\(\int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [237]

Optimal result

Integrand size = 29, antiderivative size = 66 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-((A b+a B) x)-\frac {(A b+a B) \cot (c+d x)}{d}-\frac {a A \cot ^2(c+d x)}{2 d}-\frac {(a A-b B) \log (\sin (c+d x))}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3672, 3610, 3612, 3556} \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {(a B+A b) \cot (c+d x)}{d}-\frac {(a A-b B) \log (\sin (c+d x))}{d}-x (a B+A b)-\frac {a A \cot ^2(c+d x)}{2 d} \]

[In]

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

[Out]

-((A*b + a*B)*x) - ((A*b + a*B)*Cot[c + d*x])/d - (a*A*Cot[c + d*x]^2)/(2*d) - ((a*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 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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 3672

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.50 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {a A \cot ^2(c+d x)+2 (A b+a B) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )+2 (a A-b B) (\log (\cos (c+d x))+\log (\tan (c+d x)))}{2 d} \]

[In]

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

[Out]

-1/2*(a*A*Cot[c + d*x]^2 + 2*(A*b + a*B)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 2*(a*
A - b*B)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))/d

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {\frac {\left (a A -B b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A b +B a}{\tan \left (d x +c \right )}+\left (-a A +B b \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {a A}{2 \tan \left (d x +c \right )^{2}}}{d}\)	\(89\)
default	\(\frac {\frac {\left (a A -B b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A b +B a}{\tan \left (d x +c \right )}+\left (-a A +B b \right ) \ln \left (\tan \left (d x +c \right )\right )-\frac {a A}{2 \tan \left (d x +c \right )^{2}}}{d}\)	\(89\)
parallelrisch	\(\frac {-A \left (\cot ^{2}\left (d x +c \right )\right ) a -2 A b d x -2 B x a d -2 A \cot \left (d x +c \right ) b -2 a A \ln \left (\tan \left (d x +c \right )\right )+A \ln \left (\sec ^{2}\left (d x +c \right )\right ) a -2 B \cot \left (d x +c \right ) a +2 B \ln \left (\tan \left (d x +c \right )\right ) b -B \ln \left (\sec ^{2}\left (d x +c \right )\right ) b}{2 d}\)	\(98\)
norman	\(\frac {\left (-A b -B a \right ) x \left (\tan ^{2}\left (d x +c \right )\right )-\frac {a A}{2 d}-\frac {\left (A b +B a \right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}-\frac {\left (a A -B b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (a A -B b \right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(100\)
risch	\(-A b x -B a x +i A a x -i B b x +\frac {2 i a A c}{d}-\frac {2 i B b c}{d}-\frac {2 i \left (i A a \,{\mathrm e}^{2 i \left (d x +c \right )}+A b \,{\mathrm e}^{2 i \left (d x +c \right )}+B a \,{\mathrm e}^{2 i \left (d x +c \right )}-A b -B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a A \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}\)	\(145\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.44 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {{\left (A a - B b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (2 \, {\left (B a + A b\right )} d x + A a\right )} \tan \left (d x + c\right )^{2} + A a + 2 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-1/2*((A*a - B*b)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 + (2*(B*a + A*b)*d*x + A*a)*tan(d*x
+ c)^2 + A*a + 2*(B*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. 148 vs. \(2 (56) = 112\).

Time = 0.56 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.24 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right ) \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a x & \text {for}\: c = - d x \\\frac {A a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a}{2 d \tan ^{2}{\left (c + d x \right )}} - A b x - \frac {A b}{d \tan {\left (c + d x \right )}} - B a x - \frac {B a}{d \tan {\left (c + d x \right )}} - \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.30 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (B a + A b\right )} {\left (d x + c\right )} - {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {A a + 2 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

-1/2*(2*(B*a + A*b)*(d*x + c) - (A*a - B*b)*log(tan(d*x + c)^2 + 1) + 2*(A*a - B*b)*log(tan(d*x + c)) + (A*a +
2*(B*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. 179 vs. \(2 (64) = 128\).

Time = 0.63 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.71 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, {\left (B a + A b\right )} {\left (d x + c\right )} - 8 \, {\left (A a - B b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) + 8 \, {\left (A a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.64 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,a-B\,b\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\frac {A\,a}{2}+\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b+B\,a\right )\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]

[In]

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

[Out]

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