3.42 \(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=131 \[ -\frac {(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac {3 (-B+i A) \cot (c+d x)}{2 a d}-\frac {(2 A+i B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {3 x (-B+i A)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3596, 3529, 3531, 3475} \[ -\frac {(2 A+i B) \cot ^2(c+d x)}{2 a d}+\frac {3 (-B+i A) \cot (c+d x)}{2 a d}-\frac {(2 A+i B) \log (\sin (c+d x))}{a d}+\frac {(A+i B) \cot ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {3 x (-B+i A)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps

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

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Mathematica [B]  time = 7.47, size = 902, normalized size = 6.89 \[ \frac {\left (-\frac {1}{2} A \cos (c)-\frac {1}{2} i A \sin (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \csc ^2(c+d x)}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) (\cos (d x)+i \sin (d x)) \left (\frac {1}{2} A \cos (c-d x)+\frac {1}{2} i B \cos (c-d x)-\frac {1}{2} A \cos (c+d x)-\frac {1}{2} i B \cos (c+d x)+\frac {1}{2} i A \sin (c-d x)-\frac {1}{2} B \sin (c-d x)-\frac {1}{2} i A \sin (c+d x)+\frac {1}{2} B \sin (c+d x)\right ) (A+B \tan (c+d x)) \csc (c+d x)}{2 d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\left (2 A \cos \left (\frac {c}{2}\right )+i B \cos \left (\frac {c}{2}\right )+2 i A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )\right ) \left (i \tan ^{-1}(\tan (d x)) \cos \left (\frac {c}{2}\right )-\tan ^{-1}(\tan (d x)) \sin \left (\frac {c}{2}\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\left (2 A \cos \left (\frac {c}{2}\right )+i B \cos \left (\frac {c}{2}\right )+2 i A \sin \left (\frac {c}{2}\right )-B \sin \left (\frac {c}{2}\right )\right ) \left (-\frac {1}{2} \cos \left (\frac {c}{2}\right ) \log \left (\sin ^2(c+d x)\right )-\frac {1}{2} i \sin \left (\frac {c}{2}\right ) \log \left (\sin ^2(c+d x)\right )\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {x (2 A \csc (c)+i B \csc (c)+(2 A+i B) \cot (c) (-\cos (c)-i \sin (c))) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(A+i B) \cos (2 d x) \left (\frac {1}{4} i \sin (c)-\frac {\cos (c)}{4}\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(A+i B) \left (\frac {3}{2} i d x \cos (c)-\frac {3}{2} d x \sin (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(A+i B) \left (\frac {1}{4} i \cos (c)+\frac {\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x)) \sin (2 d x) (A+B \tan (c+d x))}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*A*Cos[c/2] + I*B*Cos[c/2] + (2*I)*A*Sin[c/2] - B*Sin[c/2])*(I*ArcTan[Tan[d*x]]*Cos[c/2] - ArcTan[Tan[d*x]]
*Sin[c/2])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c +
d*x])) + ((2*A*Cos[c/2] + I*B*Cos[c/2] + (2*I)*A*Sin[c/2] - B*Sin[c/2])*(-1/2*(Cos[c/2]*Log[Sin[c + d*x]^2])
- (I/2)*Log[Sin[c + d*x]^2]*Sin[c/2])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin
[c + d*x])*(a + I*a*Tan[c + d*x])) + (x*(2*A*Csc[c] + I*B*Csc[c] + (2*A + I*B)*Cot[c]*(-Cos[c] - I*Sin[c]))*(C
os[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A +
I*B)*Cos[2*d*x]*(-1/4*Cos[c] + (I/4)*Sin[c])*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x]
+ B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c + d*x]^2*(-1/2*(A*Cos[c]) - (I/2)*A*Sin[c])*(Cos[d*x] + I*
Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*(((3
*I)/2)*d*x*Cos[c] - (3*d*x*Sin[c])/2)*(Cos[d*x] + I*Sin[d*x])*(A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin
[c + d*x])*(a + I*a*Tan[c + d*x])) + ((A + I*B)*((I/4)*Cos[c] + Sin[c]/4)*(Cos[d*x] + I*Sin[d*x])*Sin[2*d*x]*(
A + B*Tan[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + (Csc[c/2]*Csc[c + d*x]*Sec
[c/2]*(Cos[d*x] + I*Sin[d*x])*((A*Cos[c - d*x])/2 + (I/2)*B*Cos[c - d*x] - (A*Cos[c + d*x])/2 - (I/2)*B*Cos[c
+ d*x] + (I/2)*A*Sin[c - d*x] - (B*Sin[c - d*x])/2 - (I/2)*A*Sin[c + d*x] + (B*Sin[c + d*x])/2)*(A + B*Tan[c +
d*x]))/(2*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x]))

