3.8.87 \(\int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) [787]

Optimal. Leaf size=258 \[ \frac {2 \sqrt [4]{-1} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

2*(-1)^(1/4)*arctan((-1)^(3/4)*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)/a^(5/2)/d+(1/8-1/8*I)*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)/a^(5/2)/d+7/4/a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-1/5/d/cot(d*x+c)^(5/2)/(a+I*a*t
an(d*x+c))^(5/2)+1/2*I/a/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.52, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4326, 3639, 3676, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {2 \sqrt [4]{-1} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \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 )}{a^{5/2} d}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

(2*(-1)^(1/4)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sq
rt[Tan[c + d*x]])/(a^(5/2)*d) + ((1/8 - I/8)*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]])/(a^(5/2)*d) - 1/(5*d*Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^
(5/2)) + (I/2)/(a*d*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + 7/(4*a^2*d*Sqrt[Cot[c + d*x]]*Sqrt[a +
I*a*Tan[c + d*x]])

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 209

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

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 3639

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

Rule 3676

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]

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 3682

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

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 \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x) \left (-\frac {5 a}{2}+5 i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (-\frac {45 i a^2}{4}-15 a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {105 a^3}{8}-15 i a^3 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{15 a^6}\\ &=-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^4}+\frac {\left (\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}{8 a^3}\\ &=-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac {\left (i \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 )}{4 a d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \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)}}{a^{5/2} d}-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d}\\ &=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \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)}}{a^{5/2} d}-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^2 d}\\ &=\frac {2 \sqrt [4]{-1} \tan ^{-1}\left (\frac {(-1)^{3/4} \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)}}{a^{5/2} d}+\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \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)}}{a^{5/2} d}-\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 3.82, size = 241, normalized size = 0.93 \begin {gather*} \frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (1-8 e^{2 i (c+d x)}+48 e^{4 i (c+d x)}-41 e^{6 i (c+d x)}-5 e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+40 \sqrt {2} e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\cot (c+d x)}}{20 \sqrt {2} a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((I/20)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 - 8*E^((2*I)*(c + d*x)) + 48*E^((4*I)*(c +
d*x)) - 41*E^((6*I)*(c + d*x)) - 5*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/
Sqrt[-1 + E^((2*I)*(c + d*x))]] + 40*Sqrt[2]*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[
2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*a^3*d*E^((6*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. 1577 vs. \(2 (202 ) = 404\).
time = 46.56, size = 1578, normalized size = 6.12

method	result	size
default	\(\text {Expression too large to display}\)	\(1578\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-1/40-1/40*I)/d*(-15*cos(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+80*I*(
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+80*I*cos(d*x+c)^3*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)-80*I
*cos(d*x+c)^3*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+80*I*cos(d*x+c)^3*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
-1)-80*I*cos(d*x+c)^3*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)-40*I*cos(d*x+c)^2*ln(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)-I)+40*I*cos(d*x+c)^2*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-40*I*cos(d*x+c)^2*ln(((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)-1)+40*I*cos(d*x+c)^2*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)-60*I*cos(d*x+c)*ln(((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)-I)+60*I*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-60*I*cos(d*x+c)*ln(((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)-1)+60*I*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)-49*I*((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)*sin(d*x+c)-80*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)-10*cos(d*x+c)^2*2^(1/2)*arctan(
(1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+20*I*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^2-10*I*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)+20*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)*co
s(d*x+c)^3-80*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*cos(d*x+c)^2*sin(d*x+c)-80*ln(((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)+80*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)*cos(d*x+c)^2*sin(d*x+c)+80*ln
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)-84*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c
)^2*sin(d*x+c)+40*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*sin(d*x+c)+40*cos(d*x+c)*ln(((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)-1)*sin(d*x+c)-40*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)*sin(d*x+c)-40*cos(
d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*sin(d*x+c)-5*I*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+84*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)+49*sin(d*x+
c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+5*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))
-80*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+80*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+20*I*ln((
(-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)-20*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+20*I*ln(((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)-1)-20*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I)+20*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))
^(1/2)-I)-20*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+20*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))
^(1/2)-1)-20*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+I))*cos(d*x+c)^4*(a*(I*sin(d*x+c)+cos(d*x+c))/co
s(d*x+c))^(1/2)/(4*I*cos(d*x+c)^2*sin(d*x+c)+4*cos(d*x+c)^3-I*sin(d*x+c)-3*cos(d*x+c))/((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(7/2)/a^3

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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. 662 vs. \(2 (192) = 384\).
time = 0.97, size = 662, normalized size = 2.57 \begin {gather*} -\frac {{\left (10 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (2 \, \sqrt {2} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\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}} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 10 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (2 \, \sqrt {2} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\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}} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 10 \, a^{3} d \sqrt {-\frac {4 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-16 \, {\left (\sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{4} d 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}} \sqrt {-\frac {4 i}{a^{5} d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 10 \, a^{3} d \sqrt {-\frac {4 i}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (16 \, {\left (\sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} - a^{4} d 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}} \sqrt {-\frac {4 i}{a^{5} d^{2}}} - 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {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}} {\left (-41 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 48 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/40*(10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(2*sqrt(2)*(I*a^3*d*e^(2*I*d*x + 2*I*c) - I*
a^3*d)*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))*sqrt(-1/8
*I/(a^5*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 10*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*l
og(-4*(2*sqrt(2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) + I*a^3*d)*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))*sqrt(-1/8*I/(a^5*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 10
*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-16*(sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) - a^4*d*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))*sqrt(-4*I
/(a^5*d^2)) + 3*a^2*e^(2*I*d*x + 2*I*c) - a^2)*e^(-2*I*d*x - 2*I*c)) - 10*a^3*d*sqrt(-4*I/(a^5*d^2))*e^(5*I*d*
x + 5*I*c)*log(16*(sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) - a^4*d*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))*sqrt(-4*I/(a^5*d^2)) - 3*a^2*e^(2*I*d*x + 2*I*c
) + a^2)*e^(-2*I*d*x - 2*I*c)) - sqrt(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))*(-41*I*e^(6*I*d*x + 6*I*c) + 48*I*e^(4*I*d*x + 4*I*c) - 8*I*e^(2*I*d*x + 2*I*c) + I))
*e^(-5*I*d*x - 5*I*c)/(a^3*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(1/cot(d*x+c)**(7/2)/(a+I*a*tan(d*x+c))**(5/2),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(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(d*x + c) + a)^(5/2)*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 \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int(1/(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2)), x)

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