3.1.86 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [86]

Optimal. Leaf size=158 \[ -\frac {a^{5/2} (5 i A+2 B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 \sqrt {2} a^{5/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (i A-2 B) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \]

[Out]

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

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Rubi [A]
time = 0.36, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3674, 3675, 3681, 3561, 212, 3680, 65, 214} \begin {gather*} -\frac {a^{5/2} (2 B+5 i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 \sqrt {2} a^{5/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {a^2 (-2 B+i A) \sqrt {a+i a \tan (c+d x)}}{d}-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^{3/2}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^(5/2)*((5*I)*A + 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d) + (4*Sqrt[2]*a^(5/2)*(I*A + B)*ArcT
anh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (a^2*(I*A - 2*B)*Sqrt[a + I*a*Tan[c + d*x]])/d - (a*A*C
ot[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2))/d

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 214

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

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 3674

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

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3681

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

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(413\) vs. \(2(158)=316\).
time = 7.22, size = 413, normalized size = 2.61 \begin {gather*} \frac {\left (e^{-3 i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (32 \sqrt {2} (i A+B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+2 (5 i A+2 B) \left (\log \left (\left (-1+e^{i (c+d x)}\right )^2\right )-\log \left (\left (1+e^{i (c+d x)}\right )^2\right )+\log \left (3+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )-\log \left (3+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )\right )\right )-\frac {8 (A \csc (c+d x)+2 B \sec (c+d x)) (\cos (2 c)-i \sin (2 c))}{\sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x))^2}\right ) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{8 d \sec ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(32*Sqrt[2]*(I*A + B)*ArcSinh
[E^(I*(c + d*x))] + 2*((5*I)*A + 2*B)*(Log[(-1 + E^(I*(c + d*x)))^2] - Log[(1 + E^(I*(c + d*x)))^2] + Log[3 +
3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] - 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((
2*I)*(c + d*x))])] - Log[3 + 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] + 2*E^(I*(c + d*x
))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])))/E^((3*I)*(c + d*x)) - (8*(A*Csc[c + d*x] + 2*B*Sec[c + d*x]
)*(Cos[2*c] - I*Sin[2*c]))/(Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])^2))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B
*Tan[c + d*x]))/(8*d*Sec[c + d*x]^(7/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1140 vs. \(2 (132 ) = 264\).
time = 0.56, size = 1141, normalized size = 7.22

method	result	size
default	\(\text {Expression too large to display}\)	\(1141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a^2*(5*I*A*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))-8*I*B*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2
*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+8*A*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a
rctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+8*I*B*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+8*B*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))-8*I*A*2^(1/2)*arctanh(1/
2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-5*I
*A*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)+5*A*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2
*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*B*cos(d*x+c)^2*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))
-8*A*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*I
*A*cos(d*x+c)*sin(d*x+c)-2*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2))*cos(d*x+c)^2-8*B*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+4*I*B+8*I*A*2^(1/2)*cos(d*x+c)^2*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*A*cos(d*x+c)^2-5*A*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+4*B*cos(d*x+c)*sin(d*x+c)-2*B*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))-2*A
*cos(d*x+c)-4*I*B*cos(d*x+c)^2-4*B*sin(d*x+c))*((I*sin(d*x+c)+cos(d*x+c))*a/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+co
s(d*x+c)-1)/sin(d*x+c)

