\(\int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [244]

Optimal result

Integrand size = 31, antiderivative size = 72 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\left (\left (a^2 A-A b^2-2 a b B\right ) x\right )-\frac {a^2 A \cot (c+d x)}{d}-\frac {b^2 B \log (\cos (c+d x))}{d}+\frac {a (2 A b+a B) \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3685, 3705, 3556} \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-x \left (a^2 A-2 a b B-A b^2\right )-\frac {a^2 A \cot (c+d x)}{d}+\frac {a (a B+2 A b) \log (\sin (c+d x))}{d}-\frac {b^2 B \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

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

Rule 3556

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

Rule 3685

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(B*c - A*d))*(b*c - a*d)^2*((c + d*Tan[e + f*x])^(n + 1)/(f*d^2*(n +
1)*(c^2 + d^2))), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[B*(b*c - a*d)^2 + A*d*(a
^2*c - b^2*c + 2*a*b*d) + d*(B*(a^2*c - b^2*c + 2*a*b*d) + A*(2*a*b*c - a^2*d + b^2*d))*Tan[e + f*x] + b^2*B*(
c^2 + d^2)*Tan[e + f*x]^2, 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] && LtQ[n, -1]

Rule 3705

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/tan[(e_.) + (f_.)*(x_)], x_Symbol
] :> Simp[B*x, x] + (Dist[A, Int[1/Tan[e + f*x], x], x] + Dist[C, Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A,
B, C}, x] && NeQ[A, C]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {-2 a^2 A \cot (c+d x)+i (a+i b)^2 (A+i B) \log (i-\tan (c+d x))+2 a (2 A b+a B) \log (\tan (c+d x))-(a-i b)^2 (i A+B) \log (i+\tan (c+d x))}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 A a b \ln \left (\sin \left (d x +c \right )\right )+2 B a b \left (d x +c \right )+A \,b^{2} \left (d x +c \right )-B \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(84\)
default	\(\frac {A \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+B \,a^{2} \ln \left (\sin \left (d x +c \right )\right )+2 A a b \ln \left (\sin \left (d x +c \right )\right )+2 B a b \left (d x +c \right )+A \,b^{2} \left (d x +c \right )-B \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}\)	\(84\)
parallelrisch	\(\frac {\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (4 A a b +2 B \,a^{2}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 A \,a^{2} \cot \left (d x +c \right )-2 d x \left (A \,a^{2}-A \,b^{2}-2 B a b \right )}{2 d}\)	\(87\)
norman	\(\frac {\left (-A \,a^{2}+A \,b^{2}+2 B a b \right ) x \tan \left (d x +c \right )-\frac {A \,a^{2}}{d}}{\tan \left (d x +c \right )}+\frac {a \left (2 A b +B a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (2 A a b +B \,a^{2}-B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(99\)
risch	\(-\frac {2 i a^{2} B c}{d}+i B \,b^{2} x +\frac {2 i B \,b^{2} c}{d}-A \,a^{2} x +A \,b^{2} x +2 B a b x -\frac {4 i A a b c}{d}-\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-2 i A a b x -i B \,a^{2} x +\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,b^{2}}{d}\)	\(160\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {B b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} d x \tan \left (d x + c\right ) + 2 \, A a^{2} - {\left (B a^{2} + 2 \, A a b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (66) = 132\).

Time = 0.46 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.32 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{2} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{2}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{2} x & \text {for}\: c = - d x \\- A a^{2} x - \frac {A a^{2}}{d \tan {\left (c + d x \right )}} - \frac {A a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 A a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + A b^{2} x - \frac {B a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 2 B a b x + \frac {B b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.29 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (B a^{2} + 2 \, A a b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {2 \, A a^{2}}{\tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.64 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} {\left (d x + c\right )} + {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (B a^{2} \tan \left (d x + c\right ) + 2 \, A a b \tan \left (d x + c\right ) + A a^{2}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.53 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^2+2\,A\,b\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {A\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d} \]

[In]

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

[Out]

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