3.21 \(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=116 \[ \frac {(-B+i A) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {a^3 (B+3 i A) \log (\sin (c+d x))}{d}+\frac {a^3 (3 B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {3593, 3594, 3589, 3475, 3531} \[ \frac {(-B+i A) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac {a^3 (B+3 i A) \log (\sin (c+d x))}{d}+\frac {a^3 (3 B+i A) \log (\cos (c+d x))}{d}-4 a^3 x (A-i B)-\frac {a A \cot (c+d x) (a+i a \tan (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

-4*a^3*(A - I*B)*x + (a^3*(I*A + 3*B)*Log[Cos[c + d*x]])/d + (a^3*((3*I)*A + B)*Log[Sin[c + d*x]])/d - (a*A*Co
t[c + d*x]*(a + I*a*Tan[c + d*x])^2)/d + ((I*A - B)*(a^3 + I*a^3*Tan[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 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 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 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 3594

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

Rubi steps

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

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Mathematica [B]  time = 5.27, size = 291, normalized size = 2.51 \[ \frac {a^3 \csc (c) \sec (c) \csc (c+d x) \sec (c+d x) \left (4 (3 A-i B) \sin (2 c) \sin (2 (c+d x)) \tan ^{-1}(\tan (4 c+d x))+\cos (2 d x) \left ((B+3 i A) \log \left (\sin ^2(c+d x)\right )+(3 B+i A) \log \left (\cos ^2(c+d x)\right )+2 d x (-7 A+5 i B)\right )+4 A \sin (2 (c+d x))+14 A d x \cos (4 c+2 d x)-i A \cos (4 c+2 d x) \log \left (\cos ^2(c+d x)\right )-3 i A \cos (4 c+2 d x) \log \left (\sin ^2(c+d x)\right )-4 A \sin (2 c)+4 A \sin (2 d x)+4 i B \sin (2 (c+d x))-10 i B d x \cos (4 c+2 d x)-3 B \cos (4 c+2 d x) \log \left (\cos ^2(c+d x)\right )-B \cos (4 c+2 d x) \log \left (\sin ^2(c+d x)\right )-4 i B \sin (2 c)-4 i B \sin (2 d x)\right )}{16 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(a^3*Csc[c]*Csc[c + d*x]*Sec[c]*Sec[c + d*x]*(14*A*d*x*Cos[4*c + 2*d*x] - (10*I)*B*d*x*Cos[4*c + 2*d*x] - I*A*
Cos[4*c + 2*d*x]*Log[Cos[c + d*x]^2] - 3*B*Cos[4*c + 2*d*x]*Log[Cos[c + d*x]^2] - (3*I)*A*Cos[4*c + 2*d*x]*Log
[Sin[c + d*x]^2] - B*Cos[4*c + 2*d*x]*Log[Sin[c + d*x]^2] + Cos[2*d*x]*(2*(-7*A + (5*I)*B)*d*x + (I*A + 3*B)*L
og[Cos[c + d*x]^2] + ((3*I)*A + B)*Log[Sin[c + d*x]^2]) - 4*A*Sin[2*c] - (4*I)*B*Sin[2*c] + 4*A*Sin[2*d*x] - (
4*I)*B*Sin[2*d*x] + 4*A*Sin[2*(c + d*x)] + (4*I)*B*Sin[2*(c + d*x)] + 4*(3*A - I*B)*ArcTan[Tan[4*c + d*x]]*Sin
[2*c]*Sin[2*(c + d*x)]))/(16*d)

