3.8 \(\int \cot ^6(c+d x) (a+b \tan (c+d x)) (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=108 \[ -\frac {(a C+b B) \cot ^3(c+d x)}{3 d}+\frac {(a B-b C) \cot ^2(c+d x)}{2 d}+\frac {(a C+b B) \cot (c+d x)}{d}+\frac {(a B-b C) \log (\sin (c+d x))}{d}+x (a C+b B)-\frac {a B \cot ^4(c+d x)}{4 d} \]

[Out]

(B*b+C*a)*x+(B*b+C*a)*cot(d*x+c)/d+1/2*(B*a-C*b)*cot(d*x+c)^2/d-1/3*(B*b+C*a)*cot(d*x+c)^3/d-1/4*a*B*cot(d*x+c
)^4/d+(B*a-C*b)*ln(sin(d*x+c))/d

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Rubi [A]  time = 0.23, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3632, 3591, 3529, 3531, 3475} \[ -\frac {(a C+b B) \cot ^3(c+d x)}{3 d}+\frac {(a B-b C) \cot ^2(c+d x)}{2 d}+\frac {(a C+b B) \cot (c+d x)}{d}+\frac {(a B-b C) \log (\sin (c+d x))}{d}+x (a C+b B)-\frac {a B \cot ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + b*Tan[c + d*x])*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

(b*B + a*C)*x + ((b*B + a*C)*Cot[c + d*x])/d + ((a*B - b*C)*Cot[c + d*x]^2)/(2*d) - ((b*B + a*C)*Cot[c + d*x]^
3)/(3*d) - (a*B*Cot[c + d*x]^4)/(4*d) + ((a*B - b*C)*Log[Sin[c + d*x]])/d

Rule 3475

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

Rule 3529

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 3531

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 3591

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 3632

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*} \int \cot ^6(c+d x) (a+b \tan (c+d x)) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right ) \, dx &=\int \cot ^5(c+d x) (a+b \tan (c+d x)) (B+C \tan (c+d x)) \, dx\\ &=-\frac {a B \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) (b B+a C-(a B-b C) \tan (c+d x)) \, dx\\ &=-\frac {(b B+a C) \cot ^3(c+d x)}{3 d}-\frac {a B \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) (-a B+b C-(b B+a C) \tan (c+d x)) \, dx\\ &=\frac {(a B-b C) \cot ^2(c+d x)}{2 d}-\frac {(b B+a C) \cot ^3(c+d x)}{3 d}-\frac {a B \cot ^4(c+d x)}{4 d}+\int \cot ^2(c+d x) (-b B-a C+(a B-b C) \tan (c+d x)) \, dx\\ &=\frac {(b B+a C) \cot (c+d x)}{d}+\frac {(a B-b C) \cot ^2(c+d x)}{2 d}-\frac {(b B+a C) \cot ^3(c+d x)}{3 d}-\frac {a B \cot ^4(c+d x)}{4 d}+\int \cot (c+d x) (a B-b C+(b B+a C) \tan (c+d x)) \, dx\\ &=(b B+a C) x+\frac {(b B+a C) \cot (c+d x)}{d}+\frac {(a B-b C) \cot ^2(c+d x)}{2 d}-\frac {(b B+a C) \cot ^3(c+d x)}{3 d}-\frac {a B \cot ^4(c+d x)}{4 d}+(a B-b C) \int \cot (c+d x) \, dx\\ &=(b B+a C) x+\frac {(b B+a C) \cot (c+d x)}{d}+\frac {(a B-b C) \cot ^2(c+d x)}{2 d}-\frac {(b B+a C) \cot ^3(c+d x)}{3 d}-\frac {a B \cot ^4(c+d x)}{4 d}+\frac {(a B-b C) \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [C]  time = 1.15, size = 100, normalized size = 0.93 \[ -\frac {4 (a C+b B) \cot ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(c+d x)\right )+3 \left ((2 b C-2 a B) \cot ^2(c+d x)-4 (a B-b C) (\log (\tan (c+d x))+\log (\cos (c+d x)))+a B \cot ^4(c+d x)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + b*Tan[c + d*x])*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

-1/12*(4*(b*B + a*C)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2] + 3*((-2*a*B + 2*b*C)*Co
t[c + d*x]^2 + a*B*Cot[c + d*x]^4 - 4*(a*B - b*C)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]])))/d

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fricas [A]  time = 0.69, size = 138, normalized size = 1.28 \[ \frac {6 \, {\left (B a - C b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (4 \, {\left (C a + B b\right )} d x + 3 \, B a - 2 \, C b\right )} \tan \left (d x + c\right )^{4} + 12 \, {\left (C a + B b\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} - 3 \, B a - 4 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(B*a - C*b)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^4 + 3*(4*(C*a + B*b)*d*x + 3*B*a - 2
*C*b)*tan(d*x + c)^4 + 12*(C*a + B*b)*tan(d*x + c)^3 + 6*(B*a - C*b)*tan(d*x + c)^2 - 3*B*a - 4*(C*a + B*b)*ta
n(d*x + c))/(d*tan(d*x + c)^4)

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giac [B]  time = 9.32, size = 299, normalized size = 2.77 \[ -\frac {3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 192 \, {\left (C a + B b\right )} {\left (d x + c\right )} + 192 \, {\left (B a - C b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (B a - C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

