\(\int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx\) [859]

Optimal result

Integrand size = 27, antiderivative size = 143 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {39 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}-\frac {9 a \log (1+\sin (c+d x))}{16 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))} \]

[Out]

-a*csc(d*x+c)/d-1/2*a*csc(d*x+c)^2/d-39/16*a*ln(1-sin(d*x+c))/d+3*a*ln(sin(d*x+c))/d-9/16*a*ln(1+sin(d*x+c))/d
+1/8*a^3/d/(a-a*sin(d*x+c))^2+a^2/d/(a-a*sin(d*x+c))+1/8*a^2/d/(a+a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 143, 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, 90} \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x)}{d}-\frac {39 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}-\frac {9 a \log (\sin (c+d x)+1)}{16 d} \]

[In]

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

[Out]

-((a*Csc[c + d*x])/d) - (a*Csc[c + d*x]^2)/(2*d) - (39*a*Log[1 - Sin[c + d*x]])/(16*d) + (3*a*Log[Sin[c + d*x]
])/d - (9*a*Log[1 + Sin[c + d*x]])/(16*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + a^2/(d*(a - a*Sin[c + d*x])) +
a^2/(8*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 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^8 \text {Subst}\left (\int \frac {1}{(a-x)^3 x^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^8 \text {Subst}\left (\int \left (\frac {1}{4 a^5 (a-x)^3}+\frac {1}{a^6 (a-x)^2}+\frac {39}{16 a^7 (a-x)}+\frac {1}{a^5 x^3}+\frac {1}{a^6 x^2}+\frac {3}{a^7 x}-\frac {1}{8 a^6 (a+x)^2}-\frac {9}{16 a^7 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a \csc (c+d x)}{d}-\frac {a \csc ^2(c+d x)}{2 d}-\frac {39 a \log (1-\sin (c+d x))}{16 d}+\frac {3 a \log (\sin (c+d x))}{d}-\frac {9 a \log (1+\sin (c+d x))}{16 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.70 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \csc ^2(c+d x)}{2 d}-\frac {a \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},\sin ^2(c+d x)\right )}{d}-\frac {3 a \log (\cos (c+d x))}{d}+\frac {3 a \log (\sin (c+d x))}{d}+\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec ^4(c+d x)}{4 d} \]

[In]

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

[Out]

-1/2*(a*Csc[c + d*x]^2)/d - (a*Csc[c + d*x]*Hypergeometric2F1[-1/2, 3, 1/2, Sin[c + d*x]^2])/d - (3*a*Log[Cos[
c + d*x]])/d + (3*a*Log[Sin[c + d*x]])/d + (a*Sec[c + d*x]^2)/d + (a*Sec[c + d*x]^4)/(4*d)

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(129\)
default	\(\frac {a \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )}{d}\)	\(129\)
risch	\(-\frac {i \left (-6 i a \,{\mathrm e}^{8 i \left (d x +c \right )}+15 a \,{\mathrm e}^{9 i \left (d x +c \right )}-14 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+28 a \,{\mathrm e}^{7 i \left (d x +c \right )}+14 i a \,{\mathrm e}^{4 i \left (d x +c \right )}-22 a \,{\mathrm e}^{5 i \left (d x +c \right )}+6 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+28 a \,{\mathrm e}^{3 i \left (d x +c \right )}+15 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {39 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(218\)
norman	\(\frac {-\frac {a}{8 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {13 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {5 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {13 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {a \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {21 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {21 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {27 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {27 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {3 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {39 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}-\frac {9 a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(305\)
parallelrisch	\(-\frac {3 \left (26 \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+16 \left (\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )-2 \cos \left (2 d x +2 c \right )-2\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {8 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-5\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )+\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )+\frac {17 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )}{12}+\frac {17 \cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )}{12}+\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {32 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (d x +c \right )-\frac {\cos \left (2 d x +2 c \right )}{4}+\frac {1}{2}\right )}{3}\right ) a}{16 d \left (2-\sin \left (3 d x +3 c \right )-\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )\right )}\)	\(305\)

[In]

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

[Out]

1/d*(a*(1/4/sin(d*x+c)/cos(d*x+c)^4+5/8/sin(d*x+c)/cos(d*x+c)^2-15/8/sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c))
)+a*(1/4/sin(d*x+c)^2/cos(d*x+c)^4+3/4/sin(d*x+c)^2/cos(d*x+c)^2-3/2/sin(d*x+c)^2+3*ln(tan(d*x+c))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (136) = 272\).

