\(\int \cot ^5(c+d x) (a+b \tan (c+d x)) (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\) [7]

Optimal result

Integrand size = 38, antiderivative size = 87 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=(a B-b C) x+\frac {(a B-b C) \cot (c+d x)}{d}-\frac {(b B+a C) \cot ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x)}{3 d}-\frac {(b B+a C) \log (\sin (c+d x))}{d} \]

[Out]

(B*a-C*b)*x+(B*a-C*b)*cot(d*x+c)/d-1/2*(B*b+C*a)*cot(d*x+c)^2/d-1/3*a*B*cot(d*x+c)^3/d-(B*b+C*a)*ln(sin(d*x+c)
)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3713, 3672, 3610, 3612, 3556} \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {(a C+b B) \cot ^2(c+d x)}{2 d}+\frac {(a B-b C) \cot (c+d x)}{d}-\frac {(a C+b B) \log (\sin (c+d x))}{d}+x (a B-b C)-\frac {a B \cot ^3(c+d x)}{3 d} \]

[In]

Int[Cot[c + d*x]^5*(a + b*Tan[c + d*x])*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(a*B - b*C)*x + ((a*B - b*C)*Cot[c + d*x])/d - ((b*B + a*C)*Cot[c + d*x]^2)/(2*d) - (a*B*Cot[c + d*x]^3)/(3*d)
- ((b*B + a*C)*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]

Rule 3713

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

Rubi steps \begin{align*} \text {integral}& = \int \cot ^4(c+d x) (a+b \tan (c+d x)) (B+C \tan (c+d x)) \, dx \\ & = -\frac {a B \cot ^3(c+d x)}{3 d}+\int \cot ^3(c+d x) (b B+a C-(a B-b C) \tan (c+d x)) \, dx \\ & = -\frac {(b B+a C) \cot ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x)}{3 d}+\int \cot ^2(c+d x) (-a B+b C-(b B+a C) \tan (c+d x)) \, dx \\ & = \frac {(a B-b C) \cot (c+d x)}{d}-\frac {(b B+a C) \cot ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x)}{3 d}+\int \cot (c+d x) (-b B-a C+(a B-b C) \tan (c+d x)) \, dx \\ & = (a B-b C) x+\frac {(a B-b C) \cot (c+d x)}{d}-\frac {(b B+a C) \cot ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x)}{3 d}+(-b B-a C) \int \cot (c+d x) \, dx \\ & = (a B-b C) x+\frac {(a B-b C) \cot (c+d x)}{d}-\frac {(b B+a C) \cot ^2(c+d x)}{2 d}-\frac {a B \cot ^3(c+d x)}{3 d}-\frac {(b B+a C) \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 = 1.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.16 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {2 a B \cot ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(c+d x)\right )+6 b C \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )+3 (b B+a C) \left (\cot ^2(c+d x)+2 (\log (\cos (c+d x))+\log (\tan (c+d x)))\right )}{6 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + b*Tan[c + d*x])*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

-1/6*(2*a*B*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2] + 6*b*C*Cot[c + d*x]*Hypergeometr
ic2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 3*(b*B + a*C)*(Cot[c + d*x]^2 + 2*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]
])))/d

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {B b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+C b \left (-\cot \left (d x +c \right )-d x -c \right )+B a \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+C a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(95\)
default	\(\frac {B b \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+C b \left (-\cot \left (d x +c \right )-d x -c \right )+B a \left (-\frac {\cot \left (d x +c \right )^{3}}{3}+\cot \left (d x +c \right )+d x +c \right )+C a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )}{d}\)	\(95\)
parallelrisch	\(\frac {-2 B a \cot \left (d x +c \right )^{3}-3 B b \cot \left (d x +c \right )^{2}+6 B a d x -3 C a \cot \left (d x +c \right )^{2}-6 C b d x +6 B \cot \left (d x +c \right ) a -6 B \ln \left (\tan \left (d x +c \right )\right ) b +3 B \ln \left (\sec \left (d x +c \right )^{2}\right ) b -6 C b \cot \left (d x +c \right )-6 C \ln \left (\tan \left (d x +c \right )\right ) a +3 C \ln \left (\sec \left (d x +c \right )^{2}\right ) a}{6 d}\)	\(123\)
norman	\(\frac {\frac {\left (B a -C b \right ) \tan \left (d x +c \right )^{3}}{d}+\left (B a -C b \right ) x \tan \left (d x +c \right )^{4}-\frac {\left (B b +C a \right ) \tan \left (d x +c \right )^{2}}{2 d}-\frac {B a \tan \left (d x +c \right )}{3 d}}{\tan \left (d x +c \right )^{4}}-\frac {\left (B b +C a \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {\left (B b +C a \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\)	\(125\)
risch	\(i B b x +i C a x +B a x -C b x +\frac {2 i B b c}{d}+\frac {2 i C a c}{d}-\frac {2 i \left (3 i B b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 i C a \,{\mathrm e}^{4 i \left (d x +c \right )}-6 B a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 i B b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 i C a \,{\mathrm e}^{2 i \left (d x +c \right )}+6 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 B a +3 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) C a}{d}\)	\(215\)

