3.293 \(\int \frac {\cot ^5(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=234 \[ \frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sec (c+d x))}{16 d (a+b)^3}+\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (\sec (c+d x)+1)}{16 d (a-b)^3}-\frac {b^6 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )^3}-\frac {5 a+7 b}{16 d (a+b)^2 (1-\sec (c+d x))}-\frac {5 a-7 b}{16 d (a-b)^2 (\sec (c+d x)+1)}-\frac {1}{16 d (a+b) (1-\sec (c+d x))^2}-\frac {1}{16 d (a-b) (\sec (c+d x)+1)^2}+\frac {\log (\cos (c+d x))}{a d} \]

[Out]

ln(cos(d*x+c))/a/d+1/16*(8*a^2+21*a*b+15*b^2)*ln(1-sec(d*x+c))/(a+b)^3/d+1/16*(8*a^2-21*a*b+15*b^2)*ln(1+sec(d
*x+c))/(a-b)^3/d-b^6*ln(a+b*sec(d*x+c))/a/(a^2-b^2)^3/d-1/16/(a+b)/d/(1-sec(d*x+c))^2+1/16*(-5*a-7*b)/(a+b)^2/
d/(1-sec(d*x+c))-1/16/(a-b)/d/(1+sec(d*x+c))^2+1/16*(-5*a+7*b)/(a-b)^2/d/(1+sec(d*x+c))

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Rubi [A]  time = 0.29, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3885, 894} \[ -\frac {b^6 \log (a+b \sec (c+d x))}{a d \left (a^2-b^2\right )^3}+\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sec (c+d x))}{16 d (a+b)^3}+\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (\sec (c+d x)+1)}{16 d (a-b)^3}-\frac {5 a+7 b}{16 d (a+b)^2 (1-\sec (c+d x))}-\frac {5 a-7 b}{16 d (a-b)^2 (\sec (c+d x)+1)}-\frac {1}{16 d (a+b) (1-\sec (c+d x))^2}-\frac {1}{16 d (a-b) (\sec (c+d x)+1)^2}+\frac {\log (\cos (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Cos[c + d*x]]/(a*d) + ((8*a^2 + 21*a*b + 15*b^2)*Log[1 - Sec[c + d*x]])/(16*(a + b)^3*d) + ((8*a^2 - 21*a*
b + 15*b^2)*Log[1 + Sec[c + d*x]])/(16*(a - b)^3*d) - (b^6*Log[a + b*Sec[c + d*x]])/(a*(a^2 - b^2)^3*d) - 1/(1
6*(a + b)*d*(1 - Sec[c + d*x])^2) - (5*a + 7*b)/(16*(a + b)^2*d*(1 - Sec[c + d*x])) - 1/(16*(a - b)*d*(1 + Sec
[c + d*x])^2) - (5*a - 7*b)/(16*(a - b)^2*d*(1 + Sec[c + d*x]))

