\(\int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [365]

Optimal result

Integrand size = 26, antiderivative size = 55 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {4 i}{a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i \sqrt {a+i a \tan (c+d x)}}{a^3 d} \]

[Out]

4*I/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+2*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i \sqrt {a+i a \tan (c+d x)}}{a^3 d}+\frac {4 i}{a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int \frac {a-x}{(a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (\frac {2 a}{(a+x)^{3/2}}-\frac {1}{\sqrt {a+x}}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = \frac {4 i}{a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i \sqrt {a+i a \tan (c+d x)}}{a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {6 i-2 \tan (c+d x)}{a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}+\frac {2 a}{\sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\)	\(42\)
default	\(\frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}+\frac {2 a}{\sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\)	\(42\)

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3} d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**4/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i \, {\left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{a^{2}} + \frac {2}{\sqrt {i \, a \tan \left (d x + c\right ) + a} a}\right )}}{a d} \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

2*I*(sqrt(I*a*tan(d*x + c) + a)/a^2 + 2/(sqrt(I*a*tan(d*x + c) + a)*a))/(a*d)

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+2{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}}{a^3\,d} \]

[In]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

(2*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) + 2i)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c +
2*d*x) + 1))^(1/2))/(a^3*d)