3.1172 \(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=444 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f (c-i d)^{5/2}}+\frac {d \left (15 c^3+85 i c^2 d-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 f (c-i d) (c+i d)^4 (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^4+80 i c^3 d-182 c^2 d^2+1224 i c d^3+707 d^4\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 f (c-i d)^2 (c+i d)^5 \sqrt {c+d \tan (e+f x)}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 f (-d+i c)^3 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {-21 d+5 i c}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \]

[Out]

-1/8*I*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))/a^(5/2)/(c-I*d)^
(5/2)/f*2^(1/2)+1/60*d*(15*c^4+80*I*c^3*d-182*c^2*d^2+1224*I*c*d^3+707*d^4)*(a+I*a*tan(f*x+e))^(1/2)/a^3/(c-I*
d)^2/(c+I*d)^5/f/(c+d*tan(f*x+e))^(1/2)+1/20*(5*c^2+30*I*c*d-89*d^2)/a^2/(I*c-d)^3/f/(a+I*a*tan(f*x+e))^(1/2)/
(c+d*tan(f*x+e))^(3/2)+1/60*d*(15*c^3+85*I*c^2*d-221*c*d^2+361*I*d^3)*(a+I*a*tan(f*x+e))^(1/2)/a^3/(c-I*d)/(c+
I*d)^4/f/(c+d*tan(f*x+e))^(3/2)-1/5/(I*c-d)/f/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2)+1/30*(5*I*c-21*d
)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2)

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Rubi [A]  time = 1.79, antiderivative size = 444, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3559, 3596, 3598, 12, 3544, 208} \[ \frac {d \left (-182 c^2 d^2+80 i c^3 d+15 c^4+1224 i c d^3+707 d^4\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 f (c-i d)^2 (c+i d)^5 \sqrt {c+d \tan (e+f x)}}+\frac {d \left (85 i c^2 d+15 c^3-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 f (c-i d) (c+i d)^4 (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 f (-d+i c)^3 \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f (c-i d)^{5/2}}+\frac {-21 d+5 i c}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((-I/4)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[
2]*a^(5/2)*(c - I*d)^(5/2)*f) - 1/(5*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)*(c + d*Tan[e + f*x])^(3/2)) + ((
5*I)*c - 21*d)/(30*a*(c + I*d)^2*f*(a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2)) + (5*c^2 + (30*I)*
c*d - 89*d^2)/(20*a^2*(I*c - d)^3*f*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)) + (d*(15*c^3 + (85*
I)*c^2*d - 221*c*d^2 + (361*I)*d^3)*Sqrt[a + I*a*Tan[e + f*x]])/(60*a^3*(c - I*d)*(c + I*d)^4*f*(c + d*Tan[e +
f*x])^(3/2)) + (d*(15*c^4 + (80*I)*c^3*d - 182*c^2*d^2 + (1224*I)*c*d^3 + 707*d^4)*Sqrt[a + I*a*Tan[e + f*x]]
)/(60*a^3*(c - I*d)^2*(c + I*d)^5*f*Sqrt[c + d*Tan[e + f*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3596

