\(\int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\) [875]

Optimal result

Integrand size = 27, antiderivative size = 77 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {a^4}{d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 77, 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 = {2915, 12, 46} \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d} \]

[In]

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

[Out]

-((a^3*Log[1 - Sin[c + d*x]])/d) + (a^3*Log[Sin[c + d*x]])/d + a^5/(2*d*(a - a*Sin[c + d*x])^2) + 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}{(a-x)^3 x} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \frac {1}{(a-x)^3 x} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^6 \text {Subst}\left (\int \left (\frac {1}{a (a-x)^3}+\frac {1}{a^2 (a-x)^2}+\frac {1}{a^3 (a-x)}+\frac {1}{a^3 x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a^3 \log (1-\sin (c+d x))}{d}+\frac {a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {a^4}{d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-2 \log (1-\sin (c+d x))+2 \log (\sin (c+d x))+\frac {3-2 \sin (c+d x)}{(-1+\sin (c+d x))^2}\right )}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(-\frac {a^{3} \left (-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+\ln \left (\sin \left (d x +c \right )-1\right )+\frac {1}{\sin \left (d x +c \right )-1}\right )}{d}\)	\(50\)
default	\(-\frac {a^{3} \left (-\ln \left (\sin \left (d x +c \right )\right )-\frac {1}{2 \left (\sin \left (d x +c \right )-1\right )^{2}}+\ln \left (\sin \left (d x +c \right )-1\right )+\frac {1}{\sin \left (d x +c \right )-1}\right )}{d}\)	\(50\)
risch	\(-\frac {2 i \left (-a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+a^{3} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {2 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(105\)
parallelrisch	\(\frac {\left (\left (-2 \cos \left (2 d x +2 c \right )+6-8 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {3 \cos \left (2 d x +2 c \right )}{2}-4 \sin \left (d x +c \right )+\frac {3}{2}\right ) a^{3}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(114\)
norman	\(\frac {\frac {10 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {26 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {26 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {16 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {32 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {28 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {16 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(316\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.64 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3} + 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {2 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, a^{3} \sin \left (d x + c\right ) - 3 \, a^{3}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.60 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - \frac {25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 76 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 114 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 76 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(12*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*a^3*log(abs(tan(1/2*d*x + 1/2*c))) - (25*a^3*tan(1/2*d*x +
1/2*c)^4 - 76*a^3*tan(1/2*d*x + 1/2*c)^3 + 114*a^3*tan(1/2*d*x + 1/2*c)^2 - 76*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 = 9.91 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.79 \[ \int \csc (c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {2\,a^3\,\mathrm {atanh}\left (2\,\sin \left (c+d\,x\right )-1\right )}{d}-\frac {a^3\,\sin \left (c+d\,x\right )-\frac {3\,a^3}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )} \]

[In]

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

[Out]

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