + 336*Sin[2*(c + d*x)] + 40*Si
n[4*(c + d*x)]))/(1260*d*e^4*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 10.60 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.43

method	result	size
default	\(\frac {-\frac {2 i \left (\cos ^{3}\left (d x +c \right )\right )}{63}+\frac {16 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{63}-\frac {32 i \cos \left (d x +c \right )}{315}+\frac {128 \sin \left (d x +c \right )}{315}+\frac {256 i \sec \left (d x +c \right )}{315}}{d \sqrt {e \sec \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, e^{3}}\)	\(88\)

[In]

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

[Out]

2/315/d/(e*sec(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)/e^3*(-5*I*cos(d*x+c)^3+40*cos(d*x+c)^2*sin(d*x+c)-16*I
*cos(d*x+c)+64*sin(d*x+c)+128*I*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-45 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 465 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1470 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2142 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 287 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c\right )}}{2520 \, a d e^{4}} \]

[In]

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

[Out]

1/2520*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-45*I*e^(10*I*d*x + 10*I*c) - 465*
I*e^(8*I*d*x + 8*I*c) + 1470*I*e^(6*I*d*x + 6*I*c) + 2142*I*e^(4*I*d*x + 4*I*c) + 287*I*e^(2*I*d*x + 2*I*c) +
35*I)*e^(-9/2*I*d*x - 9/2*I*c)/(a*d*e^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.80 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {35 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) - 45 i \, \cos \left (\frac {7}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 252 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) - 420 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 1890 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 45 \, \sin \left (\frac {7}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 252 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 420 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 1890 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )}{2520 \, \sqrt {a} d e^{\frac {7}{2}}} \]

[In]

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

[Out]

1/2520*(35*I*cos(9/2*d*x + 9/2*c) - 45*I*cos(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*I*
cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 420*I*cos(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos
(9/2*d*x + 9/2*c))) + 1890*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 35*sin(9/2*d*x + 9
/2*c) + 45*sin(7/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 252*sin(5/9*arctan2(sin(9/2*d*x + 9/
2*c), cos(9/2*d*x + 9/2*c))) + 420*sin(1/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 1890*sin(1/9
*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))))/(sqrt(a)*d*e^(7/2))

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.94 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \sec (c+d x))^{7/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (336\,\sin \left (2\,c+2\,d\,x\right )-\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}-\cos \left (2\,c+2\,d\,x\right )\,84{}\mathrm {i}+40\,\sin \left (4\,c+4\,d\,x\right )+945{}\mathrm {i}\right )}{1260\,d\,e^4\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(336*sin(2*c + 2*d*x) - cos(4*c + 4*d*x)*5i - cos(2*c + 2*d*x)*84i + 40*sin(4*c + 4*d*
x) + 945i))/(1260*d*e^4*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))