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3598

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (5 i c-13 d)-4 i a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx}{5 a^2 (i c-d)}\\ &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}-\frac {\int \frac {-\frac {3}{4} a^2 \left (5 c^2+20 i c d-47 d^2\right )-\frac {3}{2} a^2 (5 c+21 i d) d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{5/2}} \, dx}{15 a^4 (c+i d)^2}\\ &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {\int \frac {\sqrt {a+i a \tan (e+f x)} \left (\frac {3}{8} a^3 \left (5 i c^3-35 c^2 d-135 i c d^2+361 d^3\right )+\frac {3}{2} a^3 d \left (5 i c^2-30 c d-89 i d^2\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{5/2}} \, dx}{15 a^6 (i c-d)^3}\\ &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^3+85 i c^2 d-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f (c+d \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\sqrt {a+i a \tan (e+f x)} \left (\frac {3}{16} a^4 \left (15 i c^4-90 c^3 d-260 i c^2 d^2+502 c d^3-707 i d^4\right )+\frac {3}{8} a^4 d \left (15 i c^3-85 c^2 d-221 i c d^2-361 d^3\right ) \tan (e+f x)\right )}{(c+d \tan (e+f x))^{3/2}} \, dx}{45 a^7 (i c-d)^3 \left (c^2+d^2\right )}\\ &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^3+85 i c^2 d-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^4+80 i c^3 d-182 c^2 d^2+1224 i c d^3+707 d^4\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d)^2 (c+i d)^5 f \sqrt {c+d \tan (e+f x)}}-\frac {4 \int \frac {45 a^5 (i c-d)^5 \sqrt {a+i a \tan (e+f x)}}{32 \sqrt {c+d \tan (e+f x)}} \, dx}{45 a^8 (i c-d)^3 \left (c^2+d^2\right )^2}\\ &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^3+85 i c^2 d-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^4+80 i c^3 d-182 c^2 d^2+1224 i c d^3+707 d^4\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d)^2 (c+i d)^5 f \sqrt {c+d \tan (e+f x)}}+\frac {\int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^3 (c-i d)^2}\\ &=-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^3+85 i c^2 d-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^4+80 i c^3 d-182 c^2 d^2+1224 i c d^3+707 d^4\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d)^2 (c+i d)^5 f \sqrt {c+d \tan (e+f x)}}-\frac {i \operatorname {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{4 a (c-i d)^2 f}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} (c-i d)^{5/2} f}-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 i c-21 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {5 c^2+30 i c d-89 d^2}{20 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^3+85 i c^2 d-221 c d^2+361 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f (c+d \tan (e+f x))^{3/2}}+\frac {d \left (15 c^4+80 i c^3 d-182 c^2 d^2+1224 i c d^3+707 d^4\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d)^2 (c+i d)^5 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 10.52, size = 928, normalized size = 2.09 \[ \frac {\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \sqrt {\sec (e+f x) (c \cos (e+f x)+d \sin (e+f x))} \left (\frac {\left (17 c^2+102 i d c-231 d^2\right ) \cos (2 f x) \left (\frac {1}{60} i \cos (e)-\frac {\sin (e)}{60}\right )}{(c+i d)^5}+\frac {(c+3 i d) \cos (4 f x) \left (\frac {7}{60} i \cos (e)+\frac {7 \sin (e)}{60}\right )}{(c+i d)^4}+\frac {\left (23 i \cos (e) c^5-108 d \cos (e) c^4+23 i d \sin (e) c^4-138 i d^2 \cos (e) c^3-108 d^2 \sin (e) c^3-692 d^3 \cos (e) c^2-138 i d^3 \sin (e) c^2+1623 i d^4 \cos (e) c-692 d^4 \sin (e) c+640 d^5 \cos (e)+343 i d^5 \sin (e)\right ) \left (\frac {1}{120} \cos (3 e)+\frac {1}{120} i \sin (3 e)\right )}{(c-i d)^2 (c+i d)^5 (c \cos (e)+d \sin (e))}+\frac {\cos (6 f x) \left (\frac {1}{40} i \cos (3 e)+\frac {1}{40} \sin (3 e)\right )}{(c+i d)^3}+\frac {\left (17 c^2+102 i d c-231 d^2\right ) \left (\frac {\cos (e)}{60}+\frac {1}{60} i \sin (e)\right ) \sin (2 f x)}{(c+i d)^5}+\frac {(c+3 i d) \left (\frac {7 \cos (e)}{60}-\frac {7}{60} i \sin (e)\right ) \sin (4 f x)}{(c+i d)^4}+\frac {\left (\frac {1}{40} \cos (3 e)-\frac {1}{40} i \sin (3 e)\right ) \sin (6 f x)}{(c+i d)^3}+\frac {16 \left (-\frac {1}{2} i \cos (3 e-f x) d^6+\frac {1}{2} i \cos (3 e+f x) d^6+\frac {1}{2} \sin (3 e-f x) d^6-\frac {1}{2} \sin (3 e+f x) d^6+c \cos (3 e-f x) d^5-c \cos (3 e+f x) d^5+i c \sin (3 e-f x) d^5-i c \sin (3 e+f x) d^5\right )}{3 (c-i d)^2 (c+i d)^5 (c \cos (e)+d \sin (e)) (c \cos (e+f x)+d \sin (e+f x))}+\frac {\frac {2}{3} i d^6 \cos (3 e)-\frac {2}{3} d^6 \sin (3 e)}{(c-i d)^2 (c+i d)^5 (c \cos (e+f x)+d \sin (e+f x))^2}\right )}{f (i \tan (e+f x) a+a)^{5/2}}-\frac {i e^{3 i e} \sqrt {e^{i f x}} \log \left (2 \left (e^{i (e+f x)} \sqrt {c-i d}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right ) \sec ^{\frac {5}{2}}(e+f x) (\cos (f x)+i \sin (f x))^{5/2}}{4 \sqrt {2} (c-i d)^{5/2} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} f (i \tan (e+f x) a+a)^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((-1/4*I)*E^((3*I)*e)*Sqrt[E^(I*f*x)]*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqr
t[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])]*Sec[e + f*x]^(5/2)*(Cos[f*x] + I*Sin[f*x])
^(5/2))/(Sqrt[2]*(c - I*d)^(5/2)*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]
*f*(a + I*a*Tan[e + f*x])^(5/2)) + (Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*Sqrt[Sec[e + f*x]*(c*Cos[e + f*x]
+ d*Sin[e + f*x])]*(((17*c^2 + (102*I)*c*d - 231*d^2)*Cos[2*f*x]*((I/60)*Cos[e] - Sin[e]/60))/(c + I*d)^5 + (
(c + (3*I)*d)*Cos[4*f*x]*(((7*I)/60)*Cos[e] + (7*Sin[e])/60))/(c + I*d)^4 + (((23*I)*c^5*Cos[e] - 108*c^4*d*Co
s[e] - (138*I)*c^3*d^2*Cos[e] - 692*c^2*d^3*Cos[e] + (1623*I)*c*d^4*Cos[e] + 640*d^5*Cos[e] + (23*I)*c^4*d*Sin
[e] - 108*c^3*d^2*Sin[e] - (138*I)*c^2*d^3*Sin[e] - 692*c*d^4*Sin[e] + (343*I)*d^5*Sin[e])*(Cos[3*e]/120 + (I/
120)*Sin[3*e]))/((c - I*d)^2*(c + I*d)^5*(c*Cos[e] + d*Sin[e])) + (Cos[6*f*x]*((I/40)*Cos[3*e] + Sin[3*e]/40))
/(c + I*d)^3 + ((17*c^2 + (102*I)*c*d - 231*d^2)*(Cos[e]/60 + (I/60)*Sin[e])*Sin[2*f*x])/(c + I*d)^5 + ((c + (
3*I)*d)*((7*Cos[e])/60 - ((7*I)/60)*Sin[e])*Sin[4*f*x])/(c + I*d)^4 + ((Cos[3*e]/40 - (I/40)*Sin[3*e])*Sin[6*f
*x])/(c + I*d)^3 + (((2*I)/3)*d^6*Cos[3*e] - (2*d^6*Sin[3*e])/3)/((c - I*d)^2*(c + I*d)^5*(c*Cos[e + f*x] + d*
Sin[e + f*x])^2) + (16*(c*d^5*Cos[3*e - f*x] - (I/2)*d^6*Cos[3*e - f*x] - c*d^5*Cos[3*e + f*x] + (I/2)*d^6*Cos
[3*e + f*x] + I*c*d^5*Sin[3*e - f*x] + (d^6*Sin[3*e - f*x])/2 - I*c*d^5*Sin[3*e + f*x] - (d^6*Sin[3*e + f*x])/
2))/(3*(c - I*d)^2*(c + I*d)^5*(c*Cos[e] + d*Sin[e])*(c*Cos[e + f*x] + d*Sin[e + f*x]))))/(f*(a + I*a*Tan[e +
f*x])^(5/2))