[In]

int(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(B*b*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))+C*b*(-cot(d*x+c)-d*x-c)+B*a*(-1/3*cot(d*x+c)^3+cot(d*x+c)+d*x+c)+C
*a*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c))))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {3 \, {\left (C a + B b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{3} - 3 \, {\left (2 \, {\left (B a - C b\right )} d x - C a - B b\right )} \tan \left (d x + c\right )^{3} - 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} + 2 \, B a + 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{6 \, d \tan \left (d x + c\right )^{3}} \]

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/6*(3*(C*a + B*b)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 - 3*(2*(B*a - C*b)*d*x - C*a - B*b
)*tan(d*x + c)^3 - 6*(B*a - C*b)*tan(d*x + c)^2 + 2*B*a + 3*(C*a + B*b)*tan(d*x + c))/(d*tan(d*x + c)^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (75) = 150\).

Time = 1.88 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.99 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\left (c \right )}\right ) \left (B \tan {\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\B a x + \frac {B a}{d \tan {\left (c + d x \right )}} - \frac {B a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B b}{2 d \tan ^{2}{\left (c + d x \right )}} + \frac {C a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {C a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {C a}{2 d \tan ^{2}{\left (c + d x \right )}} - C b x - \frac {C b}{d \tan {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cot(d*x+c)**5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))*(B*tan(c) + C*tan(c)**2)*cot(c)**5, Eq(d, 0)), (nan, E
q(c, -d*x)), (B*a*x + B*a/(d*tan(c + d*x)) - B*a/(3*d*tan(c + d*x)**3) + B*b*log(tan(c + d*x)**2 + 1)/(2*d) -
B*b*log(tan(c + d*x))/d - B*b/(2*d*tan(c + d*x)**2) + C*a*log(tan(c + d*x)**2 + 1)/(2*d) - C*a*log(tan(c + d*x
))/d - C*a/(2*d*tan(c + d*x)**2) - C*b*x - C*b/(d*tan(c + d*x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.20 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (B a - C b\right )} {\left (d x + c\right )} + 3 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, {\left (C a + B b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} - 2 \, B a - 3 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

1/6*(6*(B*a - C*b)*(d*x + c) + 3*(C*a + B*b)*log(tan(d*x + c)^2 + 1) - 6*(C*a + B*b)*log(tan(d*x + c)) + (6*(B
*a - C*b)*tan(d*x + c)^2 - 2*B*a - 3*(C*a + B*b)*tan(d*x + c))/tan(d*x + c)^3)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (83) = 166\).

Time = 1.61 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.72 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=\frac {B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, {\left (B a - C b\right )} {\left (d x + c\right )} + 24 \, {\left (C a + B b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 24 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {44 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 44 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(B*a*tan(1/2*d*x + 1/2*c)^3 - 3*C*a*tan(1/2*d*x + 1/2*c)^2 - 3*B*b*tan(1/2*d*x + 1/2*c)^2 - 15*B*a*tan(1/
2*d*x + 1/2*c) + 12*C*b*tan(1/2*d*x + 1/2*c) + 24*(B*a - C*b)*(d*x + c) + 24*(C*a + B*b)*log(tan(1/2*d*x + 1/2
*c)^2 + 1) - 24*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c))) + (44*C*a*tan(1/2*d*x + 1/2*c)^3 + 44*B*b*tan(1/2*d
*x + 1/2*c)^3 + 15*B*a*tan(1/2*d*x + 1/2*c)^2 - 12*C*b*tan(1/2*d*x + 1/2*c)^2 - 3*C*a*tan(1/2*d*x + 1/2*c) - 3
*B*b*tan(1/2*d*x + 1/2*c) - B*a)/tan(1/2*d*x + 1/2*c)^3)/d

Mupad [B] (verification not implemented)

Time = 7.86 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.46 \[ \int \cot ^5(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx=-\frac {{\mathrm {cot}\left (c+d\,x\right )}^3\,\left (\left (C\,b-B\,a\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {B\,b}{2}+\frac {C\,a}{2}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {B\,a}{3}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,b+C\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )}{2\,d} \]

[In]

int(cot(c + d*x)^5*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x)),x)

[Out]

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