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

Optimal. Leaf size=306 \[ -\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}-\frac {5 (-B+i A) \sqrt {\tan (c+d x)}}{2 a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]

[Out]

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

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Rubi [A]  time = 0.41, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(-B+i A) \tan ^{\frac {5}{2}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(3 A+7 i B) \tan ^{\frac {3}{2}}(c+d x)}{6 a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {5 (-B+i A) \sqrt {\tan (c+d x)}}{2 a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{8 \sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

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

Rule 617

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 628

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 1162

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 1165

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 1168

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 3528

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 3534

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 3595

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

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Mathematica [A]  time = 3.13, size = 248, normalized size = 0.81 \[ \frac {(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (\frac {2}{3} \tan (c+d x) \sec (c+d x) (\cos (d x)-i \sin (d x)) (4 (3 A+2 i B) \sin (2 (c+d x))+(11 B-15 i A) \cos (2 (c+d x))-15 i A+19 B)-(1+i) (\cos (c)+i \sin (c)) \sqrt {\sin (2 (c+d x))} \sec (c+d x) \left (((4+i) A+(1+6 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((6+i) B-(1+4 i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x])*((-1 - I)*(((4 + I)*A + (1 + 6*I)*B)*ArcSin[Cos[c + d*x] - Sin[c
+ d*x]] + ((-1 - 4*I)*A + (6 + I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*
(Cos[c] + I*Sin[c])*Sqrt[Sin[2*(c + d*x)]] + (2*Sec[c + d*x]*(Cos[d*x] - I*Sin[d*x])*((-15*I)*A + 19*B + ((-15
*I)*A + 11*B)*Cos[2*(c + d*x)] + 4*(3*A + (2*I)*B)*Sin[2*(c + d*x)])*Tan[c + d*x])/3))/(8*d*(A*Cos[c + d*x] +
B*Sin[c + d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x]))

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fricas [B]  time = 1.39, size = 706, normalized size = 2.31 \[ \frac {3 \, {\left (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 (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 {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 (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 (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 {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 (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 12 \, A B + 9 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 {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} + 2 \, A + 3 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 6 \, {\left (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 {-4 i \, A^{2} + 12 \, A B + 9 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 {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} - 2 \, A - 3 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (-27 i \, A + 19 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-30 i \, A + 38 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, 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 (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(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*((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((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^(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*((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((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^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c
))*sqrt((-4*I*A^2 + 12*A*B + 9*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((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2)) + 2*A + 3*I*B)*e^(-2*I*d*x
- 2*I*c)/(a*d)) + 6*(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2 + 12*A*B + 9*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))*sqr
t((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2)) - 2*A - 3*I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*((-27*I*A + 19*B)*e^
(4*I*d*x + 4*I*c) + (-30*I*A + 38*B)*e^(2*I*d*x + 2*I*c) - 3*I*A + 3*B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(
2*I*d*x + 2*I*c) + 1)))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{\frac {5}{2}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.44, size = 290, normalized size = 0.95 \[ -\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d a}+\frac {2 B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d a}-\frac {2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d a}-\frac {\arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) A}{d a \left (\sqrt {2}+i \sqrt {2}\right )}+\frac {i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right ) B}{d a \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \left (\sqrt {\tan }\left (d x +c \right )\right ) B}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {4 \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) A}{d a \left (\sqrt {2}-i \sqrt {2}\right )}+\frac {6 i \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right ) B}{d a \left (\sqrt {2}-i \sqrt {2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 11.33, size = 305, normalized size = 1.00 \[ \mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}\,1{}\mathrm {i}}{A}\right )\,\sqrt {-\frac {A^2\,1{}\mathrm {i}}{a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,4{}\mathrm {i}}{A}\right )\,\sqrt {\frac {A^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{4\,a^2\,d^2}}}{3\,B}\right )\,\sqrt {\frac {B^2\,9{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{a\,d}+\frac {2\,B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{a\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,a\,d}-\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {A \tan ^{\frac {5}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx + \int \frac {B \tan ^{\frac {7}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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