\(\int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [1337]

Optimal result

Integrand size = 27, antiderivative size = 132 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {b^4 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 132, 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} \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}+\frac {\left (a^2+b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac {b^4 \log (a+b \sin (c+d x))}{a^3 d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}-\frac {\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac {\csc ^2(c+d x)}{2 a d} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \left (\frac {\csc (c+d x)}{a^2}-\frac {\csc ^2(c+d x)}{2 a b}-\frac {\log (1-\sin (c+d x))}{2 b (a+b)}+\frac {\log (\sin (c+d x))}{a b}+\frac {b \log (\sin (c+d x))}{a^3}-\frac {\log (1+\sin (c+d x))}{2 (a-b) b}+\frac {b^3 \log (a+b \sin (c+d x))}{a^3 \left (a^2-b^2\right )}\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(120\)
default	\(\frac {\frac {b^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right ) \left (a -b \right ) a^{3}}-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(120\)
parallelrisch	\(\frac {8 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b^{4}-8 a^{3} \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-8 \left (\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{3}+\frac {\left (\left (-8 a^{2}-8 b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 b \right )\right ) a \right ) \left (a -b \right )}{8}\right ) \left (a +b \right )}{8 a^{5} d -8 a^{3} b^{2} d}\)	\(180\)
norman	\(\frac {-\frac {1}{8 a d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2} d}+\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}+\frac {b^{4} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{3} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}\)	\(199\)
risch	\(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}+\frac {i c}{d \left (a -b \right )}+\frac {2 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+\frac {i x}{a -b}-\frac {2 i c}{d a}-\frac {2 i b^{4} c}{a^{3} d \left (a^{2}-b^{2}\right )}-\frac {2 i b^{2} c}{a^{3} d}-\frac {2 i b^{4} x}{a^{3} \left (a^{2}-b^{2}\right )}-\frac {2 i x}{a}-\frac {2 i b^{2} x}{a^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}+\frac {b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}\)	\(330\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.70 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^{4} - a^{2} b^{2} + 2 \, {\left (b^{4} \cos \left (d x + c\right )^{2} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (a^{4} - b^{4} - {\left (a^{4} - b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (a^{4} + a^{3} b - {\left (a^{4} + a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} - a^{3} b - {\left (a^{4} - a^{3} b\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{5} - a^{3} b^{2}\right )} d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.86 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5} - a^{3} b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac {2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b - a^{3} b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} + \frac {2 \, {\left (a^{2} + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, b^{2} \sin \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) + a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.79 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\csc ^3(c+d x) \sec (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )\right )\,\left (a^2+b^2\right )}{a^3\,d}-\frac {\frac {1}{2\,a}-\frac {b\,\sin \left (c+d\,x\right )}{a^2}}{d\,{\sin \left (c+d\,x\right )}^2}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}+\frac {b^4\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^5-a^3\,b^2\right )} \]

[In]

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

[Out]

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