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cot ^5(c+d x)}{a+b \sec (c+d x)} \, dx &=-\frac {b^6 \operatorname {Subst}\left (\int \frac {1}{x (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac {b^6 \operatorname {Subst}\left (\int \left (\frac {1}{8 b^4 (a+b) (b-x)^3}+\frac {5 a+7 b}{16 b^5 (a+b)^2 (b-x)^2}+\frac {8 a^2+21 a b+15 b^2}{16 b^6 (a+b)^3 (b-x)}+\frac {1}{a b^6 x}+\frac {1}{a (a-b)^3 (a+b)^3 (a+x)}+\frac {1}{8 b^4 (-a+b) (b+x)^3}+\frac {-5 a+7 b}{16 (a-b)^2 b^5 (b+x)^2}+\frac {8 a^2-21 a b+15 b^2}{16 b^6 (-a+b)^3 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac {\log (\cos (c+d x))}{a d}+\frac {\left (8 a^2+21 a b+15 b^2\right ) \log (1-\sec (c+d x))}{16 (a+b)^3 d}+\frac {\left (8 a^2-21 a b+15 b^2\right ) \log (1+\sec (c+d x))}{16 (a-b)^3 d}-\frac {b^6 \log (a+b \sec (c+d x))}{a \left (a^2-b^2\right )^3 d}-\frac {1}{16 (a+b) d (1-\sec (c+d x))^2}-\frac {5 a+7 b}{16 (a+b)^2 d (1-\sec (c+d x))}-\frac {1}{16 (a-b) d (1+\sec (c+d x))^2}-\frac {5 a-7 b}{16 (a-b)^2 d (1+\sec (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 6.25, size = 625, normalized size = 2.67 \[ \frac {\left (-8 a^2+21 a b-15 b^2\right ) \sec (c+d x) \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{16 d (b-a)^3 (a+b \sec (c+d x))}-\frac {i \left (-8 a^2+21 a b-15 b^2\right ) \tan ^{-1}(\tan (c+d x)) \sec (c+d x) (a \cos (c+d x)+b)}{8 d (b-a)^3 (a+b \sec (c+d x))}-\frac {i \left (8 a^2+21 a b+15 b^2\right ) \tan ^{-1}(\tan (c+d x)) \sec (c+d x) (a \cos (c+d x)+b)}{8 d (a+b)^3 (a+b \sec (c+d x))}+\frac {\left (8 a^2+21 a b+15 b^2\right ) \sec (c+d x) \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{16 d (a+b)^3 (a+b \sec (c+d x))}+\frac {b^6 \sec (c+d x) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{a d \left (b^2-a^2\right )^3 (a+b \sec (c+d x))}+\frac {2 i \left (a^5-3 a^3 b^2+3 a b^4\right ) (c+d x) \sec (c+d x) (a \cos (c+d x)+b)}{d (a-b)^3 (a+b)^3 (a+b \sec (c+d x))}+\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{64 d (b-a) (a+b \sec (c+d x))}+\frac {(7 a-9 b) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{32 d (b-a)^2 (a+b \sec (c+d x))}-\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{64 d (a+b) (a+b \sec (c+d x))}+\frac {(7 a+9 b) \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (a \cos (c+d x)+b)}{32 d (a+b)^2 (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*(a^5 - 3*a^3*b^2 + 3*a*b^4)*(c + d*x)*(b + a*Cos[c + d*x])*Sec[c + d*x])/((a - b)^3*(a + b)^3*d*(a + b*
Sec[c + d*x])) - ((I/8)*(-8*a^2 + 21*a*b - 15*b^2)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x])*Sec[c + d*x])/((-
a + b)^3*d*(a + b*Sec[c + d*x])) - ((I/8)*(8*a^2 + 21*a*b + 15*b^2)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x])*
Sec[c + d*x])/((a + b)^3*d*(a + b*Sec[c + d*x])) + ((7*a + 9*b)*(b + a*Cos[c + d*x])*Csc[(c + d*x)/2]^2*Sec[c
+ d*x])/(32*(a + b)^2*d*(a + b*Sec[c + d*x])) - ((b + a*Cos[c + d*x])*Csc[(c + d*x)/2]^4*Sec[c + d*x])/(64*(a
+ b)*d*(a + b*Sec[c + d*x])) + ((-8*a^2 + 21*a*b - 15*b^2)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2]^2]*Sec[c
+ d*x])/(16*(-a + b)^3*d*(a + b*Sec[c + d*x])) + (b^6*(b + a*Cos[c + d*x])*Log[b + a*Cos[c + d*x]]*Sec[c + d*x
])/(a*(-a^2 + b^2)^3*d*(a + b*Sec[c + d*x])) + ((8*a^2 + 21*a*b + 15*b^2)*(b + a*Cos[c + d*x])*Log[Sin[(c + d*
x)/2]^2]*Sec[c + d*x])/(16*(a + b)^3*d*(a + b*Sec[c + d*x])) + ((7*a - 9*b)*(b + a*Cos[c + d*x])*Sec[(c + d*x)
/2]^2*Sec[c + d*x])/(32*(-a + b)^2*d*(a + b*Sec[c + d*x])) + ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^4*Sec[c +
d*x])/(64*(-a + b)*d*(a + b*Sec[c + d*x]))

