\(\int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [119]

Optimal result

Integrand size = 36, antiderivative size = 183 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt [4]{-1} a^2 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

[Out]

-4*(-1)^(1/4)*a^2*(I*A+B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-4*a^2*(I*A+B)*tan(d*x+c)^(1/2)/d+4/3*a^2*(A-I*
B)*tan(d*x+c)^(3/2)/d+4/5*a^2*(I*A+B)*tan(d*x+c)^(5/2)/d-2/63*a^2*(9*A-11*I*B)*tan(d*x+c)^(7/2)/d+2/9*I*B*tan(
d*x+c)^(7/2)*(a^2+I*a^2*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3675, 3673, 3609, 3614, 211} \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt [4]{-1} a^2 (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {4 a^2 (B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {4 a^2 (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \]

[In]

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

[Out]

(-4*(-1)^(1/4)*a^2*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (4*a^2*(I*A + B)*Sqrt[Tan[c + d*x]])/d
+ (4*a^2*(A - I*B)*Tan[c + d*x]^(3/2))/(3*d) + (4*a^2*(I*A + B)*Tan[c + d*x]^(5/2))/(5*d) - (2*a^2*(9*A - (11
*I)*B)*Tan[c + d*x]^(7/2))/(63*d) + (((2*I)/9)*B*Tan[c + d*x]^(7/2)*(a^2 + I*a^2*Tan[c + d*x]))/d

Rule 211

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

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 3614

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

Rule 3673

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

Rule 3675

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) \left (\frac {1}{2} a (9 A-7 i B)+\frac {1}{2} a (9 i A+11 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \tan ^{\frac {5}{2}}(c+d x) \left (9 a^2 (A-i B)+9 a^2 (i A+B) \tan (c+d x)\right ) \, dx \\ & = \frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \tan ^{\frac {3}{2}}(c+d x) \left (-9 a^2 (i A+B)+9 a^2 (A-i B) \tan (c+d x)\right ) \, dx \\ & = \frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \sqrt {\tan (c+d x)} \left (-9 a^2 (A-i B)-9 a^2 (i A+B) \tan (c+d x)\right ) \, dx \\ & = -\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {2}{9} \int \frac {9 a^2 (i A+B)-9 a^2 (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d}+\frac {\left (36 a^4 (i A+B)^2\right ) \text {Subst}\left (\int \frac {1}{9 a^2 (i A+B)+9 a^2 (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {4 \sqrt [4]{-1} a^2 (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {4 a^2 (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {4 a^2 (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 a^2 (9 A-11 i B) \tan ^{\frac {7}{2}}(c+d x)}{63 d}+\frac {2 i B \tan ^{\frac {7}{2}}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{9 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.68 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.71 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {2 a^2 \left ((315+315 i) \sqrt {2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {\tan (c+d x)}}{\sqrt {2}}\right )+\sqrt {\tan (c+d x)} \left (-630 i (A-i B)+210 (A-i B) \tan (c+d x)+126 (i A+B) \tan ^2(c+d x)-45 (A-2 i B) \tan ^3(c+d x)-35 B \tan ^4(c+d x)\right )\right )}{315 d} \]

[In]

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

[Out]

(2*a^2*((315 + 315*I)*Sqrt[2]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[Tan[c + d*x]])/Sqrt[2]] + Sqrt[Tan[c + d*x]]*((-
630*I)*(A - I*B) + 210*(A - I*B)*Tan[c + d*x] + 126*(I*A + B)*Tan[c + d*x]^2 - 45*(A - (2*I)*B)*Tan[c + d*x]^3
- 35*B*Tan[c + d*x]^4)))/(315*d)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.63

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {2 B \left (\tan ^{\frac {9}{2}}\left (d x +c \right )\right )}{9}+\frac {4 i B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 A \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {4 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {4 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {4 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {4 A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-4 i A \left (\sqrt {\tan }\left (d x +c \right )\right )-4 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (2 i A +2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (2 i B -2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(298\)
default	\(\frac {a^{2} \left (-\frac {2 B \left (\tan ^{\frac {9}{2}}\left (d x +c \right )\right )}{9}+\frac {4 i B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 A \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {4 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {4 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {4 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {4 A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-4 i A \left (\sqrt {\tan }\left (d x +c \right )\right )-4 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (2 i A +2 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (2 i B -2 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(298\)
parts	\(\frac {\left (2 i A \,a^{2}+B \,a^{2}\right ) \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {\left (2 i B \,a^{2}-A \,a^{2}\right ) \left (\frac {2 \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {A \,a^{2} \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}-\frac {B \,a^{2} \left (\frac {2 \left (\tan ^{\frac {9}{2}}\left (d x +c \right )\right )}{9}-\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(478\)

