3.97 \(\int \frac{\csc (c+d x)}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=82 \[ -\frac{7}{8 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{\tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{5}{8 a d (a \cos (c+d x)+a)^2}-\frac{1}{6 d (a \cos (c+d x)+a)^3} \]

[Out]

-ArcTanh[Cos[c + d*x]]/(8*a^3*d) - 1/(6*d*(a + a*Cos[c + d*x])^3) + 5/(8*a*d*(a + a*Cos[c + d*x])^2) - 7/(8*d*
(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.151134, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3872, 2836, 12, 88, 206} \[ -\frac{7}{8 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{\tanh ^{-1}(\cos (c+d x))}{8 a^3 d}+\frac{5}{8 a d (a \cos (c+d x)+a)^2}-\frac{1}{6 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[c + d*x]]/(8*a^3*d) - 1/(6*d*(a + a*Cos[c + d*x])^3) + 5/(8*a*d*(a + a*Cos[c + d*x])^2) - 7/(8*d*
(a^3 + a^3*Cos[c + d*x]))

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.325845, size = 97, normalized size = 1.18 \[ -\frac{\sec ^3(c+d x) \left (42 \cos ^4\left (\frac{1}{2} (c+d x)\right )-15 \cos ^2\left (\frac{1}{2} (c+d x)\right )+12 \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )+2\right )}{12 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2 - 15*Cos[(c + d*x)/2]^2 + 42*Cos[(c + d*x)/2]^4 + 12*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2]] - Log[Sin[
(c + d*x)/2]]))*Sec[c + d*x]^3)/(12*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.066, size = 90, normalized size = 1.1 \begin{align*} -{\frac{1}{6\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{3}}}+{\frac{5}{8\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{7}{8\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{16\,d{a}^{3}}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{16\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)/(a+a*sec(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.00691, size = 132, normalized size = 1.61 \begin{align*} -\frac{\frac{2 \,{\left (21 \, \cos \left (d x + c\right )^{2} + 27 \, \cos \left (d x + c\right ) + 10\right )}}{a^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}} + \frac{3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}} - \frac{3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3}}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(2*(21*cos(d*x + c)^2 + 27*cos(d*x + c) + 10)/(a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x
+ c) + a^3) + 3*log(cos(d*x + c) + 1)/a^3 - 3*log(cos(d*x + c) - 1)/a^3)/d

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Fricas [B]  time = 1.75499, size = 416, normalized size = 5.07 \begin{align*} -\frac{42 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 3 \,{\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 54 \, \cos \left (d x + c\right ) + 20}{48 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(42*cos(d*x + c)^2 + 3*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1
/2) - 3*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(-1/2*cos(d*x + c) + 1/2) + 54*cos(d*x + c
) + 20)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\csc{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.34449, size = 153, normalized size = 1.87 \begin{align*} \frac{\frac{6 \, \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3}} + \frac{\frac{18 \, a^{6}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \, a^{6}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{2 \, a^{6}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{9}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(6*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + (18*a^6*(cos(d*x + c) - 1)/(cos(d*x + c) + 1)
+ 9*a^6*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 2*a^6*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)/a^9)/d