3.542 \(\int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=201 \[ -\frac{(2+2 i) a^{3/2} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (5 B+6 i A) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}+\frac{4 a (9 A-10 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d} \]
[Out]
((-2 - 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Co
t[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (4*a*(9*A - (10*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d
) - (2*a*((6*I)*A + 5*B)*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (2*a*A*Cot[c + d*x]^(5/2)*Sqr
t[a + I*a*Tan[c + d*x]])/(5*d)
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Rubi [A] time = 0.710295, antiderivative size = 201, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used =
{4241, 3593, 3598, 12, 3544, 205} \[ -\frac{(2+2 i) a^{3/2} (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a (5 B+6 i A) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}+\frac{4 a (9 A-10 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
((-2 - 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Co
t[c + d*x]]*Sqrt[Tan[c + d*x]])/d + (4*a*(9*A - (10*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d
) - (2*a*((6*I)*A + 5*B)*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(15*d) - (2*a*A*Cot[c + d*x]^(5/2)*Sqr
t[a + I*a*Tan[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 3593
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] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
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) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{1}{5} \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 (6 i A+5 B)-\frac{1}{2} a (4 A-5 i B) \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a (6 i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a 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}{2} a^2 (9 A-10 i B)-\frac{1}{2} a^2 (6 i A+5 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a}\\ &=\frac{4 a (9 A-10 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a (6 i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a 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)}}{4 \sqrt{\tan (c+d x)}} \, dx}{15 a^2}\\ &=\frac{4 a (9 A-10 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a (6 i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}-\left (2 a (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{4 a (9 A-10 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a (6 i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}+\frac{\left (4 i a^3 (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{(2-2 i) a^{3/2} (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{4 a (9 A-10 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a (6 i A+5 B) \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{15 d}-\frac{2 a A \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{5 d}\\ \end{align*}
Mathematica [A] time = 5.24672, size = 289, normalized size = 1.44 \[ \frac{(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac{(-1+i \tan (c+d x)) \sqrt{\cot (c+d x)} \csc ^2(c+d x) ((5 B+6 i A) \sin (2 (c+d x))+(21 A-20 i B) \cos (2 (c+d x))-15 A+20 i B)}{15 \sqrt{\sec (c+d x)}}-2 \sqrt{2} (A-i B) e^{-2 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\frac{i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{d \sec ^{\frac{5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
(((-2*Sqrt[2]*(A - I*B)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[(I
*(1 + E^((2*I)*(c + d*x))))/(-1 + E^((2*I)*(c + d*x)))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]
])/E^((2*I)*(c + d*x)) + (Sqrt[Cot[c + d*x]]*Csc[c + d*x]^2*(-15*A + (20*I)*B + (21*A - (20*I)*B)*Cos[2*(c + d
*x)] + ((6*I)*A + 5*B)*Sin[2*(c + d*x)])*(-1 + I*Tan[c + d*x]))/(15*Sqrt[Sec[c + d*x]]))*(a + I*a*Tan[c + d*x]
)^(3/2)*(A + B*Tan[c + d*x]))/(d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))
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Maple [B] time = 0.585, size = 2244, normalized size = 11.2 \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))^(3/2)*(A+B*tan(d*x+c)),x)
[Out]
-1/15/d*a*2^(1/2)*(-5*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*si
n(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)+27*A*cos(d*x+c)^3*2^(1/2)-24*A*cos(d*x+c)*
2^(1/2)-25*I*B*2^(1/2)*cos(d*x+c)^3+20*I*B*2^(1/2)*cos(d*x+c)^2-18*I*A*2^(1/2)*sin(d*x+c)+25*I*B*2^(1/2)*cos(d
*x+c)-21*A*2^(1/2)*cos(d*x+c)^2-20*I*B*2^(1/2)+18*A*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(-(((c
os(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*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))-30*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/s
in(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)*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))-20*B*2^(1/2)*sin(d*x+c)+25*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+27*I*A*2^(1/2)*cos(d*x+c)^2*sin(
d*x+c)-6*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)+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(((c