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fricas [B]  time = 0.77, size = 576, normalized size = 2.46 \[ \frac {12 \, a^{6} - 32 \, a^{4} b^{2} + 20 \, a^{2} b^{4} + 2 \, {\left (5 \, a^{5} b - 14 \, a^{3} b^{3} + 9 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} - 8 \, {\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (3 \, a^{5} b - 10 \, a^{3} b^{3} + 7 \, a b^{5}\right )} \cos \left (d x + c\right ) - 16 \, {\left (b^{6} \cos \left (d x + c\right )^{4} - 2 \, b^{6} \cos \left (d x + c\right )^{2} + b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5} + {\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a^{6} + 3 \, a^{5} b - 24 \, a^{4} b^{2} - 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5} + {\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a^{6} - 3 \, a^{5} b - 24 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 24 \, a^{2} b^{4} - 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{16 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d\right )}} \]

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*(12*a^6 - 32*a^4*b^2 + 20*a^2*b^4 + 2*(5*a^5*b - 14*a^3*b^3 + 9*a*b^5)*cos(d*x + c)^3 - 8*(2*a^6 - 5*a^4*
b^2 + 3*a^2*b^4)*cos(d*x + c)^2 - 2*(3*a^5*b - 10*a^3*b^3 + 7*a*b^5)*cos(d*x + c) - 16*(b^6*cos(d*x + c)^4 - 2
*b^6*cos(d*x + c)^2 + b^6)*log(a*cos(d*x + c) + b) + (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 +
15*a*b^5 + (8*a^6 + 3*a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 + 15*a*b^5)*cos(d*x + c)^4 - 2*(8*a^6 + 3*
a^5*b - 24*a^4*b^2 - 10*a^3*b^3 + 24*a^2*b^4 + 15*a*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) + (8*a^6
- 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b^4 - 15*a*b^5 + (8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*
a^2*b^4 - 15*a*b^5)*cos(d*x + c)^4 - 2*(8*a^6 - 3*a^5*b - 24*a^4*b^2 + 10*a^3*b^3 + 24*a^2*b^4 - 15*a*b^5)*cos
(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2))/((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^4 - 2*(a^7 -
3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)^2 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d)

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giac [B]  time = 1.14, size = 649, normalized size = 2.77 \[ \frac {\frac {4 \, {\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {32 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left ({\left | a + b - \frac {2 \, 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}} \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {\frac {12 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {16 \, 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}}}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a^{2} + 2 \, a b + b^{2} + \frac {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {28 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {16 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {48 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {126 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {90 \, b^{2} {\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 (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {32 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - 2 \, b^{6}\right )} \log \left (\frac {{\left | 2 \, b + \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 2 \, {\left | a \right |} \right |}}{{\left | 2 \, b + \frac {2 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {2 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + 2 \, {\left | a \right |} \right |}}\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left | a \right |}}}{64 \, d} \]

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^2 + 21*a*b + 15*b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^3 + 3*a^2*b + 3*a*b^2 +
b^3) - 32*(a^5 - 3*a^3*b^2 + 3*a*b^4)*log(abs(a + b - 2*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))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4
- b^6) - (12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 16*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)/(a^2 - 2*a*b + b^2) - (a^2 +
2*a*b + b^2 + 12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 28*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 16
*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 48*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 126*a*b*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 90*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(cos(d*x + c) - 1)^2) - 32*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - 2*b^6)*log(abs(2*b +
2*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*abs(a))/abs(2*b + 2
*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*abs(a)))/((a^6 - 3*a^
4*b^2 + 3*a^2*b^4 - b^6)*abs(a)))/d

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maple [A]  time = 0.68, size = 308, normalized size = 1.32 \[ -\frac {b^{6} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3} a}-\frac {1}{2 d \left (8 a +8 b \right ) \left (-1+\cos \left (d x +c \right )\right )^{2}}-\frac {7 a}{16 d \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}-\frac {9 b}{16 d \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) a^{2}}{2 d \left (a +b \right )^{3}}+\frac {21 \ln \left (-1+\cos \left (d x +c \right )\right ) a b}{16 d \left (a +b \right )^{3}}+\frac {15 \ln \left (-1+\cos \left (d x +c \right )\right ) b^{2}}{16 d \left (a +b \right )^{3}}-\frac {1}{2 d \left (8 a -8 b \right ) \left (1+\cos \left (d x +c \right )\right )^{2}}+\frac {7 a}{16 d \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {9 b}{16 d \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}+\frac {\ln \left (1+\cos \left (d x +c \right )\right ) a^{2}}{2 d \left (a -b \right )^{3}}-\frac {21 \ln \left (1+\cos \left (d x +c \right )\right ) a b}{16 d \left (a -b \right )^{3}}+\frac {15 \ln \left (1+\cos \left (d x +c \right )\right ) b^{2}}{16 d \left (a -b \right )^{3}} \]

