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

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3611, 3530, 3475} \[ -\frac {b (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac {x (A b-a B)}{a^2+b^2}+\frac {A \log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[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)) + (A*Log[Sin[c + d*x]])/(a*d) - (b*(A*b - a*B)*Log[a*Cos[c + d*x] + b*Sin[c + d
*x]])/(a*(a^2 + b^2)*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 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 3611

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

Rubi steps

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

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Mathematica [C]  time = 0.39, size = 113, normalized size = 1.41 \[ -\frac {\frac {2 b (A b-a B) \log (a+b \tan (c+d x))}{a \left (a^2+b^2\right )}+\frac {(A+i B) \log (-\tan (c+d x)+i)}{a+i b}+\frac {(A-i B) \log (\tan (c+d x)+i)}{a-i b}-\frac {2 A \log (\tan (c+d x))}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(((A + I*B)*Log[I - Tan[c + d*x]])/(a + I*b) - (2*A*Log[Tan[c + d*x]])/a + ((A - I*B)*Log[I + Tan[c + d*x
]])/(a - I*b) + (2*b*(A*b - a*B)*Log[a + b*Tan[c + d*x]])/(a*(a^2 + b^2)))/d

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fricas [A]  time = 1.33, size = 118, normalized size = 1.48 \[ \frac {2 \, {\left (B a^{2} - A a b\right )} d x + {\left (A a^{2} + A b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (B a b - A b^{2}\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{3} + a b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.45, size = 113, normalized size = 1.41 \[ \frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3} b + a b^{3}} + \frac {2 \, A \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 0.68, size = 174, normalized size = 2.18 \[ -\frac {b^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d a \left (a^{2}+b^{2}\right )}+\frac {b \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )}+\frac {A \ln \left (\tan \left (d x +c \right )\right )}{d a}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a A}{2 d \left (a^{2}+b^{2}\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B b}{2 d \left (a^{2}+b^{2}\right )}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*b^2/a/(a^2+b^2)*ln(a+b*tan(d*x+c))*A+1/d*b/(a^2+b^2)*ln(a+b*tan(d*x+c))*B+1/d*A/a*ln(tan(d*x+c))-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 = 1.10, size = 107, normalized size = 1.34 \[ \frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (B a b - A b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3} + a b^{2}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, A \log \left (\tan \left (d x + c\right )\right )}{a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 7.02, size = 115, normalized size = 1.44 \[ \frac {A\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {b\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{a\,d\,\left (a^2+b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.36, size = 952, normalized size = 11.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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