\(\int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\) [245]

Optimal result

Integrand size = 31, antiderivative size = 88 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\left (b^2 B-a (2 A b+a B)\right ) x-\frac {a (2 A b+a B) \cot (c+d x)}{d}-\frac {a^2 A \cot ^2(c+d x)}{2 d}-\frac {\left (a^2 A-A b^2-2 a b B\right ) \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3685, 3709, 3612, 3556} \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {\left (a^2 A-2 a b B-A b^2\right ) \log (\sin (c+d x))}{d}-\frac {a^2 A \cot ^2(c+d x)}{2 d}+x \left (b^2 B-a (a B+2 A b)\right )-\frac {a (a B+2 A b) \cot (c+d x)}{d} \]

[In]

Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

(b^2*B - a*(2*A*b + a*B))*x - (a*(2*A*b + a*B)*Cot[c + d*x])/d - (a^2*A*Cot[c + d*x]^2)/(2*d) - ((a^2*A - A*b^
2 - 2*a*b*B)*Log[Sin[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 3612

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]

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 3709

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2)
)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e +
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2
+ b^2, 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.40 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {-2 a (2 A b+a B) \cot (c+d x)-a^2 A \cot ^2(c+d x)+(a+i b)^2 (A+i B) \log (i-\tan (c+d x))-2 \left (a^2 A-A b^2-2 a b B\right ) \log (\tan (c+d x))+(a-i b)^2 (A-i B) \log (i+\tan (c+d x))}{2 d} \]

[In]

Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2*(A + B*Tan[c + d*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.22

method	result	size
derivativedivides	\(\frac {A \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 A a b \left (-\cot \left (d x +c \right )-d x -c \right )+2 B a b \ln \left (\sin \left (d x +c \right )\right )+A \,b^{2} \ln \left (\sin \left (d x +c \right )\right )+B \,b^{2} \left (d x +c \right )}{d}\)	\(107\)
default	\(\frac {A \,a^{2} \left (-\frac {\left (\cot ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+B \,a^{2} \left (-\cot \left (d x +c \right )-d x -c \right )+2 A a b \left (-\cot \left (d x +c \right )-d x -c \right )+2 B a b \ln \left (\sin \left (d x +c \right )\right )+A \,b^{2} \ln \left (\sin \left (d x +c \right )\right )+B \,b^{2} \left (d x +c \right )}{d}\)	\(107\)
parallelrisch	\(\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-2 A \,a^{2}+2 A \,b^{2}+4 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )-A \left (\cot ^{2}\left (d x +c \right )\right ) a^{2}+\left (-4 A a b -2 B \,a^{2}\right ) \cot \left (d x +c \right )-4 d \left (A a b +\frac {1}{2} B \,a^{2}-\frac {1}{2} B \,b^{2}\right ) x}{2 d}\)	\(114\)
norman	\(\frac {\left (-2 A a b -B \,a^{2}+B \,b^{2}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )-\frac {A \,a^{2}}{2 d}-\frac {a \left (2 A b +B a \right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}-\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,a^{2}-A \,b^{2}-2 B a b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(130\)
risch	\(-2 A a b x -B \,a^{2} x +B \,b^{2} x -\frac {2 i A \,b^{2} c}{d}-i A \,b^{2} x +i A \,a^{2} x -\frac {4 i B a b c}{d}-\frac {2 i a \left (i A a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}+B a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 A b -B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-2 i B a b x +\frac {2 i a^{2} A c}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B a b}{d}\)	\(205\)

[In]

int(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.39 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {{\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + A a^{2} + {\left (A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (78) = 156\).

Time = 0.82 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.43 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{2} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{2} \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{2} x & \text {for}\: c = - d x \\\frac {A a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a^{2}}{2 d \tan ^{2}{\left (c + d x \right )}} - 2 A a b x - \frac {2 A a b}{d \tan {\left (c + d x \right )}} - \frac {A b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - B a^{2} x - \frac {B a^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 B a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{2} x & \text {otherwise} \end {cases} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )} - {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {A a^{2} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (86) = 172\).

Time = 1.10 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.69 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=-\frac {A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} {\left (d x + c\right )} - 8 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 8 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {12 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^2*tan(1/2*d*x + 1/2*c)^2 - 4*B*a^2*tan(1/2*d*x + 1/2*c) - 8*A*a*b*tan(1/2*d*x + 1/2*c) + 8*(B*a^2 +
2*A*a*b - B*b^2)*(d*x + c) - 8*(A*a^2 - 2*B*a*b - A*b^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) + 8*(A*a^2 - 2*B*a*b
- A*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) - (12*A*a^2*tan(1/2*d*x + 1/2*c)^2 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^2 -
12*A*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*B*a^2*tan(1/2*d*x + 1/2*c) - 8*A*a*b*tan(1/2*d*x + 1/2*c) - A*a^2)/tan(1/
2*d*x + 1/2*c)^2)/d

Mupad [B] (verification not implemented)

Time = 8.04 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^2+2\,B\,a\,b+A\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {A\,a^2}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^2+2\,A\,b\,a\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d} \]

[In]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^2,x)

[Out]

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