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

Optimal. Leaf size=126 \[ -\frac{1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac{1}{16 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{\tanh ^{-1}(\cos (c+d x))}{32 a^3 d}-\frac{a}{16 d (a \cos (c+d x)+a)^4}+\frac{1}{6 d (a \cos (c+d x)+a)^3}-\frac{3}{32 a d (a \cos (c+d x)+a)^2} \]

[Out]

ArcTanh[Cos[c + d*x]]/(32*a^3*d) - a/(16*d*(a + a*Cos[c + d*x])^4) + 1/(6*d*(a + a*Cos[c + d*x])^3) - 3/(32*a*
d*(a + a*Cos[c + d*x])^2) - 1/(32*d*(a^3 - a^3*Cos[c + d*x])) - 1/(16*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.134334, 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 = {3872, 2707, 88, 206} \[ -\frac{1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac{1}{16 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{\tanh ^{-1}(\cos (c+d x))}{32 a^3 d}-\frac{a}{16 d (a \cos (c+d x)+a)^4}+\frac{1}{6 d (a \cos (c+d x)+a)^3}-\frac{3}{32 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Cos[c + d*x]]/(32*a^3*d) - a/(16*d*(a + a*Cos[c + d*x])^4) + 1/(6*d*(a + a*Cos[c + d*x])^3) - 3/(32*a*
d*(a + a*Cos[c + d*x])^2) - 1/(32*d*(a^3 - a^3*Cos[c + d*x])) - 1/(16*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 2707

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

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 ^3(c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\int \frac{\cot ^3(c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(-a-x)^2 (-a+x)^5} \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a}{4 (a-x)^5}+\frac{1}{2 (a-x)^4}-\frac{3}{16 a (a-x)^3}-\frac{1}{16 a^2 (a-x)^2}+\frac{1}{32 a^2 (a+x)^2}-\frac{1}{32 a^2 \left (a^2-x^2\right )}\right ) \, dx,x,-a \cos (c+d x)\right )}{d}\\ &=-\frac{a}{16 d (a+a \cos (c+d x))^4}+\frac{1}{6 d (a+a \cos (c+d x))^3}-\frac{3}{32 a d (a+a \cos (c+d x))^2}-\frac{1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac{1}{16 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 )}{32 a^2 d}\\ &=\frac{\tanh ^{-1}(\cos (c+d x))}{32 a^3 d}-\frac{a}{16 d (a+a \cos (c+d x))^4}+\frac{1}{6 d (a+a \cos (c+d x))^3}-\frac{3}{32 a d (a+a \cos (c+d x))^2}-\frac{1}{32 d \left (a^3-a^3 \cos (c+d x)\right )}-\frac{1}{16 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.576518, size = 138, normalized size = 1.1 \[ -\frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (12 \csc ^2\left (\frac{1}{2} (c+d x)\right )+3 \sec ^8\left (\frac{1}{2} (c+d x)\right )-16 \sec ^6\left (\frac{1}{2} (c+d x)\right )+18 \sec ^4\left (\frac{1}{2} (c+d x)\right )+24 \sec ^2\left (\frac{1}{2} (c+d x)\right )+24 \left (\log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{96 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(c + d*x)/2]^6*(12*Csc[(c + d*x)/2]^2 + 24*(-Log[Cos[(c + d*x)/2]] + Log[Sin[(c + d*x)/2]]) + 24*Sec[(c
+ d*x)/2]^2 + 18*Sec[(c + d*x)/2]^4 - 16*Sec[(c + d*x)/2]^6 + 3*Sec[(c + d*x)/2]^8)*Sec[c + d*x]^3)/(96*a^3*d*
(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.079, size = 126, normalized size = 1. \begin{align*} -{\frac{1}{16\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{4}}}+{\frac{1}{6\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{3}}}-{\frac{3}{32\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}}-{\frac{1}{16\,d{a}^{3} \left ( \cos \left ( dx+c \right ) +1 \right ) }}+{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{64\,d{a}^{3}}}+{\frac{1}{32\,d{a}^{3} \left ( -1+\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{64\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/d/a^3/(cos(d*x+c)+1)^4+1/6/d/a^3/(cos(d*x+c)+1)^3-3/32/d/a^3/(cos(d*x+c)+1)^2-1/16/d/a^3/(cos(d*x+c)+1)+
1/64*ln(cos(d*x+c)+1)/a^3/d+1/32/d/a^3/(-1+cos(d*x+c))-1/64/d/a^3*ln(-1+cos(d*x+c))

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Maxima [A]  time = 1.00209, size = 197, normalized size = 1.56 \begin{align*} -\frac{\frac{2 \,{\left (3 \, \cos \left (d x + c\right )^{4} + 9 \, \cos \left (d x + c\right )^{3} - 25 \, \cos \left (d x + c\right )^{2} - 27 \, \cos \left (d x + c\right ) - 8\right )}}{a^{3} \cos \left (d x + c\right )^{5} + 3 \, a^{3} \cos \left (d x + c\right )^{4} + 2 \, a^{3} \cos \left (d x + c\right )^{3} - 2 \, 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}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(2*(3*cos(d*x + c)^4 + 9*cos(d*x + c)^3 - 25*cos(d*x + c)^2 - 27*cos(d*x + c) - 8)/(a^3*cos(d*x + c)^5
+ 3*a^3*cos(d*x + c)^4 + 2*a^3*cos(d*x + c)^3 - 2*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 = 1.77896, size = 640, normalized size = 5.08 \begin{align*} -\frac{6 \, \cos \left (d x + c\right )^{4} + 18 \, \cos \left (d x + c\right )^{3} - 50 \, \cos \left (d x + c\right )^{2} - 3 \,{\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 2 \, \cos \left (d x + c\right )^{3} - 2 \, \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 ) - 16}{192 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 2 \, 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)^3/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/192*(6*cos(d*x + c)^4 + 18*cos(d*x + c)^3 - 50*cos(d*x + c)^2 - 3*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 2*co
s(d*x + c)^3 - 2*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)^5 + 3*cos(
d*x + c)^4 + 2*cos(d*x + c)^3 - 2*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) - 16)/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 2*a^3*d*cos(d*x + c)^3 - 2*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 ^{3}{\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)**3/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(csc(c + d*x)**3/(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 = 1.34542, size = 246, normalized size = 1.95 \begin{align*} \frac{\frac{12 \,{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}}{a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}} - \frac{12 \, \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{24 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{12 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{4 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{3 \, a^{9}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}{a^{12}}}{768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(12*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)*(cos(d*x + c) + 1)/(a^3*(cos(d*x + c) - 1)) - 12*log(abs
(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^3 + (24*a^9*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 12*a^9*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 4*a^9*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 3*a^9*(cos(d*x + c) -
1)^4/(cos(d*x + c) + 1)^4)/a^12)/d