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

Optimal. Leaf size=126 \[ -\frac{\cot ^4(c+d x) \left (a^2+2 a b \sec (c+d x)+b^2\right )}{4 d}+\frac{a^2 \log (\cos (c+d x))}{d}+\frac{a (4 a+3 b) \log (1-\sec (c+d x))}{8 d}+\frac{a (4 a-3 b) \log (\sec (c+d x)+1)}{8 d}+\frac{a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d} \]

[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)

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Rubi [A]  time = 0.159482, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3885, 1805, 823, 801} \[ -\frac{\cot ^4(c+d x) \left (a^2+2 a b \sec (c+d x)+b^2\right )}{4 d}+\frac{a^2 \log (\cos (c+d x))}{d}+\frac{a (4 a+3 b) \log (1-\sec (c+d x))}{8 d}+\frac{a (4 a-3 b) \log (\sec (c+d x)+1)}{8 d}+\frac{a \cot ^2(c+d x) (2 a+3 b \sec (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[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 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]

Rule 1805

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 823

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] + Di
st[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 801

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]

Rubi steps

\begin{align*} \int \cot ^5(c+d x) (a+b \sec (c+d x))^2 \, dx &=-\frac{b^6 \operatorname{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 \operatorname{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{\operatorname{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{\operatorname{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]  time = 3.22511, size = 148, normalized size = 1.17 \[ \frac{2 \left (7 a^2+10 a b+3 b^2\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )+2 \left (7 a^2-10 a b+3 b^2\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )-(a+b)^2 \csc ^4\left (\frac{1}{2} (c+d x)\right )-(a-b)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right )+16 a \left ((4 a+3 b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+(4 a-3 b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7*a^2 + 10*a*b + 3*b^2)*Csc[(c + d*x)/2]^2 - (a + b)^2*Csc[(c + d*x)/2]^4 + 16*a*((4*a - 3*b)*Log[Cos[(c +
d*x)/2]] + (4*a + 3*b)*Log[Sin[(c + d*x)/2]]) + 2*(7*a^2 - 10*a*b + 3*b^2)*Sec[(c + d*x)/2]^2 - (a - b)^2*Sec
[(c + d*x)/2]^4)/(64*d)

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Maple [A]  time = 0.053, size = 169, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}+{\frac{ab \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,a\cos \left ( dx+c \right ) b}{4\,d}}+{\frac{3\,ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{4\,d}}-{\frac{{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 1.06366, size = 165, normalized size = 1.31 \begin{align*} \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} \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))^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

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Fricas [A]  time = 0.989525, size = 504, normalized size = 4. \begin{align*} -\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 )}} \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))^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)

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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))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.45574, size = 486, normalized size = 3.86 \begin{align*} -\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} \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))^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