\(\int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [112]

Optimal result

Integrand size = 34, antiderivative size = 130 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {2 \sqrt [4]{-1} a (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 a (i A+B) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3673, 3609, 3614, 211} \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {2 \sqrt [4]{-1} a (B+i A) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \tan ^{\frac {5}{2}}(c+d x)}{5 d}+\frac {2 a (A-i B) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2 a (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {7}{2}}(c+d x)}{7 d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.83 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 a \left (-105 \sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)} \left (-105 i (A-i B)+35 (A-i B) \tan (c+d x)+21 (i A+B) \tan ^2(c+d x)+15 i B \tan ^3(c+d x)\right )\right )}{105 d} \]

[In]

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

[Out]

(2*a*(-105*(-1)^(1/4)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]*((-105*I)*(A - I*B)
+ 35*(A - I*B)*Tan[c + d*x] + 21*(I*A + B)*Tan[c + d*x]^2 + (15*I)*B*Tan[c + d*x]^3)))/(105*d)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (105 ) = 210\).

Time = 0.09 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.09

method	result	size
derivativedivides	\(\frac {a \left (\frac {2 i B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {2 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {2 A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )-2 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (i A +B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (i B -A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(272\)
default	\(\frac {a \left (\frac {2 i B \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}+\frac {2 i A \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {2 i B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {2 A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 i A \left (\sqrt {\tan }\left (d x +c \right )\right )-2 B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (i A +B \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (i B -A \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(272\)
parts	\(\frac {\left (i a A +B a \right ) \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {a A \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {i a B \left (\frac {2 \left (\tan ^{\frac {7}{2}}\left (d x +c \right )\right )}{7}-\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(337\)

[In]

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

[Out]

1/d*a*(2/7*I*B*tan(d*x+c)^(7/2)+2/5*I*A*tan(d*x+c)^(5/2)+2/5*B*tan(d*x+c)^(5/2)-2/3*I*B*tan(d*x+c)^(3/2)+2/3*A
*tan(d*x+c)^(3/2)-2*I*A*tan(d*x+c)^(1/2)-2*B*tan(d*x+c)^(1/2)+1/4*(I*A+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)))+1/4*(-A+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. 480 vs. \(2 (100) = 200\).

Time = 0.33 (sec) , antiderivative size = 480, normalized size of antiderivative = 3.69 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \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}\right ) - 105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a^{2}}{d^{2}}} \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}\right ) - 4 \, {\left ({\left (161 i \, A + 176 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (329 i \, A + 284 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (259 i \, A + 304 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (91 i \, A + 76 \, B\right )} a\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}}}{210 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/210*(105*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A
*B + I*B^2)*a^2/d^2)*log(2*((A - I*B)*a*e^(2*I*d*x + 2*I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2*A*
B + I*B^2)*a^2/d^2)*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)) - 105*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 -
2*A*B + I*B^2)*a^2/d^2)*log(2*((A - I*B)*a*e^(2*I*d*x + 2*I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(-I*A^2 - 2
*A*B + I*B^2)*a^2/d^2)*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)) - 4*((161*I*A + 176*B)*a*e^(6*I*d*x + 6*I*c) + (329*I*A + 284*B)*a*e^(4*I*d*x + 4*I*c) + (259*I*A +
304*B)*a*e^(2*I*d*x + 2*I*c) + (91*I*A + 76*B)*a)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)
))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (100) = 200\).

Time = 0.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.55 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {-120 i \, B a \tan \left (d x + c\right )^{\frac {7}{2}} + 168 \, {\left (-i \, A - B\right )} a \tan \left (d x + c\right )^{\frac {5}{2}} - 280 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{\frac {3}{2}} + 840 \, {\left (i \, A + B\right )} a \sqrt {\tan \left (d x + c\right )} - 105 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{420 \, d} \]

[In]

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

[Out]

-1/420*(-120*I*B*a*tan(d*x + c)^(7/2) + 168*(-I*A - B)*a*tan(d*x + c)^(5/2) - 280*(A - I*B)*a*tan(d*x + c)^(3/
2) + 840*(I*A + B)*a*sqrt(tan(d*x + c)) - 105*(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)*arctan(-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)/d

Giac [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.09 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\left (i - 1\right ) \, \sqrt {2} {\left (A a - i \, B a\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (-15 i \, B a d^{6} \tan \left (d x + c\right )^{\frac {7}{2}} - 21 i \, A a d^{6} \tan \left (d x + c\right )^{\frac {5}{2}} - 21 \, B a d^{6} \tan \left (d x + c\right )^{\frac {5}{2}} - 35 \, A a d^{6} \tan \left (d x + c\right )^{\frac {3}{2}} + 35 i \, B a d^{6} \tan \left (d x + c\right )^{\frac {3}{2}} + 105 i \, A a d^{6} \sqrt {\tan \left (d x + c\right )} + 105 \, B a d^{6} \sqrt {\tan \left (d x + c\right )}\right )}}{105 \, d^{7}} \]

[In]

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

[Out]

(I - 1)*sqrt(2)*(A*a - I*B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 2/105*(-15*I*B*a*d^6*tan(d
*x + c)^(7/2) - 21*I*A*a*d^6*tan(d*x + c)^(5/2) - 21*B*a*d^6*tan(d*x + c)^(5/2) - 35*A*a*d^6*tan(d*x + c)^(3/2
) + 35*I*B*a*d^6*tan(d*x + c)^(3/2) + 105*I*A*a*d^6*sqrt(tan(d*x + c)) + 105*B*a*d^6*sqrt(tan(d*x + c)))/d^7

Mupad [B] (verification not implemented)

Time = 12.58 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.24 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2\,A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}-\frac {A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {A\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}-\frac {2\,B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}-\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {2\,B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,2{}\mathrm {i}}{7\,d}-\frac {{\left (-1\right )}^{1/4}\,A\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d} \]

[In]

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

[Out]

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