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-B*a)*x/(a^2+b^2)-B*ln(cos(d*x+c))/b/d-a*(A*b-B*a)*ln(a*cos(d*x+c)+b*sin(d*x+c))/b/(a^2+b^2)/d

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Rubi [A]  time = 0.13, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 12

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

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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]

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 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]

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*}

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Mathematica [C]  time = 0.18, 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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fricas [A]  time = 0.71, size = 110, normalized size = 1.38 \[ -\frac {2 \, {\left (B a b - A b^{2}\right )} d x - {\left (B a^{2} - A a b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} b + b^{3}\right )} d} \]

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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giac [A]  time = 0.31, size = 95, normalized size = 1.19 \[ -\frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, {\left (B a^{2} - A a b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \]

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

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maple [A]  time = 0.24, size = 159, normalized size = 1.99 \[ -\frac {a \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )}+\frac {a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right ) b}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a A}{2 d \left (a^{2}+b^{2}\right )}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B b}{2 d \left (a^{2}+b^{2}\right )}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )} \]

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/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+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

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maxima [A]  time = 1.02, size = 94, normalized size = 1.18 \[ -\frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (B a^{2} - A a b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \]

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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mupad [B]  time = 6.66, size = 100, normalized size = 1.25 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{b\,d\,\left (a^2+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.17, size = 714, normalized size = 8.92 \[ \begin {cases} \tilde {\infty } x \left (A + B \tan {\relax (c )}\right ) & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac {A d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i A d x}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {A}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {i B d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {B d x}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i B}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = - i b \\- \frac {A d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i A d x}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {A}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {i B d x \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {B d x}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i B}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = i b \\\frac {\frac {A \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - B x + \frac {B \tan {\left (c + d x \right )}}{d}}{a} & \text {for}\: b = 0 \\\frac {x \left (A + B \tan {\relax (c )}\right ) \tan {\relax (c )}}{a + b \tan {\relax (c )}} & \text {for}\: d = 0 \\- \frac {2 A a b \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} + \frac {A a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} + \frac {2 A b^{2} d x}{2 a^{2} b d + 2 b^{3} d} + \frac {2 B a^{2} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} b d + 2 b^{3} d} - \frac {2 B a b d x}{2 a^{2} b d + 2 b^{3} d} + \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} b d + 2 b^{3} d} & \text {otherwise} \end {cases} \]

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