[In]

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

[Out]

1/d*a^2*(-2/9*B*tan(d*x+c)^(9/2)+4/7*I*B*tan(d*x+c)^(7/2)-2/7*A*tan(d*x+c)^(7/2)+4/5*I*A*tan(d*x+c)^(5/2)+4/5*
B*tan(d*x+c)^(5/2)-4/3*I*B*tan(d*x+c)^(3/2)+4/3*A*tan(d*x+c)^(3/2)-4*I*A*tan(d*x+c)^(1/2)-4*B*tan(d*x+c)^(1/2)
+1/4*(2*B+2*I*A)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+
2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(-2*A+2*I*B)*2^(1/2)*(ln((1-2^
(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2
))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(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. 555 vs. \(2 (145) = 290\).

Time = 0.34 (sec) , antiderivative size = 555, normalized size of antiderivative = 3.03 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {315 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 315 \, \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, {\left ({\left (1011 i \, A + 1091 \, B\right )} a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, {\left (285 i \, A + 262 \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 42 \, {\left (84 i \, A + 89 \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, {\left (219 i \, A + 214 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (501 i \, A + 491 \, B\right )} a^{2}\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}}}{315 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

1/315*(315*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4
*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^2*e^(2*I*d*x + 2*I*c) + sqrt(-(-I*A^2 - 2*A
*B + I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))
*e^(-2*I*d*x - 2*I*c)/((-I*A - B)*a^2)) - 315*sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(8*I*d*x + 8*I*c) +
4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^2*e^(2*I
*d*x + 2*I*c) - sqrt(-(-I*A^2 - 2*A*B + I*B^2)*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-I*e^(2*I*d*x + 2*I*
c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((-I*A - B)*a^2)) - 2*((1011*I*A + 1091*B)*a^2*e^(8*I
*d*x + 8*I*c) + 10*(285*I*A + 262*B)*a^2*e^(6*I*d*x + 6*I*c) + 42*(84*I*A + 89*B)*a^2*e^(4*I*d*x + 4*I*c) + 10
*(219*I*A + 214*B)*a^2*e^(2*I*d*x + 2*I*c) + (501*I*A + 491*B)*a^2)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*
d*x + 2*I*c) + 1)))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e^(2*I*d*
x + 2*I*c) + d)

Sympy [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=- a^{2} \left (\int \left (- A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{\frac {9}{2}}{\left (c + d x \right )}\, dx + \int \left (- B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {11}{2}}{\left (c + d x \right )}\, dx + \int \left (- 2 i A \tan ^{\frac {7}{2}}{\left (c + d x \right )}\right )\, dx + \int \left (- 2 i B \tan ^{\frac {9}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

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

[Out]

-a**2*(Integral(-A*tan(c + d*x)**(5/2), x) + Integral(A*tan(c + d*x)**(9/2), x) + Integral(-B*tan(c + d*x)**(7
/2), x) + Integral(B*tan(c + d*x)**(11/2), x) + Integral(-2*I*A*tan(c + d*x)**(7/2), x) + Integral(-2*I*B*tan(
c + d*x)**(9/2), x))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.26 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {140 \, B a^{2} \tan \left (d x + c\right )^{\frac {9}{2}} + 180 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {7}{2}} + 504 \, {\left (-i \, A - B\right )} a^{2} \tan \left (d x + c\right )^{\frac {5}{2}} - 840 \, {\left (A - i \, B\right )} a^{2} \tan \left (d x + c\right )^{\frac {3}{2}} + 2520 \, {\left (i \, A + B\right )} a^{2} \sqrt {\tan \left (d x + c\right )} - 315 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2}}{630 \, d} \]

[In]

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

[Out]

-1/630*(140*B*a^2*tan(d*x + c)^(9/2) + 180*(A - 2*I*B)*a^2*tan(d*x + c)^(7/2) + 504*(-I*A - B)*a^2*tan(d*x + c
)^(5/2) - 840*(A - I*B)*a^2*tan(d*x + c)^(3/2) + 2520*(I*A + B)*a^2*sqrt(tan(d*x + c)) - 315*(2*sqrt(2)*((I -
1)*A + (I + 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arc
tan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(sqrt(2)*sqrt(tan(d*x
+ c)) + tan(d*x + c) + 1) + sqrt(2)*(-(I + 1)*A + (I - 1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) +
1))*a^2)/d

