3.97 \(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=209 \[ \frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

1/4*(A-I*B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)-2*(3*A+5*I*B)*(a+I*a*tan(d
*x+c))^(1/2)/a^2/d+1/2*(3*A+5*I*B)*tan(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^(1/2)+1/3*(I*A-B)*tan(d*x+c)^3/d/(a+I*a
*tan(d*x+c))^(3/2)+1/6*(11*A+21*I*B)*(a+I*a*tan(d*x+c))^(3/2)/a^3/d

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Rubi [A]  time = 0.54, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3595, 3592, 3527, 3480, 206} \[ \frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(-B+i A) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*a^(3/2)*d) + ((I*A - B)*Tan[c + d
*x]^3)/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((3*A + (5*I)*B)*Tan[c + d*x]^2)/(2*a*d*Sqrt[a + I*a*Tan[c + d*x]]
) - (2*(3*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d) + ((11*A + (21*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/
(6*a^3*d)

Rule 206

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

Rule 3480

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 3527

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 3592

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 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 ^3(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}-\frac {\int \frac {\tan ^2(c+d x) \left (3 a (i A-B)+\frac {3}{2} a (A+3 i B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \tan (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-3 a^2 (3 A+5 i B)+\frac {3}{4} a^2 (11 i A-21 B) \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac {\int \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (11 i A-21 B)-3 a^2 (3 A+5 i B) \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}+\frac {(A-i B) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{2 a d}\\ &=\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(i A-B) \tan ^3(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(3 A+5 i B) \tan ^2(c+d x)}{2 a d \sqrt {a+i a \tan (c+d x)}}-\frac {2 (3 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{a^2 d}+\frac {(11 A+21 i B) (a+i a \tan (c+d x))^{3/2}}{6 a^3 d}\\ \end {align*}

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Mathematica [A]  time = 4.77, size = 176, normalized size = 0.84 \[ \frac {i \sec ^3(c+d x) (21 (3 A+5 i B) \cos (c+d x)+(37 A+51 i B) \cos (3 (c+d x))+2 i \sin (c+d x) ((39 A+53 i B) \cos (2 (c+d x))+39 A+61 i B))-\frac {24 i (A-i B) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{3/2}}}{24 a d (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-24*I)*(A - I*B)*E^((3*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^(3/2) + I*Sec[c +
d*x]^3*(21*(3*A + (5*I)*B)*Cos[c + d*x] + (37*A + (51*I)*B)*Cos[3*(c + d*x)] + (2*I)*(39*A + (61*I)*B + (39*A
+ (53*I)*B)*Cos[2*(c + d*x)])*Sin[c + d*x]))/(24*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.61, size = 439, normalized size = 2.10 \[ \frac {3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (4 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-4 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left (2 \, {\left (19 \, A + 26 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (17 \, A + 29 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, {\left (2 \, A + 3 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(1/2)*(a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2)
)*log((sqrt(2)*sqrt(1/2)*(4*I*a^2*d*e^(2*I*d*x + 2*I*c) + 4*I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A
^2 - 2*I*A*B - B^2)/(a^3*d^2)) + (4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 3*sqrt(1/2)*(a
^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))*log((sqrt(2)*sqrt(
1/2)*(-4*I*a^2*d*e^(2*I*d*x + 2*I*c) - 4*I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)
/(a^3*d^2)) + (4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - sqrt(2)*(2*(19*A + 26*I*B)*e^(6*I
*d*x + 6*I*c) + 3*(17*A + 29*I*B)*e^(4*I*d*x + 4*I*c) + 6*(2*A + 3*I*B)*e^(2*I*d*x + 2*I*c) - A - I*B)*sqrt(a/
(e^(2*I*d*x + 2*I*c) + 1)))/(a^2*d*e^(5*I*d*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 153, normalized size = 0.73 \[ -\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 i B a \sqrt {a +i a \tan \left (d x +c \right )}+A \sqrt {a +i a \tan \left (d x +c \right )}\, a -\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8}+\frac {a^{2} \left (7 i B +5 A \right )}{4 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a^{3} \left (i B +A \right )}{6 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.02, size = 160, normalized size = 0.77 \[ -\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 16 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 48 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{2} + \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (5 \, A + 7 i \, B\right )} a^{3} - 2 \, {\left (A + i \, B\right )} a^{4}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}}{24 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*sqrt(2)*(A - I*B)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt
(I*a*tan(d*x + c) + a))) - 16*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a + 48*sqrt(I*a*tan(d*x + c) + a)*(A + 2*I*B)*a
^2 + 4*(3*(I*a*tan(d*x + c) + a)*(5*A + 7*I*B)*a^3 - 2*(A + I*B)*a^4)/(I*a*tan(d*x + c) + a)^(3/2))/(a^4*d)

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mupad [B]  time = 7.44, size = 233, normalized size = 1.11 \[ \frac {\frac {B\,1{}\mathrm {i}}{3\,d}-\frac {B\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\frac {A\,a}{3}-\frac {5\,A\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2}}{a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2\,d}-\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,a^3\,d}-\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/2}\,d}+\frac {\sqrt {2}\,A\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{4\,a^{3/2}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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