3.536 \(\int \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=198 \[ -\frac{2 (5 B+i A) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}+\frac{2 (13 A-5 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{(1+i) \sqrt{a} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d} \]
[Out]
((-1 - I)*Sqrt[a]*(A - I*B)*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*(13*A - (5*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) -
(2*(I*A + 5*B)*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (2*A*Cot[c + d*x]^(5/2)*Sqrt[a + I*a*Ta
n[c + d*x]])/(5*d)
________________________________________________________________________________________
Rubi [A] time = 0.668029, antiderivative size = 198, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used =
{4241, 3598, 12, 3544, 205} \[ -\frac{2 (5 B+i A) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}+\frac{2 (13 A-5 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{(1+i) \sqrt{a} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]
[Out]
((-1 - I)*Sqrt[a]*(A - I*B)*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*(13*A - (5*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) -
(2*(I*A + 5*B)*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (2*A*Cot[c + d*x]^(5/2)*Sqrt[a + I*a*Ta
n[c + d*x]])/(5*d)
Rule 4241
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]
Rule 3598
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 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3544
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 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \cot ^{\frac{7}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\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 (\frac{1}{2} a (i A+5 B)-2 a A \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{5 a}\\ &=-\frac{2 (i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 A \cot ^{\frac{5}{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{1}{4} a^2 (13 A-5 i B)-\frac{1}{2} a^2 (i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^2}\\ &=\frac{2 (13 A-5 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 (i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 A \cot ^{\frac{5}{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 a^3 (i A+B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^3}\\ &=\frac{2 (13 A-5 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 (i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\left ((i A+B) \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 (13 A-5 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 (i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{\left (2 i a^2 (i A+B) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{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{(1-i) \sqrt{a} (i A+B) \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 (13 A-5 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 (i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}\\ \end{align*}
Mathematica [A] time = 2.98304, size = 188, normalized size = 0.95 \[ \frac{e^{-i (c+d x)} \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-15 (A-i B) \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+2 A e^{i (c+d x)} \left (-20 e^{2 i (c+d x)}+17 e^{4 i (c+d x)}+15\right )-20 i B e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )}{15 d \left (-1+e^{2 i (c+d x)}\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]
[Out]
(((-20*I)*B*E^((3*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x))) + 2*A*E^(I*(c + d*x))*(15 - 20*E^((2*I)*(c + d*x))
+ 17*E^((4*I)*(c + d*x))) - 15*(A - I*B)*(-1 + E^((2*I)*(c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^
((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*E^(I*(c + d*x))*(-1 + E^((2*I)*(c +
d*x)))^2)
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Maple [B] time = 0.667, size = 2243, normalized size = 11.3 \begin{align*} \text{result too large to display} \end{align*}
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))^(1/2)*(A+B*tan(d*x+c)),x)
[Out]
-1/30/d*2^(1/2)*(-10*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)+30*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin
(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)+30*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c
)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)+34*A*cos(d*x+c)^3*2^(1/2)-28*A*cos(d*x+c)*2
^(1/2)-32*A*2^(1/2)*cos(d*x+c)^2+26*A*2^(1/2)-20*I*B*cos(d*x+c)^3*2^(1/2)+10*I*B*cos(d*x+c)^2*2^(1/2)-26*I*A*s
in(d*x+c)*2^(1/2)+20*I*B*cos(d*x+c)*2^(1/2)+30*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x
+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)+15*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+
c)-cos(d*x+c)-sin(d*x+c)+1))-10*I*B*2^(1/2)-30*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-15*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*ln(-(((cos(d*x+c)-1)
/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(
d*x+c)+cos(d*x+c)+sin(d*x+c)-1))-2*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)-30*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*
sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-30*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*
x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-15*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*l
n(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))-10*B*2^(1/2)*sin(d*x+c)+20*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/
2)+15*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(
1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(
d*x+c)+1))+15*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x
+c)+sin(d*x+c)-1))+30*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)*2^(1/2)+1)+30*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)
-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-30*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(
d*x+c))^(1/2)*2^(1/2)+1)-30*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^
(1/2)*2^(1/2)-1)-15*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1
/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d
*x+c)-1))+30*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1
)-30*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1
/2)+1)-30*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)
*2^(1/2)-1)-15*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+
