\(\int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\) [1333]

Optimal result

Integrand size = 25, antiderivative size = 80 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {a^2 \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2916, 12, 1643} \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^2 \log (a+b \sin (c+d x))}{b d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}+\frac {\log (\sin (c+d x)+1)}{2 d (a-b)} \]

[In]

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

[Out]

-1/2*Log[1 - Sin[c + d*x]]/((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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {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 {\text {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 {\text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.90 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-((a-b) b \log (1-\sin (c+d x)))+b (a+b) \log (1+\sin (c+d x))-2 a^2 \log (a+b \sin (c+d x))}{2 (a-b) b (a+b) d} \]

[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)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) b}}{d}\)	\(76\)
default	\(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) b}}{d}\)	\(76\)
parallelrisch	\(\frac {-a^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )+\left (a^{2}-b^{2}\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\left (-a +b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \left (a +b \right )\right ) b}{a^{2} b d -b^{3} d}\)	\(109\)
norman	\(\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}-\frac {a^{2} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d b \left (a^{2}-b^{2}\right )}\)	\(118\)
risch	\(-\frac {i x}{b}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {2 i a^{2} x}{b \left (a^{2}-b^{2}\right )}+\frac {2 i a^{2} c}{d b \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d b \left (a^{2}-b^{2}\right )}\)	\(197\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.92 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a b - b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d} \]

[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)

Sympy [F]

\[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, a^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b - b^{3}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, a^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 13.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.46 \[ \int \frac {\sin (c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )}+\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b\,d}-\frac {a^2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{b\,d\,\left (a^2-b^2\right )} \]

[In]

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

[Out]

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