\(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\) [877]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {6 a^3 \log (1-\sin (c+d x))}{d}+\frac {6 a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {3 a^4}{d (a-a \sin (c+d x))} \]

[Out]

-3*a^3*csc(d*x+c)/d-1/2*a^3*csc(d*x+c)^2/d-6*a^3*ln(1-sin(d*x+c))/d+6*a^3*ln(sin(d*x+c))/d+1/2*a^5/d/(a-a*sin(
d*x+c))^2+3*a^4/d/(a-a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 46} \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {3 a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {3 a^3 \csc (c+d x)}{d}-\frac {6 a^3 \log (1-\sin (c+d x))}{d}+\frac {6 a^3 \log (\sin (c+d x))}{d} \]

[In]

Int[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

(-3*a^3*Csc[c + d*x])/d - (a^3*Csc[c + d*x]^2)/(2*d) - (6*a^3*Log[1 - Sin[c + d*x]])/d + (6*a^3*Log[Sin[c + d*
x]])/d + a^5/(2*d*(a - a*Sin[c + d*x])^2) + (3*a^4)/(d*(a - a*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 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2915

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {a^5 \text {Subst}\left (\int \frac {a^3}{(a-x)^3 x^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^8 \text {Subst}\left (\int \frac {1}{(a-x)^3 x^3} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^8 \text {Subst}\left (\int \left (\frac {1}{a^3 (a-x)^3}+\frac {3}{a^4 (a-x)^2}+\frac {6}{a^5 (a-x)}+\frac {1}{a^3 x^3}+\frac {3}{a^4 x^2}+\frac {6}{a^5 x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {3 a^3 \csc (c+d x)}{d}-\frac {a^3 \csc ^2(c+d x)}{2 d}-\frac {6 a^3 \log (1-\sin (c+d x))}{d}+\frac {6 a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {3 a^4}{d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.66 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \left (6 \csc (c+d x)+\csc ^2(c+d x)+12 \log (1-\sin (c+d x))-12 \log (\sin (c+d x))-\frac {1}{(-1+\sin (c+d x))^2}+\frac {6}{-1+\sin (c+d x)}\right )}{2 d} \]

[In]

Integrate[Csc[c + d*x]^3*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.60

method	result	size
risch	\(-\frac {4 i \left (-9 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+16 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-13 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-9 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+13 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-3 a^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {12 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {6 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(178\)
parallelrisch	\(\frac {\left (\left (-48 \cos \left (2 d x +2 c \right )-192 \sin \left (d x +c \right )+144\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (24 \cos \left (2 d x +2 c \right )+96 \sin \left (d x +c \right )-72\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-136 \cos \left (d x +c \right )+34 \cos \left (2 d x +2 c \right )+106\right ) \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (88 \cos \left (d x +c \right )-88\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-4\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{4 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(195\)
derivativedivides	\(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(215\)
default	\(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(215\)

[In]

int(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-4*I*(-9*I*a^3*exp(6*I*(d*x+c))+3*a^3*exp(7*I*(d*x+c))+16*I*a^3*exp(4*I*(d*x+c))-13*a^3*exp(5*I*(d*x+c))-9*I*a
^3*exp(2*I*(d*x+c))+13*a^3*exp(3*I*(d*x+c))-3*a^3*exp(I*(d*x+c)))/(exp(2*I*(d*x+c))-1)^2/(exp(I*(d*x+c))-I)^4/
d-12*a^3/d*ln(exp(I*(d*x+c))-I)+6*a^3/d*ln(exp(2*I*(d*x+c))-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (109) = 218\).

Time = 0.29 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.12 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {18 \, a^{3} \cos \left (d x + c\right )^{2} - 17 \, a^{3} - 12 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 3 \, a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (a^{3} \cos \left (d x + c\right )^{4} - 3 \, a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, {\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right )^{4} - 3 \, d \cos \left (d x + c\right )^{2} + 2 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right ) + 2 \, d\right )}} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(18*a^3*cos(d*x + c)^2 - 17*a^3 - 12*(a^3*cos(d*x + c)^4 - 3*a^3*cos(d*x + c)^2 + 2*a^3 + 2*(a^3*cos(d*x
+ c)^2 - a^3)*sin(d*x + c))*log(1/2*sin(d*x + c)) + 12*(a^3*cos(d*x + c)^4 - 3*a^3*cos(d*x + c)^2 + 2*a^3 + 2*
(a^3*cos(d*x + c)^2 - a^3)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 4*(3*a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c
))/(d*cos(d*x + c)^4 - 3*d*cos(d*x + c)^2 + 2*(d*cos(d*x + c)^2 - d)*sin(d*x + c) + 2*d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**5*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {12 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 12 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac {12 \, a^{3} \sin \left (d x + c\right )^{3} - 18 \, a^{3} \sin \left (d x + c\right )^{2} + 4 \, a^{3} \sin \left (d x + c\right ) + a^{3}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )^{2}}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(12*a^3*log(sin(d*x + c) - 1) - 12*a^3*log(sin(d*x + c)) + (12*a^3*sin(d*x + c)^3 - 18*a^3*sin(d*x + c)^2
+ 4*a^3*sin(d*x + c) + a^3)/(sin(d*x + c)^4 - 2*sin(d*x + c)^3 + sin(d*x + c)^2))/d

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.78 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 96 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 48 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {72 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} - \frac {8 \, {\left (25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 92 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 92 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{8 \, d} \]

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 + 96*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 48*a^3*log(abs(tan(1/2*d*x + 1/
2*c))) + 12*a^3*tan(1/2*d*x + 1/2*c) + (72*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan
(1/2*d*x + 1/2*c)^2 - 8*(25*a^3*tan(1/2*d*x + 1/2*c)^4 - 92*a^3*tan(1/2*d*x + 1/2*c)^3 + 136*a^3*tan(1/2*d*x +
1/2*c)^2 - 92*a^3*tan(1/2*d*x + 1/2*c) + 25*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {12\,a^3\,\mathrm {atanh}\left (2\,\sin \left (c+d\,x\right )-1\right )}{d}-\frac {6\,a^3\,{\sin \left (c+d\,x\right )}^3-9\,a^3\,{\sin \left (c+d\,x\right )}^2+2\,a^3\,\sin \left (c+d\,x\right )+\frac {a^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2\right )} \]

[In]

int((a + a*sin(c + d*x))^3/(cos(c + d*x)^5*sin(c + d*x)^3),x)

[Out]

(12*a^3*atanh(2*sin(c + d*x) - 1))/d - (2*a^3*sin(c + d*x) + a^3/2 - 9*a^3*sin(c + d*x)^2 + 6*a^3*sin(c + d*x)
^3)/(d*(sin(c + d*x)^2 - 2*sin(c + d*x)^3 + sin(c + d*x)^4))