3.1333 \(\int \frac{\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=80 \[ -\frac{a^2 \log (a+b \sin (c+d x))}{b d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.155777, antiderivative size = 80, 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, 1629} \[ -\frac{a^2 \log (a+b \sin (c+d x))}{b d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0654118, size = 72, normalized size = 0.9 \[ \frac{-2 a^2 \log (a+b \sin (c+d x))-b (a-b) \log (1-\sin (c+d x))+b (a+b) \log (\sin (c+d x)+1)}{2 b d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.062, size = 81, normalized size = 1. \begin{align*} -{\frac{{a}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) \left ( a-b \right ) b}}-{\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 ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*a^2/(a+b)/(a-b)/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))

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Maxima [A]  time = 0.988743, size = 92, normalized size = 1.15 \begin{align*} -\frac{\frac{2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b - b^{3}} - \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}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.32669, size = 174, normalized size = 2.17 \begin{align*} -\frac{2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left (a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a b - b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left (a^{2} b - b^{3}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.21428, size = 96, normalized size = 1.2 \begin{align*} -\frac{\frac{2 \, a^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - 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}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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