\(\int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [1349]

Optimal result

Integrand size = 27, antiderivative size = 156 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {\log (\sin (c+d x))}{a d}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {b^4 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}+\frac {1}{4 (a-b) d (1+\sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 156, 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} \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b^4 \log (a+b \sin (c+d x))}{a d \left (a^2-b^2\right )^2}+\frac {1}{4 d (a+b) (1-\sin (c+d x))}+\frac {1}{4 d (a-b) (\sin (c+d x)+1)}-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 d (a+b)^2}-\frac {(2 a-3 b) \log (\sin (c+d x)+1)}{4 d (a-b)^2}+\frac {\log (\sin (c+d x))}{a d} \]

[In]

Int[(Csc[c + d*x]*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

-1/4*((2*a + 3*b)*Log[1 - Sin[c + d*x]])/((a + b)^2*d) + Log[Sin[c + d*x]]/(a*d) - ((2*a - 3*b)*Log[1 + Sin[c
+ d*x]])/(4*(a - b)^2*d) - (b^4*Log[a + b*Sin[c + d*x]])/(a*(a^2 - b^2)^2*d) + 1/(4*(a + b)*d*(1 - Sin[c + d*x
])) + 1/(4*(a - b)*d*(1 + Sin[c + d*x]))

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^3 \text {Subst}\left (\int \frac {b}{x (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^4 \text {Subst}\left (\int \frac {1}{x (a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^4 \text {Subst}\left (\int \left (\frac {1}{4 b^3 (a+b) (b-x)^2}+\frac {2 a+3 b}{4 b^4 (a+b)^2 (b-x)}+\frac {1}{a b^4 x}-\frac {1}{a (a-b)^2 (a+b)^2 (a+x)}-\frac {1}{4 (a-b) b^3 (b+x)^2}+\frac {-2 a+3 b}{4 (a-b)^2 b^4 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d} \\ & = -\frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 (a+b)^2 d}+\frac {\log (\sin (c+d x))}{a d}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{4 (a-b)^2 d}-\frac {b^4 \log (a+b \sin (c+d x))}{a \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\sin (c+d x))}+\frac {1}{4 (a-b) d (1+\sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b^4 \left (-\frac {(2 a+3 b) \log (1-\sin (c+d x))}{b^4 (a+b)^2}+\frac {4 \log (\sin (c+d x))}{a b^4}-\frac {(2 a-3 b) \log (1+\sin (c+d x))}{(a-b)^2 b^4}-\frac {4 \log (a+b \sin (c+d x))}{a (a-b)^2 (a+b)^2}-\frac {1}{b^4 (a+b) (-1+\sin (c+d x))}+\frac {1}{(a-b) b^4 (1+\sin (c+d x))}\right )}{4 d} \]

[In]

Integrate[(Csc[c + d*x]*Sec[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

(b^4*(-(((2*a + 3*b)*Log[1 - Sin[c + d*x]])/(b^4*(a + b)^2)) + (4*Log[Sin[c + d*x]])/(a*b^4) - ((2*a - 3*b)*Lo
g[1 + Sin[c + d*x]])/((a - b)^2*b^4) - (4*Log[a + b*Sin[c + d*x]])/(a*(a - b)^2*(a + b)^2) - 1/(b^4*(a + b)*(-
1 + Sin[c + d*x])) + 1/((a - b)*b^4*(1 + Sin[c + d*x]))))/(4*d)