-1/192*(3*B*a*tan(1/2*d*x + 1/2*c)^4 - 8*C*a*tan(1/2*d*x + 1/2*c)^3 - 8*B*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*ta
n(1/2*d*x + 1/2*c)^2 + 24*C*b*tan(1/2*d*x + 1/2*c)^2 + 120*C*a*tan(1/2*d*x + 1/2*c) + 120*B*b*tan(1/2*d*x + 1/
2*c) - 192*(C*a + B*b)*(d*x + c) + 192*(B*a - C*b)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 192*(B*a - C*b)*log(abs(t
an(1/2*d*x + 1/2*c))) + (400*B*a*tan(1/2*d*x + 1/2*c)^4 - 400*C*b*tan(1/2*d*x + 1/2*c)^4 - 120*C*a*tan(1/2*d*x
+ 1/2*c)^3 - 120*B*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*tan(1/2*d*x + 1/2*c)^2 + 24*C*b*tan(1/2*d*x + 1/2*c)^2 +
8*C*a*tan(1/2*d*x + 1/2*c) + 8*B*b*tan(1/2*d*x + 1/2*c) + 3*B*a)/tan(1/2*d*x + 1/2*c)^4)/d

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maple [A]  time = 0.53, size = 150, normalized size = 1.39 \[ -\frac {a B \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a B \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a B \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a C \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {C \cot \left (d x +c \right ) a}{d}+a C x +\frac {C a c}{d}-\frac {B b \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {B \cot \left (d x +c \right ) b}{d}+B x b +\frac {B b c}{d}-\frac {C b \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {C b \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

-1/4*a*B*cot(d*x+c)^4/d+1/2*a*B*cot(d*x+c)^2/d+a*B*ln(sin(d*x+c))/d-1/3/d*a*C*cot(d*x+c)^3+1/d*C*cot(d*x+c)*a+
a*C*x+1/d*C*a*c-1/3/d*B*b*cot(d*x+c)^3+1/d*B*cot(d*x+c)*b+B*x*b+1/d*B*b*c-1/2/d*C*b*cot(d*x+c)^2-1/d*C*b*ln(si
n(d*x+c))

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maxima [A]  time = 0.93, size = 122, normalized size = 1.13 \[ \frac {12 \, {\left (C a + B b\right )} {\left (d x + c\right )} - 6 \, {\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, {\left (C a + B b\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (B a - C b\right )} \tan \left (d x + c\right )^{2} - 3 \, B a - 4 \, {\left (C a + B b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+b*tan(d*x+c))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(12*(C*a + B*b)*(d*x + c) - 6*(B*a - C*b)*log(tan(d*x + c)^2 + 1) + 12*(B*a - C*b)*log(tan(d*x + c)) + (1
2*(C*a + B*b)*tan(d*x + c)^3 + 6*(B*a - C*b)*tan(d*x + c)^2 - 3*B*a - 4*(C*a + B*b)*tan(d*x + c))/tan(d*x + c)
^4)/d

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mupad [B]  time = 8.82, size = 145, normalized size = 1.34 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a-C\,b\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\left (-B\,b-C\,a\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {C\,b}{2}-\frac {B\,a}{2}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {B\,b}{3}+\frac {C\,a}{3}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {B\,a}{4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(B*tan(c + d*x) + C*tan(c + d*x)^2)*(a + b*tan(c + d*x)),x)

[Out]

(log(tan(c + d*x))*(B*a - C*b))/d - (cot(c + d*x)^4*((B*a)/4 + tan(c + d*x)*((B*b)/3 + (C*a)/3) - tan(c + d*x)
^3*(B*b + C*a) - tan(c + d*x)^2*((B*a)/2 - (C*b)/2)))/d - (log(tan(c + d*x) - 1i)*(B + C*1i)*(a + b*1i))/(2*d)
+ (log(tan(c + d*x) + 1i)*(B - C*1i)*(a*1i + b)*1i)/(2*d)

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sympy [A]  time = 5.87, size = 211, normalized size = 1.95 \[ \begin {cases} \text {NaN} & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (a + b \tan {\relax (c )}\right ) \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \cot ^{6}{\relax (c )} & \text {for}\: d = 0 \\- \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {B a}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {B a}{4 d \tan ^{4}{\left (c + d x \right )}} + B b x + \frac {B b}{d \tan {\left (c + d x \right )}} - \frac {B b}{3 d \tan ^{3}{\left (c + d x \right )}} + C a x + \frac {C a}{d \tan {\left (c + d x \right )}} - \frac {C a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {C b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {C b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {C b}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(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(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(a + b*tan(c))*(B*tan(c) + C*tan(c)**
2)*cot(c)**6, Eq(d, 0)), (-B*a*log(tan(c + d*x)**2 + 1)/(2*d) + B*a*log(tan(c + d*x))/d + B*a/(2*d*tan(c + d*x
)**2) - B*a/(4*d*tan(c + d*x)**4) + B*b*x + B*b/(d*tan(c + d*x)) - B*b/(3*d*tan(c + d*x)**3) + C*a*x + C*a/(d*
tan(c + d*x)) - C*a/(3*d*tan(c + d*x)**3) + C*b*log(tan(c + d*x)**2 + 1)/(2*d) - C*b*log(tan(c + d*x))/d - C*b
/(2*d*tan(c + d*x)**2), True))

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