3.311 \(\int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=69 \[ -\frac {b^2 B \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac {b B x}{a^2+b^2}+\frac {B \log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {21, 3571, 3530, 3475} \[ -\frac {b^2 B \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac {b B x}{a^2+b^2}+\frac {B \log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cot[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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 3571

Int[1/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[((a*
c - b*d)*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[b^2/((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^2/((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}, 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+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac {\cot (c+d x)}{a+b \tan (c+d x)} \, dx\\ &=-\frac {b B x}{a^2+b^2}+\frac {B \int \cot (c+d x) \, dx}{a}-\frac {\left (b^2 B\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac {b B x}{a^2+b^2}+\frac {B \log (\sin (c+d x))}{a d}-\frac {b^2 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 = 0.12, size = 79, normalized size = 1.14 \[ -\frac {B \left (2 b^2 \log (a \cot (c+d x)+b)+a (a+i b) \log (-\cot (c+d x)+i)+a (a-i b) \log (\cot (c+d x)+i)\right )}{2 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cot[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.72, size = 104, normalized size = 1.51 \[ -\frac {2 \, B a b d x + B b^{2} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{3} + a b^{2}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.45, size = 92, normalized size = 1.33 \[ -\frac {\frac {2 \, B b^{3} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3} b + a b^{3}} + \frac {2 \, {\left (d x + c\right )} B b}{a^{2} + b^{2}} + \frac {B a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, B \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.58, size = 99, normalized size = 1.43 \[ -\frac {b^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d a \left (a^{2}+b^{2}\right )}+\frac {\ln \left (\tan \left (d x +c \right )\right ) B}{d a}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a B}{2 d \left (a^{2}+b^{2}\right )}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.59, size = 88, normalized size = 1.28 \[ -\frac {\frac {2 \, B b^{2} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3} + a b^{2}} + \frac {2 \, {\left (d x + c\right )} B b}{a^{2} + b^{2}} + \frac {B a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, B \log \left (\tan \left (d x + c\right )\right )}{a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.36, size = 99, normalized size = 1.43 \[ \frac {B\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{a\,d}-\frac {B\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {B\,b^2\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{a\,d\,\left (a^2+b^2\right )}-\frac {B\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x))^2,x)

[Out]

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

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sympy [A]  time = 3.32, size = 683, normalized size = 9.90 \[ \begin {cases} \frac {\tilde {\infty } B x \cot {\relax (c )}}{\tan {\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {B \left (- x - \frac {1}{d \tan {\left (c + d x \right )}}\right )}{b} & \text {for}\: a = 0 \\- \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 {i 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 {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 {2 i B \log {\left (\tan {\left (c + d x \right )} \right )} \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {2 B \log {\left (\tan {\left (c + d x \right )} \right )}}{- 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 {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 {i 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 {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 {2 i B \log {\left (\tan {\left (c + d x \right )} \right )} \tan {\left (c + d x \right )}}{- 2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {2 B \log {\left (\tan {\left (c + d x \right )} \right )}}{- 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 (B a + B b \tan {\relax (c )}\right ) \cot {\relax (c )}}{\left (a + b \tan {\relax (c )}\right )^{2}} & \text {for}\: d = 0 \\\frac {B \left (- \frac {\log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {\log {\left (\tan {\left (c + d x \right )} \right )}}{d}\right )}{a} & \text {for}\: b = 0 \\- \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d + 2 a b^{2} d} + \frac {2 B a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} - \frac {2 B a b d x}{2 a^{3} d + 2 a b^{2} d} - \frac {2 B b^{2} \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} + \frac {2 B b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)

[Out]

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

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