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}}{d}\)	\(137\)
default	\(\frac {\frac {\ln \left (\sin \left (d x +c \right )\right )}{a}+\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (-2 a +3 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-2 a -3 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a +b \right )^{2}}-\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2} a}}{d}\)	\(137\)
parallelrisch	\(\frac {-b^{4} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a -b \right )^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (a +\frac {3 b}{2}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\left (a -\frac {3 b}{2}\right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (a -b \right ) \left (a +b \right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a \left (a \cos \left (2 d x +2 c \right )+2 b \sin \left (d x +c \right )-a \right )}{2}\right ) \left (a -b \right )\right ) \left (a +b \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} a d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(218\)
norman	\(\frac {-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (a^{2}-b^{2}\right )}-\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{2}-b^{2}\right )}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a^{2}-b^{2}\right ) d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (2 a -3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {\left (2 a +3 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{4} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\)	\(248\)
risch	\(\frac {i a c}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {i a c}{d \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 i b c}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {3 i b c}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b^{4} x}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i x}{a}+\frac {i \left (-2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 i b x}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 i b x}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{4} c}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 i c}{d a}+\frac {i a x}{a^{2}+2 a b +b^{2}}+\frac {i a x}{a^{2}-2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{\left (a^{2}+2 a b +b^{2}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{\left (a^{2}-2 a b +b^{2}\right ) d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 \left (a^{2}-2 a b +b^{2}\right ) d}-\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}\)	\(502\)

[In]

int(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.37 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {4 \, b^{4} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, a^{4} + 2 \, a^{2} b^{2} - 4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (2 \, a^{4} + a^{3} b - 4 \, a^{2} b^{2} - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{4} - a^{3} b - 4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(4*b^4*cos(d*x + c)^2*log(b*sin(d*x + c) + a) - 2*a^4 + 2*a^2*b^2 - 4*(a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c
)^2*log(-1/2*sin(d*x + c)) + (2*a^4 + a^3*b - 4*a^2*b^2 - 3*a*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*a
^4 - a^3*b - 4*a^2*b^2 + 3*a*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(a^3*b - a*b^3)*sin(d*x + c))/((a^
5 - 2*a^3*b^2 + a*b^4)*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\csc {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Integral(csc(c + d*x)*sec(c + d*x)**3/(a + b*sin(c + d*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}} - \frac {4 \, \log \left (\sin \left (d x + c\right )\right )}{a}}{4 \, d} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(4*b^4*log(b*sin(d*x + c) + a)/(a^5 - 2*a^3*b^2 + a*b^4) + (2*a - 3*b)*log(sin(d*x + c) + 1)/(a^2 - 2*a*b
+ b^2) + (2*a + 3*b)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(b*sin(d*x + c) - a)/((a^2 - b^2)*sin(d*x
+ c)^2 - a^2 + b^2) - 4*log(sin(d*x + c))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {4 \, b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b - 2 \, a^{3} b^{3} + a b^{5}} + \frac {{\left (2 \, a - 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (2 \, a + 3 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {4 \, \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a} - \frac {2 \, {\left (a^{3} \sin \left (d x + c\right )^{2} - 2 \, a b^{2} \sin \left (d x + c\right )^{2} + a^{2} b \sin \left (d x + c\right ) - b^{3} \sin \left (d x + c\right ) - 2 \, a^{3} + 3 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/4*(4*b^5*log(abs(b*sin(d*x + c) + a))/(a^5*b - 2*a^3*b^3 + a*b^5) + (2*a - 3*b)*log(abs(sin(d*x + c) + 1))/
(a^2 - 2*a*b + b^2) + (2*a + 3*b)*log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) - 4*log(abs(sin(d*x + c)))/a
- 2*(a^3*sin(d*x + c)^2 - 2*a*b^2*sin(d*x + c)^2 + a^2*b*sin(d*x + c) - b^3*sin(d*x + c) - 2*a^3 + 3*a*b^2)/((
a^4 - 2*a^2*b^2 + b^4)*(sin(d*x + c)^2 - 1)))/d

Mupad [B] (verification not implemented)

Time = 13.00 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09 \[ \int \frac {\csc (c+d x) \sec ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {b}{4\,{\left (a-b\right )}^2}-\frac {1}{2\,\left (a-b\right )}\right )}{d}-\frac {\frac {a}{2\,\left (a^2-b^2\right )}-\frac {b\,\sin \left (c+d\,x\right )}{2\,\left (a^2-b^2\right )}}{d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )}+\frac {\ln \left (\sin \left (c+d\,x\right )\right )}{a\,d}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{2\,\left (a+b\right )}\right )}{d}-\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{a\,d\,{\left (a^2-b^2\right )}^2} \]

[In]

int(1/(cos(c + d*x)^3*sin(c + d*x)*(a + b*sin(c + d*x))),x)

[Out]

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