\(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx\) [876]

Optimal result

Integrand size = 29, antiderivative size = 93 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a^3 \log (1-\sin (c+d x))}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {2 a^4}{d (a-a \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 93, 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} \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {2 a^4}{d (a-a \sin (c+d x))}-\frac {a^3 \csc (c+d x)}{d}-\frac {3 a^3 \log (1-\sin (c+d x))}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d} \]

[In]

Int[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

-((a^3*Csc[c + d*x])/d) - (3*a^3*Log[1 - Sin[c + d*x]])/d + (3*a^3*Log[Sin[c + d*x]])/d + a^5/(2*d*(a - a*Sin[
c + d*x])^2) + (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^2}{(a-x)^3 x^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \frac {1}{(a-x)^3 x^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {1}{a^2 (a-x)^3}+\frac {2}{a^3 (a-x)^2}+\frac {3}{a^4 (a-x)}+\frac {1}{a^3 x^2}+\frac {3}{a^4 x}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a^3 \csc (c+d x)}{d}-\frac {3 a^3 \log (1-\sin (c+d x))}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}+\frac {2 a^4}{d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \left (-2 \csc (c+d x)-6 \log (1-\sin (c+d x))+6 \log (\sin (c+d x))+\frac {1}{(-1+\sin (c+d x))^2}-\frac {4}{-1+\sin (c+d x)}\right )}{2 d} \]

[In]

Integrate[Csc[c + d*x]^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3,x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.47

method	result	size
risch	\(-\frac {2 i a^{3} \left (-9 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}+9 i {\mathrm e}^{2 i \left (d x +c \right )}-10 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {6 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(137\)
parallelrisch	\(\frac {\left (\frac {9}{2}+6 \left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \left (\cos \left (d x +c \right )-1\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {9 \cos \left (2 d x +2 c \right )}{2}\right ) a^{3}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(146\)
derivativedivides	\(\frac {\frac {a^{3}}{4 \cos \left (d x +c \right )^{4}}+3 a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(168\)
default	\(\frac {\frac {a^{3}}{4 \cos \left (d x +c \right )^{4}}+3 a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{3} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(168\)

[In]

int(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-2*I*a^3/(exp(2*I*(d*x+c))-1)/(exp(I*(d*x+c))-I)^4/d*(-9*I*exp(4*I*(d*x+c))+3*exp(5*I*(d*x+c))+9*I*exp(2*I*(d*
x+c))-10*exp(3*I*(d*x+c))+3*exp(I*(d*x+c)))-6*a^3/d*ln(exp(I*(d*x+c))-I)+3*a^3/d*ln(exp(2*I*(d*x+c))-1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.99 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) - 8 \, a^{3} + 6 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 6 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} - {\left (a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (2 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(6*a^3*cos(d*x + c)^2 + 9*a^3*sin(d*x + c) - 8*a^3 + 6*(2*a^3*cos(d*x + c)^2 - 2*a^3 - (a^3*cos(d*x + c)^2
- 2*a^3)*sin(d*x + c))*log(1/2*sin(d*x + c)) - 6*(2*a^3*cos(d*x + c)^2 - 2*a^3 - (a^3*cos(d*x + c)^2 - 2*a^3)
*sin(d*x + c))*log(-sin(d*x + c) + 1))/(2*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 2*d)*sin(d*x + c) - 2*d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**2*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 = 90, normalized size of antiderivative = 0.97 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {6 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + \frac {6 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}}{\sin \left (d x + c\right )^{3} - 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.78 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {12 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {25 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 88 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 130 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 88 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 25 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{2 \, d} \]

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^5*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(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))) + a^3*tan(1/2*d*x + 1/2
*c) + (6*a^3*tan(1/2*d*x + 1/2*c) + a^3)/tan(1/2*d*x + 1/2*c) - (25*a^3*tan(1/2*d*x + 1/2*c)^4 - 88*a^3*tan(1/
2*d*x + 1/2*c)^3 + 130*a^3*tan(1/2*d*x + 1/2*c)^2 - 88*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 = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6\,a^3\,\mathrm {atanh}\left (2\,\sin \left (c+d\,x\right )-1\right )}{d}-\frac {3\,a^3\,{\sin \left (c+d\,x\right )}^2-\frac {9\,a^3\,\sin \left (c+d\,x\right )}{2}+a^3}{d\,\left ({\sin \left (c+d\,x\right )}^3-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\right )} \]

[In]

int((a + a*sin(c + d*x))^3/(cos(c + d*x)^5*sin(c + d*x)^2),x)

[Out]

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