\(\int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) [252]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\left (\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x\right )-\frac {b^2 (A b+3 a B) \log (\cos (c+d x))}{d}+\frac {a^2 (3 A b+a B) \log (\sin (c+d x))}{d}+\frac {b^2 (a A+b B) \tan (c+d x)}{d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^2}{d} \]

[Out]

-(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*x-b^2*(A*b+3*B*a)*ln(cos(d*x+c))/d+a^2*(3*A*b+B*a)*ln(sin(d*x+c))/d+b^2*(A*
a+B*b)*tan(d*x+c)/d-a*A*cot(d*x+c)*(a+b*tan(d*x+c))^2/d

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3686, 3718, 3705, 3556} \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {a^2 (a B+3 A b) \log (\sin (c+d x))}{d}-x \left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right )+\frac {b^2 (a A+b B) \tan (c+d x)}{d}-\frac {b^2 (3 a B+A b) \log (\cos (c+d x))}{d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^2}{d} \]

[In]

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

[Out]

-((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*x) - (b^2*(A*b + 3*a*B)*Log[Cos[c + d*x]])/d + (a^2*(3*A*b + a*B)*Lo
g[Sin[c + d*x]])/d + (b^2*(a*A + b*B)*Tan[c + d*x])/d - (a*A*Cot[c + d*x]*(a + b*Tan[c + d*x])^2)/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 3686

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*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

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]

Rule 3718

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

Rubi steps \begin{align*} \text {integral}& = -\frac {a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}+\int \cot (c+d x) (a+b \tan (c+d x)) \left (a (3 A b+a B)-\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+b (a A+b B) \tan ^2(c+d x)\right ) \, dx \\ & = \frac {b^2 (a A+b B) \tan (c+d x)}{d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}-\int \cot (c+d x) \left (-a^2 (3 A b+a B)+\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)-b^2 (A b+3 a B) \tan ^2(c+d x)\right ) \, dx \\ & = -\left (\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x\right )+\frac {b^2 (a A+b B) \tan (c+d x)}{d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^2}{d}+\left (a^2 (3 A b+a B)\right ) \int \cot (c+d x) \, dx+\left (b^2 (A b+3 a B)\right ) \int \tan (c+d x) \, dx \\ & = -\left (\left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) x\right )-\frac {b^2 (A b+3 a B) \log (\cos (c+d x))}{d}+\frac {a^2 (3 A b+a B) \log (\sin (c+d x))}{d}+\frac {b^2 (a A+b B) \tan (c+d x)}{d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^2}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {-2 a^3 A \cot (c+d x)+i (a+i b)^3 (A+i B) \log (i-\tan (c+d x))+2 a^2 (3 A b+a B) \log (\tan (c+d x))+(i a+b)^3 (A-i B) \log (i+\tan (c+d x))+2 b^3 B \tan (c+d x)}{2 d} \]

[In]

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

[Out]

(-2*a^3*A*Cot[c + d*x] + I*(a + I*b)^3*(A + I*B)*Log[I - Tan[c + d*x]] + 2*a^2*(3*A*b + a*B)*Log[Tan[c + d*x]]
+ (I*a + b)^3*(A - I*B)*Log[I + Tan[c + d*x]] + 2*b^3*B*Tan[c + d*x])/(2*d)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99

method	result	size
parallelrisch	\(\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (6 A \,a^{2} b +2 B \,a^{3}\right ) \ln \left (\tan \left (d x +c \right )\right )-2 A \cot \left (d x +c \right ) a^{3}+2 B \,b^{3} \tan \left (d x +c \right )-2 d x \left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right )}{2 d}\)	\(118\)
derivativedivides	\(\frac {B \,b^{3} \tan \left (d x +c \right )+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{\tan \left (d x +c \right )}+a^{2} \left (3 A b +B a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(124\)
default	\(\frac {B \,b^{3} \tan \left (d x +c \right )+\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{\tan \left (d x +c \right )}+a^{2} \left (3 A b +B a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\)	\(124\)
norman	\(\frac {\left (-A \,a^{3}+3 A a \,b^{2}+3 B \,a^{2} b -B \,b^{3}\right ) x \tan \left (d x +c \right )+\frac {B \,b^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {A \,a^{3}}{d}}{\tan \left (d x +c \right )}+\frac {a^{2} \left (3 A b +B a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}-\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(136\)
risch	\(-A \,a^{3} x +3 A a \,b^{2} x +3 B \,a^{2} b x -B \,b^{3} x -3 i A \,a^{2} b x -i B \,a^{3} x +3 i B a \,b^{2} x +i A \,b^{3} x -\frac {2 i a^{3} B c}{d}-\frac {2 i \left (A \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-B \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+A \,a^{3}+B \,b^{3}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 i A \,b^{3} c}{d}+\frac {6 i B a \,b^{2} c}{d}-\frac {6 i A \,a^{2} b c}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A \,b^{3}}{d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B a \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A b}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B}{d}\)	\(269\)

[In]

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

[Out]

1/2*((-3*A*a^2*b+A*b^3-B*a^3+3*B*a*b^2)*ln(sec(d*x+c)^2)+(6*A*a^2*b+2*B*a^3)*ln(tan(d*x+c))-2*A*cot(d*x+c)*a^3
+2*B*b^3*tan(d*x+c)-2*d*x*(A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.22 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, B b^{3} \tan \left (d x + c\right )^{2} - 2 \, A a^{3} - 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} d x \tan \left (d x + c\right ) + {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) - {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.77 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.87 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{3} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{3} \cot ^{2}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{3} x & \text {for}\: c = - d x \\- A a^{3} x - \frac {A a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 A a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 A a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 A a b^{2} x + \frac {A b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 B a^{2} b x + \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - B b^{3} x + \frac {B b^{3} \tan {\left (c + d x \right )}}{d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.05 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, B b^{3} \tan \left (d x + c\right ) - \frac {2 \, A a^{3}}{\tan \left (d x + c\right )} - 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left (\tan \left (d x + c\right )\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.44 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.28 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {2 \, B b^{3} \tan \left (d x + c\right ) - 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (B a^{3} \tan \left (d x + c\right ) + 3 \, A a^{2} b \tan \left (d x + c\right ) + A a^{3}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.70 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \cot ^2(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3+3\,A\,b\,a^2\right )}{d}-\frac {A\,a^3\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {B\,b^3\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (a+b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (a-b\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]

[In]

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

[Out]

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