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fricas [B]  time = 0.71, size = 1770, normalized size = 3.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

(sqrt(2)*(3*c^6 + 6*I*c^5*d + 3*c^4*d^2 + 12*I*c^3*d^3 - 3*c^2*d^4 + 6*I*c*d^5 - 3*d^6 + (23*c^6 + 62*I*c^5*d
+ 55*c^4*d^2 + 860*I*c^3*d^3 + 3145*c^2*d^4 - 3298*I*c*d^5 - 983*d^6)*e^(10*I*f*x + 10*I*e) + (80*c^6 + 284*I*
c^5*d - 80*c^4*d^2 + 2360*I*c^3*d^3 + 4960*c^2*d^4 - 1540*I*c*d^5 + 544*d^6)*e^(8*I*f*x + 8*I*e) + (105*c^6 +
426*I*c^5*d - 387*c^4*d^2 + 1908*I*c^3*d^3 + 1167*c^2*d^4 + 1962*I*c*d^5 + 1179*d^6)*e^(6*I*f*x + 6*I*e) + (65
*c^6 + 254*I*c^5*d - 251*c^4*d^2 + 508*I*c^3*d^3 - 697*c^2*d^4 + 254*I*c*d^5 - 381*d^6)*e^(4*I*f*x + 4*I*e) +
(20*c^6 + 56*I*c^5*d + 4*c^4*d^2 + 112*I*c^3*d^3 - 52*c^2*d^4 + 56*I*c*d^5 - 36*d^6)*e^(2*I*f*x + 2*I*e))*sqrt
(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) + ((-3
0*I*a^3*c^9 + 30*a^3*c^8*d - 120*I*a^3*c^7*d^2 + 120*a^3*c^6*d^3 - 180*I*a^3*c^5*d^4 + 180*a^3*c^4*d^5 - 120*I
*a^3*c^3*d^6 + 120*a^3*c^2*d^7 - 30*I*a^3*c*d^8 + 30*a^3*d^9)*f*e^(9*I*f*x + 9*I*e) + (-60*I*a^3*c^9 + 180*a^3
*c^8*d + 480*a^3*c^6*d^3 + 360*I*a^3*c^5*d^4 + 360*a^3*c^4*d^5 + 480*I*a^3*c^3*d^6 + 180*I*a^3*c*d^8 - 60*a^3*
d^9)*f*e^(7*I*f*x + 7*I*e) + (-30*I*a^3*c^9 + 150*a^3*c^8*d + 240*I*a^3*c^7*d^2 + 420*I*a^3*c^5*d^4 - 420*a^3*
c^4*d^5 - 240*a^3*c^2*d^7 - 150*I*a^3*c*d^8 + 30*a^3*d^9)*f*e^(5*I*f*x + 5*I*e))*sqrt(-I/((8*I*a^5*c^5 + 40*a^
5*c^4*d - 80*I*a^5*c^3*d^2 - 80*a^5*c^2*d^3 + 40*I*a^5*c*d^4 + 8*a^5*d^5)*f^2))*log((4*I*a^3*c^3 + 12*a^3*c^2*
d - 12*I*a^3*c*d^2 - 4*a^3*d^3)*f*sqrt(-I/((8*I*a^5*c^5 + 40*a^5*c^4*d - 80*I*a^5*c^3*d^2 - 80*a^5*c^2*d^3 + 4
0*I*a^5*c*d^4 + 8*a^5*d^5)*f^2))*e^(I*f*x + I*e) + sqrt(2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(
2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)) + ((30*I*a^3*c^9 - 30*a^3*
c^8*d + 120*I*a^3*c^7*d^2 - 120*a^3*c^6*d^3 + 180*I*a^3*c^5*d^4 - 180*a^3*c^4*d^5 + 120*I*a^3*c^3*d^6 - 120*a^
3*c^2*d^7 + 30*I*a^3*c*d^8 - 30*a^3*d^9)*f*e^(9*I*f*x + 9*I*e) + (60*I*a^3*c^9 - 180*a^3*c^8*d - 480*a^3*c^6*d
^3 - 360*I*a^3*c^5*d^4 - 360*a^3*c^4*d^5 - 480*I*a^3*c^3*d^6 - 180*I*a^3*c*d^8 + 60*a^3*d^9)*f*e^(7*I*f*x + 7*
I*e) + (30*I*a^3*c^9 - 150*a^3*c^8*d - 240*I*a^3*c^7*d^2 - 420*I*a^3*c^5*d^4 + 420*a^3*c^4*d^5 + 240*a^3*c^2*d
^7 + 150*I*a^3*c*d^8 - 30*a^3*d^9)*f*e^(5*I*f*x + 5*I*e))*sqrt(-I/((8*I*a^5*c^5 + 40*a^5*c^4*d - 80*I*a^5*c^3*
d^2 - 80*a^5*c^2*d^3 + 40*I*a^5*c*d^4 + 8*a^5*d^5)*f^2))*log((-4*I*a^3*c^3 - 12*a^3*c^2*d + 12*I*a^3*c*d^2 + 4
*a^3*d^3)*f*sqrt(-I/((8*I*a^5*c^5 + 40*a^5*c^4*d - 80*I*a^5*c^3*d^2 - 80*a^5*c^2*d^3 + 40*I*a^5*c*d^4 + 8*a^5*
d^5)*f^2))*e^(I*f*x + I*e) + sqrt(2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))
*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)))/((-120*I*a^3*c^9 + 120*a^3*c^8*d - 480*I*a^3*c^
7*d^2 + 480*a^3*c^6*d^3 - 720*I*a^3*c^5*d^4 + 720*a^3*c^4*d^5 - 480*I*a^3*c^3*d^6 + 480*a^3*c^2*d^7 - 120*I*a^
3*c*d^8 + 120*a^3*d^9)*f*e^(9*I*f*x + 9*I*e) + (-240*I*a^3*c^9 + 720*a^3*c^8*d + 1920*a^3*c^6*d^3 + 1440*I*a^3
*c^5*d^4 + 1440*a^3*c^4*d^5 + 1920*I*a^3*c^3*d^6 + 720*I*a^3*c*d^8 - 240*a^3*d^9)*f*e^(7*I*f*x + 7*I*e) + (-12
0*I*a^3*c^9 + 600*a^3*c^8*d + 960*I*a^3*c^7*d^2 + 1680*I*a^3*c^5*d^4 - 1680*a^3*c^4*d^5 - 960*a^3*c^2*d^7 - 60
0*I*a^3*c*d^8 + 120*a^3*d^9)*f*e^(5*I*f*x + 5*I*e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Evaluation time: 8.04Error: Bad Argument Type

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maple [B]  time = 0.37, size = 10145, normalized size = 22.85 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(5/2),x)

[Out]

Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*(c + d*tan(e + f*x))**(5/2)), x)

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