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

Optimal. Leaf size=42 \[ -\frac {(a B+A b) \log (\cos (c+d x))}{d}+x (a A-b B)+\frac {b B \tan (c+d x)}{d} \]

[Out]

(A*a-B*b)*x-(A*b+B*a)*ln(cos(d*x+c))/d+b*B*tan(d*x+c)/d

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3525, 3475} \[ -\frac {(a B+A b) \log (\cos (c+d x))}{d}+x (a A-b B)+\frac {b B \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 59, normalized size = 1.40 \[ a A x-\frac {a B \log (\cos (c+d x))}{d}-\frac {A b \log (\cos (c+d x))}{d}-\frac {b B \tan ^{-1}(\tan (c+d x))}{d}+\frac {b B \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 50, normalized size = 1.19 \[ \frac {2 \, {\left (A a - B b\right )} d x + 2 \, B b \tan \left (d x + c\right ) - {\left (B a + A b\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.33, size = 329, normalized size = 7.83 \[ \frac {2 \, A a d x \tan \left (d x\right ) \tan \relax (c) - 2 \, B b d x \tan \left (d x\right ) \tan \relax (c) - B a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right ) \tan \relax (c) - A b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) \tan \left (d x\right ) \tan \relax (c) - 2 \, A a d x + 2 \, B b d x + B a \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) + A b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{4} \tan \relax (c)^{2} - 2 \, \tan \left (d x\right )^{3} \tan \relax (c) + \tan \left (d x\right )^{2} \tan \relax (c)^{2} + \tan \left (d x\right )^{2} - 2 \, \tan \left (d x\right ) \tan \relax (c) + 1\right )}}{\tan \relax (c)^{2} + 1}\right ) - 2 \, B b \tan \left (d x\right ) - 2 \, B b \tan \relax (c)}{2 \, {\left (d \tan \left (d x\right ) \tan \relax (c) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 77, normalized size = 1.83 \[ \frac {b B \tan \left (d x +c \right )}{d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A b}{2 d}+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B}{2 d}+\frac {a A \arctan \left (\tan \left (d x +c \right )\right )}{d}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.91, size = 50, normalized size = 1.19 \[ \frac {2 \, B b \tan \left (d x + c\right ) + 2 \, {\left (A a - B b\right )} {\left (d x + c\right )} + {\left (B a + A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.32, size = 55, normalized size = 1.31 \[ \frac {B\,b\,\mathrm {tan}\left (c+d\,x\right )+\frac {A\,b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}+\frac {B\,a\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}+A\,a\,d\,x-B\,b\,d\,x}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.18, size = 73, normalized size = 1.74 \[ \begin {cases} A a x + \frac {A b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - B b x + \frac {B b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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