3.277 \(\int \cot ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=92 \[ -\frac{\cot ^2(c+d x) \left (a^2+2 a b \sec (c+d x)+b^2\right )}{2 d}-\frac{a^2 \log (\cos (c+d x))}{d}-\frac{a (a+b) \log (1-\sec (c+d x))}{2 d}-\frac{a (a-b) \log (\sec (c+d x)+1)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.129499, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3885, 1805, 801} \[ -\frac{\cot ^2(c+d x) \left (a^2+2 a b \sec (c+d x)+b^2\right )}{2 d}-\frac{a^2 \log (\cos (c+d x))}{d}-\frac{a (a+b) \log (1-\sec (c+d x))}{2 d}-\frac{a (a-b) \log (\sec (c+d x)+1)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.458545, size = 82, normalized size = 0.89 \[ -\frac{(a+b)^2 \csc ^2\left (\frac{1}{2} (c+d x)\right )+(a-b)^2 \sec ^2\left (\frac{1}{2} (c+d x)\right )+8 a \left ((a+b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+(a-b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)^2*Csc[(c + d*x)/2]^2 + 8*a*((a - b)*Log[Cos[(c + d*x)/2]] + (a + b)*Log[Sin[(c + d*x)/2]]) + (a - b)
^2*Sec[(c + d*x)/2]^2)/(8*d)

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Maple [A]  time = 0.052, size = 108, normalized size = 1.2 \begin{align*} -{\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 ) ^{3}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a\cos \left ( dx+c \right ) b}{d}}-{\frac{ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

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]

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

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Maxima [A]  time = 0.987193, size = 97, normalized size = 1.05 \begin{align*} -\frac{{\left (a^{2} - a b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) +{\left (a^{2} + a b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac{2 \, a b \cos \left (d x + c\right ) + a^{2} + b^{2}}{\cos \left (d x + c\right )^{2} - 1}}{2 \, d} \end{align*}

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

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Fricas [A]  time = 0.790496, size = 275, normalized size = 2.99 \begin{align*} \frac{2 \, a b \cos \left (d x + c\right ) + a^{2} + b^{2} -{\left ({\left (a^{2} - a b\right )} \cos \left (d x + c\right )^{2} - a^{2} + a b\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left ({\left (a^{2} + a b\right )} \cos \left (d x + c\right )^{2} - a^{2} - a b\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left (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)^3*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{2} \cot ^{3}{\left (c + d x \right )}\, dx \end{align*}

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

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Giac [B]  time = 1.3528, size = 282, normalized size = 3.07 \begin{align*} \frac{8 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) + \frac{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 4 \,{\left (a^{2} + 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{4 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{4 \, a b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{\cos \left (d x + c\right ) - 1}}{8 \, d} \end{align*}

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*(8*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) + a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) -
2*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 4*(a^2 + a*b)*log(ab
s(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) + (a^2 + 2*a*b + b^2 + 4*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
+ 4*a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))*(cos(d*x + c) + 1)/(cos(d*x + c) - 1))/d