\(\int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [147]

Optimal result

Integrand size = 36, antiderivative size = 364 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+6 i) A-(29+i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {((5-7 i) A+(28+30 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((6+i) A+(1+29 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((6+i) A+(1+29 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a^3 d}+\frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

(1/32+1/32*I)*((1+6*I)*A-(29+I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)-1/32*((5-7*I)*A+(28+30*I)
*B)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/a^3/d*2^(1/2)+(1/64+1/64*I)*((6+I)*A+(1+29*I)*B)*ln(1-2^(1/2)*tan(d*x+c
)^(1/2)+tan(d*x+c))/a^3/d*2^(1/2)-(1/64+1/64*I)*((6+I)*A+(1+29*I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))
/a^3/d*2^(1/2)+5/8*(A+6*I*B)*tan(d*x+c)^(1/2)/a^3/d+1/6*(I*A-B)*tan(d*x+c)^(7/2)/d/(a+I*a*tan(d*x+c))^3+1/12*(
2*A+5*I*B)*tan(d*x+c)^(5/2)/a/d/(a+I*a*tan(d*x+c))^2-7/24*(I*A-4*B)*tan(d*x+c)^(3/2)/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3676, 3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+6 i) A-(29+i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {((5-7 i) A+(28+30 i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {7 (-4 B+i A) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((6+i) A+(1+29 i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((6+i) A+(1+29 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(-B+i A) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2} \]

[In]

Int[(Tan[c + d*x]^(7/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((-1/16 - I/16)*((1 + 6*I)*A - (29 + I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a^3*d) - (((5 - 7*
I)*A + (28 + 30*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(16*Sqrt[2]*a^3*d) + ((1/32 + I/32)*((6 + I)*A +
(1 + 29*I)*B)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a^3*d) - ((1/32 + I/32)*((6 + I)*A
+ (1 + 29*I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a^3*d) + (5*(A + (6*I)*B)*Sqrt[T
an[c + d*x]])/(8*a^3*d) + ((I*A - B)*Tan[c + d*x]^(7/2))/(6*d*(a + I*a*Tan[c + d*x])^3) + ((2*A + (5*I)*B)*Tan
[c + d*x]^(5/2))/(12*a*d*(a + I*a*Tan[c + d*x])^2) - (7*(I*A - 4*B)*Tan[c + d*x]^(3/2))/(24*d*(a^3 + I*a^3*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 3609

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

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 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, 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 {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^{\frac {5}{2}}(c+d x) \left (\frac {7}{2} a (i A-B)+\frac {1}{2} a (A+13 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (-5 a^2 (2 A+5 i B)+a^2 (4 i A-31 B) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4} \\ & = \frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \sqrt {\tan (c+d x)} \left (-21 a^3 (i A-4 B)-15 a^3 (A+6 i B) \tan (c+d x)\right ) \, dx}{48 a^6} \\ & = \frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\int \frac {15 a^3 (A+6 i B)-21 a^3 (i A-4 B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{48 a^6} \\ & = \frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {\text {Subst}\left (\int \frac {15 a^3 (A+6 i B)-21 a^3 (i A-4 B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{24 a^6 d} \\ & = \frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((5+7 i) A-(28-30 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d}-\frac {((5-7 i) A+(28+30 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{16 a^3 d} \\ & = \frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {((5+7 i) A-(28-30 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 )}{32 \sqrt {2} a^3 d}+\frac {((5+7 i) A-(28-30 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 )}{32 \sqrt {2} a^3 d}-\frac {((5-7 i) A+(28+30 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d}-\frac {((5-7 i) A+(28+30 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{32 a^3 d} \\ & = \frac {((5+7 i) A-(28-30 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {((5+7 i) A-(28-30 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {((5-7 i) A+(28+30 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((5-7 i) A+(28+30 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d} \\ & = \frac {((5-7 i) A+(28+30 i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((5-7 i) A+(28+30 i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((5+7 i) A-(28-30 i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}-\frac {((5+7 i) A-(28-30 i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {5 (A+6 i B) \sqrt {\tan (c+d x)}}{8 a^3 d}+\frac {(i A-B) \tan ^{\frac {7}{2}}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {(2 A+5 i B) \tan ^{\frac {5}{2}}(c+d x)}{12 a d (a+i a \tan (c+d x))^2}-\frac {7 (i A-4 B) \tan ^{\frac {3}{2}}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.60 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) \left (-12 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+12 (-1)^{3/4} (6 A+29 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+2 i ((9 A+33 i B) \cos (c+d x)+21 (A+7 i B) \cos (3 (c+d x))+2 i (19 A+97 i B+(19 A+145 i B) \cos (2 (c+d x))) \sin (c+d x)) \sqrt {\tan (c+d x)}\right )}{96 a^3 d (-i+\tan (c+d x))^3} \]

