\(\int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [67]

Optimal result

Integrand size = 36, antiderivative size = 194 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 (7 A-i B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d} \]

[Out]

(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d-8/35*(7*A-I*B)*(a+I*a*tan(d*x+
c))^(1/2)/d+2/35*(7*A-I*B)*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/d+2/7*B*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3
/d-2/105*(7*A-31*I*B)*(a+I*a*tan(d*x+c))^(3/2)/a/d

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3678, 3673, 3608, 3561, 212} \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}-\frac {2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}-\frac {8 (7 A-i B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \]

[In]

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

[Out]

(Sqrt[2]*Sqrt[a]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (8*(7*A - I*B)*Sqrt[a +
I*a*Tan[c + d*x]])/(35*d) + (2*(7*A - I*B)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(35*d) + (2*B*Tan[c + d*
x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(7*d) - (2*(7*A - (31*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/(105*a*d)

Rule 212

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

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3608

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 3678

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*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {2 \int \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-3 a B+\frac {1}{2} a (7 A-i B) \tan (c+d x)\right ) \, dx}{7 a} \\ & = \frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {4 \int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-a^2 (7 A-i B)-\frac {1}{4} a^2 (7 i A+31 B) \tan (c+d x)\right ) \, dx}{35 a^2} \\ & = \frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}+\frac {4 \int \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{4} a^2 (7 i A+31 B)-a^2 (7 A-i B) \tan (c+d x)\right ) \, dx}{35 a^2} \\ & = -\frac {8 (7 A-i B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}+(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = -\frac {8 (7 A-i B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d}+\frac {(2 a (A-i B)) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = \frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {8 (7 A-i B) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 (7 A-i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{35 d}+\frac {2 B \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}-\frac {2 (7 A-31 i B) (a+i a \tan (c+d x))^{3/2}}{105 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.67 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (-91 A+43 i B+(-7 i A-31 B) \tan (c+d x)+3 (7 A-i B) \tan ^2(c+d x)+15 B \tan ^3(c+d x)\right )}{105 d} \]

[In]

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

[Out]

(Sqrt[2]*Sqrt[a]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (2*Sqrt[a + I*a*Tan[c +
d*x]]*(-91*A + (43*I)*B + ((-7*I)*A - 31*B)*Tan[c + d*x] + 3*(7*A - I*B)*Tan[c + d*x]^2 + 15*B*Tan[c + d*x]^3)
)/(105*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {4 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {4 i B \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {2 A \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{3} d}\)	\(163\)
default	\(\frac {\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {4 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {4 i B \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {2 A \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{3} d}\)	\(163\)
parts	\(\frac {2 A \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-a^{2} \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{2}}+\frac {2 i B \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}} a}{5}+\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}}{3}-\frac {a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d \,a^{3}}\)	\(189\)

[In]

int((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/d/a^3*(1/7*I*B*(a+I*a*tan(d*x+c))^(7/2)-2/5*I*B*a*(a+I*a*tan(d*x+c))^(5/2)-1/5*A*a*(a+I*a*tan(d*x+c))^(5/2)+
2/3*I*B*a^2*(a+I*a*tan(d*x+c))^(3/2)+1/3*A*a^2*(a+I*a*tan(d*x+c))^(3/2)-A*a^3*(a+I*a*tan(d*x+c))^(1/2)+1/2*a^(
7/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(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. 440 vs. \(2 (151) = 302\).

Time = 0.28 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.27 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {105 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 105 \, \sqrt {2} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 4 \, \sqrt {2} {\left ({\left (119 \, A - 92 i \, B\right )} e^{\left (7 i \, d x + 7 i \, c\right )} + 7 \, {\left (37 \, A - 16 i \, B\right )} e^{\left (5 i \, d x + 5 i \, c\right )} + 35 \, {\left (7 \, A - 4 i \, B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + 105 \, A e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{210 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/210*(105*sqrt(2)*(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((A^2 -
2*I*A*B - B^2)*a/d^2)*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) - (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((A^2 - 2*I*
A*B - B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 105*sqrt(2)*(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((A^2 - 2*I*A*B - B^2)*a/d^2)*log(-4*((-I
*A - B)*a*e^(I*d*x + I*c) - (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((A^2 - 2*I*A*B - B^2)*a/d^2)*sqrt(a/(e^(2*I*
d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 4*sqrt(2)*((119*A - 92*I*B)*e^(7*I*d*x + 7*I*c) + 7*(37*A -
16*I*B)*e^(5*I*d*x + 5*I*c) + 35*(7*A - 4*I*B)*e^(3*I*d*x + 3*I*c) + 105*A*e^(I*d*x + I*c))*sqrt(a/(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]

\[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)*tan(d*x+c)**3*(A+B*tan(d*x+c)),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*(A + B*tan(c + d*x))*tan(c + d*x)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.79 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {105 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 60 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} B a + 84 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (A + 2 i \, B\right )} a^{2} - 140 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A + 2 i \, B\right )} a^{3} + 420 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A a^{4}}{210 \, a^{4} d} \]

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/210*(105*sqrt(2)*(A - I*B)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + s
qrt(I*a*tan(d*x + c) + a))) - 60*I*(I*a*tan(d*x + c) + a)^(7/2)*B*a + 84*(I*a*tan(d*x + c) + a)^(5/2)*(A + 2*I
*B)*a^2 - 140*(I*a*tan(d*x + c) + a)^(3/2)*(A + 2*I*B)*a^3 + 420*sqrt(I*a*tan(d*x + c) + a)*A*a^4)/(a^4*d)

Giac [F(-1)]

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

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.11 \[ \int \tan ^3(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a\,d}-\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a\,d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,4{}\mathrm {i}}{5\,a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a^3\,d}-\frac {\sqrt {2}\,B\,\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{d}-\frac {\sqrt {2}\,A\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{d} \]

[In]

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

[Out]

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