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

Optimal. Leaf size=103 \[ \frac {b^2 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac {x (a A+b B)}{a^2+b^2}-\frac {(A b-a B) \log (\sin (c+d x))}{a^2 d}-\frac {A \cot (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3609, 3651, 3530, 3475} \[ \frac {b^2 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac {x (a A+b B)}{a^2+b^2}-\frac {(A b-a B) \log (\sin (c+d x))}{a^2 d}-\frac {A \cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a*A + b*B)*x)/(a^2 + b^2)) - (A*Cot[c + d*x])/(a*d) - ((A*b - a*B)*Log[Sin[c + d*x]])/(a^2*d) + (b^2*(A*b
- a*B)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/(a^2*(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 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ
[a, 0])))

Rule 3651

Int[((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*tan[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[((a*(A*c - c*C + B*d) + b*(B*c - A*d + C*d
))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[
e + f*x])/(a + b*Tan[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Ta
n[e + f*x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, 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 ^2(c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx &=-\frac {A \cot (c+d x)}{a d}-\frac {\int \frac {\cot (c+d x) \left (A b-a B+a A \tan (c+d x)+A b \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a}\\ &=-\frac {(a A+b B) x}{a^2+b^2}-\frac {A \cot (c+d x)}{a d}-\frac {(A b-a B) \int \cot (c+d x) \, dx}{a^2}+\frac {\left (b^2 (A b-a B)\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2 \left (a^2+b^2\right )}\\ &=-\frac {(a A+b B) x}{a^2+b^2}-\frac {A \cot (c+d x)}{a d}-\frac {(A b-a B) \log (\sin (c+d x))}{a^2 d}+\frac {b^2 (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right ) d}\\ \end {align*}

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Mathematica [C]  time = 0.88, size = 138, normalized size = 1.34 \[ \frac {\frac {2 b^2 (A b-a B) \log (a+b \tan (c+d x))}{a^2 \left (a^2+b^2\right )}+\frac {2 (a B-A b) \log (\tan (c+d x))}{a^2}+\frac {i (A+i B) \log (-\tan (c+d x)+i)}{a+i b}-\frac {(B+i A) \log (\tan (c+d x)+i)}{a-i b}-\frac {2 A \cot (c+d x)}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 177, normalized size = 1.72 \[ -\frac {2 \, A a^{3} + 2 \, A a b^{2} + 2 \, {\left (A a^{3} + B a^{2} b\right )} d x \tan \left (d x + c\right ) - {\left (B a^{3} - A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + {\left (B a b^{2} - A b^{3}\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, {\left (a^{4} + a^{2} b^{2}\right )} d \tan \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*A*a^3 + 2*A*a*b^2 + 2*(A*a^3 + B*a^2*b)*d*x*tan(d*x + c) - (B*a^3 - A*a^2*b + B*a*b^2 - A*b^3)*log(tan
(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c) + (B*a*b^2 - A*b^3)*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c
) + a^2)/(tan(d*x + c)^2 + 1))*tan(d*x + c))/((a^4 + a^2*b^2)*d*tan(d*x + c))

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giac [A]  time = 0.68, size = 157, normalized size = 1.52 \[ -\frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac {2 \, {\left (B a b^{3} - A b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + a^{2} b^{3}} - \frac {2 \, {\left (B a - A b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac {2 \, {\left (B a \tan \left (d x + c\right ) - A b \tan \left (d x + c\right ) + A a\right )}}{a^{2} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(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^3 - A*b
^4)*log(abs(b*tan(d*x + c) + a))/(a^4*b + a^2*b^3) - 2*(B*a - A*b)*log(abs(tan(d*x + c)))/a^2 + 2*(B*a*tan(d*x
+ c) - A*b*tan(d*x + c) + A*a)/(a^2*tan(d*x + c)))/d

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maple [B]  time = 0.60, size = 214, normalized size = 2.08 \[ \frac {b^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \,a^{2} \left (a^{2}+b^{2}\right )}-\frac {b^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d a \left (a^{2}+b^{2}\right )}-\frac {A}{d a \tan \left (d x +c \right )}-\frac {\ln \left (\tan \left (d x +c \right )\right ) A b}{d \,a^{2}}+\frac {\ln \left (\tan \left (d x +c \right )\right ) B}{d a}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A b}{2 d \left (a^{2}+b^{2}\right )}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a B}{2 d \left (a^{2}+b^{2}\right )}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a}{d \left (a^{2}+b^{2}\right )}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b}{d \left (a^{2}+b^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*b^3/a^2/(a^2+b^2)*ln(a+b*tan(d*x+c))*A-1/d*b^2/a/(a^2+b^2)*ln(a+b*tan(d*x+c))*B-1/d*A/a/tan(d*x+c)-1/d/a^2
*ln(tan(d*x+c))*A*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-1/2/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 = 0.75, size = 131, normalized size = 1.27 \[ -\frac {\frac {2 \, {\left (A a + B b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} + \frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + a^{2} b^{2}} + \frac {{\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac {2 \, {\left (B a - A b\right )} \log \left (\tan \left (d x + c\right )\right )}{a^{2}} + \frac {2 \, A}{a \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(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*b^2 - A*b^3)*log(b*tan(d*x + c) + a)/(a^4 + a^2*b^2) + (B*a
- A*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2) - 2*(B*a - A*b)*log(tan(d*x + c))/a^2 + 2*A/(a*tan(d*x + c)))/d

