\(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [390]

Optimal result

Integrand size = 26, antiderivative size = 125 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}}-\frac {i \sec (c+d x)}{8 a^2 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

-1/16*I*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(7/2)/d*2^(1/2)+1/2*I*sec(d*x+c)/a/
d/(a+I*a*tan(d*x+c))^(5/2)-1/8*I*sec(d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3582, 3583, 3570, 212} \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{8 \sqrt {2} a^{7/2} d}-\frac {i \sec (c+d x)}{8 a^2 d (a+i a \tan (c+d x))^{3/2}}+\frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((-1/8*I)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(7/2)*d) + ((I/2)*S
ec[c + d*x])/(a*d*(a + I*a*Tan[c + d*x])^(5/2)) - ((I/8)*Sec[c + d*x])/(a^2*d*(a + I*a*Tan[c + d*x])^(3/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3570

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 3582

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d^2*(
d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Dist[d^2*((m - 2)/(a*(m + n - 1
))), 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] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && 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 \sec (c+d x)}{3 a d (a+i a \tan (c+d x))^{5/2}}-\frac {2 \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx}{3 a} \\ & = \frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}}-\frac {i \sec (c+d x)}{8 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{16 a^3} \\ & = \frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}}-\frac {i \sec (c+d x)}{8 a^2 d (a+i a \tan (c+d x))^{3/2}}-\frac {i \text {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 )}{8 a^3 d} \\ & = -\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{8 \sqrt {2} a^{7/2} d}+\frac {i \sec (c+d x)}{2 a d (a+i a \tan (c+d x))^{5/2}}-\frac {i \sec (c+d x)}{8 a^2 d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {i \sec ^3(c+d x) \left (-3+e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-3 \cos (2 (c+d x))+i \sin (2 (c+d x))\right )}{16 a^3 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

((I/16)*Sec[c + d*x]^3*(-3 + E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c +
d*x))]] - 3*Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]))/(a^3*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 794 vs. \(2 (100 ) = 200\).

Time = 10.65 (sec) , antiderivative size = 795, normalized size of antiderivative = 6.36

method	result	size
default	\(\frac {8 i \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )+4 i \tan \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-4 i \tan \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+8 \cos \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-4 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-4 i \tan \left (d x +c \right ) \sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+4 \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-4 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-i \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-8 \sec \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-4 \sec \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-3 \left (\sec ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-2 \left (\sec ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (\sec ^{3}\left (d x +c \right )\right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-2 \left (\sec ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{16 d \left (\tan \left (d x +c \right )-i\right )^{3} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{3} \left (\cos \left (d x +c \right )+1\right )}\)	\(795\)

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/16/d/(tan(d*x+c)-I)^3/(a*(1+I*tan(d*x+c)))^(1/2)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a^3/(cos(d*x+c)+1)*(8*I*
arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+4*I*tan(d
*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-4*I*tan(d*x+c)
*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+8*cos(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d
*x+c)/(cos(d*x+c)+1))^(1/2))-4*I*tan(d*x+c)*sec(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(
-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-4*I*tan(d*x+c)*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+4*arctan(1/2*(
I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-4*(-cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)-I*tan(d*x+c)*sec(d*x+c)^2*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+
1))^(1/2))-8*sec(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/
2))-4*sec(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*sec(d*x+c)^2*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos
(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2*sec(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+sec(d*x+c)^3*
arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2*sec(d*x+c)^3*(-cos
(d*x+c)/(cos(d*x+c)+1))^(1/2))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (94) = 188\).

Time = 0.26 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.14 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{3} d}\right ) + i \, \sqrt {\frac {1}{2}} a^{4} d \sqrt {\frac {1}{a^{7} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{7} d^{2}}} + i\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{3} d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Integral(sec(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(7/2), 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. 977 vs. \(2 (94) = 188\).

Time = 0.46 (sec) , antiderivative size = 977, normalized size of antiderivative = 7.82 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-1/64*(4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*((-I*sqrt(2)*cos(4*d*x + 4*c
) - sqrt(2)*sin(4*d*x + 4*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(4*d*x +
4*c) - I*sqrt(2)*sin(4*d*x + 4*c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(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)*((-I*sqrt(2)*cos(4*d*x + 4*c) - sqrt(2)*si
n(4*d*x + 4*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(4*d*x + 4*c) - I*sqrt(
2)*sin(4*d*x + 4*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - (2*sqrt(2)*arctan2((c
os(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(si
n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 2*sqrt(2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*c
os(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)
- I*sqrt(2)*log(sqrt(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)*s
in(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) + I*sqrt(2)*log(sqrt(co
s(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(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))*sqrt(a))/(a^4*d)

Giac [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(I*a*tan(d*x + c) + a)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(7/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(7/2)), x)