c)-sin(d*x+c)+1))+30*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*2^(1/2)+1)+30*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)*2^(1/2)-1)+15*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ln(-(((cos(d*x+c)-1
)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin
(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+34*I*A*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)-30*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1))*(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
________________________________________________________________________________________
Maxima [B] time = 3.90849, size = 1875, normalized size = 9.47 \begin{align*} \text{result too large to display} \end{align*}
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))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
-1/900*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((-(900*I + 900)*A + (900*I -
900)*B)*cos(3*d*x + 3*c) + ((1170*I + 1170)*A - (750*I - 750)*B)*cos(d*x + c) + (-(900*I - 900)*A - (900*I + 9
00)*B)*sin(3*d*x + 3*c) + ((1170*I - 1170)*A + (750*I + 750)*B)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c) - 1)) + (((900*I - 900)*A + (900*I + 900)*B)*cos(3*d*x + 3*c) + (-(1170*I - 1170)*A - (750*
I + 750)*B)*cos(d*x + c) + (-(900*I + 900)*A + (900*I - 900)*B)*sin(3*d*x + 3*c) + ((1170*I + 1170)*A - (750*I
- 750)*B)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((((900*I - 900)*
A + (900*I + 900)*B)*cos(2*d*x + 2*c)^2 + ((900*I - 900)*A + (900*I + 900)*B)*sin(2*d*x + 2*c)^2 + (-(1800*I -
1800)*A - (1800*I + 1800)*B)*cos(2*d*x + 2*c) + (900*I - 900)*A + (900*I + 900)*B)*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*arctan2(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*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + (((450*I + 450)*A - (450*I - 450)*B)*cos(2*d*x
+ 2*c)^2 + ((450*I + 450)*A - (450*I - 450)*B)*sin(2*d*x + 2*c)^2 + (-(900*I + 900)*A + (900*I - 900)*B)*cos(
2*d*x + 2*c) + (450*I + 450)*A - (450*I - 450)*B)*log(4*cos(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)*(cos(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*arctan2(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) + (((-(900*I + 900)*A + (900*I - 900)*B)*cos(5*d*x + 5*c) + (-(150*I
+ 150)*A - (750*I - 750)*B)*cos(3*d*x + 3*c) + (-(390*I + 390)*A - (150*I - 150)*B)*cos(d*x + c) + (-(900*I -
900)*A - (900*I + 900)*B)*sin(5*d*x + 5*c) + (-(150*I - 150)*A + (750*I + 750)*B)*sin(3*d*x + 3*c) + (-(390*I
- 390)*A + (150*I + 150)*B)*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (((-(240*
I + 240)*A - (600*I - 600)*B)*cos(d*x + c) + (-(240*I - 240)*A + (600*I + 600)*B)*sin(d*x + c))*cos(2*d*x + 2*
c)^2 + ((-(240*I + 240)*A - (600*I - 600)*B)*cos(d*x + c) + (-(240*I - 240)*A + (600*I + 600)*B)*sin(d*x + c))
*sin(2*d*x + 2*c)^2 + (((480*I + 480)*A + (1200*I - 1200)*B)*cos(d*x + c) + ((480*I - 480)*A - (1200*I + 1200)
*B)*sin(d*x + c))*cos(2*d*x + 2*c) + (-(240*I + 240)*A - (600*I - 600)*B)*cos(d*x + c) + (-(240*I - 240)*A + (
600*I + 600)*B)*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (((900*I - 900)*A + (
900*I + 900)*B)*cos(5*d*x + 5*c) + ((150*I - 150)*A - (750*I + 750)*B)*cos(3*d*x + 3*c) + ((390*I - 390)*A - (
150*I + 150)*B)*cos(d*x + c) + (-(900*I + 900)*A + (900*I - 900)*B)*sin(5*d*x + 5*c) + (-(150*I + 150)*A - (75
0*I - 750)*B)*sin(3*d*x + 3*c) + (-(390*I + 390)*A - (150*I - 150)*B)*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) - 1)) + ((((240*I - 240)*A - (600*I + 600)*B)*cos(d*x + c) + (-(240*I + 240)*A - (600
*I - 600)*B)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (((240*I - 240)*A - (600*I + 600)*B)*cos(d*x + c) + (-(240*I +
240)*A - (600*I - 600)*B)*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((-(480*I - 480)*A + (1200*I + 1200)*B)*cos(d*x
+ c) + ((480*I + 480)*A + (1200*I - 1200)*B)*sin(d*x + c))*cos(2*d*x + 2*c) + ((240*I - 240)*A - (600*I + 600)
*B)*cos(d*x + c) + (-(240*I + 240)*A - (600*I - 600)*B)*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(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] time = 1.53342, size = 1337, normalized size = 6.75 \begin{align*} \frac{4 \, \sqrt{2}{\left ({\left (17 \, A - 10 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 10 \,{\left (2 \, A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, A\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}} e^{\left (i \, d x + i \, c\right )} - 15 \,{\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{{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac{{\left (\sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - B\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}} e^{\left (i \, d x + i \, c\right )} + i \, d \sqrt{\frac{{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + 15 \,{\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{{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac{{\left (\sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - B\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}} e^{\left (i \, d x + i \, c\right )} - i \, d \sqrt{\frac{{\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right )}{30 \,{\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{align*}
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))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
1/30*(4*sqrt(2)*((17*A - 10*I*B)*e^(4*I*d*x + 4*I*c) - 10*(2*A - I*B)*e^(2*I*d*x + 2*I*c) + 15*A)*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))*e^(I*d*x + I*c) - 15*(d*e^(4
*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)*a/d^2)*log((sqrt(2)*((I*A + B)
*e^(2*I*d*x + 2*I*c) - I*A - B)*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))*e^(I*d*x + I*c) + I*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x
- 2*I*c)/(I*A + B)) + 15*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((2*I*A^2 + 4*A*B - 2*I*B^
2)*a/d^2)*log((sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) - I*A - B)*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))*e^(I*d*x + I*c) - I*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)*a/d^2)
*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)))/(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)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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))**(1/2)*(A+B*tan(d*x+c)),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt{i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}
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))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c)^(7/2), x)