\(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [374]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} a^{5/2} d}+\frac {i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3582, 3583, 3570, 212} \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} a^{5/2} d} \]

[In]

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-I)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*a^(5/2)*d) + (I*Sec[c + d
*x])/(a*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)}{a d (a+i a \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\sec (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{a} \\ & = \frac {i \sec (c+d x)}{a 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}{2 a^2} \\ & = \frac {i \sec (c+d x)}{a 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 )}{a^2 d} \\ & = -\frac {i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {2} a^{5/2} d}+\frac {i \sec (c+d x)}{a d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.73 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i e^{-\frac {1}{2} i (2 c+d x)} \left (-1-e^{2 i (c+d x)}+e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sec ^3(c+d x) \left (\cos \left (c+\frac {d x}{2}\right )+i \sin \left (c+\frac {d x}{2}\right )\right )}{2 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

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. 395 vs. \(2 (71 ) = 142\).

Time = 9.18 (sec) , antiderivative size = 396, normalized size of antiderivative = 4.60

method	result	size
default	\(\frac {\left (\sqrt {2}\, \arctan \left (\frac {\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right ) \sqrt {2}}{2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right )+2 i \arctan \left (\frac {\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right ) \sqrt {2}}{2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right ) \sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\left (\csc ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right ) \sqrt {2}}{2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right ) \sqrt {2}\, \left (1-\cos \left (d x +c \right )\right )^{2}-2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}+2 i \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right ) \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )+i\right )^{3}}{2 d {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )}^{\frac {5}{2}} {\left (-\frac {a \left (2 i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )}^{\frac {5}{2}}}\)	\(396\)

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/2/d*(2^(1/2)*arctan(1/2*(I*(csc(d*x+c)-cot(d*x+c))-1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/2))+2*I*a
rctan(1/2*(I*(csc(d*x+c)-cot(d*x+c))-1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/2))*2^(1/2)*(csc(d*x+c)-c
ot(d*x+c))-csc(d*x+c)^2*arctan(1/2*(I*(csc(d*x+c)-cot(d*x+c))-1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/
2))*2^(1/2)*(1-cos(d*x+c))^2-2*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/2)+2*I*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(
1/2)*(csc(d*x+c)-cot(d*x+c)))*(-csc(d*x+c)+cot(d*x+c)+I)^3/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(5/2)/(-a*(2*I*(c
sc(d*x+c)-cot(d*x+c))-csc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(5/2)

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. 245 vs. \(2 (67) = 134\).

Time = 0.25 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.85 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (i \, \sqrt {2} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) - i \, \sqrt {2} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d}\right ) - 2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/4*(I*sqrt(2)*a^3*d*sqrt(1/(a^5*d^2))*e^(2*I*d*x + 2*I*c)*log(2*((I*a^2*d*e^(2*I*d*x + 2*I*c) + I*a^2*d)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) - I*sqrt(2)*a^3*d*sqrt(1/(a^5*d
^2))*e^(2*I*d*x + 2*I*c)*log(2*((-I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqr
t(1/(a^5*d^2)) - I)*e^(-I*d*x - I*c)/(a^2*d)) - 2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-I*e^(2*I*d*x + 2
*I*c) - I))*e^(-2*I*d*x - 2*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(5/2), 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. 827 vs. \(2 (67) = 134\).

Time = 0.70 (sec) , antiderivative size = 827, normalized size of antiderivative = 9.62 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/8*(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(2*d*x + 2*c)
- sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2)*cos(2*d*x + 2
*c) - I*sqrt(2)*sin(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a) - (2*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) - 2*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) - 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*ar
ctan2(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) + 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)*sin(1/2*arc
tan2(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^3*d)

Giac [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(5/2)), x)