\(\int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [83]

Optimal result

Integrand size = 34, antiderivative size = 171 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3673, 3608, 3559, 3561, 212} \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d} \]

[In]

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

[Out]

(-4*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (4*a^2*(A - I*B)*Sqrt
[a + I*a*Tan[c + d*x]])/d + (2*a*(A - I*B)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d) + (2*A*(a + I*a*Tan[c + d*x])^(
5/2))/(5*d) - (((2*I)/7)*B*(a + I*a*Tan[c + d*x])^(7/2))/(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 3559

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

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}+\int (a+i a \tan (c+d x))^{5/2} (-B+A \tan (c+d x)) \, dx \\ & = \frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-(i A+B) \int (a+i a \tan (c+d x))^{5/2} \, dx \\ & = \frac {2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-(2 a (i A+B)) \int (a+i a \tan (c+d x))^{3/2} \, dx \\ & = \frac {4 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-\left (4 a^2 (i A+B)\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {4 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {\left (8 a^3 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = -\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (A-i B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 A (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac {2 i B (a+i a \tan (c+d x))^{7/2}}{7 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.77 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-420 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^2 \sqrt {a+i a \tan (c+d x)} \left (266 A-260 i B+(77 i A+80 B) \tan (c+d x)+(-21 A+45 i B) \tan ^2(c+d x)-15 B \tan ^3(c+d x)\right )}{105 d} \]

[In]

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

[Out]

(-420*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + 2*a^2*Sqrt[a + I*a*Tan
[c + d*x]]*(266*A - (260*I)*B + ((77*I)*A + 80*B)*Tan[c + d*x] + (-21*A + (45*I)*B)*Tan[c + d*x]^2 - 15*B*Tan[
c + d*x]^3))/(105*d)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 i B \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {2 A \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 i a^{3} B \sqrt {a +i a \tan \left (d x +c \right )}+4 A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a d}\)	\(165\)
default	\(\frac {-\frac {2 i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}+\frac {2 A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {2 i B \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {2 A \,a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-4 i a^{3} B \sqrt {a +i a \tan \left (d x +c \right )}+4 A \,a^{3} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {7}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a d}\)	\(165\)
parts	\(\frac {A \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}+\frac {2 i B \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} a^{2}}{3}-2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+2 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d a}\)	\(187\)

[In]

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

[Out]

2/d/a*(-1/7*I*B*(a+I*a*tan(d*x+c))^(7/2)+1/5*A*a*(a+I*a*tan(d*x+c))^(5/2)-1/3*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)-2*I*B*a^3*(a+I*a*tan(d*x+c))^(1/2)+2*A*a^3*(a+I*a*tan(d*x+c))^(1/2)-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. 477 vs. \(2 (130) = 260\).

Time = 0.26 (sec) , antiderivative size = 477, normalized size of antiderivative = 2.79 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (105 \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{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 )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 105 \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{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 )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, \sqrt {2} {\left (2 \, {\left (91 \, A - 100 i \, B\right )} a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 7 \, {\left (61 \, A - 55 i \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 350 \, {\left (A - i \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 105 \, {\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\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 )}} \]

[In]

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

[Out]

-2/105*(105*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^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)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) - sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^2)*(I*d*e^(
2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 105*sqrt(2)*sq
rt((A^2 - 2*I*A*B - B^2)*a^5/d^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)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) - sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*
d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 2*sqrt(2)*(2*(91*A - 100*I*B)*a^2*e
^(7*I*d*x + 7*I*c) + 7*(61*A - 55*I*B)*a^2*e^(5*I*d*x + 5*I*c) + 350*(A - I*B)*a^2*e^(3*I*d*x + 3*I*c) + 105*(
A - I*B)*a^2*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]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.89 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (105 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 15 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} B a + 21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} A a^{2} + 35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{3} + 210 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{4}\right )}}{105 \, a^{2} d} \]

[In]

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

[Out]

2/105*(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) + sq
rt(I*a*tan(d*x + c) + a))) - 15*I*(I*a*tan(d*x + c) + a)^(7/2)*B*a + 21*(I*a*tan(d*x + c) + a)^(5/2)*A*a^2 + 3
5*(I*a*tan(d*x + c) + a)^(3/2)*(A - I*B)*a^3 + 210*sqrt(I*a*tan(d*x + c) + a)*(A - I*B)*a^4)/(a^2*d)

Giac [F(-1)]

Timed out. \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 1.67 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.24 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}+\frac {2\,A\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}-\frac {B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {4\,A\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {B\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a\,d}+\frac {\sqrt {2}\,B\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d}+\frac {\sqrt {2}\,A\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \]

[In]

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

[Out]

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