3.543 \(\int \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=157 \[ \frac {(2+2 i) a^{3/2} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a (3 B+4 i A) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]
[Out]
(2+2*I)*a^(3/2)*(I*A+B)*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-2/3*a*A*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/3*a*(4*I*A+3*B)*cot(d*x+c)^(1/2)*(a+I*a*t
an(d*x+c))^(1/2)/d
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Rubi [A] time = 0.51, antiderivative size = 157, normalized size of antiderivative = 1.00,
number of steps used = 6, 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} \[ \frac {(2+2 i) a^{3/2} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a (3 B+4 i A) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
((2 + 2*I)*a^(3/2)*(I*A + 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*a*((4*I)*A + 3*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(3*d) -
(2*a*A*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(3*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 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]
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 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 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]
Rubi steps
\begin {align*} \int \cot ^{\frac {5}{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 {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {1}{3} \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 (4 i A+3 B)-\frac {1}{2} a (2 A-3 i B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a (4 i A+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {3 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{2 \sqrt {\tan (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 a (4 i A+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\left (2 a (A-i 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 a (4 i A+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (4 i a^3 (A-i 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 {2 a (4 i A+3 B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}
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Mathematica [A] time = 5.18, size = 259, normalized size = 1.65 \[ \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (2 \sqrt {2} (B+i A) e^{-2 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\frac {i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-\frac {2 (\cot (c+d x)-i) (A \csc (c+d x)+(3 B+4 i A) \sec (c+d x))}{3 \sqrt {\cot (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}\right )}{d \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
(((2*Sqrt[2]*(I*A + 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)) - (2*(-I + Cot[c + d*x])*(A*Csc[c + d*x] + ((4*I)*A + 3*B)*Sec[c + d*x]))/(3*Sqrt[Cot[c
+ d*x]]*Sec[c + d*x]^(3/2)))*(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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fricas [B] time = 0.58, size = 456, normalized size = 2.90 \[ \frac {\sqrt {2} {\left ({\left (-40 i \, A - 24 \, B\right )} a e^{\left (3 i \, d x + 3 i \, c\right )} + {\left (24 i \, A + 24 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 3 \, \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) + 3 \, \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {{\left (-32 i \, A^{2} - 64 \, A B + 32 i \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
1/12*(sqrt(2)*((-40*I*A - 24*B)*a*e^(3*I*d*x + 3*I*c) + (24*I*A + 24*B)*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)) - 3*sqrt((-32*I*A^2 - 64*A*B + 32*I
*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-(8*(A - I*B)*a^2*e^(I*d*x + I*c) + sqrt(2)*sqrt((-32*I*A^2 - 6
4*A*B + 32*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) + 3*sqrt((-32*I*A^2 - 64*A*B + 32*
I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(-(8*(A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(2)*sqrt((-32*I*A^2 -
64*A*B + 32*I*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)))/(d*e^(2*I*d*x + 2*I*c) - d)
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
Timed out
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maple [B] time = 4.23, size = 2017, normalized size = 12.85 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)
[Out]
-1/3/d*(5*A*2^(1/2)*cos(d*x+c)^2-A*2^(1/2)*cos(d*x+c)-3*B*sin(d*x+c)*2^(1/2)-4*A*2^(1/2)+6*A*((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+6*B*arctan(2^(1/2)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+6*B*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+3*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))*cos(d*x+c)^2+6*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin
(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+3*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(
2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)^2-3*I*B*2^(1/2)*cos(d*x+c)^2-4*I*A*sin(d*x+c)*2^(
1/2)+3*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)-6*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)-3*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*si
n(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)
-1))-6*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-6*B*arctan(2^
(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-6*B*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-3*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/
2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))-6*I*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)-6*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)-1)-3*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+
c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))+6
*I*B*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+3*I*B*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2
^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+6*I*B*((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+3*I*B*2^(1/2)+3*I*A*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(
d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))*cos(d*x+c)^2-6*I*B*arctan(2^(1/2)*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-3*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1
/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)^2-6*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+6*I*A*arctan(2^(1/2)*((-1+cos(d*x+c))/sin
(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+6*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arc
tan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)^2+5*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2))*(cos(d*x+c
)/sin(d*x+c))^(5/2)*(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)^2*a*2^(1/2)
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maxima [B] time = 1.23, size = 1101, normalized size = 7.01 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
-1/9*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((((18*I - 18)*A + (18*I + 18)*B)
*a*cos(3*d*x + 3*c) + (-(6*I - 6)*A - (18*I + 18)*B)*a*cos(d*x + c) + (-(18*I + 18)*A + (18*I - 18)*B)*a*sin(3
*d*x + 3*c) + ((6*I + 6)*A - (18*I - 18)*B)*a*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
- 1)) + (((18*I + 18)*A - (18*I - 18)*B)*a*cos(3*d*x + 3*c) + (-(6*I + 6)*A + (18*I - 18)*B)*a*cos(d*x + c) +
((18*I - 18)*A + (18*I + 18)*B)*a*sin(3*d*x + 3*c) + (-(6*I - 6)*A - (18*I + 18)*B)*a*sin(d*x + c))*sin(3/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((((18*I + 18)*A - (18*I - 18)*B)*a*cos(2*d*x + 2*c
)^2 + ((18*I + 18)*A - (18*I - 18)*B)*a*sin(2*d*x + 2*c)^2 + (-(36*I + 36)*A + (36*I - 36)*B)*a*cos(2*d*x + 2*
c) + ((18*I + 18)*A - (18*I - 18)*B)*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*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)) + ((-(9*I - 9)*A - (9*I + 9)*B)*a*cos(2*d*x + 2*c)^2 + (-(9*I - 9)*A - (9*I + 9)*B)*a*sin(2*
d*x + 2*c)^2 + ((18*I - 18)*A + (18*I + 18)*B)*a*cos(2*d*x + 2*c) + (-(9*I - 9)*A - (9*I + 9)*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*arct
an2(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) + ((((12*I - 12)*A*
a*cos(d*x + c) - (12*I + 12)*A*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (12*I - 12)*A*a*cos(d*x + c) + ((12*I - 12
)*A*a*cos(d*x + c) - (12*I + 12)*A*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 - (12*I + 12)*A*a*sin(d*x + c) + (-(24*I
- 24)*A*a*cos(d*x + c) + (24*I + 24)*A*a*sin(d*x + c))*cos(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c) - 1)) + (((12*I + 12)*A*a*cos(d*x + c) + (12*I - 12)*A*a*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (12
*I + 12)*A*a*cos(d*x + c) + ((12*I + 12)*A*a*cos(d*x + c) + (12*I - 12)*A*a*sin(d*x + c))*sin(2*d*x + 2*c)^2 +
(12*I - 12)*A*a*sin(d*x + c) + (-(24*I + 24)*A*a*cos(d*x + c) - (24*I - 24)*A*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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)^(5/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)
[Out]
int(cot(c + d*x)^(5/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**(5/2)*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)
[Out]
Timed out
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