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mupad [B]  time = 7.70, size = 140, normalized size = 1.36 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b^3-B\,a\,b^2\right )}{d\,\left (a^4+a^2\,b^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (b+a\,1{}\mathrm {i}\right )}-\frac {A\,\mathrm {cot}\left (c+d\,x\right )}{a\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (a+b\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a + b*tan(c + d*x))*(A*b^3 - B*a*b^2))/(d*(a^4 + a^2*b^2)) - (log(tan(c + d*x))*(A*b - B*a))/(a^2*d) + (l
og(tan(c + d*x) + 1i)*(A - B*1i))/(2*d*(a*1i + b)) - (A*cot(c + d*x))/(a*d) + (log(tan(c + d*x) - 1i)*(A*1i -
B))/(2*d*(a + b*1i))

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sympy [A]  time = 4.50, size = 2067, normalized size = 20.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*A*x, Eq(a, 0) & Eq(b, 0) & Eq(c, 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)/a, Eq(b, 0)), ((A*log(tan(c + d*x)**2 + 1)/(2*d) - A*log(tan(c + d*x
))/d - A/(2*d*tan(c + d*x)**2) - B*x - B/(d*tan(c + d*x)))/b, Eq(a, 0)), (-3*I*A*d*x*tan(c + d*x)**2/(2*b*d*ta
n(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - 3*A*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) -
A*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) + I*A*log(tan(c + d*
x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) + 2*A*log(tan(c + d*x))*tan(c + d*x)**2
/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - 2*I*A*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2
- 2*I*b*d*tan(c + d*x)) - 3*I*A*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - 2*A/(2*b*d*tan(c
+ d*x)**2 - 2*I*b*d*tan(c + d*x)) + B*d*x*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - I*
B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - I*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)*
*2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) - B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x
)**2 - 2*I*b*d*tan(c + d*x)) + 2*I*B*log(tan(c + d*x))*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c
+ d*x)) + 2*B*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)) + B*tan(c + d*x)/(
2*b*d*tan(c + d*x)**2 - 2*I*b*d*tan(c + d*x)), Eq(a, -I*b)), (3*I*A*d*x*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2
+ 2*I*b*d*tan(c + d*x)) - 3*A*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - A*log(tan(c +
d*x)**2 + 1)*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - I*A*log(tan(c + d*x)**2 + 1)*ta
n(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + 2*A*log(tan(c + d*x))*tan(c + d*x)**2/(2*b*d*tan(c
+ d*x)**2 + 2*I*b*d*tan(c + d*x)) + 2*I*A*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan
(c + d*x)) + 3*I*A*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - 2*A/(2*b*d*tan(c + d*x)**2 +
2*I*b*d*tan(c + d*x)) + B*d*x*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + I*B*d*x*tan(c +
d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + I*B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2/(2*b*d*tan
(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) - B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*
d*tan(c + d*x)) - 2*I*B*log(tan(c + d*x))*tan(c + d*x)**2/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + 2*B
*log(tan(c + d*x))*tan(c + d*x)/(2*b*d*tan(c + d*x)**2 + 2*I*b*d*tan(c + d*x)) + B*tan(c + d*x)/(2*b*d*tan(c +
d*x)**2 + 2*I*b*d*tan(c + d*x)), Eq(a, I*b)), (zoo*A*x/a, Eq(c, -d*x)), (x*(A + B*tan(c))*cot(c)**2/(a + b*ta
n(c)), Eq(d, 0)), (-2*A*a**3*d*x*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*A*a**3/
(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) + A*a**2*b*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*a**4*
d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*A*a**2*b*log(tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x
) + 2*a**2*b**2*d*tan(c + d*x)) - 2*A*a*b**2/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) + 2*A*b**3*l
og(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*A*b**3*log(tan(c
+ d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - B*a**3*log(tan(c + d*x)**2 + 1)*ta
n(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) + 2*B*a**3*log(tan(c + d*x))*tan(c + d*x)/(2*a
**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)) - 2*B*a**2*b*d*x*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2
*b**2*d*tan(c + d*x)) - 2*B*a*b**2*log(a/b + tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d
*tan(c + d*x)) + 2*B*a*b**2*log(tan(c + d*x))*tan(c + d*x)/(2*a**4*d*tan(c + d*x) + 2*a**2*b**2*d*tan(c + d*x)
), True))

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