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Maxima [A]
time = 0.49, size = 163, normalized size = 1.03 \begin {gather*} -\frac {i \, {\left (4 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{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 ) - {\left (5 \, A - 2 i \, B\right )} a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - 4 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A a}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(4*sqrt(2)*(A - I*B)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqr
t(I*a*tan(d*x + c) + a))) - (5*A - 2*I*B)*a^(3/2)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x
+ c) + a) + sqrt(a))) - 4*I*sqrt(I*a*tan(d*x + c) + a)*B*a - 2*I*sqrt(I*a*tan(d*x + c) + a)*A*a/tan(d*x + c))
*a/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 705 vs. \(2 (125) = 250\).
time = 2.79, size = 705, normalized size = 4.46 \begin {gather*} -\frac {8 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - 8 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) - \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-5 i \, A - 2 \, B\right )} a^{3} + 2 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (5 i \, A + 2 \, B\right )} a}\right ) + \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {16 \, {\left (3 \, {\left (-5 i \, A - 2 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-5 i \, A - 2 \, B\right )} a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {{\left (25 \, A^{2} - 20 i \, A B - 4 \, B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (5 i \, A + 2 \, B\right )} a}\right ) + 4 \, \sqrt {2} {\left ({\left (i \, A + 2 \, B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (i \, A - 2 \, 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}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(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)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))
*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 8*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(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)*(d*e^(2*I*d*x + 2*I*c) + d)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^
2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-16*(3*(-5*I*A - 2*B)*a^3*e^(2*I*d*x + 2*I*c) + (-5*I*A - 2*B)*a^3 + 2*sqrt
(2)*sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((5*I*A + 2*B)*a)) + sqrt(-(25*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(2*I
*d*x + 2*I*c) - d)*log(-16*(3*(-5*I*A - 2*B)*a^3*e^(2*I*d*x + 2*I*c) + (-5*I*A - 2*B)*a^3 - 2*sqrt(2)*sqrt(-(2
5*A^2 - 20*I*A*B - 4*B^2)*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
)))*e^(-2*I*d*x - 2*I*c)/((5*I*A + 2*B)*a)) + 4*sqrt(2)*((I*A + 2*B)*a^2*e^(3*I*d*x + 3*I*c) + (I*A - 2*B)*a^2
*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 8.38, size = 2500, normalized size = 15.82 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((d^4*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (
A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A
*B*a^5*21i)/(2*d^2))^(1/2)*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*246
4i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*6i)/(A^3*a^14*d*126i - 336*B^3*a^14*d - A*B^2*a^14*d*1032i + 876*A^2*B*
a^14*d + A*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4
- (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/
d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)) + (A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((4
9*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^
4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d^2))^(1/2)*14i)/(A^3*a^11*d*
126i - 336*B^3*a^11*d + A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*
2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2
*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) -
(B^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4
+ (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) +
(A*B*a^5*21i)/(2*d^2))^(1/2)*56i)/(A^3*a^11*d*126i - 336*B^3*a^11*d + A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^2
2)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4
*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1
/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) + (88*A*B*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^22)/d
^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a
^6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d^2))^(1/2))/(A^3*a^11*d*126i - 336*B^3*a^
11*d + A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3
*B*a^22*616i)/d^4)^(1/2)*2i - 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B
^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d))*(((49*A^4*a^22)/d^
4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^
6) - (57*A^2*a^5)/(8*d^2) + (9*B^2*a^5)/(2*d^2) + (A*B*a^5*21i)/(2*d^2))^(1/2)*2i - atan((d^4*(a + a*tan(c + d
*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2) - ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^
2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2)
*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i
)/d^4)^(1/2)*6i)/(A^3*a^14*d*126i - 336*B^3*a^14*d - A*B^2*a^14*d*1032i + 876*A^2*B*a^14*d - A*a^3*d^3*((49*A^
4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(
1/2)*2i + 4*B*a^3*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d
^4 - (A^3*B*a^22*616i)/d^4)^(1/2)) - (A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2
*a^5)/(8*d^2) - ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (
A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2)*14i)/(A^3*a^11*d*126i - 336*B^3*a^11*d - A*
d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*6
16i)/d^4)^(1/2)*2i + 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2
464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i + 876*A^2*B*a^11*d) + (B^2*a^8*d^2*(a + a*tan(c
+ d*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2) - ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328
*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1
/2)*56i)/(A^3*a^11*d*126i - 336*B^3*a^11*d - A*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^2
2)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2)*2i + 4*B*d^3*((49*A^4*a^22)/d^4 + (784*B^4*a^22
)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*616i)/d^4)^(1/2) - A*B^2*a^11*d*1032i +
876*A^2*B*a^11*d) - (88*A*B*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((9*B^2*a^5)/(2*d^2) - (57*A^2*a^5)/(8*d^2)
- ((49*A^4*a^22)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*2464i)/d^4 - (A^3*B*a^22*61
6i)/d^4)^(1/2)/(8*a^6) + (A*B*a^5*21i)/(2*d^2))^(1/2))/(A^3*a^11*d*126i - 336*B^3*a^11*d - A*d^3*((49*A^4*a^22
)/d^4 + (784*B^4*a^22)/d^4 - (2328*A^2*B^2*a^22...

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