3.8 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=111 \[ -\frac {a (B+i A) \cot ^3(c+d x)}{3 d}+\frac {a (A-i B) \cot ^2(c+d x)}{2 d}+\frac {a (B+i A) \cot (c+d x)}{d}+\frac {a (A-i B) \log (\sin (c+d x))}{d}+a x (B+i A)-\frac {a A \cot ^4(c+d x)}{4 d} \]

[Out]

a*(I*A+B)*x+a*(I*A+B)*cot(d*x+c)/d+1/2*a*(A-I*B)*cot(d*x+c)^2/d-1/3*a*(I*A+B)*cot(d*x+c)^3/d-1/4*a*A*cot(d*x+c
)^4/d+a*(A-I*B)*ln(sin(d*x+c))/d

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Rubi [A]  time = 0.19, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3591, 3529, 3531, 3475} \[ -\frac {a (B+i A) \cot ^3(c+d x)}{3 d}+\frac {a (A-i B) \cot ^2(c+d x)}{2 d}+\frac {a (B+i A) \cot (c+d x)}{d}+\frac {a (A-i B) \log (\sin (c+d x))}{d}+a x (B+i A)-\frac {a A \cot ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

a*(I*A + B)*x + (a*(I*A + B)*Cot[c + d*x])/d + (a*(A - I*B)*Cot[c + d*x]^2)/(2*d) - (a*(I*A + B)*Cot[c + d*x]^
3)/(3*d) - (a*A*Cot[c + d*x]^4)/(4*d) + (a*(A - I*B)*Log[Sin[c + d*x]])/d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2
+ b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) (a (i A+B)-a (A-i B) \tan (c+d x)) \, dx\\ &=-\frac {a (i A+B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) (-a (A-i B)-a (i A+B) \tan (c+d x)) \, dx\\ &=\frac {a (A-i B) \cot ^2(c+d x)}{2 d}-\frac {a (i A+B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot ^2(c+d x) (-a (i A+B)+a (A-i B) \tan (c+d x)) \, dx\\ &=\frac {a (i A+B) \cot (c+d x)}{d}+\frac {a (A-i B) \cot ^2(c+d x)}{2 d}-\frac {a (i A+B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\int \cot (c+d x) (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=a (i A+B) x+\frac {a (i A+B) \cot (c+d x)}{d}+\frac {a (A-i B) \cot ^2(c+d x)}{2 d}-\frac {a (i A+B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+(a (A-i B)) \int \cot (c+d x) \, dx\\ &=a (i A+B) x+\frac {a (i A+B) \cot (c+d x)}{d}+\frac {a (A-i B) \cot ^2(c+d x)}{2 d}-\frac {a (i A+B) \cot ^3(c+d x)}{3 d}-\frac {a A \cot ^4(c+d x)}{4 d}+\frac {a (A-i B) \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [C]  time = 0.88, size = 96, normalized size = 0.86 \[ -\frac {a \left (4 (B+i A) \cot ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(c+d x)\right )-6 (A-i B) \cot ^2(c+d x)-12 (A-i B) (\log (\tan (c+d x))+\log (\cos (c+d x)))+3 A \cot ^4(c+d x)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

-1/12*(a*(-6*(A - I*B)*Cot[c + d*x]^2 + 3*A*Cot[c + d*x]^4 + 4*(I*A + B)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2
, 1, -1/2, -Tan[c + d*x]^2] - 12*(A - I*B)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/d

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fricas [B]  time = 0.42, size = 206, normalized size = 1.86 \[ -\frac {6 \, {\left (4 \, A - 3 i \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} - 36 \, {\left (A - i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (16 \, A - 13 i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \, {\left (A - i \, B\right )} a - 3 \, {\left ({\left (A - i \, B\right )} a e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (A - i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*(6*(4*A - 3*I*B)*a*e^(6*I*d*x + 6*I*c) - 36*(A - I*B)*a*e^(4*I*d*x + 4*I*c) + 2*(16*A - 13*I*B)*a*e^(2*I*
d*x + 2*I*c) - 8*(A - I*B)*a - 3*((A - I*B)*a*e^(8*I*d*x + 8*I*c) - 4*(A - I*B)*a*e^(6*I*d*x + 6*I*c) + 6*(A -
I*B)*a*e^(4*I*d*x + 4*I*c) - 4*(A - I*B)*a*e^(2*I*d*x + 2*I*c) + (A - I*B)*a)*log(e^(2*I*d*x + 2*I*c) - 1))/(
d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*e^(2*I*d*x + 2*I*c) + d)