Time = 0.31 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.06 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {30 \, a \cos \left (d x + c\right )^{4} - 16 \, a \cos \left (d x + c\right )^{2} + 48 \, {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 9 \, {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 39 \, {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2} - {\left (a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

[In]

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

[Out]

1/16*(30*a*cos(d*x + c)^4 - 16*a*cos(d*x + c)^2 + 48*(a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - (a*cos(d*x + c)^4
- a*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) - 9*(a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - (a*cos(d*x
+ c)^4 - a*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 39*(a*cos(d*x + c)^4 - a*cos(d*x + c)^2 - (a*
cos(d*x + c)^4 - a*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 2*(3*a*cos(d*x + c)^2 + a)*sin(d*x +
c) - 6*a)/(d*cos(d*x + c)^4 - d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - d*cos(d*x + c)^2)*sin(d*x + c))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {9 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 39 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 48 \, a \log \left (\sin \left (d x + c\right )\right ) + \frac {2 \, {\left (15 \, a \sin \left (d x + c\right )^{4} - 3 \, a \sin \left (d x + c\right )^{3} - 22 \, a \sin \left (d x + c\right )^{2} + 4 \, a \sin \left (d x + c\right ) + 4 \, a\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{4} - \sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )^{2}}}{16 \, d} \]

[In]

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

[Out]

-1/16*(9*a*log(sin(d*x + c) + 1) + 39*a*log(sin(d*x + c) - 1) - 48*a*log(sin(d*x + c)) + 2*(15*a*sin(d*x + c)^
4 - 3*a*sin(d*x + c)^3 - 22*a*sin(d*x + c)^2 + 4*a*sin(d*x + c) + 4*a)/(sin(d*x + c)^5 - sin(d*x + c)^4 - sin(
d*x + c)^3 + sin(d*x + c)^2))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {36 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 156 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 192 \, a \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {4 \, {\left (9 \, a \sin \left (d x + c\right ) + 11 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {27 \, a \sin \left (d x + c\right )^{4} + 74 \, a \sin \left (d x + c\right )^{3} - 141 \, a \sin \left (d x + c\right )^{2} + 32 \, a}{{\left (\sin \left (d x + c\right )^{2} - \sin \left (d x + c\right )\right )}^{2}}}{64 \, d} \]

[In]

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

[Out]

-1/64*(36*a*log(abs(sin(d*x + c) + 1)) + 156*a*log(abs(sin(d*x + c) - 1)) - 192*a*log(abs(sin(d*x + c))) - 4*(
9*a*sin(d*x + c) + 11*a)/(sin(d*x + c) + 1) + (27*a*sin(d*x + c)^4 + 74*a*sin(d*x + c)^3 - 141*a*sin(d*x + c)^
2 + 32*a)/(sin(d*x + c)^2 - sin(d*x + c))^2)/d

Mupad [B] (verification not implemented)

Time = 10.72 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int \csc ^3(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {3\,a\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {39\,a\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{16\,d}-\frac {9\,a\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{16\,d}-\frac {\frac {15\,a\,{\sin \left (c+d\,x\right )}^4}{8}-\frac {3\,a\,{\sin \left (c+d\,x\right )}^3}{8}-\frac {11\,a\,{\sin \left (c+d\,x\right )}^2}{4}+\frac {a\,\sin \left (c+d\,x\right )}{2}+\frac {a}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^5-{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2\right )} \]

[In]

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

[Out]

(3*a*log(sin(c + d*x)))/d - (39*a*log(sin(c + d*x) - 1))/(16*d) - (9*a*log(sin(c + d*x) + 1))/(16*d) - (a/2 +
(a*sin(c + d*x))/2 - (11*a*sin(c + d*x)^2)/4 - (3*a*sin(c + d*x)^3)/8 + (15*a*sin(c + d*x)^4)/8)/(d*(sin(c + d
*x)^2 - sin(c + d*x)^3 - sin(c + d*x)^4 + sin(c + d*x)^5))