3.239 \(\int \cot ^5(c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {3591, 3529, 3531, 3475} \[ -\frac {(a B+A b) \cot ^3(c+d x)}{3 d}+\frac {(a A-b B) \cot ^2(c+d x)}{2 d}+\frac {(a B+A b) \cot (c+d x)}{d}+\frac {(a A-b B) \log (\sin (c+d x))}{d}+x (a B+A b)-\frac {a A \cot ^4(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(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 - b*B)*Cot[c + d*x]^2)/(2*d) - ((A*b + a*B)*Cot[c + d*x]^
3)/(3*d) - (a*A*Cot[c + d*x]^4)/(4*d) + ((a*A - b*B)*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]

Rubi steps

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

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Mathematica [C]  time = 1.17, size = 100, normalized size = 0.93 \[ -\frac {4 (a B+A b) \cot ^3(c+d x) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\tan ^2(c+d x)\right )+3 \left ((2 b B-2 a A) \cot ^2(c+d x)-4 (a A-b B) (\log (\tan (c+d x))+\log (\cos (c+d x)))+a A \cot ^4(c+d x)\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.59, size = 138, normalized size = 1.28 \[ \frac {6 \, {\left (A a - B b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{4} + 3 \, {\left (4 \, {\left (B a + A b\right )} d x + 3 \, A a - 2 \, B b\right )} \tan \left (d x + c\right )^{4} + 12 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (A a - B b\right )} \tan \left (d x + c\right )^{2} - 3 \, A a - 4 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{12 \, d \tan \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.33, size = 299, normalized size = 2.77 \[ -\frac {3 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 8 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 192 \, {\left (B a + A b\right )} {\left (d x + c\right )} + 192 \, {\left (A a - B b\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 192 \, {\left (A a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 400 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.41, size = 150, normalized size = 1.39 \[ -\frac {a A \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {a A \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a A \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a B \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {B \cot \left (d x +c \right ) a}{d}+a B x +\frac {B a c}{d}-\frac {A b \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}+\frac {A \cot \left (d x +c \right ) b}{d}+A x b +\frac {A b c}{d}-\frac {B b \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}-\frac {B b \ln \left (\sin \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 122, normalized size = 1.13 \[ \frac {12 \, {\left (B a + A b\right )} {\left (d x + c\right )} - 6 \, {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {12 \, {\left (B a + A b\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (A a - B b\right )} \tan \left (d x + c\right )^{2} - 3 \, A a - 4 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.39, size = 145, normalized size = 1.34 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,a-B\,b\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^4\,\left (\left (-A\,b-B\,a\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {B\,b}{2}-\frac {A\,a}{2}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^2+\left (\frac {A\,b}{3}+\frac {B\,a}{3}\right )\,\mathrm {tan}\left (c+d\,x\right )+\frac {A\,a}{4}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.61, size = 211, normalized size = 1.95 \[ \begin {cases} \tilde {\infty } A a x & \text {for}\: \left (c = 0 \vee c = - d x\right ) \wedge \left (c = - d x \vee d = 0\right ) \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right ) \cot ^{5}{\relax (c )} & \text {for}\: d = 0 \\- \frac {A a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {A a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + \frac {A a}{2 d \tan ^{2}{\left (c + d x \right )}} - \frac {A a}{4 d \tan ^{4}{\left (c + d x \right )}} + A b x + \frac {A b}{d \tan {\left (c + d x \right )}} - \frac {A b}{3 d \tan ^{3}{\left (c + d x \right )}} + B a x + \frac {B a}{d \tan {\left (c + d x \right )}} - \frac {B a}{3 d \tan ^{3}{\left (c + d x \right )}} + \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {B b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {B b}{2 d \tan ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*A*a*x, (Eq(c, 0) | Eq(c, -d*x)) & (Eq(d, 0) | Eq(c, -d*x))), (x*(A + B*tan(c))*(a + b*tan(c))*c
ot(c)**5, Eq(d, 0)), (-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*a/(4*d*tan(c + d*x)**4) + A*b*x + A*b/(d*tan(c + d*x)) - A*b/(3*d*tan(c + d*x)**3) + 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), True))

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