[In]

Integrate[(Tan[c + d*x]^(7/2)*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(-12*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(Cos[3*(c + d*x)] + I*Sin[3*(c
+ d*x)]) + 12*(-1)^(3/4)*(6*A + (29*I)*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(Cos[3*(c + d*x)] + I*Sin[3*
(c + d*x)]) + (2*I)*((9*A + (33*I)*B)*Cos[c + d*x] + 21*(A + (7*I)*B)*Cos[3*(c + d*x)] + (2*I)*(19*A + (97*I)*
B + (19*A + (145*I)*B)*Cos[2*(c + d*x)])*Sin[c + d*x])*Sqrt[Tan[c + d*x]]))/(96*a^3*d*(-I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.50

method	result	size
derivativedivides	\(\frac {2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {i \left (\frac {-i \left (9 i A -20 B \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {38 i A}{3}+\frac {98 B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-14 i B -5 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {2 \left (29 i B +6 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {i A}{16}+\frac {B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(181\)
default	\(\frac {2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {i \left (\frac {-i \left (9 i A -20 B \right ) \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )+\left (-\frac {38 i A}{3}+\frac {98 B}{3}\right ) \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+\left (-14 i B -5 A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {2 \left (29 i B +6 A \right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{8}+\frac {4 \left (\frac {i A}{16}+\frac {B}{16}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d \,a^{3}}\)	\(181\)

[In]

int(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d/a^3*(2*I*B*tan(d*x+c)^(1/2)-1/8*I*((-I*(9*I*A-20*B)*tan(d*x+c)^(5/2)+(-38/3*I*A+98/3*B)*tan(d*x+c)^(3/2)+(
-5*A-14*I*B)*tan(d*x+c)^(1/2))/(tan(d*x+c)-I)^3-2*(29*I*B+6*A)/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(
2^(1/2)-I*2^(1/2))))+4*(1/16*I*A+1/16*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. 685 vs. \(2 (271) = 542\).

Time = 0.28 (sec) , antiderivative size = 685, normalized size of antiderivative = 1.88 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} 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^{6} 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 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} 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^{6} 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 \, a^{3} d \sqrt {\frac {36 i \, A^{2} - 348 \, A B - 841 i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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 {36 i \, A^{2} - 348 \, A B - 841 i \, B^{2}}{a^{6} d^{2}}} + 6 i \, A - 29 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 3 \, a^{3} d \sqrt {\frac {36 i \, A^{2} - 348 \, A B - 841 i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} 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 {36 i \, A^{2} - 348 \, A B - 841 i \, B^{2}}{a^{6} d^{2}}} - 6 i \, A + 29 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (10 \, A + 73 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (14 \, A + 41 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (5 \, A + 8 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + 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}}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/96*(3*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*((I*a^3*d*e^(2*I*d*x + 2*I*
c) + I*a^3*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^6*
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^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)
/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*((-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*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^6*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^3*d*sqrt((36*I*A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log
(-1/8*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((
36*I*A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2)) + 6*I*A - 29*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*a^3*d*sqrt((36*I*
A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt((-
I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((36*I*A^2 - 348*A*B - 841*I*B^2)/(a^6*d^2)) - 6*I*A
+ 29*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(10*A + 73*I*B)*e^(6*I*d*x + 6*I*c) + (14*A + 41*I*B)*e^(4*I*d*x
+ 4*I*c) - (5*A + 8*I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c
) + 1)))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)**(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.95 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.46 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\left (i + 1\right ) \, \sqrt {2} {\left (-6 i \, A + 29 \, B\right )} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {\left (i - 1\right ) \, \sqrt {2} {\left (-i \, A - B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{16 \, a^{3} d} + \frac {2 i \, B \sqrt {\tan \left (d x + c\right )}}{a^{3} d} + \frac {-27 i \, A \tan \left (d x + c\right )^{\frac {5}{2}} + 60 \, B \tan \left (d x + c\right )^{\frac {5}{2}} - 38 \, A \tan \left (d x + c\right )^{\frac {3}{2}} - 98 i \, B \tan \left (d x + c\right )^{\frac {3}{2}} + 15 i \, A \sqrt {\tan \left (d x + c\right )} - 42 \, B \sqrt {\tan \left (d x + c\right )}}{24 \, a^{3} d {\left (\tan \left (d x + c\right ) - i\right )}^{3}} \]

