3.269 \(\int \frac{\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\)

Optimal. Leaf size=80 \[ -\frac{a (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{b d \left (a^2+b^2\right )}+\frac{x (A b-a B)}{a^2+b^2}-\frac{B \log (\cos (c+d x))}{b d} \]

[Out]

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

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Rubi [A]  time = 0.126949, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3589, 3475, 12, 3531, 3530} \[ -\frac{a (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{b d \left (a^2+b^2\right )}+\frac{x (A b-a B)}{a^2+b^2}-\frac{B \log (\cos (c+d x))}{b d} \]

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

Rule 3589

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*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] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [C]  time = 0.155029, size = 98, normalized size = 1.22 \[ \frac{b (a-i b) (A+i B) \log (-\tan (c+d x)+i)+b (a+i b) (A-i B) \log (\tan (c+d x)+i)+2 a (a B-A b) \log (a+b \tan (c+d x))}{2 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 159, normalized size = 2. \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Aa}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bb}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{a\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{{a}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{d \left ({a}^{2}+{b}^{2} \right ) b}} \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/d/(a^2+b^2)*ln(1+tan(d*x+c)^2)*A*a+1/2/d/(a^2+b^2)*ln(1+tan(d*x+c)^2)*B*b+1/d/(a^2+b^2)*A*arctan(tan(d*x+c
))*b-1/d/(a^2+b^2)*B*arctan(tan(d*x+c))*a-1/d*a/(a^2+b^2)*ln(a+b*tan(d*x+c))*A+1/d*a^2/(a^2+b^2)/b*ln(a+b*tan(
d*x+c))*B

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Maxima [A]  time = 1.52825, size = 127, normalized size = 1.59 \begin{align*} -\frac{\frac{2 \,{\left (B a - A b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{2 \,{\left (B a^{2} - A a b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac{{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{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*(2*(B*a - A*b)*(d*x + c)/(a^2 + b^2) - 2*(B*a^2 - A*a*b)*log(b*tan(d*x + c) + a)/(a^2*b + b^3) - (A*a + B
*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2))/d

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Fricas [A]  time = 1.86233, size = 251, normalized size = 3.14 \begin{align*} -\frac{2 \,{\left (B a b - A b^{2}\right )} d x -{\left (B a^{2} - A a b\right )} \log \left (\frac{b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) +{\left (B a^{2} + B b^{2}\right )} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \,{\left (a^{2} b + b^{3}\right )} 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*(2*(B*a*b - A*b^2)*d*x - (B*a^2 - A*a*b)*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c
)^2 + 1)) + (B*a^2 + B*b^2)*log(1/(tan(d*x + c)^2 + 1)))/((a^2*b + b^3)*d)

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Sympy [A]  time = 4.40048, size = 700, normalized size = 8.75 \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)

[Out]

Piecewise((zoo*x*(A + B*tan(c)), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((A*log(tan(c + d*x)**2 + 1)/(2*d) - B*x + B
*tan(c + d*x)/d)/a, Eq(b, 0)), (-A*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) + I*A*d*x/(-2*b*d*tan(c +
d*x) + 2*I*b*d) + A/(-2*b*d*tan(c + d*x) + 2*I*b*d) - I*B*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d) - B
*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) - B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d)
+ I*B*log(tan(c + d*x)**2 + 1)/(-2*b*d*tan(c + d*x) + 2*I*b*d) + I*B/(-2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, -
I*b)), (A*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*A*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - A/(2*b*d*
tan(c + d*x) + 2*I*b*d) - I*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + B*d*x/(2*b*d*tan(c + d*x) + 2*
I*b*d) + B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*B*log(tan(c + d*x)**2 + 1)
/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*B/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), (x*(A + B*tan(c))*tan(c)/(a
+ b*tan(c)), Eq(d, 0)), (-2*A*a*b*log(a/b + tan(c + d*x))/(2*a**2*b*d + 2*b**3*d) + A*a*b*log(tan(c + d*x)**2
+ 1)/(2*a**2*b*d + 2*b**3*d) + 2*A*b**2*d*x/(2*a**2*b*d + 2*b**3*d) + 2*B*a**2*log(a/b + tan(c + d*x))/(2*a**
2*b*d + 2*b**3*d) - 2*B*a*b*d*x/(2*a**2*b*d + 2*b**3*d) + B*b**2*log(tan(c + d*x)**2 + 1)/(2*a**2*b*d + 2*b**3
*d), True))

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Giac [A]  time = 1.21786, size = 128, normalized size = 1.6 \begin{align*} -\frac{\frac{2 \,{\left (B a - A b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \,{\left (B a^{2} - A a b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{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="giac")

[Out]

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