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 + 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

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Rubi [A]  time = 0.18726, antiderivative size = 108, normalized size of antiderivative = 1., 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 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 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 3475

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

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*}

Mathematica [C]  time = 1.16954, size = 100, normalized size = 0.93 \[ -\frac{4 (a B+A b) \cot ^3(c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},-\tan ^2(c+d x)\right )+3 \left ((2 b B-2 a A) \cot ^2(c+d x)-4 (a A-b B) (\log (\tan (c+d x))+\log (\cos (c+d x)))+a A \cot ^4(c+d x)\right )}{12 d} \]

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]

-(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)*Cot[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]])))/(12*d)

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Maple [A]  time = 0.066, size = 150, normalized size = 1.4 \begin{align*} -{\frac{Ab \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{A\cot \left ( dx+c \right ) b}{d}}+Axb+{\frac{Abc}{d}}-{\frac{Bb \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{Bb\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{Aa \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{A \left ( \cot \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}+{\frac{Aa\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{aB \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}+{\frac{B\cot \left ( dx+c \right ) a}{d}}+aBx+{\frac{aBc}{d}} \end{align*}

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]

-1/3/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(sin(d*x+c))-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

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Maxima [A]  time = 1.47806, size = 165, normalized size = 1.53 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 2.06044, size = 340, normalized size = 3.15 \begin{align*} \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}} \end{align*}

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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Sympy [A]  time = 8.45852, size = 211, normalized size = 1.95 \begin{align*} \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{\left (c \right )}\right ) \left (a + b \tan{\left (c \right )}\right ) \cot ^{5}{\left (c \right )} & \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} \end{align*}

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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Giac [B]  time = 1.36492, size = 404, normalized size = 3.74 \begin{align*} -\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} \end{align*}

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