\(\int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\) [1447]

Optimal result

Integrand size = 25, antiderivative size = 36 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {b \tan (c+d x)}{d} \]

[Out]

-a*arctanh(cos(d*x+c))/d+a*sec(d*x+c)/d+b*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2917, 2702, 327, 213, 3852, 8} \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {b \tan (c+d x)}{d} \]

[In]

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

[Out]

-((a*ArcTanh[Cos[c + d*x]])/d) + (a*Sec[c + d*x])/d + (b*Tan[c + d*x])/d

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2702

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = a \int \csc (c+d x) \sec ^2(c+d x) \, dx+b \int \sec ^2(c+d x) \, dx \\ & = \frac {a \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d}-\frac {b \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d} \\ & = \frac {a \sec (c+d x)}{d}+\frac {b \tan (c+d x)}{d}+\frac {a \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (c+d x)\right )}{d} \\ & = -\frac {a \text {arctanh}(\cos (c+d x))}{d}+\frac {a \sec (c+d x)}{d}+\frac {b \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.56 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a \sec (c+d x)}{d}+\frac {b \tan (c+d x)}{d} \]

[In]

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

[Out]

-((a*Log[Cos[(c + d*x)/2]])/d) + (a*Log[Sin[(c + d*x)/2]])/d + (a*Sec[c + d*x])/d + (b*Tan[c + d*x])/d

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b \tan \left (d x +c \right )}{d}\)	\(41\)
default	\(\frac {a \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+b \tan \left (d x +c \right )}{d}\)	\(41\)
parallelrisch	\(\frac {a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{d \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(61\)
risch	\(\frac {2 i \left (-i a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}\)	\(71\)
norman	\(\frac {-\frac {2 a}{d}-\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(104\)

[In]

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

[Out]

1/d*(a*(1/cos(d*x+c)+ln(csc(d*x+c)-cot(d*x+c)))+b*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.81 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - a \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, b \sin \left (d x + c\right ) - 2 \, a}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(csc(d*x+c)*sec(d*x+c)**2*(a+b*sin(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 2 \, b \tan \left (d x + c\right )}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.44 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

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