\(\int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [762]

Optimal result

Integrand size = 27, antiderivative size = 44 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \sec (c+d x)}{d}+\frac {2 a^2 \tan (c+d x)}{d} \]

[Out]

-a^2*arctanh(cos(d*x+c))/d+2*a^2*sec(d*x+c)/d+2*a^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2952, 3852, 8, 2702, 327, 213, 2686} \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {2 a^2 \sec (c+d x)}{d} \]

[In]

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

[Out]

-((a^2*ArcTanh[Cos[c + d*x]])/d) + (2*a^2*Sec[c + d*x])/d + (2*a^2*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 2686

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

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 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.57 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}\right )}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {\left (-4+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )\right ) a^{2}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(45\)
derivativedivides	\(\frac {\frac {a^{2}}{\cos \left (d x +c \right )}+2 a^{2} \tan \left (d x +c \right )+a^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(58\)
default	\(\frac {\frac {a^{2}}{\cos \left (d x +c \right )}+2 a^{2} \tan \left (d x +c \right )+a^{2} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(58\)
risch	\(\frac {4 a^{2}}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(63\)
norman	\(\frac {-\frac {4 a^{2}}{d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(152\)

[In]

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

[Out]

(-4+ln(tan(1/2*d*x+1/2*c))*(tan(1/2*d*x+1/2*c)-1))*a^2/d/(tan(1/2*d*x+1/2*c)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (44) = 88\).

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.86 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2} \sin \left (d x + c\right ) + 4 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right ) - a^{2} \sin \left (d x + c\right ) + a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (a^{2} \cos \left (d x + c\right ) - a^{2} \sin \left (d x + c\right ) + a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} {\left (\frac {2}{\cos \left (d x + c\right )} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 4 \, a^{2} \tan \left (d x + c\right ) + \frac {2 \, a^{2}}{\cos \left (d x + c\right )}}{2 \, d} \]

[In]

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

[Out]

1/2*(a^2*(2/cos(d*x + c) - log(cos(d*x + c) + 1) + log(cos(d*x + c) - 1)) + 4*a^2*tan(d*x + c) + 2*a^2/cos(d*x
+ c))/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {4 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

[In]

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

[Out]

(a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 4*a^2/(tan(1/2*d*x + 1/2*c) - 1))/d

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \csc (c+d x) \sec ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {4\,a^2}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

[In]

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

[Out]

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