\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx\) [139]

Optimal result

Integrand size = 36, antiderivative size = 296 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {((7-5 i) A+(5+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((6+i) A+(1+4 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((-7-5 i) A+(5-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \]

[Out]

-1/8*((7-5*I)*A+(5+3*I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(-1/8+1/8*I)*((6+I)*A+(1+4*I)*B)*ar
ctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+1/16*((7+5*I)*A+(-5+3*I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x
+c))/a/d*2^(1/2)+1/16*((-7-5*I)*A+(5-3*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/2)+5/2*(I*A-B)
/a/d/tan(d*x+c)^(1/2)+1/6*(-7*A-3*I*B)/a/d/tan(d*x+c)^(3/2)+1/2*(A+I*B)/d/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {((7-5 i) A+(5+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((6+i) A+(1+4 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (-B+i A)}{2 a d \sqrt {\tan (c+d x)}}+\frac {((7+5 i) A-(5-3 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d}+\frac {((5-3 i) B-(7+5 i) A) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]

[In]

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

[Out]

(((7 - 5*I)*A + (5 + 3*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(4*Sqrt[2]*a*d) - ((1/4 - I/4)*((6 + I)*A
+ (1 + 4*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) + (((7 + 5*I)*A - (5 - 3*I)*B)*Log[1 - S
qrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(8*Sqrt[2]*a*d) + (((-7 - 5*I)*A + (5 - 3*I)*B)*Log[1 + Sqrt[2]*Sqr
t[Tan[c + d*x]] + Tan[c + d*x]])/(8*Sqrt[2]*a*d) - (7*A + (3*I)*B)/(6*a*d*Tan[c + d*x]^(3/2)) + (5*(I*A - B))/
(2*a*d*Sqrt[Tan[c + d*x]]) + (A + I*B)/(2*d*Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

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

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3677

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]

Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac {\int \frac {\frac {1}{2} a (7 A+3 i B)-\frac {5}{2} a (i A-B) \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{2 a^2} \\ & = -\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac {\int \frac {-\frac {5}{2} a (i A-B)-\frac {1}{2} a (7 A+3 i B) \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2} \\ & = -\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac {\int \frac {-\frac {1}{2} a (7 A+3 i B)+\frac {5}{2} a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} a (7 A+3 i B)+\frac {5}{2} a (i A-B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = -\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac {((7+5 i) A-(5-3 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a d}-\frac {((7-5 i) A+(5+3 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{4 a d} \\ & = -\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}+\frac {((7+5 i) A-(5-3 i) B) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 \sqrt {2} a d}+\frac {((7+5 i) A-(5-3 i) B) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 \sqrt {2} a d}-\frac {((7-5 i) A+(5+3 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 a d}-\frac {((7-5 i) A+(5+3 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{8 a d} \\ & = \frac {((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((7+5 i) A-(5-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))}-\frac {((7-5 i) A+(5+3 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((7-5 i) A+(5+3 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d} \\ & = \frac {((7-5 i) A+(5+3 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {((7-5 i) A+(5+3 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((7+5 i) A-(5-3 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{8 \sqrt {2} a d}-\frac {7 A+3 i B}{6 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {5 (i A-B)}{2 a d \sqrt {\tan (c+d x)}}+\frac {A+i B}{2 d \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.56 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.40 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {3 (-i A+B)-2 (3 A+2 i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-i \tan (c+d x)\right ) (-i+\tan (c+d x))-(A-i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},i \tan (c+d x)\right ) (-i+\tan (c+d x))}{6 a d \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))} \]

[In]

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

[Out]

(3*((-I)*A + B) - 2*(3*A + (2*I)*B)*Hypergeometric2F1[-3/2, 1, -1/2, (-I)*Tan[c + d*x]]*(-I + Tan[c + d*x]) -
(A - I*B)*Hypergeometric2F1[-3/2, 1, -1/2, I*Tan[c + d*x]]*(-I + Tan[c + d*x]))/(6*a*d*Tan[c + d*x]^(3/2)*(-I
+ Tan[c + d*x]))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.54

method	result	size
derivativedivides	\(\frac {-\frac {i \left (\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}-\frac {4 \left (2 i B +3 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(160\)
default	\(\frac {-\frac {i \left (\frac {i \left (i A -B \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}-\frac {4 \left (2 i B +3 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}-\frac {2 A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 \left (-i A +B \right )}{\sqrt {\tan \left (d x +c \right )}}+\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(160\)

[In]

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

[Out]

1/d/a*(-1/2*I*(I*(I*A-B)*tan(d*x+c)^(1/2)/(tan(d*x+c)-I)-4*(3*A+2*I*B)/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)
^(1/2)/(2^(1/2)-I*2^(1/2))))-2/3*A/tan(d*x+c)^(3/2)-2*(-I*A+B)/tan(d*x+c)^(1/2)+4*(-1/4*I*A-1/4*B)/(2^(1/2)+I*
2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*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. 795 vs. \(2 (221) = 442\).

