3.305 \(\int \frac {\cot ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=197 \[ \frac {b^4}{a d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {b^4 \left (5 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 d \left (a^2-b^2\right )^3}-\frac {\log (\cos (c+d x))}{a^2 d}+\frac {1}{4 d (a+b)^2 (1-\sec (c+d x))}+\frac {1}{4 d (a-b)^2 (\sec (c+d x)+1)}-\frac {(a+2 b) \log (1-\sec (c+d x))}{2 d (a+b)^3}-\frac {(a-2 b) \log (\sec (c+d x)+1)}{2 d (a-b)^3} \]

[Out]

-ln(cos(d*x+c))/a^2/d-1/2*(a+2*b)*ln(1-sec(d*x+c))/(a+b)^3/d-1/2*(a-2*b)*ln(1+sec(d*x+c))/(a-b)^3/d-b^4*(5*a^2
-b^2)*ln(a+b*sec(d*x+c))/a^2/(a^2-b^2)^3/d+1/4/(a+b)^2/d/(1-sec(d*x+c))+1/4/(a-b)^2/d/(1+sec(d*x+c))+b^4/a/(a^
2-b^2)^2/d/(a+b*sec(d*x+c))

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Rubi [A]  time = 0.23, antiderivative size = 197, 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} \[ \frac {b^4}{a d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))}-\frac {b^4 \left (5 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 d \left (a^2-b^2\right )^3}-\frac {\log (\cos (c+d x))}{a^2 d}+\frac {1}{4 d (a+b)^2 (1-\sec (c+d x))}+\frac {1}{4 d (a-b)^2 (\sec (c+d x)+1)}-\frac {(a+2 b) \log (1-\sec (c+d x))}{2 d (a+b)^3}-\frac {(a-2 b) \log (\sec (c+d x)+1)}{2 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]

[Out]

-(Log[Cos[c + d*x]]/(a^2*d)) - ((a + 2*b)*Log[1 - Sec[c + d*x]])/(2*(a + b)^3*d) - ((a - 2*b)*Log[1 + Sec[c +
d*x]])/(2*(a - b)^3*d) - (b^4*(5*a^2 - b^2)*Log[a + b*Sec[c + d*x]])/(a^2*(a^2 - b^2)^3*d) + 1/(4*(a + b)^2*d*
(1 - Sec[c + d*x])) + 1/(4*(a - b)^2*d*(1 + Sec[c + d*x])) + b^4/(a*(a^2 - b^2)^2*d*(a + b*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 ^3(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac {b^4 \operatorname {Subst}\left (\int \frac {1}{x (a+x)^2 \left (b^2-x^2\right )^2} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac {b^4 \operatorname {Subst}\left (\int \left (\frac {1}{4 b^3 (a+b)^2 (b-x)^2}+\frac {a+2 b}{2 b^4 (a+b)^3 (b-x)}+\frac {1}{a^2 b^4 x}-\frac {1}{a (a-b)^2 (a+b)^2 (a+x)^2}+\frac {-5 a^2+b^2}{a^2 (a-b)^3 (a+b)^3 (a+x)}-\frac {1}{4 (a-b)^2 b^3 (b+x)^2}+\frac {-a+2 b}{2 (a-b)^3 b^4 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac {\log (\cos (c+d x))}{a^2 d}-\frac {(a+2 b) \log (1-\sec (c+d x))}{2 (a+b)^3 d}-\frac {(a-2 b) \log (1+\sec (c+d x))}{2 (a-b)^3 d}-\frac {b^4 \left (5 a^2-b^2\right ) \log (a+b \sec (c+d x))}{a^2 \left (a^2-b^2\right )^3 d}+\frac {1}{4 (a+b)^2 d (1-\sec (c+d x))}+\frac {1}{4 (a-b)^2 d (1+\sec (c+d x))}+\frac {b^4}{a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 2.27, size = 351, normalized size = 1.78 \[ \frac {\sec ^2(c+d x) (a \cos (c+d x)+b) \left (-\frac {8 b^5}{a^2 (a-b)^2 (a+b)^2}+\frac {8 b^4 \left (b^2-5 a^2\right ) (a \cos (c+d x)+b) \log (a \cos (c+d x)+b)}{a^2 \left (a^2-b^2\right )^3}-\frac {16 i \left (a^4-3 a^2 b^2-2 b^4\right ) (c+d x) (a \cos (c+d x)+b)}{(a-b)^3 (a+b)^3}+\frac {4 (a-2 b) \log \left (\cos ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{(b-a)^3}+\frac {8 i (a+2 b) \tan ^{-1}(\tan (c+d x)) (a \cos (c+d x)+b)}{(a+b)^3}+\frac {8 i (a-2 b) \tan ^{-1}(\tan (c+d x)) (a \cos (c+d x)+b)}{(a-b)^3}-\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a+b)^2}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{(a-b)^2}-\frac {4 (a+2 b) \log \left (\sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)}{(a+b)^3}\right )}{8 d (a+b \sec (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + b*Sec[c + d*x])^2,x]