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fricas [A]  time = 0.62, size = 189, normalized size = 1.44 \[ \frac {{\left (14 i \, A - 10 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + {\left ({\left (-28 i \, A + 20 \, B\right )} d x - A - 9 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left ({\left (14 i \, A - 10 \, B\right )} d x + 10 \, A + 10 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left ({\left (2 \, A + i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (2 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (2 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - A - i \, B}{4 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((14*I*A - 10*B)*d*x*e^(6*I*d*x + 6*I*c) + ((-28*I*A + 20*B)*d*x - A - 9*I*B)*e^(4*I*d*x + 4*I*c) + ((14*I
*A - 10*B)*d*x + 10*A + 10*I*B)*e^(2*I*d*x + 2*I*c) - 4*((2*A + I*B)*e^(6*I*d*x + 6*I*c) - 2*(2*A + I*B)*e^(4*
I*d*x + 4*I*c) + (2*A + I*B)*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - A - I*B)/(a*d*e^(6*I*d*x + 6*
I*c) - 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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giac [A]  time = 0.95, size = 165, normalized size = 1.26 \[ -\frac {\frac {4 \, {\left (2 \, A + i \, B\right )} \log \left (-i \, \tan \left (d x + c\right )\right )}{a} - \frac {{\left (7 \, A + 5 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac {{\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac {7 \, A \tan \left (d x + c\right ) + 5 i \, B \tan \left (d x + c\right ) - 9 i \, A + 7 \, B}{a {\left (\tan \left (d x + c\right ) - i\right )}} - \frac {2 \, {\left (6 \, A \tan \left (d x + c\right )^{2} + 3 i \, B \tan \left (d x + c\right )^{2} + 2 i \, A \tan \left (d x + c\right ) - 2 \, B \tan \left (d x + c\right ) - A\right )}}{a \tan \left (d x + c\right )^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(2*A + I*B)*log(-I*tan(d*x + c))/a - (7*A + 5*I*B)*log(tan(d*x + c) - I)/a - (A - I*B)*log(-I*tan(d*x
+ c) + 1)/a + (7*A*tan(d*x + c) + 5*I*B*tan(d*x + c) - 9*I*A + 7*B)/(a*(tan(d*x + c) - I)) - 2*(6*A*tan(d*x +
c)^2 + 3*I*B*tan(d*x + c)^2 + 2*I*A*tan(d*x + c) - 2*B*tan(d*x + c) - A)/(a*tan(d*x + c)^2))/d

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maple [A]  time = 0.72, size = 206, normalized size = 1.57 \[ \frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}-\frac {A}{2 a d \tan \left (d x +c \right )^{2}}-\frac {i B \ln \left (\tan \left (d x +c \right )\right )}{a d}-\frac {2 A \ln \left (\tan \left (d x +c \right )\right )}{d a}+\frac {i A}{a d \tan \left (d x +c \right )}-\frac {B}{a d \tan \left (d x +c \right )}+\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}-\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {7 \ln \left (\tan \left (d x +c \right )-i\right ) A}{4 d a}+\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right ) B}{4 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/d/a*A*ln(tan(d*x+c)+I)-1/4*I/d/a*B*ln(tan(d*x+c)+I)-1/2/a/d*A/tan(d*x+c)^2-I/a/d*B*ln(tan(d*x+c))-2/d/a*A*
ln(tan(d*x+c))+I/a/d/tan(d*x+c)*A-1/a/d/tan(d*x+c)*B+1/2*I/d/a/(tan(d*x+c)-I)*A-1/2/d/a/(tan(d*x+c)-I)*B+7/4/d
/a*ln(tan(d*x+c)-I)*A+5/4*I/a/d*ln(tan(d*x+c)-I)*B

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 6.51, size = 153, normalized size = 1.17 \[ -\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {3\,A}{2\,a}+\frac {B\,3{}\mathrm {i}}{2\,a}\right )+\frac {A}{2\,a}-\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a}+\frac {A\,1{}\mathrm {i}}{2\,a}\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}+{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (2\,A+B\,1{}\mathrm {i}\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (7\,A+B\,5{}\mathrm {i}\right )}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c + d*x) + 1i)*(A - B*1i))/(4*a*d) - (log(tan(c + d*x))*(2*A + B*1i))/(a*d) - (tan(c + d*x)^2*((3*A)/
(2*a) + (B*3i)/(2*a)) + A/(2*a) - tan(c + d*x)*((A*1i)/(2*a) - B/a))/(d*(tan(c + d*x)^2 + tan(c + d*x)^3*1i))
+ (log(tan(c + d*x) - 1i)*(7*A + B*5i))/(4*a*d)

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sympy [A]  time = 1.33, size = 201, normalized size = 1.53 \[ \frac {2 A - 2 i B e^{2 i c} e^{2 i d x} + 2 i B}{a d e^{4 i c} e^{4 i d x} - 2 a d e^{2 i c} e^{2 i d x} + a d} + \begin {cases} - \frac {\left (A + i B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac {7 i A - 5 B}{2 a} + \frac {\left (7 i A e^{2 i c} + i A - 5 B e^{2 i c} - B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 7 i A + 5 B\right )}{2 a} - \frac {\left (2 A + i B\right ) \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*A - 2*I*B*exp(2*I*c)*exp(2*I*d*x) + 2*I*B)/(a*d*exp(4*I*c)*exp(4*I*d*x) - 2*a*d*exp(2*I*c)*exp(2*I*d*x) + a
*d) + Piecewise((-(A + I*B)*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(4*a*d*exp(2*I*c), 0)), (x*(-(7*I*A - 5*B)/(2
*a) + (7*I*A*exp(2*I*c) + I*A - 5*B*exp(2*I*c) - B)*exp(-2*I*c)/(2*a)), True)) - x*(-7*I*A + 5*B)/(2*a) - (2*A
+ I*B)*log(exp(2*I*d*x) - exp(-2*I*c))/(a*d)

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