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)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)*cot(d*x+c)^(1/2)*tan(d
*x+c)^(1/2)/d-2/15*(I*A+5*B)*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/5*A*cot(d*x+c)^(5/2)*(a+I*a*tan(d*x
+c))^(1/2)/d+2/15*(13*A-5*I*B)*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.67, antiderivative size = 198, normalized size of antiderivative = 1.00, 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 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 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 {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*}

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Mathematica [A]  time = 3.43, 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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fricas [B]  time = 0.52, size = 475, normalized size = 2.40 \[ \frac {8 \, \sqrt {2} {\left ({\left (17 \, A - 10 i \, B\right )} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 \, {\left (2 \, A - i \, B\right )} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, A e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 4 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (8 i \, A^{2} + 16 \, A B - 8 i \, B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} - 4 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right )}{60 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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]

1/60*(8*sqrt(2)*((17*A - 10*I*B)*e^(5*I*d*x + 5*I*c) - 10*(2*A - I*B)*e^(3*I*d*x + 3*I*c) + 15*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)) - 15*(d*e^(4
*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a/d^2)*log((sqrt(2)*(I*d*e^(2
*I*d*x + 2*I*c) - I*d)*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a/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)) - 4*(A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - 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((8*I*A^2 + 16*A*B - 8*I*B^2)*a/d^2)*log((sqrt(2
)*(-I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((8*I*A^2 + 16*A*B - 8*I*B^2)*a/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)) - 4*(A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]

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)

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maple [B]  time = 4.44, size = 2243, normalized size = 11.33 \[ \text {Expression too large to display} \]

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*(15*I*B*cos(d*x+c)^2*sin(d*x+c)*((-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))+34*A*2^(1/2)*cos(d*x+c)^3-32*A*2^(1/2)*cos(d*x+c)^2-28*A*2^(1/2)*cos(d*x+c)-10*B*sin(d*x+c
)*2^(1/2)+30*I*B*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)-1)+26*A*2^(1/2)-15*A*cos(d*x+c)^2*sin(d*x+c)*((-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))-30*A*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arcta
n(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)+30*B*cos(d*x+c)^2*sin(d*x+c)*((-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)+30*B*cos(d*x+c)^2*sin(d*x+c)*((-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)+15*B*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/si
n(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))+34*I*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-
2*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)-30*I*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)+1)-30*I*A*sin(d*x+c)*((-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)-15*I*A*sin(d*x+c)*((-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))-30*I*B*sin(d*x+c)*((-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)-15*I*B*sin(d*x+c)*((-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))-30*I*B*sin(d*x+c)*((-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)-30*A*cos(d*x+c)^2*sin(d*x+c)*((-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)+30*A*sin(d*x+c)*((-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)+15*A*sin(d*x+c)*((-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)-si
n(d*x+c)+1))+30*A*sin(d*x+c)*((-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)-30*B*sin(d*x+c)*((-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)
-30*B*sin(d*x+c)*((-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)-15*B*
sin(d*x+c)*((-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))-20*I*B*co
s(d*x+c)^3*2^(1/2)+10*I*B*cos(d*x+c)^2*2^(1/2)-26*I*A*sin(d*x+c)*2^(1/2)+20*I*B*2^(1/2)*cos(d*x+c)-10*I*B*2^(1
/2)-10*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)+20*B*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+30*I*A*cos(d*x+c)^2*sin(d*x+c)*((-
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)+30*I*A*cos(d*x+c)^2*sin(d
*x+c)*((-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)+15*I*A*cos(d*x+c
)^2*sin(d*x+c)*((-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)-c
os(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))+30*I*
B*cos(d*x+c)^2*sin(d*x+c)*((-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)/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)

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maxima [B]  time = 2.23, size = 1389, normalized size = 7.02 \[ \text {result too large to display} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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