[Out]

((b + a*Cos[c + d*x])*((-8*b^5)/(a^2*(a - b)^2*(a + b)^2) - ((16*I)*(a^4 - 3*a^2*b^2 - 2*b^4)*(c + d*x)*(b + a
*Cos[c + d*x]))/((a - b)^3*(a + b)^3) + ((8*I)*(a - 2*b)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x]))/(a - b)^3
+ ((8*I)*(a + 2*b)*ArcTan[Tan[c + d*x]]*(b + a*Cos[c + d*x]))/(a + b)^3 - ((b + a*Cos[c + d*x])*Csc[(c + d*x)/
2]^2)/(a + b)^2 + (4*(a - 2*b)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2]^2])/(-a + b)^3 + (8*b^4*(-5*a^2 + b^2
)*(b + a*Cos[c + d*x])*Log[b + a*Cos[c + d*x]])/(a^2*(a^2 - b^2)^3) - (4*(a + 2*b)*(b + a*Cos[c + d*x])*Log[Si
n[(c + d*x)/2]^2])/(a + b)^3 - ((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a - b)^2)*Sec[c + d*x]^2)/(8*d*(a +
b*Sec[c + d*x])^2)

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fricas [B]  time = 0.77, size = 693, normalized size = 3.52 \[ \frac {a^{6} b + a^{2} b^{5} - 2 \, b^{7} - 2 \, {\left (a^{6} b - a^{4} b^{3} + a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right ) + 2 \, {\left (5 \, a^{2} b^{5} - b^{7} - {\left (5 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )^{3} - {\left (5 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, a^{3} b^{4} - a b^{6}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) + {\left (a^{6} b + a^{5} b^{2} - 3 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5} - {\left (a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{6} b + a^{5} b^{2} - 3 \, a^{4} b^{3} - 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} + a^{6} b - 3 \, a^{5} b^{2} - 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{6} b - a^{5} b^{2} - 3 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5} - {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{6} b - a^{5} b^{2} - 3 \, a^{4} b^{3} + 5 \, a^{3} b^{4} - 2 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} - a^{6} b - 3 \, a^{5} b^{2} + 5 \, a^{4} b^{3} - 2 \, a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left ({\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} d \cos \left (d x + c\right )^{3} + {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} d \cos \left (d x + c\right ) - {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(a^6*b + a^2*b^5 - 2*b^7 - 2*(a^6*b - a^4*b^3 + a^2*b^5 - b^7)*cos(d*x + c)^2 + (a^7 - 2*a^5*b^2 + a^3*b^4
)*cos(d*x + c) + 2*(5*a^2*b^5 - b^7 - (5*a^3*b^4 - a*b^6)*cos(d*x + c)^3 - (5*a^2*b^5 - b^7)*cos(d*x + c)^2 +
(5*a^3*b^4 - a*b^6)*cos(d*x + c))*log(a*cos(d*x + c) + b) + (a^6*b + a^5*b^2 - 3*a^4*b^3 - 5*a^3*b^4 - 2*a^2*b
^5 - (a^7 + a^6*b - 3*a^5*b^2 - 5*a^4*b^3 - 2*a^3*b^4)*cos(d*x + c)^3 - (a^6*b + a^5*b^2 - 3*a^4*b^3 - 5*a^3*b
^4 - 2*a^2*b^5)*cos(d*x + c)^2 + (a^7 + a^6*b - 3*a^5*b^2 - 5*a^4*b^3 - 2*a^3*b^4)*cos(d*x + c))*log(1/2*cos(d
*x + c) + 1/2) + (a^6*b - a^5*b^2 - 3*a^4*b^3 + 5*a^3*b^4 - 2*a^2*b^5 - (a^7 - a^6*b - 3*a^5*b^2 + 5*a^4*b^3 -
2*a^3*b^4)*cos(d*x + c)^3 - (a^6*b - a^5*b^2 - 3*a^4*b^3 + 5*a^3*b^4 - 2*a^2*b^5)*cos(d*x + c)^2 + (a^7 - a^6
*b - 3*a^5*b^2 + 5*a^4*b^3 - 2*a^3*b^4)*cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/((a^9 - 3*a^7*b^2 + 3*a^5*
b^4 - a^3*b^6)*d*cos(d*x + c)^3 + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d*cos(d*x + c)^2 - (a^9 - 3*a^7*b^
2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c) - (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d)