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/d*b^6/(a+b)^3/(a-b)^3/a*ln(b+a*cos(d*x+c))-1/2/d/(8*a+8*b)/(-1+cos(d*x+c))^2-7/16/d/(a+b)^2/(-1+cos(d*x+c))
*a-9/16/d/(a+b)^2/(-1+cos(d*x+c))*b+1/2/d/(a+b)^3*ln(-1+cos(d*x+c))*a^2+21/16/d/(a+b)^3*ln(-1+cos(d*x+c))*a*b+
15/16/d/(a+b)^3*ln(-1+cos(d*x+c))*b^2-1/2/d/(8*a-8*b)/(1+cos(d*x+c))^2+7/16/d/(a-b)^2/(1+cos(d*x+c))*a-9/16/d/
(a-b)^2/(1+cos(d*x+c))*b+1/2/d/(a-b)^3*ln(1+cos(d*x+c))*a^2-21/16/d/(a-b)^3*ln(1+cos(d*x+c))*a*b+15/16/d/(a-b)
^3*ln(1+cos(d*x+c))*b^2

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maxima [A]  time = 0.52, size = 289, normalized size = 1.24 \[ -\frac {\frac {16 \, b^{6} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}} - \frac {{\left (8 \, a^{2} - 21 \, a b + 15 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (8 \, a^{2} + 21 \, a b + 15 \, b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (5 \, a^{2} b - 9 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \, a^{3} - 10 \, a b^{2} - 4 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} - {\left (3 \, a^{2} b - 7 \, b^{3}\right )} \cos \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}}}{16 \, d} \]

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*(16*b^6*log(a*cos(d*x + c) + b)/(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6) - (8*a^2 - 21*a*b + 15*b^2)*log(co
s(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - (8*a^2 + 21*a*b + 15*b^2)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2
*b + 3*a*b^2 + b^3) - 2*((5*a^2*b - 9*b^3)*cos(d*x + c)^3 + 6*a^3 - 10*a*b^2 - 4*(2*a^3 - 3*a*b^2)*cos(d*x + c
)^2 - (3*a^2*b - 7*b^3)*cos(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4
- 2*a^2*b^2 + b^4)*cos(d*x + c)^2))/d

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mupad [B]  time = 2.27, size = 290, normalized size = 1.24 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (8\,a^2+21\,a\,b+15\,b^2\right )}{d\,\left (8\,a^3+24\,a^2\,b+24\,a\,b^2+8\,b^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4\,d\,\left (16\,a-16\,b\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {16\,b}{{\left (16\,a-16\,b\right )}^2}-\frac {3}{16\,a-16\,b}\right )}{d}-\frac {\frac {a^2-2\,a\,b+b^2}{4\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3+2\,a^2\,b+5\,a\,b^2-4\,b^3\right )}{{\left (a+b\right )}^2}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (16\,a^2-32\,a\,b+16\,b^2\right )}-\frac {b^6\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{a\,d\,{\left (a^2-b^2\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(tan(c/2 + (d*x)/2))*(21*a*b + 8*a^2 + 15*b^2))/(d*(24*a*b^2 + 24*a^2*b + 8*a^3 + 8*b^3)) - tan(c/2 + (d*x
)/2)^4/(4*d*(16*a - 16*b)) - log(tan(c/2 + (d*x)/2)^2 + 1)/(a*d) - (tan(c/2 + (d*x)/2)^2*((16*b)/(16*a - 16*b)
^2 - 3/(16*a - 16*b)))/d - ((a^2 - 2*a*b + b^2)/(4*(a + b)) + (tan(c/2 + (d*x)/2)^2*(5*a*b^2 + 2*a^2*b - 3*a^3
- 4*b^3))/(a + b)^2)/(d*tan(c/2 + (d*x)/2)^4*(16*a^2 - 32*a*b + 16*b^2)) - (b^6*log(a + b - a*tan(c/2 + (d*x)
/2)^2 + b*tan(c/2 + (d*x)/2)^2))/(a*d*(a^2 - b^2)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]

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]

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

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