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

Optimal. Leaf size=91 \[ \frac {a (B+i A) \tan ^2(c+d x)}{2 d}+\frac {a (A-i B) \tan (c+d x)}{d}+\frac {a (B+i A) \log (\cos (c+d x))}{d}-a x (A-i B)+\frac {i a B \tan ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 91, 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, 3528, 3525, 3475} \[ \frac {a (B+i A) \tan ^2(c+d x)}{2 d}+\frac {a (A-i B) \tan (c+d x)}{d}+\frac {a (B+i A) \log (\cos (c+d x))}{d}-a x (A-i B)+\frac {i a B \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(A - I*B)*x) + (a*(I*A + B)*Log[Cos[c + d*x]])/d + (a*(A - I*B)*Tan[c + d*x])/d + (a*(I*A + B)*Tan[c + d*x
]^2)/(2*d) + ((I/3)*a*B*Tan[c + d*x]^3)/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 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 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 ^2(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\frac {i a B \tan ^3(c+d x)}{3 d}+\int \tan ^2(c+d x) (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=\frac {a (i A+B) \tan ^2(c+d x)}{2 d}+\frac {i a B \tan ^3(c+d x)}{3 d}+\int \tan (c+d x) (-a (i A+B)+a (A-i B) \tan (c+d x)) \, dx\\ &=-a (A-i B) x+\frac {a (A-i B) \tan (c+d x)}{d}+\frac {a (i A+B) \tan ^2(c+d x)}{2 d}+\frac {i a B \tan ^3(c+d x)}{3 d}-(a (i A+B)) \int \tan (c+d x) \, dx\\ &=-a (A-i B) x+\frac {a (i A+B) \log (\cos (c+d x))}{d}+\frac {a (A-i B) \tan (c+d x)}{d}+\frac {a (i A+B) \tan ^2(c+d x)}{2 d}+\frac {i a B \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.88, size = 86, normalized size = 0.95 \[ \frac {a \left (-6 (A-i B) \tan ^{-1}(\tan (c+d x))+3 (B+i A) \tan ^2(c+d x)+6 (A-i B) \tan (c+d x)+6 (B+i A) \log (\cos (c+d x))+2 i B \tan ^3(c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.51, size = 164, normalized size = 1.80 \[ \frac {{\left (12 i \, A + 18 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (18 i \, A + 18 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (6 i \, A + 8 \, B\right )} a + {\left ({\left (3 i \, A + 3 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (9 i \, A + 9 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (9 i \, A + 9 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (3 i \, A + 3 \, B\right )} a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((12*I*A + 18*B)*a*e^(4*I*d*x + 4*I*c) + (18*I*A + 18*B)*a*e^(2*I*d*x + 2*I*c) + (6*I*A + 8*B)*a + ((3*I*A
+ 3*B)*a*e^(6*I*d*x + 6*I*c) + (9*I*A + 9*B)*a*e^(4*I*d*x + 4*I*c) + (9*I*A + 9*B)*a*e^(2*I*d*x + 2*I*c) + (3
*I*A + 3*B)*a)*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 = 0.60, size = 284, normalized size = 3.12 \[ \frac {3 i \, A a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 \, B a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 i \, A a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, B a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 i \, A a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, B a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 i \, A a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, B a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, A a e^{\left (2 i \, d x + 2 i \, c\right )} + 18 \, B a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 3 \, B a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 6 i \, A a + 8 \, B a}{3 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*I*A*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 3*B*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I
*c) + 1) + 9*I*A*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*B*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 9*I*A*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*B*a*e^(2*I*d*x + 2*I*c)*log(e^(2
*I*d*x + 2*I*c) + 1) + 12*I*A*a*e^(4*I*d*x + 4*I*c) + 18*B*a*e^(4*I*d*x + 4*I*c) + 18*I*A*a*e^(2*I*d*x + 2*I*c
) + 18*B*a*e^(2*I*d*x + 2*I*c) + 3*I*A*a*log(e^(2*I*d*x + 2*I*c) + 1) + 3*B*a*log(e^(2*I*d*x + 2*I*c) + 1) + 6
*I*A*a + 8*B*a)/(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 = 141, normalized size = 1.55 \[ \frac {i a B \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {i a A \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {i a B \tan \left (d x +c \right )}{d}+\frac {a B \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a A \tan \left (d x +c \right )}{d}-\frac {i a \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A}{2 d}-\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B}{2 d}+\frac {i a B \arctan \left (\tan \left (d x +c \right )\right )}{d}-\frac {a A \arctan \left (\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 82, normalized size = 0.90 \[ -\frac {-2 i \, B a \tan \left (d x + c\right )^{3} - {\left (3 i \, A + 3 \, B\right )} a \tan \left (d x + c\right )^{2} + 6 \, {\left (d x + c\right )} {\left (A - i \, B\right )} a + 3 \, {\left (i \, A + B\right )} a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (A - i \, B\right )} a \tan \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.11, size = 82, normalized size = 0.90 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {B\,a}{2}+\frac {A\,a\,1{}\mathrm {i}}{2}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B\,a+A\,a\,1{}\mathrm {i}\right )}{d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,a-B\,a\,1{}\mathrm {i}\right )}{d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.66, size = 175, normalized size = 1.92 \[ \frac {i a \left (A - i B\right ) \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {6 A a - 8 i B a + \left (18 A a e^{2 i c} - 18 i B a e^{2 i c}\right ) e^{2 i d x} + \left (12 A a e^{4 i c} - 18 i B a e^{4 i c}\right ) e^{4 i d x}}{- 3 i d e^{6 i c} e^{6 i d x} - 9 i d e^{4 i c} e^{4 i d x} - 9 i d e^{2 i c} e^{2 i d x} - 3 i d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*a*(A - I*B)*log(exp(2*I*d*x) + exp(-2*I*c))/d + (6*A*a - 8*I*B*a + (18*A*a*exp(2*I*c) - 18*I*B*a*exp(2*I*c))
*exp(2*I*d*x) + (12*A*a*exp(4*I*c) - 18*I*B*a*exp(4*I*c))*exp(4*I*d*x))/(-3*I*d*exp(6*I*c)*exp(6*I*d*x) - 9*I*
d*exp(4*I*c)*exp(4*I*d*x) - 9*I*d*exp(2*I*c)*exp(2*I*d*x) - 3*I*d)

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