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

Optimal. Leaf size=107 \[ \frac {a^2 (B+i A) \tan (c+d x)}{d}-\frac {2 a^2 (A-i B) \log (\cos (c+d x))}{d}-2 a^2 x (B+i A)+\frac {A (a+i a \tan (c+d x))^2}{2 d}-\frac {i B (a+i a \tan (c+d x))^3}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3592, 3527, 3477, 3475} \[ \frac {a^2 (B+i A) \tan (c+d x)}{d}-\frac {2 a^2 (A-i B) \log (\cos (c+d x))}{d}-2 a^2 x (B+i A)+\frac {A (a+i a \tan (c+d x))^2}{2 d}-\frac {i B (a+i a \tan (c+d x))^3}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*a^2*(I*A + B)*x - (2*a^2*(A - I*B)*Log[Cos[c + d*x]])/d + (a^2*(I*A + B)*Tan[c + d*x])/d + (A*(a + I*a*Tan[
c + d*x])^2)/(2*d) - ((I/3)*B*(a + I*a*Tan[c + d*x])^3)/(a*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 3477

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

Rule 3527

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

Rule 3592

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

Rubi steps

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

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Mathematica [B]  time = 4.34, size = 273, normalized size = 2.55 \[ \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \left ((A-i B) \cos ^3(c+d x) (-4 d x \sin (2 c)-4 i d x \cos (2 c))+2 (B+i A) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \tan ^{-1}(\tan (3 c+d x))+\frac {1}{3} (6 A-7 i B) \sec (c) (\sin (2 c)+i \cos (2 c)) \sin (d x) \cos ^2(c+d x)-(A-i B) (\cos (2 c)-i \sin (2 c)) \cos ^3(c+d x) \log \left (\cos ^2(c+d x)\right )-\frac {1}{6} (\cos (2 c)-i \sin (2 c)) (3 A+2 B \tan (c)-6 i B) \cos (c+d x)+\frac {1}{3} B \cos (c) (\tan (c)+i)^2 \sin (d x)\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(I*A + B)*ArcTan[Tan[3*c + d*x]]*Cos[c + d*x]^3*(Cos[2*c] - I*Sin[2*c]) - (A - I*B)*Cos[c + d*x]^3*Log[Cos
[c + d*x]^2]*(Cos[2*c] - I*Sin[2*c]) + (A - I*B)*Cos[c + d*x]^3*((-4*I)*d*x*Cos[2*c] - 4*d*x*Sin[2*c]) + ((6*A
- (7*I)*B)*Cos[c + d*x]^2*Sec[c]*(I*Cos[2*c] + Sin[2*c])*Sin[d*x])/3 + (B*Cos[c]*Sin[d*x]*(I + Tan[c])^2)/3 -
(Cos[c + d*x]*(Cos[2*c] - I*Sin[2*c])*(3*A - (6*I)*B + 2*B*Tan[c]))/6)*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c
+ d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [A]  time = 0.41, size = 175, normalized size = 1.64 \[ -\frac {2 \, {\left (3 \, {\left (3 \, A - 5 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (5 \, A - 6 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (6 \, A - 7 i \, B\right )} a^{2} + 3 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (A - i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (A - i \, B\right )} a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.52, size = 312, normalized size = 2.92 \[ -\frac {6 \, A a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, B a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 18 i \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 18 i \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, A a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 30 i \, B a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 36 i \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, A a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, B a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, A a^{2} - 14 i \, B a^{2}}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*A*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*B*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 18*A*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 18*I*B*a^2*e^(4*I*d*x + 4*I*c)*log
(e^(2*I*d*x + 2*I*c) + 1) + 18*A*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 18*I*B*a^2*e^(2*I*d*x
+ 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*A*a^2*e^(4*I*d*x + 4*I*c) - 30*I*B*a^2*e^(4*I*d*x + 4*I*c) + 30*A*a
^2*e^(2*I*d*x + 2*I*c) - 36*I*B*a^2*e^(2*I*d*x + 2*I*c) + 6*A*a^2*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*B*a^2*log
(e^(2*I*d*x + 2*I*c) + 1) + 12*A*a^2 - 14*I*B*a^2)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2
*I*d*x + 2*I*c) + d)

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maple [A]  time = 0.02, size = 158, normalized size = 1.48 \[ \frac {i a^{2} B \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {a^{2} B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {2 i a^{2} A \tan \left (d x +c \right )}{d}-\frac {a^{2} A \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} B \tan \left (d x +c \right )}{d}-\frac {i a^{2} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}+\frac {a^{2} A \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {2 i a^{2} A \arctan \left (\tan \left (d x +c \right )\right )}{d}-\frac {2 a^{2} B \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 94, normalized size = 0.88 \[ -\frac {2 \, B a^{2} \tan \left (d x + c\right )^{3} + 3 \, {\left (A - 2 i \, B\right )} a^{2} \tan \left (d x + c\right )^{2} - 6 \, {\left (d x + c\right )} {\left (-2 i \, A - 2 \, B\right )} a^{2} - 6 \, {\left (A - i \, B\right )} a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - {\left (12 i \, A + 12 \, B\right )} a^{2} \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.13, size = 111, normalized size = 1.04 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {a^2\,\left (B+A\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {B\,a^2\,1{}\mathrm {i}}{2}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (a^2\,\left (B+A\,1{}\mathrm {i}\right )+B\,a^2+A\,a^2\,1{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (2\,A\,a^2-B\,a^2\,2{}\mathrm {i}\right )}{d}-\frac {B\,a^2\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.70, size = 178, normalized size = 1.66 \[ - \frac {2 a^{2} \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 12 A a^{2} + 14 i B a^{2} + \left (- 30 A a^{2} e^{2 i c} + 36 i B a^{2} e^{2 i c}\right ) e^{2 i d x} + \left (- 18 A a^{2} e^{4 i c} + 30 i B a^{2} e^{4 i c}\right ) e^{4 i d x}}{3 d e^{6 i c} e^{6 i d x} + 9 d e^{4 i c} e^{4 i d x} + 9 d e^{2 i c} e^{2 i d x} + 3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**2*(A - I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-12*A*a**2 + 14*I*B*a**2 + (-30*A*a**2*exp(2*I*c) + 36*
I*B*a**2*exp(2*I*c))*exp(2*I*d*x) + (-18*A*a**2*exp(4*I*c) + 30*I*B*a**2*exp(4*I*c))*exp(4*I*d*x))/(3*d*exp(6*
I*c)*exp(6*I*d*x) + 9*d*exp(4*I*c)*exp(4*I*d*x) + 9*d*exp(2*I*c)*exp(2*I*d*x) + 3*d)

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