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

Optimal. Leaf size=58 \[ \frac {(A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac {x (a A+b B)}{a^2+b^2} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3531, 3530} \[ \frac {(A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac {x (a A+b B)}{a^2+b^2} \]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 66, normalized size = 1.14 \[ \frac {2 (a A+b B) \tan ^{-1}(\tan (c+d x))-(A b-a B) \left (\log \left (\sec ^2(c+d x)\right )-2 \log (a+b \tan (c+d x))\right )}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 76, normalized size = 1.31 \[ \frac {2 \, {\left (A a + B b\right )} d x - {\left (B a - A b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} + b^{2}\right )} d} \]

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 - (B*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))
)/((a^2 + b^2)*d)

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giac [A]  time = 0.29, size = 94, normalized size = 1.62 \[ \frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, {\left (B a b - A b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.25, size = 153, normalized size = 2.64 \[ \frac {\ln \left (a +b \tan \left (d x +c \right )\right ) A b}{d \left (a^{2}+b^{2}\right )}-\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) a B}{d \left (a^{2}+b^{2}\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A b}{2 d \left (a^{2}+b^{2}\right )}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a B}{2 d \left (a^{2}+b^{2}\right )}+\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )} \]

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]

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

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maxima [A]  time = 0.49, size = 88, normalized size = 1.52 \[ \frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (B a - A b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.67, size = 93, normalized size = 1.60 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{d\,\left (a^2+b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \]

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]

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

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sympy [A]  time = 0.95, size = 524, normalized size = 9.03 \[ \begin {cases} \frac {\tilde {\infty } x \left (A + B \tan {\relax (c )}\right )}{\tan {\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {i A d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {A d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {i A}{2 b d \tan {\left (c + d x \right )} - 2 i b d} + \frac {B d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i B d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {B}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\- \frac {i A d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {A d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {i A}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {B d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i B d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {B}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x \left (A + B \tan {\relax (c )}\right )}{a + b \tan {\relax (c )}} & \text {for}\: d = 0 \\\frac {A x + \frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d}}{a} & \text {for}\: b = 0 \\\frac {2 A a d x}{2 a^{2} d + 2 b^{2} d} + \frac {2 A b \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} - \frac {A b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} - \frac {2 B a \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} + \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} + \frac {2 B b d x}{2 a^{2} d + 2 b^{2} d} & \text {otherwise} \end {cases} \]

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

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