[In]

integrate(tan(d*x+c)^(7/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-(1/16*I + 1/16)*sqrt(2)*(-6*I*A + 29*B)*arctan((1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) + (1/16*I -
1/16)*sqrt(2)*(-I*A - B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^3*d) + 2*I*B*sqrt(tan(d*x + c))/
(a^3*d) + 1/24*(-27*I*A*tan(d*x + c)^(5/2) + 60*B*tan(d*x + c)^(5/2) - 38*A*tan(d*x + c)^(3/2) - 98*I*B*tan(d*
x + c)^(3/2) + 15*I*A*sqrt(tan(d*x + c)) - 42*B*sqrt(tan(d*x + c)))/(a^3*d*(tan(d*x + c) - I)^3)

Mupad [B] (verification not implemented)

Time = 11.53 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\mathrm {atan}\left (\frac {8\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}}{3\,A}\right )\,\sqrt {\frac {A^2\,9{}\mathrm {i}}{64\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {16\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,16{}\mathrm {i}}{29\,B}\right )\,\sqrt {-\frac {B^2\,841{}\mathrm {i}}{256\,a^6\,d^2}}\,2{}\mathrm {i}+\frac {\frac {5\,A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{8\,a^3\,d}-\frac {9\,A\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{8\,a^3\,d}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,19{}\mathrm {i}}{12\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}-\frac {\frac {49\,B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{12\,a^3\,d}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,7{}\mathrm {i}}{4\,a^3\,d}+\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,5{}\mathrm {i}}{2\,a^3\,d}}{-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a^3\,d} \]

[In]

int((tan(c + d*x)^(7/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

atan((8*a^3*d*tan(c + d*x)^(1/2)*((A^2*9i)/(64*a^6*d^2))^(1/2))/(3*A))*((A^2*9i)/(64*a^6*d^2))^(1/2)*2i + atan
((16*a^3*d*tan(c + d*x)^(1/2)*(-(A^2*1i)/(256*a^6*d^2))^(1/2))/A)*(-(A^2*1i)/(256*a^6*d^2))^(1/2)*2i + atan((a
^3*d*tan(c + d*x)^(1/2)*((B^2*1i)/(256*a^6*d^2))^(1/2)*16i)/B)*((B^2*1i)/(256*a^6*d^2))^(1/2)*2i - atan((a^3*d
*tan(c + d*x)^(1/2)*(-(B^2*841i)/(256*a^6*d^2))^(1/2)*16i)/(29*B))*(-(B^2*841i)/(256*a^6*d^2))^(1/2)*2i + ((5*
A*tan(c + d*x)^(1/2))/(8*a^3*d) + (A*tan(c + d*x)^(3/2)*19i)/(12*a^3*d) - (9*A*tan(c + d*x)^(5/2))/(8*a^3*d))/
(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(c + d*x)^3*1i + 1) - ((49*B*tan(c + d*x)^(3/2))/(12*a^3*d) - (B*tan(
c + d*x)^(1/2)*7i)/(4*a^3*d) + (B*tan(c + d*x)^(5/2)*5i)/(2*a^3*d))/(tan(c + d*x)*3i - 3*tan(c + d*x)^2 - tan(
c + d*x)^3*1i + 1) + (B*tan(c + d*x)^(1/2)*2i)/(a^3*d)