\(\int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [1336]

Optimal result

Integrand size = 27, antiderivative size = 110 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\csc (c+d x)}{a d}-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {b \log (\sin (c+d x))}{a^2 d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b^3 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2916, 12, 908} \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^3 \log (a+b \sin (c+d x))}{a^2 d \left (a^2-b^2\right )}-\frac {b \log (\sin (c+d x))}{a^2 d}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}+\frac {\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac {\csc (c+d x)}{a d} \]

[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 12

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

Rule 908

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]))

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 {b^2}{x^2 (a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^3 \text {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 \text {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] (verified)

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \left (-\frac {\csc (c+d x)}{a b}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)}-\frac {\log (\sin (c+d x))}{a^2}+\frac {\log (1+\sin (c+d x))}{2 (a-b) b}-\frac {b^2 \log (a+b \sin (c+d x))}{a^2 \left (a^2-b^2\right )}\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}-\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{2}}}{d}\)	\(102\)
default	\(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}-\frac {1}{a \sin \left (d x +c \right )}-\frac {b \ln \left (\sin \left (d x +c \right )\right )}{a^{2}}-\frac {b^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{2}}}{d}\)	\(102\)
parallelrisch	\(\frac {-\ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b^{3}-a^{2} \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\left (a -b \right ) \left (a \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b \right )}{2}\right ) \left (a +b \right )}{d \left (a^{4}-a^{2} b^{2}\right )}\)	\(136\)
norman	\(\frac {-\frac {1}{2 a d}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}-\frac {b^{3} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} d \left (a^{2}-b^{2}\right )}\)	\(156\)
risch	\(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b^{3} x}{a^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i b^{3} c}{a^{2} d \left (a^{2}-b^{2}\right )}+\frac {2 i b x}{a^{2}}+\frac {2 i b c}{a^{2} d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{2} d \left (a^{2}-b^{2}\right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}\)	\(261\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) \sin \left (d x + c\right ) + 2 \, a^{3} - 2 \, a b^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - {\left (a^{3} + a^{2} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + {\left (a^{3} - a^{2} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right )}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} d \sin \left (d x + c\right )} \]

[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))

Sympy [F]

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

[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)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - a^{2} b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{2}} + \frac {2}{a \sin \left (d x + c\right )}}{2 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.03 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, b^{4} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - a^{2} b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac {2 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{a^{2} \sin \left (d x + c\right )}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 13.38 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^2(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,\left (a+b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,\left (a-b\right )}+\frac {1}{a\,\sin \left (c+d\,x\right )}+\frac {b\,\ln \left (\sin \left (c+d\,x\right )\right )}{a^2}+\frac {b^3\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a^4-a^2\,b^2}}{d} \]

[In]

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

[Out]

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