\(\int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\) [808]

Optimal result

Integrand size = 27, antiderivative size = 73 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2948, 2845, 3057, 12, 3855} \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))} \]

[In]

Int[Csc[c + d*x]*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

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

Rule 12

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

Rule 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2948

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

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = a^4 \int \frac {\csc (c+d x)}{(a-a \sin (c+d x))^2} \, dx \\ & = \frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} a^2 \int \frac {\csc (c+d x) (3 a+a \sin (c+d x))}{a-a \sin (c+d x)} \, dx \\ & = \frac {4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac {1}{3} \int 3 a^2 \csc (c+d x) \, dx \\ & = \frac {4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}+a^2 \int \csc (c+d x) \, dx \\ & = -\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {4 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.45 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (1+\sin (c+d x))^2 \left (-6 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \sec (c+d x)+4 \sec ^3(c+d x)+12 \tan (c+d x)+4 \tan ^3(c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

[In]

Integrate[Csc[c + d*x]*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*(1 + Sin[c + d*x])^2*(-6*Log[Cos[(c + d*x)/2]] + 6*Log[Sin[(c + d*x)/2]] + 6*Sec[c + d*x] + 4*Sec[c + d*x
]^3 + 12*Tan[c + d*x] + 4*Tan[c + d*x]^3))/(6*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\(\frac {\left (\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {10}{3}\right ) a^{2}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(71\)
derivativedivides	\(\frac {\frac {a^{2}}{3 \cos \left (d x +c \right )^{3}}-2 a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(81\)
default	\(\frac {\frac {a^{2}}{3 \cos \left (d x +c \right )^{3}}-2 a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} \left (\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {1}{\cos \left (d x +c \right )}+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\)	\(81\)
risch	\(\frac {2 a^{2} \left (-9 i {\mathrm e}^{i \left (d x +c \right )}+3 \,{\mathrm e}^{2 i \left (d x +c \right )}-4\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}\)	\(88\)
norman	\(\frac {-\frac {10 a^{2}}{3 d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {16 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\)	\(228\)

[In]

int(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

((tan(1/2*d*x+1/2*c)-1)^3*ln(tan(1/2*d*x+1/2*c))-4*tan(1/2*d*x+1/2*c)^2+6*tan(1/2*d*x+1/2*c)-10/3)*a^2/d/(tan(
1/2*d*x+1/2*c)-1)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (68) = 136\).

Time = 0.28 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.16 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {8 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 2 \, a^{2} + 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2} \cos \left (d x + c\right ) - 2 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right ) + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right ) - a^{2}\right )} \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(8*a^2*cos(d*x + c)^2 + 10*a^2*cos(d*x + c) + 2*a^2 + 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c) - 2*a^2 +
(a^2*cos(d*x + c) + 2*a^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*(a^2*cos(d*x + c)^2 - a^2*cos(d*x + c
) - 2*a^2 + (a^2*cos(d*x + c) + 2*a^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) - 2*(4*a^2*cos(d*x + c) - a^
2)*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)**4*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.23 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} + 1\right )}}{\cos \left (d x + c\right )^{3}} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, a^{2}}{\cos \left (d x + c\right )^{3}}}{6 \, d} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2 + a^2*(2*(3*cos(d*x + c)^2 + 1)/cos(d*x + c)^3 - 3*log(cos(d*x +
c) + 1) + 3*log(cos(d*x + c) - 1)) + 2*a^2/cos(d*x + c)^3)/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {2 \, {\left (6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

[In]

integrate(csc(d*x+c)*sec(d*x+c)^4*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(3*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 2*(6*a^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*tan(1/2*d*x + 1/2*c) + 5*a
^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 10.63 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.34 \[ \int \csc (c+d x) \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-6\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {10\,a^2}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

[In]

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^4*sin(c + d*x)),x)

[Out]

(a^2*log(tan(c/2 + (d*x)/2)))/d - (4*a^2*tan(c/2 + (d*x)/2)^2 + (10*a^2)/3 - 6*a^2*tan(c/2 + (d*x)/2))/(d*(3*t
an(c/2 + (d*x)/2) - 3*tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^3 - 1))