3.3.13 \(\int \frac {\csc (c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [213]

Optimal. Leaf size=109 \[ \frac {b^2}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\log (1-\cos (c+d x))}{2 (a+b)^2 d}-\frac {\log (1+\cos (c+d x))}{2 (a-b)^2 d}+\frac {2 a b \log (b+a \cos (c+d x))}{\left (a^2-b^2\right )^2 d} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3957, 2916, 12, 1643} \begin {gather*} \frac {b^2}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {2 a b \log (a \cos (c+d x)+b)}{d \left (a^2-b^2\right )^2}+\frac {\log (1-\cos (c+d x))}{2 d (a+b)^2}-\frac {\log (\cos (c+d x)+1)}{2 d (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 1643

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]

Rule 2916

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/b)*x)^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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

Rubi steps

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

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Mathematica [A]
time = 0.20, size = 165, normalized size = 1.51 \begin {gather*} \frac {-a^2 \cos (c+d x) \left ((a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-2 a b \log (b+a \cos (c+d x))-(a-b)^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+b \left (-a (a+b)^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 b \log (b+a \cos (c+d x))+(a-b) \left (b (a+b)+a (a-b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{a (a-b)^2 (a+b)^2 d (b+a \cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 98, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {b^{2}}{\left (a +b \right ) \left (a -b \right ) a \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 a b \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(98\)
default	\(\frac {-\frac {\ln \left (1+\cos \left (d x +c \right )\right )}{2 \left (a -b \right )^{2}}+\frac {\ln \left (-1+\cos \left (d x +c \right )\right )}{2 \left (a +b \right )^{2}}+\frac {b^{2}}{\left (a +b \right ) \left (a -b \right ) a \left (b +a \cos \left (d x +c \right )\right )}+\frac {2 a b \ln \left (b +a \cos \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}}{d}\)	\(98\)
norman	\(-\frac {2 b^{2}}{d \left (a^{3}-b \,a^{2}-b^{2} a +b^{3}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}+2 b a +b^{2}\right )}+\frac {2 b a \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\)	\(153\)
risch	\(\frac {i x}{a^{2}-2 b a +b^{2}}+\frac {i c}{d \left (a^{2}-2 b a +b^{2}\right )}-\frac {i x}{a^{2}+2 b a +b^{2}}-\frac {i c}{d \left (a^{2}+2 b a +b^{2}\right )}-\frac {4 i a b x}{a^{4}-2 b^{2} a^{2}+b^{4}}-\frac {4 i a b c}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}-\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{a d \left (-a^{2}+b^{2}\right ) \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \left (a^{2}-2 b a +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \left (a^{2}+2 b a +b^{2}\right )}+\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{d \left (a^{4}-2 b^{2} a^{2}+b^{4}\right )}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 123, normalized size = 1.13 \begin {gather*} \frac {\frac {4 \, a b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, b^{2}}{a^{3} b - a b^{3} + {\left (a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )} - \frac {\log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {\log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.11, size = 210, normalized size = 1.93 \begin {gather*} \frac {2 \, a^{2} b^{2} - 2 \, b^{4} + 4 \, {\left (a^{3} b \cos \left (d x + c\right ) + a^{2} b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right ) - {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3} + {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\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^{3} b - 2 \, a^{2} b^{2} + a b^{3} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\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^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)/(a+b*sec(d*x+c))**2,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (105) = 210\).
time = 0.50, size = 213, normalized size = 1.95 \begin {gather*} \frac {\frac {4 \, a b \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^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac {\log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, {\left (a b + b^{2} + \frac {a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\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 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(4*a*b*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^4 - 2*a^2*b^2 + b^4) + log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/(a^2 + 2*a*b + b^2) - 4*(a*b + b
^2 + a*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((a^3 + a^2*b - a*b^2 - b^3)*(a + b + a*(cos(d*x + c) - 1)/(co
s(d*x + c) + 1) - b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))))/d

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Mupad [B]
time = 0.23, size = 103, normalized size = 0.94 \begin {gather*} \frac {\ln \left (\cos \left (c+d\,x\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}-\frac {\ln \left (\cos \left (c+d\,x\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {b^2}{a\,d\,\left (a^2-b^2\right )\,\left (b+a\,\cos \left (c+d\,x\right )\right )}+\frac {2\,a\,b\,\ln \left (b+a\,\cos \left (c+d\,x\right )\right )}{d\,{\left (a^2-b^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)*(a + b/cos(c + d*x))^2),x)

[Out]

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

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