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giac [B]  time = 1.93, size = 656, normalized size = 3.33 \[ -\frac {\frac {4 \, {\left (a + 2 \, b\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 {8 \, {\left (5 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | -a - b - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}} - \frac {a^{5} - a^{4} b - a^{3} b^{2} + a^{2} b^{3} + \frac {3 \, a^{5} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, a^{4} b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, a^{3} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {20 \, a b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4 \, b^{5} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{5} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, a^{4} b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a^{3} b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {12 \, a^{2} b^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a b^{4} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, b^{5} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (\frac {a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {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 )}} - \frac {\cos \left (d x + c\right ) - 1}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {8 \, \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right )}{a^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/8*(4*(a + 2*b)*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) + 8*(5*a^2
*b^4 - b^6)*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
))/(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6) - (a^5 - a^4*b - a^3*b^2 + a^2*b^3 + 3*a^5*(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - 3*a^4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*a^3*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1
) + 3*a^2*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 20*a*b^4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4*b^5*(
cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*a^5*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 4*a^4*b*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 - 2*a^3*b^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 12*a^2*b^3*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2 + 4*a*b^4*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 4*b^5*(cos(d*x + c) - 1)^2/(
cos(d*x + c) + 1)^2)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 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)) - (cos(d*x + c) - 1)/((a^2 - 2*a*b + b^2)*(cos(d*x + c) + 1)) - 8*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) + 1))/a^2)/d

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maple [A]  time = 0.72, size = 226, normalized size = 1.15 \[ -\frac {b^{5}}{d \,a^{2} \left (a +b \right )^{2} \left (a -b \right )^{2} \left (b +a \cos \left (d x +c \right )\right )}-\frac {5 b^{4} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3}}+\frac {b^{6} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \left (a +b \right )^{3} \left (a -b \right )^{3} a^{2}}+\frac {1}{4 d \left (a +b \right )^{2} \left (-1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) a}{2 d \left (a +b \right )^{3}}-\frac {\ln \left (-1+\cos \left (d x +c \right )\right ) b}{d \left (a +b \right )^{3}}-\frac {1}{4 d \left (a -b \right )^{2} \left (1+\cos \left (d x +c \right )\right )}-\frac {\ln \left (1+\cos \left (d x +c \right )\right ) a}{2 d \left (a -b \right )^{3}}+\frac {\ln \left (1+\cos \left (d x +c \right )\right ) b}{d \left (a -b \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x)