Time = 0.29 (sec) , antiderivative size = 795, normalized size of antiderivative = 2.69 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {3 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {9 i \, A^{2} - 12 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {9 i \, A^{2} - 12 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} + 3 i \, A - 2 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {9 i \, A^{2} - 12 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {9 i \, A^{2} - 12 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} - 3 i \, A + 2 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left ({\left (19 \, A + 27 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (19 \, A + 3 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (35 \, A + 27 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{24 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

[In]

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

[Out]

1/24*(3*(a*d*e^(6*I*d*x + 6*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((-I*A^2 - 2*A*B +
I*B^2)/(a^2*d^2))*log(-2*((I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(
I*A + B)) - 3*(a*d*e^(6*I*d*x + 6*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((-I*A^2 - 2
*A*B + I*B^2)/(a^2*d^2))*log(-2*((-I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*
I*d*x + 2*I*c) + 1))*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2
*I*c)/(I*A + B)) - 6*(a*d*e^(6*I*d*x + 6*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((9*I
*A^2 - 12*A*B - 4*I*B^2)/(a^2*d^2))*log(-((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e
^(2*I*d*x + 2*I*c) + 1))*sqrt((9*I*A^2 - 12*A*B - 4*I*B^2)/(a^2*d^2)) + 3*I*A - 2*B)*e^(-2*I*d*x - 2*I*c)/(a*d
)) + 6*(a*d*e^(6*I*d*x + 6*I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((9*I*A^2 - 12*A*B
- 4*I*B^2)/(a^2*d^2))*log(((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I
*c) + 1))*sqrt((9*I*A^2 - 12*A*B - 4*I*B^2)/(a^2*d^2)) - 3*I*A + 2*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2*((19*A +
27*I*B)*e^(6*I*d*x + 6*I*c) - (19*A + 3*I*B)*e^(4*I*d*x + 4*I*c) - (35*A + 27*I*B)*e^(2*I*d*x + 2*I*c) + 3*A
+ 3*I*B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(a*d*e^(6*I*d*x + 6*I*c) - 2*a*d*e^(4*I
*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

Sympy [F]

\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=- \frac {i \left (\int \frac {A}{\tan ^{\frac {7}{2}}{\left (c + d x \right )} - i \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \tan {\left (c + d x \right )}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )} - i \tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx\right )}{a} \]

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(5/2)/(a+I*a*tan(d*x+c)),x)

[Out]

-I*(Integral(A/(tan(c + d*x)**(7/2) - I*tan(c + d*x)**(5/2)), x) + Integral(B*tan(c + d*x)/(tan(c + d*x)**(7/2
) - I*tan(c + d*x)**(5/2)), x))/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.90 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.48 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (-3 i \, A + 2 \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{2 \, a d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (i \, A + B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{4 \, a d} - \frac {-i \, A \sqrt {\tan \left (d x + c\right )} + B \sqrt {\tan \left (d x + c\right )}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} + \frac {2 i \, {\left (3 \, A \tan \left (d x + c\right ) + 3 i \, B \tan \left (d x + c\right ) + i \, A\right )}}{3 \, a d \tan \left (d x + c\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

-(1/2*I + 1/2)*sqrt(2)*(-3*I*A + 2*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a*d) + (1/4*I - 1/4)*s
qrt(2)*(I*A + B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a*d) - 1/2*(-I*A*sqrt(tan(d*x + c)) + B*sq
rt(tan(d*x + c)))/(a*d*(tan(d*x + c) - I)) + 2/3*I*(3*A*tan(d*x + c) + 3*I*B*tan(d*x + c) + I*A)/(a*d*tan(d*x
+ c)^(3/2))

Mupad [B] (verification not implemented)

Time = 11.37 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.02 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{4\,a^2\,d^2}}}{3\,A}\right )\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}+2\,\mathrm {atanh}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{a^2\,d^2}}+2\,\mathrm {atanh}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}-\frac {\frac {2\,A}{3\,a\,d}+\frac {5\,A\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,a\,d}-\frac {A\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}}{3\,a\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}+{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,1{}\mathrm {i}}-\frac {\frac {2\,B}{a\,d}+\frac {B\,\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}}{2\,a\,d}}{\sqrt {\mathrm {tan}\left (c+d\,x\right )}+{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,1{}\mathrm {i}} \]

[In]

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

[Out]

atan((2*a*d*tan(c + d*x)^(1/2)*((A^2*9i)/(4*a^2*d^2))^(1/2))/(3*A))*((A^2*9i)/(4*a^2*d^2))^(1/2)*2i - atan((4*
a*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(16*a^2*d^2))^(1/2))/A)*(-(A^2*1i)/(16*a^2*d^2))^(1/2)*2i + 2*atanh((a*d*tan
(c + d*x)^(1/2)*(-(B^2*1i)/(a^2*d^2))^(1/2))/B)*(-(B^2*1i)/(a^2*d^2))^(1/2) + 2*atanh((4*a*d*tan(c + d*x)^(1/2
)*((B^2*1i)/(16*a^2*d^2))^(1/2))/B)*((B^2*1i)/(16*a^2*d^2))^(1/2) - ((2*A)/(3*a*d) - (A*tan(c + d*x)*4i)/(3*a*
d) + (5*A*tan(c + d*x)^2)/(2*a*d))/(tan(c + d*x)^(3/2) + tan(c + d*x)^(5/2)*1i) - ((2*B)/(a*d) + (B*tan(c + d*
x)*5i)/(2*a*d))/(tan(c + d*x)^(1/2) + tan(c + d*x)^(3/2)*1i)