\(\int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\) [867]

Optimal result

Integrand size = 29, antiderivative size = 116 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \csc (c+d x)}{d}-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {5 a^3}{4 d (a-a \sin (c+d x))} \]

[Out]

-a^2*csc(d*x+c)/d-17/8*a^2*ln(1-sin(d*x+c))/d+2*a^2*ln(sin(d*x+c))/d+1/8*a^2*ln(1+sin(d*x+c))/d+1/4*a^4/d/(a-a
*sin(d*x+c))^2+5/4*a^3/d/(a-a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 116, 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, 90} \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {5 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc (c+d x)}{d}-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[In]

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

[Out]

-((a^2*Csc[c + d*x])/d) - (17*a^2*Log[1 - Sin[c + d*x]])/(8*d) + (2*a^2*Log[Sin[c + d*x]])/d + (a^2*Log[1 + Si
n[c + d*x]])/(8*d) + a^4/(4*d*(a - a*Sin[c + d*x])^2) + (5*a^3)/(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 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^2}{(a-x)^3 x^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \frac {1}{(a-x)^3 x^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^7 \text {Subst}\left (\int \left (\frac {1}{2 a^3 (a-x)^3}+\frac {5}{4 a^4 (a-x)^2}+\frac {17}{8 a^5 (a-x)}+\frac {1}{a^4 x^2}+\frac {2}{a^5 x}+\frac {1}{8 a^5 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {a^2 \csc (c+d x)}{d}-\frac {17 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {2 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {5 a^3}{4 d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.64 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-8 \csc (c+d x)-17 \log (1-\sin (c+d x))+16 \log (\sin (c+d x))+\log (1+\sin (c+d x))+\frac {2}{(-1+\sin (c+d x))^2}-\frac {10}{-1+\sin (c+d x)}\right )}{8 d} \]

[In]

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

[Out]

(a^2*(-8*Csc[c + d*x] - 17*Log[1 - Sin[c + d*x]] + 16*Log[Sin[c + d*x]] + Log[1 + Sin[c + d*x]] + 2/(-1 + Sin[
c + d*x])^2 - 10/(-1 + Sin[c + d*x])))/(8*d)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(154\)
default	\(\frac {a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (\frac {1}{4 \cos \left (d x +c \right )^{4}}+\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(154\)
risch	\(-\frac {i a^{2} \left (-28 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{5 i \left (d x +c \right )}+28 i {\mathrm e}^{2 i \left (d x +c \right )}-34 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{2 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {17 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(158\)
parallelrisch	\(\frac {\left (\frac {7}{2}+17 \left (\frac {3}{4}-\frac {\cos \left (2 d x +2 c \right )}{4}-\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-\frac {3}{4}+\frac {\cos \left (2 d x +2 c \right )}{4}+\sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {19 \left (\cos \left (d x +c \right )-1\right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+\sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {7 \cos \left (2 d x +2 c \right )}{2}\right ) a^{2}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(178\)
norman	\(\frac {\frac {8 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2}}{2 d}+\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {9 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {7 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {7 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {9 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {8 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {17 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(320\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (110) = 220\).

Time = 0.28 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.07 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {18 \, a^{2} \cos \left (d x + c\right )^{2} + 28 \, a^{2} \sin \left (d x + c\right ) - 26 \, a^{2} + 16 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 17 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (2 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

[In]

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

[Out]

1/8*(18*a^2*cos(d*x + c)^2 + 28*a^2*sin(d*x + c) - 26*a^2 + 16*(2*a^2*cos(d*x + c)^2 - 2*a^2 - (a^2*cos(d*x +
c)^2 - 2*a^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) + (2*a^2*cos(d*x + c)^2 - 2*a^2 - (a^2*cos(d*x + c)^2 - 2*a^
2)*sin(d*x + c))*log(sin(d*x + c) + 1) - 17*(2*a^2*cos(d*x + c)^2 - 2*a^2 - (a^2*cos(d*x + c)^2 - 2*a^2)*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. \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**2*sec(d*x+c)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 17 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (9 \, a^{2} \sin \left (d x + c\right )^{2} - 14 \, a^{2} \sin \left (d x + c\right ) + 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{3} - 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right )}}{8 \, d} \]

[In]

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

[Out]

1/8*(a^2*log(sin(d*x + c) + 1) - 17*a^2*log(sin(d*x + c) - 1) + 16*a^2*log(sin(d*x + c)) - 2*(9*a^2*sin(d*x +
c)^2 - 14*a^2*sin(d*x + c) + 4*a^2)/(sin(d*x + c)^3 - 2*sin(d*x + c)^2 + sin(d*x + c)))/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.99 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 34 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 32 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {16 \, {\left (2 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )} + \frac {51 \, a^{2} \sin \left (d x + c\right )^{2} - 122 \, a^{2} \sin \left (d x + c\right ) + 75 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

[In]

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

[Out]

1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) - 34*a^2*log(abs(sin(d*x + c) - 1)) + 32*a^2*log(abs(sin(d*x + c))) - 1
6*(2*a^2*sin(d*x + c) + a^2)/sin(d*x + c) + (51*a^2*sin(d*x + c)^2 - 122*a^2*sin(d*x + c) + 75*a^2)/(sin(d*x +
c) - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 9.86 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.95 \[ \int \csc ^2(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {17\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d}+\frac {2\,a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d}-\frac {\frac {9\,a^2\,{\sin \left (c+d\,x\right )}^2}{4}-\frac {7\,a^2\,\sin \left (c+d\,x\right )}{2}+a^2}{d\,\left ({\sin \left (c+d\,x\right )}^3-2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )\right )} \]

[In]

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

[Out]

(a^2*log(sin(c + d*x) + 1))/(8*d) - (17*a^2*log(sin(c + d*x) - 1))/(8*d) + (2*a^2*log(sin(c + d*x)))/d - (a^2
- (7*a^2*sin(c + d*x))/2 + (9*a^2*sin(c + d*x)^2)/4)/(d*(sin(c + d*x) - 2*sin(c + d*x)^2 + sin(c + d*x)^3))