3.8.57 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [757]

Optimal. Leaf size=218 \[ -\frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {12 a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d} \]

[Out]

(-2-2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^
(1/2)/d-2/5*I*a^2*cot(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)-2/5*a^2*cot(d*x+c)^(5/2)/d/(a+I*a*tan(d*x+c))^(1
/2)-4/5*I*a*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d+12/5*a*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.43, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4326, 3634, 3677, 3679, 12, 3625, 211} \begin {gather*} -\frac {(2+2 i) a^{3/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {12 a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-2 - 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]
]*Sqrt[Tan[c + d*x]])/d - (((2*I)/5)*a^2*Cot[c + d*x]^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (2*a^2*Cot[c + d
*x]^(5/2))/(5*d*Sqrt[a + I*a*Tan[c + d*x]]) + (12*a*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(5*d) - (((
4*I)/5)*a*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3625

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

Rule 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((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 - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3677

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3679

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*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] && LtQ[n, -1]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

\begin {align*} \int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {1}{5} \left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {9 i a^2}{2}+\frac {11}{2} a^2 \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\\ &=-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-3 i a^3+2 a^3 \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{5 a^2}\\ &=-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {9 a^4}{2}+3 i a^4 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^3}\\ &=-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {12 a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\left (8 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {15 i a^5 \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {12 a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\left (2 i a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {12 a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {\left (4 a^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac {(2+2 i) a^{3/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 i a^2 \cot ^{\frac {3}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 a^2 \cot ^{\frac {5}{2}}(c+d x)}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {12 a \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {4 i a \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 1.92, size = 161, normalized size = 0.74 \begin {gather*} \frac {4 a \left (e^{i (c+d x)} \left (5-10 e^{2 i (c+d x)}+9 e^{4 i (c+d x)}\right )-5 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \cos (c+d x) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{5 d \left (-1+e^{2 i (c+d x)}\right )^2 \left (1+e^{2 i (c+d x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(4*a*(E^(I*(c + d*x))*(5 - 10*E^((2*I)*(c + d*x)) + 9*E^((4*I)*(c + d*x))) - 5*(-1 + E^((2*I)*(c + d*x)))^(5/2
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Cos[c + d*x]*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c
+ d*x]])/(5*d*(-1 + E^((2*I)*(c + d*x)))^2*(1 + E^((2*I)*(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. 1146 vs. \(2 (175 ) = 350\).
time = 47.15, size = 1147, normalized size = 5.26

method	result	size
default	\(\text {Expression too large to display}\)	\(1147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/d*(-5*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin
(d*x+c)-sin(d*x+c)-cos(d*x+c)+1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-
1))-2*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-10*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)*2^(1/2)-1)+10*I*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)-10*sin(d*x+c)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^
(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-5*sin(d*x+c)*cos(d*x+c)^2*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-
cos(d*x+c)+1))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-10*sin(d*x+c)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*2^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+10*I*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)-10*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)+9*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+5*I*cos(d*x+c)^2*s
in(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*
x+c)-cos(d*x+c)+1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))+9*cos(d*x+
c)^3*2^(1/2)-6*I*2^(1/2)*sin(d*x+c)-7*2^(1/2)*cos(d*x+c)^2+10*sin(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*2^(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+5*sin(d*x+c)*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2
)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-cos(d*
x+c)+1))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+10*sin(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-8*cos(d*x+c)*2^(1/2)+6*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(7/2)*(a*(I*sin(d*
x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3*2^(1/2)*a

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1181 vs. \(2 (164) = 328\).
time = 0.65, size = 1181, normalized size = 5.42 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-1/15*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(15*I + 15)*a*cos(3*d*x + 3
*c) + (16*I + 16)*a*cos(d*x + c) - (15*I - 15)*a*sin(3*d*x + 3*c) + (16*I - 16)*a*sin(d*x + c))*cos(3/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + ((15*I - 15)*a*cos(3*d*x + 3*c) - (16*I - 16)*a*cos(d*x + c) - (1
5*I + 15)*a*sin(3*d*x + 3*c) + (16*I + 16)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
- 1)))*sqrt(a) + 15*(2*((I - 1)*a*cos(2*d*x + 2*c)^2 + (I - 1)*a*sin(2*d*x + 2*c)^2 - (2*I - 2)*a*cos(2*d*x +
2*c) + (I - 1)*a)*arctan2(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 -
2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + ((
I + 1)*a*cos(2*d*x + 2*c)^2 + (I + 1)*a*sin(2*d*x + 2*c)^2 - (2*I + 2)*a*cos(2*d*x + 2*c) + (I + 1)*a)*log(4*c
os(d*x + c)^2 + 4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(c
os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
- 1))^2) + 8*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c) - 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 2*((-(15*I + 1
5)*a*cos(5*d*x + 5*c) + (5*I + 5)*a*cos(3*d*x + 3*c) - (2*I + 2)*a*cos(d*x + c) - (15*I - 15)*a*sin(5*d*x + 5*
c) + (5*I - 5)*a*sin(3*d*x + 3*c) - (2*I - 2)*a*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*
c) - 1)) + 3*(((I + 1)*a*cos(d*x + c) + (I - 1)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((I + 1)*a*cos(d*x + c) +
(I - 1)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 2*(-(I + 1)*a*cos(d*x + c) - (I - 1)*a*sin(d*x + c))*cos(2*d*x +
2*c) + (I + 1)*a*cos(d*x + c) + (I - 1)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) -
1)) + ((15*I - 15)*a*cos(5*d*x + 5*c) - (5*I - 5)*a*cos(3*d*x + 3*c) + (2*I - 2)*a*cos(d*x + c) - (15*I + 15)*
a*sin(5*d*x + 5*c) + (5*I + 5)*a*sin(3*d*x + 3*c) - (2*I + 2)*a*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c) - 1)) + 3*((-(I - 1)*a*cos(d*x + c) + (I + 1)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(I - 1
)*a*cos(d*x + c) + (I + 1)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + 2*((I - 1)*a*cos(d*x + c) - (I + 1)*a*sin(d*x
+ c))*cos(2*d*x + 2*c) - (I - 1)*a*cos(d*x + c) + (I + 1)*a*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c) - 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*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. 402 vs. \(2 (164) = 328\).
time = 0.74, size = 402, normalized size = 1.84 \begin {gather*} \frac {8 \, \sqrt {2} {\left (9 \, a e^{\left (5 i \, d x + 5 i \, c\right )} - 10 \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 5 \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} - 5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} + 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) + 5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \log \left (-\frac {{\left (\sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {32 i \, a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} - 8 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{20 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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)^(7/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/20*(8*sqrt(2)*(9*a*e^(5*I*d*x + 5*I*c) - 10*a*e^(3*I*d*x + 3*I*c) + 5*a*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 5*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^
(2*I*d*x + 2*I*c) + d)*sqrt(32*I*a^3/d^2)*log(1/2*(sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(32*I*a^3/d^2)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) + 8*I*a^2*e^(I*d*x +
I*c))*e^(-I*d*x - I*c)/a) + 5*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(32*I*a^3/d^2)*log(-1
/2*(sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(32*I*a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*
x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 8*I*a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a))/(d*e^(4*I*d*x + 4*I
*c) - 2*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)**(7/2)*(a+I*a*tan(d*x+c))**(3/2),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)^(7/2)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(3/2), x)

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