3.1335 \(\int \frac{\csc (c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=93 \[ \frac{b^2 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}+\frac{\log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.145838, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2837, 12, 894} \[ \frac{b^2 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}+\frac{\log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.106112, size = 84, normalized size = 0.9 \[ -\frac{-\frac{2 b^2 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )}+\frac{\log (1-\sin (c+d x))}{a+b}+\frac{\log (\sin (c+d x)+1)}{a-b}-\frac{2 \log (\sin (c+d x))}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 95, normalized size = 1. \begin{align*}{\frac{{b}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{da \left ( a+b \right ) \left ( a-b \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }}+{\frac{\ln \left ( \sin \left ( dx+c \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01384, size = 108, normalized size = 1.16 \begin{align*} \frac{\frac{2 \, b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3} - a b^{2}} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac{2 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72232, size = 225, normalized size = 2.42 \begin{align*} \frac{2 \, b^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \,{\left (a^{2} - b^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (d x + c\right )\right ) -{\left (a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{2} - a b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{3} - a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23994, size = 116, normalized size = 1.25 \begin{align*} \frac{\frac{2 \, b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b - a b^{3}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac{2 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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