[Out]

-1/d/a^2*b^5/(a+b)^2/(a-b)^2/(b+a*cos(d*x+c))-5/d*b^4/(a+b)^3/(a-b)^3*ln(b+a*cos(d*x+c))+1/d*b^6/(a+b)^3/(a-b)
^3/a^2*ln(b+a*cos(d*x+c))+1/4/d/(a+b)^2/(-1+cos(d*x+c))-1/2/d/(a+b)^3*ln(-1+cos(d*x+c))*a-1/d/(a+b)^3*ln(-1+co
s(d*x+c))*b-1/4/d/(a-b)^2/(1+cos(d*x+c))-1/2/d/(a-b)^3*ln(1+cos(d*x+c))*a+1/d/(a-b)^3*ln(1+cos(d*x+c))*b

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maxima [A]  time = 0.54, size = 303, normalized size = 1.54 \[ -\frac {\frac {2 \, {\left (5 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}} + \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (a + 2 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {a^{4} b + a^{2} b^{3} + 2 \, b^{5} - 2 \, {\left (a^{4} b + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - a^{3} b^{2}\right )} \cos \left (d x + c\right )}{a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(2*(5*a^2*b^4 - b^6)*log(a*cos(d*x + c) + b)/(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6) + (a - 2*b)*log(cos(
d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (a + 2*b)*log(cos(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3
) + (a^4*b + a^2*b^3 + 2*b^5 - 2*(a^4*b + b^5)*cos(d*x + c)^2 + (a^5 - a^3*b^2)*cos(d*x + c))/(a^6*b - 2*a^4*b
^3 + a^2*b^5 - (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)^3 - (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c)^2 + (a^
7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)))/d

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mupad [B]  time = 2.48, size = 313, normalized size = 1.59 \[ \frac {\frac {a^2-2\,a\,b+b^2}{2\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^5-4\,a^4\,b+6\,a^3\,b^2-4\,a^2\,b^3+a\,b^4-16\,b^5\right )}{2\,a\,{\left (a+b\right )}^2\,\left (a-b\right )}}{d\,\left (\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-4\,a^3+4\,a^2\,b+4\,a\,b^2-4\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d\,{\left (a-b\right )}^2}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+2\,b\right )}{d\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}-\frac {b^4\,\ln \left (a+b-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )\,\left (5\,a^2-b^2\right )}{a^2\,d\,{\left (a^2-b^2\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3/(a + b/cos(c + d*x))^2,x)

[Out]

((a^2 - 2*a*b + b^2)/(2*(a + b)) - (tan(c/2 + (d*x)/2)^2*(a*b^4 - 4*a^4*b + a^5 - 16*b^5 - 4*a^2*b^3 + 6*a^3*b
^2))/(2*a*(a + b)^2*(a - b)))/(d*(tan(c/2 + (d*x)/2)^2*(4*a*b^2 + 4*a^2*b - 4*a^3 - 4*b^3) + tan(c/2 + (d*x)/2
)^4*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3))) - tan(c/2 + (d*x)/2)^2/(8*d*(a - b)^2) + log(tan(c/2 + (d*x)/2)^2
+ 1)/(a^2*d) - (log(tan(c/2 + (d*x)/2))*(a + 2*b))/(d*(3*a*b^2 + 3*a^2*b + a^3 + b^3)) - (b^4*log(a + b - a*ta
n(c/2 + (d*x)/2)^2 + b*tan(c/2 + (d*x)/2)^2)*(5*a^2 - b^2))/(a^2*d*(a^2 - b^2)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+b*sec(d*x+c))**2,x)

[Out]

Integral(cot(c + d*x)**3/(a + b*sec(c + d*x))**2, x)

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