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

Optimal. Leaf size=160 \[ -\frac{4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{8 i \sec (c+d x)}{a^3 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((8*I)*Sqrt[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(a^(7/2)*d) - (((2*I)/5)*
Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) - (((4*I)/3)*Sec[c + d*x]^3)/(a^2*d*(a + I*a*Tan[c + d*x])^
(3/2)) - ((8*I)*Sec[c + d*x])/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.251263, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3491, 3489, 206} \[ -\frac{4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{8 i \sec (c+d x)}{a^3 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*I)*Sqrt[2]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(a^(7/2)*d) - (((2*I)/5)*
Sec[c + d*x]^5)/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) - (((4*I)/3)*Sec[c + d*x]^3)/(a^2*d*(a + I*a*Tan[c + d*x])^
(3/2)) - ((8*I)*Sec[c + d*x])/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3491

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

Rule 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^7(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}+\frac{2 \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx}{a}\\ &=-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}-\frac{4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac{4 \int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{a^2}\\ &=-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}-\frac{4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{8 i \sec (c+d x)}{a^3 d \sqrt{a+i a \tan (c+d x)}}+\frac{8 \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{a^3}\\ &=-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}-\frac{4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{8 i \sec (c+d x)}{a^3 d \sqrt{a+i a \tan (c+d x)}}+\frac{(16 i) \operatorname{Subst}\left (\int \frac{1}{2-a x^2} \, dx,x,\frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^3 d}\\ &=\frac{8 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{7/2} d}-\frac{2 i \sec ^5(c+d x)}{5 a d (a+i a \tan (c+d x))^{5/2}}-\frac{4 i \sec ^3(c+d x)}{3 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac{8 i \sec (c+d x)}{a^3 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.11821, size = 130, normalized size = 0.81 \[ -\frac{128 e^{7 i (c+d x)} \left (-35 e^{2 i (c+d x)}-15 e^{4 i (c+d x)}+15 \left (1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-23\right )}{15 a^3 d \left (1+e^{2 i (c+d x)}\right )^6 (\tan (c+d x)-i)^3 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-128*E^((7*I)*(c + d*x))*(-23 - 35*E^((2*I)*(c + d*x)) - 15*E^((4*I)*(c + d*x)) + 15*(1 + E^((2*I)*(c + d*x))
)^(5/2)*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(15*a^3*d*(1 + E^((2*I)*(c + d*x)))^6*(-I + Tan[c + d*x])^3*S
qrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.302, size = 399, normalized size = 2.5 \begin{align*}{\frac{2}{15\,{a}^{4}d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( -15\,\sqrt{2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}-30\,\sqrt{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}-15\,\sqrt{2}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{5/2}\sin \left ( dx+c \right ) +92\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-76\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+92\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -19\,i\cos \left ( dx+c \right ) -16\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +3\,i-3\,\sin \left ( dx+c \right ) \right ) \sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15/d/a^4*(-15*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-30*2^(1/2)*sin(d*x+c)*cos(d*x+c)*arctan
(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(5/2)-15*2^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*sin(d*x+c)+92*I*cos(d*x+c)^3-76*I*cos(d*x+c)^2+92*cos(d*x+c
)^2*sin(d*x+c)-19*I*cos(d*x+c)-16*cos(d*x+c)*sin(d*x+c)+3*I-3*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x
+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^2

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Maxima [B]  time = 2.35029, size = 1574, normalized size = 9.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*((60*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*ar
ctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 60*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c
)^2 + 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2
*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - (30*I*sqrt
(2)*cos(2*d*x + 2*c)^2 + 30*I*sqrt(2)*sin(2*d*x + 2*c)^2 + 60*I*sqrt(2)*cos(2*d*x + 2*c) + 30*I*sqrt(2))*log(s
qrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1
)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - (-30*I*sqrt(2)*cos(2*d*x + 2*c)^2 - 30
*I*sqrt(2)*sin(2*d*x + 2*c)^2 - 60*I*sqrt(2)*cos(2*d*x + 2*c) - 30*I*sqrt(2))*log(sqrt(cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(c
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1))^2 - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
^(1/4)*sqrt(a) - ((-120*I*sqrt(2)*cos(4*d*x + 4*c) - 280*I*sqrt(2)*cos(2*d*x + 2*c) + 120*sqrt(2)*sin(4*d*x +
4*c) + 280*sqrt(2)*sin(2*d*x + 2*c) - 184*I*sqrt(2))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
- (120*sqrt(2)*cos(4*d*x + 4*c) + 280*sqrt(2)*cos(2*d*x + 2*c) + 120*I*sqrt(2)*sin(4*d*x + 4*c) + 280*I*sqrt(2
)*sin(2*d*x + 2*c) + 184*sqrt(2))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a))/((a^4*cos
(2*d*x + 2*c)^2 + a^4*sin(2*d*x + 2*c)^2 + 2*a^4*cos(2*d*x + 2*c) + a^4)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c
)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*d)

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Fricas [B]  time = 2.25258, size = 1052, normalized size = 6.58 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-120 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 280 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 184 i\right )} e^{\left (i \, d x + i \, c\right )} + \sqrt{2}{\left (60 i \, a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 120 i \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + 60 i \, a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt{\frac{1}{a^{7} d^{2}}} \log \left ({\left (\sqrt{2} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt{2}{\left (-60 i \, a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} - 120 i \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - 60 i \, a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt{\frac{1}{a^{7} d^{2}}} \log \left (-{\left (\sqrt{2} a^{4} d \sqrt{\frac{1}{a^{7} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{15 \,{\left (a^{4} d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-120*I*e^(4*I*d*x + 4*I*c) - 280*I*e^(2*I*d*x + 2*I*c) - 184*
I)*e^(I*d*x + I*c) + sqrt(2)*(60*I*a^4*d*e^(5*I*d*x + 5*I*c) + 120*I*a^4*d*e^(3*I*d*x + 3*I*c) + 60*I*a^4*d*e^
(I*d*x + I*c))*sqrt(1/(a^7*d^2))*log((sqrt(2)*a^4*d*sqrt(1/(a^7*d^2))*e^(I*d*x + I*c) + sqrt(2)*sqrt(a/(e^(2*I
*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*(-60*I*a^4*d*e^(5*I
*d*x + 5*I*c) - 120*I*a^4*d*e^(3*I*d*x + 3*I*c) - 60*I*a^4*d*e^(I*d*x + I*c))*sqrt(1/(a^7*d^2))*log(-(sqrt(2)*
a^4*d*sqrt(1/(a^7*d^2))*e^(I*d*x + I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*
e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a^4*d*e^(5*I*d*x + 5*I*c) + 2*a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x
+ I*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{7}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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