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

Optimal. Leaf size=129 \[ \frac {(-B+i A) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}-\frac {3 (-B+i A) \tan (c+d x)}{2 a d}-\frac {(A+2 i B) \log (\cos (c+d x))}{a d}+\frac {3 x (-B+i A)}{2 a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3595, 3528, 3525, 3475} \[ \frac {(-B+i A) \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(A+2 i B) \tan ^2(c+d x)}{2 a d}-\frac {3 (-B+i A) \tan (c+d x)}{2 a d}-\frac {(A+2 i B) \log (\cos (c+d x))}{a d}+\frac {3 x (-B+i A)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[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) - ((A + (2*I)*B)*Log[Cos[c + d*x]])/(a*d) - (3*(I*A - B)*Tan[c + d*x])/(2*a*d) - ((A + (
2*I)*B)*Tan[c + d*x]^2)/(2*a*d) + ((I*A - B)*Tan[c + d*x]^3)/(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 3525

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

Rule 3528

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

Rule 3595

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 7.31, size = 898, normalized size = 6.96 \[ \frac {\left (\frac {1}{2} B \sin (c)-\frac {1}{2} i B \cos (c)\right ) (\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \sec ^2(c+d x)}{d (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {(\cos (d x)+i \sin (d x)) (A \cos (c-d x)+i B \cos (c-d x)-A \cos (c+d x)-i B \cos (c+d x)+i A \sin (c-d x)-B \sin (c-d x)-i A \sin (c+d x)+B \sin (c+d x)) (A+B \tan (c+d x)) \sec (c+d x)}{2 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) (A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {x (\cos (d x)+i \sin (d x)) (-i A \sec (c)+2 B \sec (c)+(A+2 i B) (\cos (c)+i \sin (c)) \tan (c)) (A+B \tan (c+d x))}{(A \cos (c+d x)+B \sin (c+d x)) (i \tan (c+d x) a+a)}+\frac {\left (A \cos \left (\frac {c}{2}\right )+2 i B \cos \left (\frac {c}{2}\right )+i A \sin \left (\frac {c}{2}\right )-2 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 (A \cos \left (\frac {c}{2}\right )+2 i B \cos \left (\frac {c}{2}\right )+i A \sin \left (\frac {c}{2}\right )-2 B \sin \left (\frac {c}{2}\right )\right ) \left (-\frac {1}{2} \cos \left (\frac {c}{2}\right ) \log \left (\cos ^2(c+d x)\right )-\frac {1}{2} i \sin \left (\frac {c}{2}\right ) \log \left (\cos ^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 {(A+i B) \cos (2 d x) \left (\frac {\cos (c)}{4}-\frac {1}{4} i \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 {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 {(B-i A) \left (\frac {\cos (c)}{4}-\frac {1}{4} i \sin (c)\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[(Tan[c + d*x]^3*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x]),x]

[Out]

((A*Cos[c/2] + (2*I)*B*Cos[c/2] + I*A*Sin[c/2] - 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])) + ((A*Cos[c/2] + (2*I)*B*Cos[c/2] + I*A*Sin[c/2] - 2*B*Sin[c/2])*(-1/2*(Cos[c/2]*Log[Cos[c + d*x]^2])
- (I/2)*Log[Cos[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])) + ((A + I*B)*Cos[2*d*x]*(Cos[c]/4 - (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])) + (Sec[c + d*x]^2*((-1/2*I)*
B*Cos[c] + (B*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)*(((3*I)/2)*d*x*Cos[c] - (3*d*x*Sin[c])/2)*(Cos[d*x] + I*Sin[d*x])*(A + B*Ta
n[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + (((-I)*A + B)*(Cos[c]/4 - (I/4)*Si
n[c])*(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*T
an[c + d*x])) + (Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])*(A*Cos[c - d*x] + I*B*Cos[c - d*x] - A*Cos[c + d*x] - I*
B*Cos[c + d*x] + I*A*Sin[c - d*x] - B*Sin[c - d*x] - I*A*Sin[c + d*x] + B*Sin[c + d*x])*(A + B*Tan[c + d*x]))/
(2*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])) + (
x*(Cos[d*x] + I*Sin[d*x])*((-I)*A*Sec[c] + 2*B*Sec[c] + (A + (2*I)*B)*(Cos[c] + I*Sin[c])*Tan[c])*(A + B*Tan[c
+ d*x]))/((A*Cos[c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.82, size = 181, normalized size = 1.40 \[ \frac {{\left (10 i \, A - 14 \, B\right )} d x e^{\left (6 i \, d x + 6 i \, c\right )} + {\left ({\left (20 i \, A - 28 \, B\right )} d x + 9 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left ({\left (10 i \, A - 14 \, B\right )} d x + 10 \, A + 10 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, {\left ({\left (A + 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (A + 2 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + A + i \, B}{4 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.94, size = 125, normalized size = 0.97 \[ \frac {\frac {{\left (5 \, A + 7 i \, B\right )} \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac {{\left (A - i \, B\right )} \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} - \frac {2 \, {\left (i \, B a \tan \left (d x + c\right )^{2} + 2 i \, A a \tan \left (d x + c\right ) - 2 \, B a \tan \left (d x + c\right )\right )}}{a^{2}} - \frac {5 \, A \tan \left (d x + c\right ) + 7 i \, B \tan \left (d x + c\right ) - 3 i \, A + 5 \, B}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.19, size = 169, normalized size = 1.31 \[ \frac {B \tan \left (d x +c \right )}{d a}-\frac {i B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d a}-\frac {i A \tan \left (d x +c \right )}{d a}-\frac {A \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}+\frac {i B \ln \left (\tan \left (d x +c \right )+i\right )}{4 d a}+\frac {5 \ln \left (\tan \left (d x +c \right )-i\right ) A}{4 d a}+\frac {7 i \ln \left (\tan \left (d x +c \right )-i\right ) B}{4 d a}-\frac {i A}{2 d a \left (\tan \left (d x +c \right )-i\right )}+\frac {B}{2 d a \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(tan(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.

________________________________________________________________________________________

mupad [B]  time = 6.38, size = 141, normalized size = 1.09 \[ \frac {\frac {A}{2\,a}-\frac {A+B\,1{}\mathrm {i}}{2\,a}+\frac {A+B\,2{}\mathrm {i}}{2\,a}}{d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-\frac {B}{a}+\frac {A\,1{}\mathrm {i}}{a}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (5\,A+B\,7{}\mathrm {i}\right )}{4\,a\,d}-\frac {B\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.28, size = 206, normalized size = 1.60 \[ \frac {- 2 A e^{2 i c} e^{2 i d x} - 2 A - 2 i B}{- a d e^{4 i c} e^{4 i d x} - 2 a d e^{2 i c} e^{2 i d x} - a d} + \begin {cases} - \frac {\left (- A - i B\right ) e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: 4 a d e^{2 i c} \neq 0 \\x \left (- \frac {5 i A - 7 B}{2 a} - \frac {\left (- 5 i A e^{2 i c} + i A + 7 B e^{2 i c} - B\right ) e^{- 2 i c}}{2 a}\right ) & \text {otherwise} \end {cases} - \frac {x \left (- 5 i A + 7 B\right )}{2 a} - \frac {\left (A + 2 i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________