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fricas [A]  time = 0.64, size = 138, normalized size = 1.19 \[ \frac {{\left (-2 i \, A + 2 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-2 i \, A - 2 \, B\right )} a^{3} + {\left ({\left (i \, A + 3 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-i \, A - 3 \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + {\left ({\left (3 i \, A + B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-3 i \, A - B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} - d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.97, size = 258, normalized size = 2.22 \[ \frac {3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, {\left (i \, A a^{3} + 3 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 12 \, {\left (-4 i \, A a^{3} - 4 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 6 \, {\left (i \, A a^{3} + 3 \, B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - 6 \, {\left (-3 i \, A a^{3} - B a^{3}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {-10 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 14 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 i \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 i \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 14 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*A*a^3*tan(1/2*d*x + 1/2*c) + 6*(I*A*a^3 + 3*B*a^3)*log(tan(1/2*d*x + 1/2*c) + 1) + 12*(-4*I*A*a^3 - 4*B
*a^3)*log(tan(1/2*d*x + 1/2*c) + I) + 6*(I*A*a^3 + 3*B*a^3)*log(tan(1/2*d*x + 1/2*c) - 1) - 6*(-3*I*A*a^3 - B*
a^3)*log(tan(1/2*d*x + 1/2*c)) + (-10*I*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 14*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 3*A*a
^3*tan(1/2*d*x + 1/2*c)^2 + 12*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 + 10*I*A*a^3*tan(1/2*d*x + 1/2*c) + 14*B*a^3*tan
(1/2*d*x + 1/2*c) + 3*A*a^3)/(tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)))/d

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maple [A]  time = 0.39, size = 134, normalized size = 1.16 \[ 4 i B x \,a^{3}+\frac {i A \,a^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {3 i A \,a^{3} \ln \left (\sin \left (d x +c \right )\right )}{d}-4 A \,a^{3} x -\frac {i a^{3} B \tan \left (d x +c \right )}{d}+\frac {4 i B \,a^{3} c}{d}-\frac {A \cot \left (d x +c \right ) a^{3}}{d}-\frac {4 A \,a^{3} c}{d}+\frac {3 a^{3} B \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{3} B \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.83, size = 86, normalized size = 0.74 \[ -\frac {4 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a^{3} - {\left (-2 i \, A - 2 \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - {\left (3 i \, A + B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) + i \, B a^{3} \tan \left (d x + c\right ) + \frac {A a^{3}}{\tan \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.43, size = 76, normalized size = 0.66 \[ \frac {a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B+A\,3{}\mathrm {i}\right )}{d}-\frac {4\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )}{d}-\frac {A\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}-\frac {B\,a^3\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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sympy [B]  time = 2.00, size = 223, normalized size = 1.92 \[ \frac {i a^{3} \left (A - 3 i B\right ) \log {\left (e^{2 i d x} + \frac {- 2 i A a^{3} - 2 B a^{3} + i a^{3} \left (A - 3 i B\right )}{- i A a^{3} e^{2 i c} + B a^{3} e^{2 i c}} \right )}}{d} + \frac {i a^{3} \left (3 A - i B\right ) \log {\left (e^{2 i d x} + \frac {- 2 i A a^{3} - 2 B a^{3} + i a^{3} \left (3 A - i B\right )}{- i A a^{3} e^{2 i c} + B a^{3} e^{2 i c}} \right )}}{d} + \frac {2 i A a^{3} + 2 B a^{3} + \left (2 i A a^{3} e^{2 i c} - 2 B a^{3} e^{2 i c}\right ) e^{2 i d x}}{- d e^{4 i c} e^{4 i d x} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

I*a**3*(A - 3*I*B)*log(exp(2*I*d*x) + (-2*I*A*a**3 - 2*B*a**3 + I*a**3*(A - 3*I*B))/(-I*A*a**3*exp(2*I*c) + B*
a**3*exp(2*I*c)))/d + I*a**3*(3*A - I*B)*log(exp(2*I*d*x) + (-2*I*A*a**3 - 2*B*a**3 + I*a**3*(3*A - I*B))/(-I*
A*a**3*exp(2*I*c) + B*a**3*exp(2*I*c)))/d + (2*I*A*a**3 + 2*B*a**3 + (2*I*A*a**3*exp(2*I*c) - 2*B*a**3*exp(2*I
*c))*exp(2*I*d*x))/(-d*exp(4*I*c)*exp(4*I*d*x) + d)

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