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*(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]))

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Rubi [A]  time = 0.173158, antiderivative size = 129, normalized size of antiderivative = 1., 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 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]

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 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 3475

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

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 = 7.1989, 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])*(-(Cos[c/2]*Log[Cos[c + d*x]^2])/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*((-I/2)*B*Co
s[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*Tan[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)*Sin[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*Tan[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*Co
s[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*(C
os[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]))

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Maple [A]  time = 0.033, size = 169, normalized size = 1.3 \begin{align*}{\frac{B\tan \left ( dx+c \right ) }{ad}}-{\frac{{\frac{i}{2}}B \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{ad}}-{\frac{iA\tan \left ( dx+c \right ) }{ad}}+{\frac{5\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) A}{4\,ad}}+{\frac{{\frac{7\,i}{4}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) B}{ad}}-{\frac{{\frac{i}{2}}A}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{B}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{A\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{4\,ad}}+{\frac{{\frac{i}{4}}B\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [A]  time = 1.46816, size = 527, normalized size = 4.09 \begin{align*} \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 )}} \end{align*}

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))

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Sympy [A]  time = 13.3631, size = 196, normalized size = 1.52 \begin{align*} \frac{\frac{2 A e^{- 2 i c} e^{2 i d x}}{a d} + \frac{\left (2 A + 2 i B\right ) e^{- 4 i c}}{a d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \frac{\left (\begin{cases} 5 i A x e^{2 i c} + \frac{A e^{- 2 i d x}}{2 d} - 7 B x e^{2 i c} + \frac{i B e^{- 2 i d x}}{2 d} & \text{for}\: d \neq 0 \\x \left (5 i A e^{2 i c} - i A - 7 B e^{2 i c} + B\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i c}}{2 a} - \frac{\left (A + 2 i B\right ) \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \end{align*}

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

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Giac [A]  time = 2.01044, size = 169, normalized size = 1.31 \begin{align*} \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} \end{align*}

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