Giac [A] (verification not implemented)

none

Time = 1.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.06 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {\left (2 i - 2\right ) \, \sqrt {2} {\left (A a^{2} - i \, B a^{2}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (35 \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac {9}{2}} + 45 \, A a^{2} d^{8} \tan \left (d x + c\right )^{\frac {7}{2}} - 90 i \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac {7}{2}} - 126 i \, A a^{2} d^{8} \tan \left (d x + c\right )^{\frac {5}{2}} - 126 \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac {5}{2}} - 210 \, A a^{2} d^{8} \tan \left (d x + c\right )^{\frac {3}{2}} + 210 i \, B a^{2} d^{8} \tan \left (d x + c\right )^{\frac {3}{2}} + 630 i \, A a^{2} d^{8} \sqrt {\tan \left (d x + c\right )} + 630 \, B a^{2} d^{8} \sqrt {\tan \left (d x + c\right )}\right )}}{315 \, d^{9}} \]

[In]

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

[Out]

(2*I - 2)*sqrt(2)*(A*a^2 - I*B*a^2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 2/315*(35*B*a^2*d^8*
tan(d*x + c)^(9/2) + 45*A*a^2*d^8*tan(d*x + c)^(7/2) - 90*I*B*a^2*d^8*tan(d*x + c)^(7/2) - 126*I*A*a^2*d^8*tan
(d*x + c)^(5/2) - 126*B*a^2*d^8*tan(d*x + c)^(5/2) - 210*A*a^2*d^8*tan(d*x + c)^(3/2) + 210*I*B*a^2*d^8*tan(d*
x + c)^(3/2) + 630*I*A*a^2*d^8*sqrt(tan(d*x + c)) + 630*B*a^2*d^8*sqrt(tan(d*x + c)))/d^9

Mupad [B] (verification not implemented)

Time = 13.16 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.78 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {4\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}-\frac {A\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,4{}\mathrm {i}}{d}+\frac {A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,4{}\mathrm {i}}{5\,d}-\frac {2\,A\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{7\,d}-\frac {4\,B\,a^2\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}-\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,4{}\mathrm {i}}{3\,d}+\frac {4\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}+\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,4{}\mathrm {i}}{7\,d}-\frac {2\,B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^{9/2}}{9\,d}+\frac {\sqrt {2}\,A\,a^2\,\ln \left (-4\,A\,a^2\,d+\sqrt {2}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d}-\frac {\sqrt {4{}\mathrm {i}}\,A\,a^2\,\ln \left (-4\,A\,a^2\,d+2\,\sqrt {4{}\mathrm {i}}\,A\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^2\,\ln \left (B\,a^2\,d\,4{}\mathrm {i}+\sqrt {2}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (1-\mathrm {i}\right )}{d}-\frac {\sqrt {-4{}\mathrm {i}}\,B\,a^2\,\ln \left (B\,a^2\,d\,4{}\mathrm {i}+2\,\sqrt {-4{}\mathrm {i}}\,B\,a^2\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

[In]

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

[Out]

(4*A*a^2*tan(c + d*x)^(3/2))/(3*d) - (A*a^2*tan(c + d*x)^(1/2)*4i)/d + (A*a^2*tan(c + d*x)^(5/2)*4i)/(5*d) - (
2*A*a^2*tan(c + d*x)^(7/2))/(7*d) - (4*B*a^2*tan(c + d*x)^(1/2))/d - (B*a^2*tan(c + d*x)^(3/2)*4i)/(3*d) + (4*
B*a^2*tan(c + d*x)^(5/2))/(5*d) + (B*a^2*tan(c + d*x)^(7/2)*4i)/(7*d) - (2*B*a^2*tan(c + d*x)^(9/2))/(9*d) + (
2^(1/2)*A*a^2*log(- 4*A*a^2*d - 2^(1/2)*A*a^2*d*tan(c + d*x)^(1/2)*(2 + 2i))*(1 + 1i))/d - (4i^(1/2)*A*a^2*log
(2*4i^(1/2)*A*a^2*d*tan(c + d*x)^(1/2) - 4*A*a^2*d))/d + (2^(1/2)*B*a^2*log(B*a^2*d*4i - 2^(1/2)*B*a^2*d*tan(c
+ d*x)^(1/2)*(2 - 2i))*(1 - 1i))/d - ((-4i)^(1/2)*B*a^2*log(B*a^2*d*4i + 2*(-4i)^(1/2)*B*a^2*d*tan(c + d*x)^(
1/2)))/d