3.262 \(\int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx\)

Optimal. Leaf size=102 \[ \frac{(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac{(8 a-3 b) \log (\cos (c+d x)+1)}{16 d}-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d} \]

[Out]

((8*a + 3*b)*Log[1 - Cos[c + d*x]])/(16*d) + ((8*a - 3*b)*Log[1 + Cos[c + d*x]])/(16*d) - (Cot[c + d*x]^4*(a +
b*Sec[c + d*x]))/(4*d) + (Cot[c + d*x]^2*(4*a + 3*b*Sec[c + d*x]))/(8*d)

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Rubi [A]  time = 0.131437, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3882, 3883, 2668, 633, 31} \[ \frac{(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac{(8 a-3 b) \log (\cos (c+d x)+1)}{16 d}-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + b*Sec[c + d*x]),x]

[Out]

((8*a + 3*b)*Log[1 - Cos[c + d*x]])/(16*d) + ((8*a - 3*b)*Log[1 + Cos[c + d*x]])/(16*d) - (Cot[c + d*x]^4*(a +
b*Sec[c + d*x]))/(4*d) + (Cot[c + d*x]^2*(4*a + 3*b*Sec[c + d*x]))/(8*d)

Rule 3882

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[((e*Cot[c
+ d*x])^(m + 1)*(a + b*Csc[c + d*x]))/(d*e*(m + 1)), x] - Dist[1/(e^2*(m + 1)), Int[(e*Cot[c + d*x])^(m + 2)*(
a*(m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[m, -1]

Rule 3883

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))/cot[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[(b + a*Sin[c + d*x])/Cos[
c + d*x], x] /; FreeQ[{a, b, c, d}, x]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx &=-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{1}{4} \int \cot ^3(c+d x) (-4 a-3 b \sec (c+d x)) \, dx\\ &=-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}+\frac{1}{8} \int \cot (c+d x) (8 a+3 b \sec (c+d x)) \, dx\\ &=-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}+\frac{1}{8} \int (3 b+8 a \cos (c+d x)) \csc (c+d x) \, dx\\ &=-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}-\frac{a \operatorname{Subst}\left (\int \frac{3 b+x}{64 a^2-x^2} \, dx,x,8 a \cos (c+d x)\right )}{d}\\ &=-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}-\frac{(8 a-3 b) \operatorname{Subst}\left (\int \frac{1}{-8 a-x} \, dx,x,8 a \cos (c+d x)\right )}{16 d}-\frac{(8 a+3 b) \operatorname{Subst}\left (\int \frac{1}{8 a-x} \, dx,x,8 a \cos (c+d x)\right )}{16 d}\\ &=\frac{(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac{(8 a-3 b) \log (1+\cos (c+d x))}{16 d}-\frac{\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac{\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.3295, size = 166, normalized size = 1.63 \[ \frac{a \left (-\cot ^4(c+d x)+2 \cot ^2(c+d x)+4 \log (\tan (c+d x))+4 \log (\cos (c+d x))\right )}{4 d}-\frac{b \csc ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}+\frac{5 b \csc ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{b \sec ^4\left (\frac{1}{2} (c+d x)\right )}{64 d}-\frac{5 b \sec ^2\left (\frac{1}{2} (c+d x)\right )}{32 d}+\frac{3 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}-\frac{3 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*Csc[(c + d*x)/2]^2)/(32*d) - (b*Csc[(c + d*x)/2]^4)/(64*d) - (3*b*Log[Cos[(c + d*x)/2]])/(8*d) + (3*b*Log
[Sin[(c + d*x)/2]])/(8*d) + (a*(2*Cot[c + d*x]^2 - Cot[c + d*x]^4 + 4*Log[Cos[c + d*x]] + 4*Log[Tan[c + d*x]])
)/(4*d) - (5*b*Sec[(c + d*x)/2]^2)/(32*d) + (b*Sec[(c + d*x)/2]^4)/(64*d)

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Maple [A]  time = 0.044, size = 134, normalized size = 1.3 \begin{align*} -{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{ \left ( \cot \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}+{\frac{a\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{8\,d}}+{\frac{3\,b\cos \left ( dx+c \right ) }{8\,d}}+{\frac{3\,b\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+b*sec(d*x+c)),x)

[Out]

-1/4/d*a*cot(d*x+c)^4+1/2/d*a*cot(d*x+c)^2+1/d*a*ln(sin(d*x+c))-1/4/d*b/sin(d*x+c)^4*cos(d*x+c)^5+1/8/d*b/sin(
d*x+c)^2*cos(d*x+c)^5+1/8/d*b*cos(d*x+c)^3+3/8/d*b*cos(d*x+c)+3/8/d*b*ln(csc(d*x+c)-cot(d*x+c))

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Maxima [A]  time = 0.981967, size = 134, normalized size = 1.31 \begin{align*} \frac{{\left (8 \, a - 3 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) +{\left (8 \, a + 3 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (5 \, b \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} - 3 \, b \cos \left (d x + c\right ) - 6 \, a\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((8*a - 3*b)*log(cos(d*x + c) + 1) + (8*a + 3*b)*log(cos(d*x + c) - 1) - 2*(5*b*cos(d*x + c)^3 + 8*a*cos(
d*x + c)^2 - 3*b*cos(d*x + c) - 6*a)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))/d

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Fricas [A]  time = 0.834091, size = 440, normalized size = 4.31 \begin{align*} -\frac{10 \, b \cos \left (d x + c\right )^{3} + 16 \, a \cos \left (d x + c\right )^{2} - 6 \, b \cos \left (d x + c\right ) -{\left ({\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} + 8 \, a - 3 \, b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left ({\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{2} + 8 \, a + 3 \, b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 12 \, a}{16 \,{\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(10*b*cos(d*x + c)^3 + 16*a*cos(d*x + c)^2 - 6*b*cos(d*x + c) - ((8*a - 3*b)*cos(d*x + c)^4 - 2*(8*a - 3
*b)*cos(d*x + c)^2 + 8*a - 3*b)*log(1/2*cos(d*x + c) + 1/2) - ((8*a + 3*b)*cos(d*x + c)^4 - 2*(8*a + 3*b)*cos(
d*x + c)^2 + 8*a + 3*b)*log(-1/2*cos(d*x + c) + 1/2) - 12*a)/(d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^2 + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+b*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.40751, size = 359, normalized size = 3.52 \begin{align*} \frac{4 \,{\left (8 \, a + 3 \, b\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 64 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac{{\left (a + b + \frac{12 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{8 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{48 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{18 \, b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac{12 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{8 \, b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(4*(8*a + 3*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 64*a*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) + 1)) - (a + b + 12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b*(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) + 48*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(co
s(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 - 12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b*(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)
/d