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

Optimal. Leaf size=163 \[ -\frac{b^3}{2 a^2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{b^2 \left (3 a^2-b^2\right )}{a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{b \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)^3}-\frac{\log (\cos (c+d x)+1)}{2 d (a-b)^3} \]

[Out]

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

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Rubi [A]  time = 0.31935, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3872, 2837, 12, 1629} \[ -\frac{b^3}{2 a^2 d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{b^2 \left (3 a^2-b^2\right )}{a^2 d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{b \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^3}+\frac{\log (1-\cos (c+d x))}{2 d (a+b)^3}-\frac{\log (\cos (c+d x)+1)}{2 d (a-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[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 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.554559, size = 203, normalized size = 1.25 \[ \frac{\sec ^3(c+d x) (a \cos (c+d x)+b) \left (-\frac{2 b^2 \left (b^2-3 a^2\right ) (a \cos (c+d x)+b)}{a^2 (a-b)^2 (a+b)^2}+\frac{2 b \left (3 a^2+b^2\right ) (a \cos (c+d x)+b)^2 \log (a \cos (c+d x)+b)}{\left (a^2-b^2\right )^3}+\frac{b^3}{a^2 \left (b^2-a^2\right )}+\frac{2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^2}{(b-a)^3}+\frac{2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b)^2}{(a+b)^3}\right )}{2 d (a+b \sec (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cos[c + d*x])*(b^3/(a^2*(-a^2 + b^2)) - (2*b^2*(-3*a^2 + b^2)*(b + a*Cos[c + d*x]))/(a^2*(a - b)^2*(a
+ b)^2) + (2*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2]])/(-a + b)^3 + (2*b*(3*a^2 + b^2)*(b + a*Cos[c + d*x]
)^2*Log[b + a*Cos[c + d*x]])/(a^2 - b^2)^3 + (2*(b + a*Cos[c + d*x])^2*Log[Sin[(c + d*x)/2]])/(a + b)^3)*Sec[c
+ d*x]^3)/(2*d*(a + b*Sec[c + d*x])^3)

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Maple [A]  time = 0.071, size = 206, normalized size = 1.3 \begin{align*} -{\frac{{b}^{3}}{2\,d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \left ( b+a\cos \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b\ln \left ( b+a\cos \left ( dx+c \right ) \right ){a}^{2}}{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+{\frac{{b}^{3}\ln \left ( b+a\cos \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{3} \left ( a-b \right ) ^{3}}}+3\,{\frac{{b}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) }}-{\frac{{b}^{4}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{2} \left ( b+a\cos \left ( dx+c \right ) \right ) }}-{\frac{\ln \left ( \cos \left ( dx+c \right ) +1 \right ) }{2\, \left ( a-b \right ) ^{3}d}}+{\frac{\ln \left ( -1+\cos \left ( dx+c \right ) \right ) }{2\,d \left ( a+b \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.979614, size = 325, normalized size = 1.99 \begin{align*} \frac{\frac{2 \,{\left (3 \, a^{2} b + b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{5 \, a^{2} b^{3} - b^{5} + 2 \,{\left (3 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )}{a^{6} b^{2} - 2 \, a^{4} b^{4} + a^{2} b^{6} +{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \cos \left (d x + c\right )} - \frac{\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac{\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(3*a^2*b + b^3)*log(a*cos(d*x + c) + b)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + (5*a^2*b^3 - b^5 + 2*(3*a
^3*b^2 - a*b^4)*cos(d*x + c))/(a^6*b^2 - 2*a^4*b^4 + a^2*b^6 + (a^8 - 2*a^6*b^2 + a^4*b^4)*cos(d*x + c)^2 + 2*
(a^7*b - 2*a^5*b^3 + a^3*b^5)*cos(d*x + c)) - log(cos(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + log(cos(
d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/d

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Fricas [B]  time = 2.60537, size = 1021, normalized size = 6.26 \begin{align*} \frac{5 \, a^{4} b^{3} - 6 \, a^{2} b^{5} + b^{7} + 2 \,{\left (3 \, a^{5} b^{2} - 4 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right ) + 2 \,{\left (3 \, a^{4} b^{3} + a^{2} b^{5} +{\left (3 \, a^{6} b + a^{4} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) -{\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5} +{\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) +{\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5} +{\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \,{\left ({\left (a^{10} - 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} - a^{4} b^{6}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{9} b - 3 \, a^{7} b^{3} + 3 \, a^{5} b^{5} - a^{3} b^{7}\right )} d \cos \left (d x + c\right ) +{\left (a^{8} b^{2} - 3 \, a^{6} b^{4} + 3 \, a^{4} b^{6} - a^{2} b^{8}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*a^4*b^3 - 6*a^2*b^5 + b^7 + 2*(3*a^5*b^2 - 4*a^3*b^4 + a*b^6)*cos(d*x + c) + 2*(3*a^4*b^3 + a^2*b^5 + (
3*a^6*b + a^4*b^3)*cos(d*x + c)^2 + 2*(3*a^5*b^2 + a^3*b^4)*cos(d*x + c))*log(a*cos(d*x + c) + b) - (a^5*b^2 +
3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5 + (a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*cos(d*x + c)^2 + 2*(a^6*b + 3*a^5*b^2
+ 3*a^4*b^3 + a^3*b^4)*cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + (a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5
+ (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*cos(d*x + c)^2 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*cos(d*x
+ c))*log(-1/2*cos(d*x + c) + 1/2))/((a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*d*cos(d*x + c)^2 + 2*(a^9*b - 3*
a^7*b^3 + 3*a^5*b^5 - a^3*b^7)*d*cos(d*x + c) + (a^8*b^2 - 3*a^6*b^4 + 3*a^4*b^6 - a^2*b^8)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.50738, size = 610, normalized size = 3.74 \begin{align*} \frac{\frac{2 \,{\left (3 \, a^{2} b + b^{3}\right )} \log \left ({\left | -a - b - \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{\log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{9 \, a^{3} b + 15 \, a^{2} b^{2} + 3 \, a b^{3} - 3 \, b^{4} + \frac{18 \, a^{3} b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{6 \, a^{2} b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{10 \, a b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{2 \, b^{4}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{9 \, a^{3} b{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{9 \, a^{2} b^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a b^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{3 \, b^{4}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )}{\left (a + b + \frac{a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{b{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(3*a^2*b + b^3)*log(abs(-a - b - a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + b*(cos(d*x + c) - 1)/(cos(d*
x + c) + 1)))/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^3 + 3
*a^2*b + 3*a*b^2 + b^3) - (9*a^3*b + 15*a^2*b^2 + 3*a*b^3 - 3*b^4 + 18*a^3*b*(cos(d*x + c) - 1)/(cos(d*x + c)
+ 1) + 6*a^2*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 10*a*b^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2*b^
4*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 9*a^3*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 9*a^2*b^2*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 3*a*b^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 3*b^4*(cos(d*x + c)
- 1)^2/(cos(d*x + c) + 1)^2)/((a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*(a + b + a*(cos(d*x + c) - 1
)/(cos(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))^2))/d