3.233 \(\int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=65 \[ \frac{(a B+A b) \tan (c+d x)}{d}-\frac{(a A-b B) \log (\cos (c+d x))}{d}-x (a B+A b)+\frac{b B \tan ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0585006, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3592, 3525, 3475} \[ \frac{(a B+A b) \tan (c+d x)}{d}-\frac{(a A-b B) \log (\cos (c+d x))}{d}-x (a B+A b)+\frac{b B \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + b*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

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

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]

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

Mathematica [A]  time = 0.271831, size = 67, normalized size = 1.03 \[ \frac{-2 (a B+A b) \tan ^{-1}(\tan (c+d x))+2 (a B+A b) \tan (c+d x)+2 (b B-a A) \log (\cos (c+d x))+b B \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

(-2*(A*b + a*B)*ArcTan[Tan[c + d*x]] + 2*(-(a*A) + b*B)*Log[Cos[c + d*x]] + 2*(A*b + a*B)*Tan[c + d*x] + b*B*T
an[c + d*x]^2)/(2*d)

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Maple [A]  time = 0.013, size = 105, normalized size = 1.6 \begin{align*}{\frac{B \left ( \tan \left ( dx+c \right ) \right ) ^{2}b}{2\,d}}+{\frac{A\tan \left ( dx+c \right ) b}{d}}+{\frac{B\tan \left ( dx+c \right ) a}{d}}+{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) A}{2\,d}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bb}{2\,d}}-{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d}}-{\frac{aB\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.49162, size = 89, normalized size = 1.37 \begin{align*} \frac{B b \tan \left (d x + c\right )^{2} - 2 \,{\left (B a + A b\right )}{\left (d x + c\right )} +{\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (B a + A b\right )} \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93071, size = 161, normalized size = 2.48 \begin{align*} \frac{B b \tan \left (d x + c\right )^{2} - 2 \,{\left (B a + A b\right )} d x -{\left (A a - B b\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \,{\left (B a + A b\right )} \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.297419, size = 104, normalized size = 1.6 \begin{align*} \begin{cases} \frac{A a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - A b x + \frac{A b \tan{\left (c + d x \right )}}{d} - B a x + \frac{B a \tan{\left (c + d x \right )}}{d} - \frac{B b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{B b \tan ^{2}{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (A + B \tan{\left (c \right )}\right ) \left (a + b \tan{\left (c \right )}\right ) \tan{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a*log(tan(c + d*x)**2 + 1)/(2*d) - A*b*x + A*b*tan(c + d*x)/d - B*a*x + B*a*tan(c + d*x)/d - B*b*
log(tan(c + d*x)**2 + 1)/(2*d) + B*b*tan(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*tan(c))*(a + b*tan(c))*tan(c)
, True))

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Giac [B]  time = 1.6015, size = 832, normalized size = 12.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*B*a*d*x*tan(d*x)^2*tan(c)^2 + 2*A*b*d*x*tan(d*x)^2*tan(c)^2 + A*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan
(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 -
B*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*ta
n(d*x)*tan(c) + 1))*tan(d*x)^2*tan(c)^2 - 4*B*a*d*x*tan(d*x)*tan(c) - 4*A*b*d*x*tan(d*x)*tan(c) - B*b*tan(d*x)
^2*tan(c)^2 - 2*A*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + ta
n(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) + 2*B*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d
*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1))*tan(d*x)*tan(c) + 2*B*a*tan(d*x)^2*t
an(c) + 2*A*b*tan(d*x)^2*tan(c) + 2*B*a*tan(d*x)*tan(c)^2 + 2*A*b*tan(d*x)*tan(c)^2 + 2*B*a*d*x + 2*A*b*d*x -
B*b*tan(d*x)^2 - B*b*tan(c)^2 + A*a*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)
^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) - B*b*log(4*(tan(c)^2 + 1)/(tan(d*x)^4*tan(c)^2 - 2*tan(d*x
)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)) - 2*B*a*tan(d*x) - 2*A*b*tan(d*x) - 2*
B*a*tan(c) - 2*A*b*tan(c) - B*b)/(d*tan(d*x)^2*tan(c)^2 - 2*d*tan(d*x)*tan(c) + d)