\(\int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [988]

Optimal result

Integrand size = 31, antiderivative size = 27 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 i a}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

-2/5*I*a/f/(c-I*c*tan(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 32} \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 i a}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[In]

Int[(a + I*a*Tan[e + f*x])/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(((-2*I)/5)*a)/(f*(c - I*c*Tan[e + f*x])^(5/2))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = (a c) \int \frac {\sec ^2(e+f x)}{(c-i c \tan (e+f x))^{7/2}} \, dx \\ & = \frac {(i a) \text {Subst}\left (\int \frac {1}{(c+x)^{7/2}} \, dx,x,-i c \tan (e+f x)\right )}{f} \\ & = -\frac {2 i a}{5 f (c-i c \tan (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 i a}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

(((-2*I)/5)*a)/(f*(c - I*c*Tan[e + f*x])^(5/2))

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {2 i a}{5 f \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(22\)
default	\(-\frac {2 i a}{5 f \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(22\)
risch	\(-\frac {i a \left ({\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {2}}{20 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(53\)
parts	\(\frac {2 i a c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 c^{\frac {7}{2}}}-\frac {1}{8 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{10 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}+\frac {i a \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}-\frac {1}{5 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f}\)	\(192\)

[In]

int((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/5*I*a/f/(c-I*c*tan(f*x+e))^(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. 68 vs. \(2 (19) = 38\).

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (-i \, a e^{\left (6 i \, f x + 6 i \, e\right )} - 3 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 3 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{20 \, c^{3} f} \]

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/20*sqrt(2)*(-I*a*e^(6*I*f*x + 6*I*e) - 3*I*a*e^(4*I*f*x + 4*I*e) - 3*I*a*e^(2*I*f*x + 2*I*e) - I*a)*sqrt(c/(
e^(2*I*f*x + 2*I*e) + 1))/(c^3*f)

Sympy [A] (verification not implemented)

Time = 8.45 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=\begin {cases} - \frac {2 i a}{5 f \left (- i c \tan {\left (e + f x \right )} + c\right )^{\frac {5}{2}}} & \text {for}\: f \neq 0 \\\frac {x \left (i a \tan {\left (e \right )} + a\right )}{\left (- i c \tan {\left (e \right )} + c\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Piecewise((-2*I*a/(5*f*(-I*c*tan(e + f*x) + c)**(5/2)), Ne(f, 0)), (x*(I*a*tan(e) + a)/(-I*c*tan(e) + c)**(5/2
), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {2 i \, a}{5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} f} \]

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2/5*I*a/((-I*c*tan(f*x + e) + c)^(5/2)*f)

Giac [F]

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

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)/(-I*c*tan(f*x + e) + c)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 6.95 (sec) , antiderivative size = 118, normalized size of antiderivative = 4.37 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (-{\cos \left (e+f\,x\right )}^2\,6{}\mathrm {i}-{\cos \left (2\,e+2\,f\,x\right )}^2\,6{}\mathrm {i}-{\cos \left (3\,e+3\,f\,x\right )}^2\,2{}\mathrm {i}+3\,\sin \left (2\,e+2\,f\,x\right )+3\,\sin \left (4\,e+4\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )+6{}\mathrm {i}\right )}{20\,c^3\,f} \]

[In]

int((a + a*tan(e + f*x)*1i)/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

(a*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/2)*(3*sin(2*e + 2*f*x) + 3*sin(4*e +
4*f*x) + sin(6*e + 6*f*x) - cos(2*e + 2*f*x)^2*6i - cos(3*e + 3*f*x)^2*2i - cos(e + f*x)^2*6i + 6i))/(20*c^3*f
)