\(\int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [358]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {3572, 3570, 212} \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

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

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 3572

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {2 \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{a} \\ & = -\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}}+\frac {(4 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 d} \\ & = \frac {2 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {2 i \sec (c+d x)}{a d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {8 e^{3 i (c+d x)} \left (-1+\sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{a d \left (1+e^{2 i (c+d x)}\right )^2 (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(8*E^((3*I)*(c + d*x))*(-1 + Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(a*d*(1 +
E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*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. 222 vs. \(2 (71 ) = 142\).

Time = 8.94 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.59

method	result	size
default	\(-\frac {2 \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )+i\right )^{3} \left (-\sqrt {2}\, \arctan \left (\frac {\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right ) \sqrt {2}}{2 \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}}\right ) \sqrt {\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}+i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-1\right )}{d {\left (-\frac {a \left (2 i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1}\right )}^{\frac {3}{2}} {\left (\left (\csc ^{2}\left (d x +c \right )\right ) \left (1-\cos \left (d x +c \right )\right )^{2}-1\right )}^{2}}\)	\(223\)

[In]

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

[Out]

-2/d*(-csc(d*x+c)+cot(d*x+c)+I)^3*(-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))*(csc(d*x+c)^2*(1-cos(d*x+c))^2-1)^(1/2)+I*(csc(d*x+c)-cot(d*x+c))-1)/(-a*(2*I*(csc(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))^(3/2)/(csc(d*x+c)^2*(1-cos
(d*x+c))^2-1)^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. 196 vs. \(2 (67) = 134\).

Time = 0.24 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.28 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-\frac {8 \, {\left ({\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) + i \, \sqrt {2} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-\frac {8 \, {\left ({\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a d}\right ) - 2 i \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{a^{2} d} \]

[In]

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

[Out]

(-I*sqrt(2)*a^2*d*sqrt(1/(a^3*d^2))*log(-8*((I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt(1/(a^3*d^2)) - I)*e^(-I*d*x - I*c)/(a*d)) + I*sqrt(2)*a^2*d*sqrt(1/(a^3*d^2))*log(-8*((-I*a*d*e^(2*I*
d*x + 2*I*c) - I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) - I)*e^(-I*d*x - I*c)/(a*d)) - 2*I*s
qrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(a^2*d)

Sympy [F]

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

[In]

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

[Out]

Integral(sec(c + d*x)**3/(I*a*(tan(c + d*x) - I))**(3/2), 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. 813 vs. \(2 (67) = 134\).

Time = 0.45 (sec) , antiderivative size = 813, normalized size of antiderivative = 9.45 \[ \int \frac {\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/2*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(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*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) + 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*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) + 4*(I*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c) + 1)) + sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*
d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*a^2*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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