\(\int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx\) [278]

Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {a^2 \log (\cos (c+d x))}{d}+\frac {a (4 a+3 b) \log (1-\sec (c+d x))}{8 d}+\frac {a (4 a-3 b) \log (1+\sec (c+d x))}{8 d}+\frac {a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d}-\frac {\cot ^4(c+d x) \left (a^2+b^2+2 a b \sec (c+d x)\right )}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3970, 1819, 837, 815} \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {\cot ^4(c+d x) \left (a^2+2 a b \sec (c+d x)+b^2\right )}{4 d}+\frac {a^2 \log (\cos (c+d x))}{d}+\frac {a (4 a+3 b) \log (1-\sec (c+d x))}{8 d}+\frac {a (4 a-3 b) \log (\sec (c+d x)+1)}{8 d}+\frac {a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d} \]

[In]

Int[Cot[c + d*x]^5*(a + b*Sec[c + d*x])^2,x]

[Out]

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

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 3970

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*} \text {integral}& = -\frac {b^6 \text {Subst}\left (\int \frac {(a+x)^2}{x \left (b^2-x^2\right )^3} \, dx,x,b \sec (c+d x)\right )}{d} \\ & = -\frac {\cot ^4(c+d x) \left (a^2+b^2+2 a b \sec (c+d x)\right )}{4 d}+\frac {b^4 \text {Subst}\left (\int \frac {-4 a^2-6 a x}{x \left (b^2-x^2\right )^2} \, dx,x,b \sec (c+d x)\right )}{4 d} \\ & = \frac {a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d}-\frac {\cot ^4(c+d x) \left (a^2+b^2+2 a b \sec (c+d x)\right )}{4 d}+\frac {\text {Subst}\left (\int \frac {-8 a^2 b^2-6 a b^2 x}{x \left (b^2-x^2\right )} \, dx,x,b \sec (c+d x)\right )}{8 d} \\ & = \frac {a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d}-\frac {\cot ^4(c+d x) \left (a^2+b^2+2 a b \sec (c+d x)\right )}{4 d}+\frac {\text {Subst}\left (\int \left (-\frac {a (4 a+3 b)}{b-x}-\frac {8 a^2}{x}+\frac {a (4 a-3 b)}{b+x}\right ) \, dx,x,b \sec (c+d x)\right )}{8 d} \\ & = \frac {a^2 \log (\cos (c+d x))}{d}+\frac {a (4 a+3 b) \log (1-\sec (c+d x))}{8 d}+\frac {a (4 a-3 b) \log (1+\sec (c+d x))}{8 d}+\frac {a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d}-\frac {\cot ^4(c+d x) \left (a^2+b^2+2 a b \sec (c+d x)\right )}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.25 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.10 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {16 a^2 \log (\cos (c+d x))+2 a (4 a+3 b) \log (1-\sec (c+d x))+2 a (4 a-3 b) \log (1+\sec (c+d x))-\frac {(a+b)^2}{(-1+\sec (c+d x))^2}+\frac {(a+b) (5 a+b)}{-1+\sec (c+d x)}-\frac {(a-b)^2}{(1+\sec (c+d x))^2}-\frac {(a-b) (5 a-b)}{1+\sec (c+d x)}}{16 d} \]

[In]

Integrate[Cot[c + d*x]^5*(a + b*Sec[c + d*x])^2,x]

[Out]

(16*a^2*Log[Cos[c + d*x]] + 2*a*(4*a + 3*b)*Log[1 - Sec[c + d*x]] + 2*a*(4*a - 3*b)*Log[1 + Sec[c + d*x]] - (a
+ b)^2/(-1 + Sec[c + d*x])^2 + ((a + b)*(5*a + b))/(-1 + Sec[c + d*x]) - (a - b)^2/(1 + Sec[c + d*x])^2 - ((a
- b)*(5*a - b))/(1 + Sec[c + d*x]))/(16*d)

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )-\frac {b^{2} \cos \left (d x +c \right )^{4}}{4 \sin \left (d x +c \right )^{4}}}{d}\)	\(136\)
default	\(\frac {a^{2} \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )-\frac {b^{2} \cos \left (d x +c \right )^{4}}{4 \sin \left (d x +c \right )^{4}}}{d}\)	\(136\)
risch	\(-i a^{2} x -\frac {2 i a^{2} c}{d}-\frac {5 a b \,{\mathrm e}^{7 i \left (d x +c \right )}+8 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+4 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-8 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 a b \,{\mathrm e}^{3 i \left (d x +c \right )}+8 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5 a b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) a b}{4 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) a b}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(236\)

[In]

int(cot(d*x+c)^5*(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.61 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {10 \, a b \cos \left (d x + c\right )^{3} - 6 \, a b \cos \left (d x + c\right ) + 4 \, {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 6 \, a^{2} - 2 \, b^{2} - {\left ({\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} - 3 \, a b\right )} \cos \left (d x + c\right )^{2} + 4 \, a^{2} - 3 \, a b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 3 \, a b\right )} \cos \left (d x + c\right )^{2} + 4 \, a^{2} + 3 \, a b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]

[In]

integrate(cot(d*x+c)^5*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cot ^{5}{\left (c + d x \right )}\, dx \]

[In]

integrate(cot(d*x+c)**5*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*cot(c + d*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.97 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\left (4 \, a^{2} - 3 \, a b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (4 \, a^{2} + 3 \, a b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a b \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right ) + 2 \, {\left (2 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - b^{2}\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (118) = 236\).

Time = 0.37 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.86 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=-\frac {64 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) + \frac {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {16 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {2 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 8 \, {\left (4 \, a^{2} + 3 \, a b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) + \frac {{\left (a^{2} + 2 \, a b + b^{2} + \frac {12 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {16 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {4 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {48 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {36 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}}{64 \, d} \]

[In]

integrate(cot(d*x+c)^5*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/64*(64*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) + 12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) +
1) - 16*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + a^2*(cos(d*
x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 2*a*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b^2*(cos(d*x + c) - 1)^
2/(cos(d*x + c) + 1)^2 - 8*(4*a^2 + 3*a*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) + (a^2 + 2*a*b +
b^2 + 12*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 16*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 4*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 + 36*a*b*(cos(d*x + c) -
1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 14.34 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.30 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {5\,a^2}{32}-\frac {3\,a\,b}{16}+\frac {b^2}{32}+\frac {{\left (a-b\right )}^2}{32}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\left (a-b\right )}^2}{64\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+\frac {3\,b\,a}{4}\right )}{d}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a\,b}{2}+\frac {a^2}{4}+\frac {b^2}{4}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^2+4\,a\,b+b^2\right )\right )}{16\,d} \]

[In]

int(cot(c + d*x)^5*(a + b/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)^2*((5*a^2)/32 - (3*a*b)/16 + b^2/32 + (a - b)^2/32))/d - (tan(c/2 + (d*x)/2)^4*(a - b)^2)/
(64*d) + (log(tan(c/2 + (d*x)/2))*((3*a*b)/4 + a^2))/d - (a^2*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (cot(c/2 + (d
*x)/2)^4*((a*b)/2 + a^2/4 + b^2/4 - tan(c/2 + (d*x)/2)^2*(4*a*b + 3*a^2 + b^2)))/(16*d)