3.1336 \(\int \frac{\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=110 \[ -\frac{b^3 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac{\csc (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.186549, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 894} \[ -\frac{b^3 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac{\csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[c + d*x]/(a*d)) - Log[1 - Sin[c + d*x]]/(2*(a + b)*d) - (b*Log[Sin[c + d*x]])/(a^2*d) + Log[1 + Sin[c +
d*x]]/(2*(a - b)*d) - (b^3*Log[a + b*Sin[c + d*x]])/(a^2*(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 ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{b^2}{x^2 (a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \frac{1}{x^2 (a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{b^3 \operatorname{Subst}\left (\int \left (\frac{1}{2 b^3 (a+b) (b-x)}+\frac{1}{a b^2 x^2}-\frac{1}{a^2 b^2 x}-\frac{1}{a^2 (a-b) (a+b) (a+x)}-\frac{1}{2 b^3 (-a+b) (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{\csc (c+d x)}{a d}-\frac{\log (1-\sin (c+d x))}{2 (a+b) d}-\frac{b \log (\sin (c+d x))}{a^2 d}+\frac{\log (1+\sin (c+d x))}{2 (a-b) d}-\frac{b^3 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 0.234847, size = 97, normalized size = 0.88 \[ \frac{\frac{2 b^3 \log (a+b \sin (c+d x))}{a^2 \left (b^2-a^2\right )}-\frac{2 b \log (\sin (c+d x))}{a^2}-\frac{\log (1-\sin (c+d x))}{a+b}+\frac{\log (\sin (c+d x)+1)}{a-b}-\frac{2 \csc (c+d x)}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.082, size = 113, normalized size = 1. \begin{align*} -{\frac{{b}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) \left ( a-b \right ){a}^{2}}}-{\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{1}{da\sin \left ( dx+c \right ) }}-{\frac{b\ln \left ( \sin \left ( dx+c \right ) \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.982456, size = 128, normalized size = 1.16 \begin{align*} -\frac{\frac{2 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - a^{2} 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 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac{2}{a \sin \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.15817, size = 348, normalized size = 3.16 \begin{align*} -\frac{2 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{3} - 2 \, a b^{2} + 2 \,{\left (a^{2} b - b^{3}\right )} \log \left (\frac{1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) -{\left (a^{3} + a^{2} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) +{\left (a^{3} - a^{2} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{2 \,{\left (a^{4} - a^{2} b^{2}\right )} d \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\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)**2*sec(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.20944, size = 153, normalized size = 1.39 \begin{align*} -\frac{\frac{2 \, b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - a^{2} 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 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac{2 \,{\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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