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giac [B]  time = 1.21, size = 282, normalized size = 2.54 \[ -\frac {3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 i \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 384 \, {\left (A a - i \, B a\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) - 192 \, {\left (A a - i \, B a\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 i \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 i \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 i \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/192*(3*A*a*tan(1/2*d*x + 1/2*c)^4 - 8*I*A*a*tan(1/2*d*x + 1/2*c)^3 - 8*B*a*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*
tan(1/2*d*x + 1/2*c)^2 + 24*I*B*a*tan(1/2*d*x + 1/2*c)^2 + 120*I*A*a*tan(1/2*d*x + 1/2*c) + 120*B*a*tan(1/2*d*
x + 1/2*c) + 384*(A*a - I*B*a)*log(tan(1/2*d*x + 1/2*c) + I) - 192*(A*a - I*B*a)*log(tan(1/2*d*x + 1/2*c)) + (
400*A*a*tan(1/2*d*x + 1/2*c)^4 - 400*I*B*a*tan(1/2*d*x + 1/2*c)^4 - 120*I*A*a*tan(1/2*d*x + 1/2*c)^3 - 120*B*a
*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x + 1/2*c)^2 + 24*I*B*a*tan(1/2*d*x + 1/2*c)^2 + 8*I*A*a*tan(1/2*d*
x + 1/2*c) + 8*B*a*tan(1/2*d*x + 1/2*c) + 3*A*a)/tan(1/2*d*x + 1/2*c)^4)/d

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maple [A]  time = 0.43, size = 159, normalized size = 1.43 \[ -\frac {i A a \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {i A \cot \left (d x +c \right ) a}{d}+i A a x +\frac {i A a c}{d}-\frac {i B a \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {i B a \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a A \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a A \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a A \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a B \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {B \cot \left (d x +c \right ) a}{d}+a B x +\frac {B a c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

-1/3*I/d*A*a*cot(d*x+c)^3+I/d*A*a*cot(d*x+c)+I*A*a*x+I/d*A*a*c-1/2*I/d*B*a*cot(d*x+c)^2-I/d*B*a*ln(sin(d*x+c))
-1/4*a*A*cot(d*x+c)^4/d+1/2*a*A*cot(d*x+c)^2/d+a*A*ln(sin(d*x+c))/d-1/3/d*a*B*cot(d*x+c)^3+1/d*B*cot(d*x+c)*a+
a*B*x+1/d*B*a*c

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maxima [A]  time = 0.86, size = 115, normalized size = 1.04 \[ \frac {12 \, {\left (d x + c\right )} {\left (i \, A + B\right )} a - 6 \, {\left (A - i \, B\right )} a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (A - i \, B\right )} a \log \left (\tan \left (d x + c\right )\right ) + \frac {{\left (12 i \, A + 12 \, B\right )} a \tan \left (d x + c\right )^{3} + 6 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )^{2} + {\left (-4 i \, A - 4 \, B\right )} a \tan \left (d x + c\right ) - 3 \, A a}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*(I*A + B)*a - 6*(A - I*B)*a*log(tan(d*x + c)^2 + 1) + 12*(A - I*B)*a*log(tan(d*x + c)) + ((
12*I*A + 12*B)*a*tan(d*x + c)^3 + 6*(A - I*B)*a*tan(d*x + c)^2 + (-4*I*A - 4*B)*a*tan(d*x + c) - 3*A*a)/tan(d*
x + c)^4)/d

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mupad [B]  time = 6.48, size = 100, normalized size = 0.90 \[ -\frac {\left (-B\,a-A\,a\,1{}\mathrm {i}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (-\frac {A\,a}{2}+\frac {B\,a\,1{}\mathrm {i}}{2}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {B\,a}{3}+\frac {A\,a\,1{}\mathrm {i}}{3}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {A\,a}{4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4}+\frac {a\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^5*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i),x)

[Out]

(a*atan(2*tan(c + d*x) + 1i)*(A - B*1i)*2i)/d - ((A*a)/4 + tan(c + d*x)*((A*a*1i)/3 + (B*a)/3) - tan(c + d*x)^
3*(A*a*1i + B*a) - tan(c + d*x)^2*((A*a)/2 - (B*a*1i)/2))/(d*tan(c + d*x)^4)

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sympy [B]  time = 3.77, size = 218, normalized size = 1.96 \[ \frac {a \left (A - i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 8 A a + 8 i B a + \left (32 A a e^{2 i c} - 26 i B a e^{2 i c}\right ) e^{2 i d x} + \left (- 36 A a e^{4 i c} + 36 i B a e^{4 i c}\right ) e^{4 i d x} + \left (24 A a e^{6 i c} - 18 i B a e^{6 i c}\right ) e^{6 i d x}}{- 3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} - 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} - 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

a*(A - I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-8*A*a + 8*I*B*a + (32*A*a*exp(2*I*c) - 26*I*B*a*exp(2*I*c))*
exp(2*I*d*x) + (-36*A*a*exp(4*I*c) + 36*I*B*a*exp(4*I*c))*exp(4*I*d*x) + (24*A*a*exp(6*I*c) - 18*I*B*a*exp(6*I
*c))*exp(6*I*d*x))/(-3*d*exp(8*I*c)*exp(8*I*d*x) + 12*d*exp(6*I*c)*exp(6*I*d*x) - 18*d*exp(4*I*c)*exp(4*I*d*x)
+ 12*d*exp(2*I*c)*exp(2*I*d*x) - 3*d)

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