os(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)*arcta
n(((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)/sin(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))-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
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Maxima [B] time = 3.37154, size = 1940, normalized size = 9.65 \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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
-1/225*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((-(450*I + 450)*A + (450*I -
450)*B)*a*cos(3*d*x + 3*c) + ((480*I + 480)*A - (450*I - 450)*B)*a*cos(d*x + c) + (-(450*I - 450)*A - (450*I +
450)*B)*a*sin(3*d*x + 3*c) + ((480*I - 480)*A + (450*I + 450)*B)*a*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) - 1)) + (((450*I - 450)*A + (450*I + 450)*B)*a*cos(3*d*x + 3*c) + (-(480*I - 480)*A - (
450*I + 450)*B)*a*cos(d*x + c) + (-(450*I + 450)*A + (450*I - 450)*B)*a*sin(3*d*x + 3*c) + ((480*I + 480)*A -
(450*I - 450)*B)*a*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((((450*I
- 450)*A + (450*I + 450)*B)*a*cos(2*d*x + 2*c)^2 + ((450*I - 450)*A + (450*I + 450)*B)*a*sin(2*d*x + 2*c)^2 +
(-(900*I - 900)*A - (900*I + 900)*B)*a*cos(2*d*x + 2*c) + ((450*I - 450)*A + (450*I + 450)*B)*a)*arctan2(2*(c
os(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*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + (((225*I + 225)*A - (225*I - 225
)*B)*a*cos(2*d*x + 2*c)^2 + ((225*I + 225)*A - (225*I - 225)*B)*a*sin(2*d*x + 2*c)^2 + (-(450*I + 450)*A + (45
0*I - 450)*B)*a*cos(2*d*x + 2*c) + ((225*I + 225)*A - (225*I - 225)*B)*a)*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) + (((-(450*I + 450)*A + (450*I - 450)*B)*a*co
s(5*d*x + 5*c) + ((150*I + 150)*A - (600*I - 600)*B)*a*cos(3*d*x + 3*c) + (-(60*I + 60)*A + (150*I - 150)*B)*a
*cos(d*x + c) + (-(450*I - 450)*A - (450*I + 450)*B)*a*sin(5*d*x + 5*c) + ((150*I - 150)*A + (600*I + 600)*B)*
a*sin(3*d*x + 3*c) + (-(60*I - 60)*A - (150*I + 150)*B)*a*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c) - 1)) + ((((90*I + 90)*A - (150*I - 150)*B)*a*cos(d*x + c) + ((90*I - 90)*A + (150*I + 150)*B)*a*
sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((90*I + 90)*A - (150*I - 150)*B)*a*cos(d*x + c) + (((90*I + 90)*A - (150*I
- 150)*B)*a*cos(d*x + c) + ((90*I - 90)*A + (150*I + 150)*B)*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((90*I - 90
)*A + (150*I + 150)*B)*a*sin(d*x + c) + ((-(180*I + 180)*A + (300*I - 300)*B)*a*cos(d*x + c) + (-(180*I - 180)
*A - (300*I + 300)*B)*a*sin(d*x + c))*cos(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1
)) + (((450*I - 450)*A + (450*I + 450)*B)*a*cos(5*d*x + 5*c) + (-(150*I - 150)*A - (600*I + 600)*B)*a*cos(3*d*
x + 3*c) + ((60*I - 60)*A + (150*I + 150)*B)*a*cos(d*x + c) + (-(450*I + 450)*A + (450*I - 450)*B)*a*sin(5*d*x
+ 5*c) + ((150*I + 150)*A - (600*I - 600)*B)*a*sin(3*d*x + 3*c) + (-(60*I + 60)*A + (150*I - 150)*B)*a*sin(d*
x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (((-(90*I - 90)*A - (150*I + 150)*B)*a*cos(
d*x + c) + ((90*I + 90)*A - (150*I - 150)*B)*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (-(90*I - 90)*A - (150*I + 1
50)*B)*a*cos(d*x + c) + ((-(90*I - 90)*A - (150*I + 150)*B)*a*cos(d*x + c) + ((90*I + 90)*A - (150*I - 150)*B)
*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 + ((90*I + 90)*A - (150*I - 150)*B)*a*sin(d*x + c) + (((180*I - 180)*A + (
300*I + 300)*B)*a*cos(d*x + c) + (-(180*I + 180)*A + (300*I - 300)*B)*a*sin(d*x + c))*cos(2*d*x + 2*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)
________________________________________________________________________________________
Fricas [B] time = 1.44653, size = 1436, normalized size = 7.14 \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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
1/30*(4*sqrt(2)*((27*A - 25*I*B)*a*e^(4*I*d*x + 4*I*c) - 10*(3*A - 4*I*B)*a*e^(2*I*d*x + 2*I*c) + 15*(A - I*B)
*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*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(
(sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) + (-2*I*A - 2*B)*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) + I*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^
2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((2*I*A + 2*B)*a)) + 15*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d
^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((2*I*A + 2*B)*a*e^(2*I*d*x + 2*I*c) +
(-2*I*A - 2*B)*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) - I*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(
(2*I*A + 2*B)*a)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
________________________________________________________________________________________
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))**(3/2)*(A+B*tan(d*x+c)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \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))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^(7/2), x)