\(\int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\) [869]

Optimal result

Integrand size = 29, antiderivative size = 150 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {4 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {49 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {6 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (1+\sin (c+d x))}{8 d}+\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {9 a^3}{4 d (a-a \sin (c+d x))} \]

[Out]

-4*a^2*csc(d*x+c)/d-a^2*csc(d*x+c)^2/d-1/3*a^2*csc(d*x+c)^3/d-49/8*a^2*ln(1-sin(d*x+c))/d+6*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+9/4*a^3/d/(a-a*sin(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 150, 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} \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {9 a^3}{4 d (a-a \sin (c+d x))}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {4 a^2 \csc (c+d x)}{d}-\frac {49 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {6 a^2 \log (\sin (c+d x))}{d}+\frac {a^2 \log (\sin (c+d x)+1)}{8 d} \]

[In]

Int[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(-4*a^2*Csc[c + d*x])/d - (a^2*Csc[c + d*x]^2)/d - (a^2*Csc[c + d*x]^3)/(3*d) - (49*a^2*Log[1 - Sin[c + d*x]])
/(8*d) + (6*a^2*Log[Sin[c + d*x]])/d + (a^2*Log[1 + Sin[c + d*x]])/(8*d) + a^4/(4*d*(a - a*Sin[c + d*x])^2) +
(9*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^4}{(a-x)^3 x^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^9 \text {Subst}\left (\int \frac {1}{(a-x)^3 x^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {a^9 \text {Subst}\left (\int \left (\frac {1}{2 a^5 (a-x)^3}+\frac {9}{4 a^6 (a-x)^2}+\frac {49}{8 a^7 (a-x)}+\frac {1}{a^4 x^4}+\frac {2}{a^5 x^3}+\frac {4}{a^6 x^2}+\frac {6}{a^7 x}+\frac {1}{8 a^7 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d} \\ & = -\frac {4 a^2 \csc (c+d x)}{d}-\frac {a^2 \csc ^2(c+d x)}{d}-\frac {a^2 \csc ^3(c+d x)}{3 d}-\frac {49 a^2 \log (1-\sin (c+d x))}{8 d}+\frac {6 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 {9 a^3}{4 d (a-a \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.04 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.89 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^5 \left (-\frac {4 \csc (c+d x)}{a^3}-\frac {\csc ^2(c+d x)}{a^3}-\frac {\csc ^3(c+d x)}{3 a^3}-\frac {49 \log (1-\sin (c+d x))}{8 a^3}+\frac {6 \log (\sin (c+d x))}{a^3}+\frac {\log (1+\sin (c+d x))}{8 a^3}+\frac {1}{4 a (a-a \sin (c+d x))^2}+\frac {9}{4 a^2 (a-a \sin (c+d x))}\right )}{d} \]

[In]

Integrate[Csc[c + d*x]^4*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^5*((-4*Csc[c + d*x])/a^3 - Csc[c + d*x]^2/a^3 - Csc[c + d*x]^3/(3*a^3) - (49*Log[1 - Sin[c + d*x]])/(8*a^3)
+ (6*Log[Sin[c + d*x]])/a^3 + Log[1 + Sin[c + d*x]]/(8*a^3) + 1/(4*a*(a - a*Sin[c + d*x])^2) + 9/(4*a^2*(a -
a*Sin[c + d*x]))))/d

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.36

method	result	size
risch	\(-\frac {i a^{2} \left (-228 i {\mathrm e}^{8 i \left (d x +c \right )}+75 \,{\mathrm e}^{9 i \left (d x +c \right )}+652 i {\mathrm e}^{6 i \left (d x +c \right )}-412 \,{\mathrm e}^{7 i \left (d x +c \right )}-652 i {\mathrm e}^{4 i \left (d x +c \right )}+738 \,{\mathrm e}^{5 i \left (d x +c \right )}+228 i {\mathrm e}^{2 i \left (d x +c \right )}-412 \,{\mathrm e}^{3 i \left (d x +c \right )}+75 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{6 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {49 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}+\frac {6 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\)	\(204\)
derivativedivides	\(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7}{12 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {35}{24 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {35}{8 \sin \left (d x +c \right )}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(219\)
default	\(\frac {a^{2} \left (\frac {1}{4 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}+\frac {5}{8 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {15}{8 \sin \left (d x +c \right )}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{4}}+\frac {3}{4 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right )^{2}}-\frac {3}{2 \sin \left (d x +c \right )^{2}}+3 \ln \left (\tan \left (d x +c \right )\right )\right )+a^{2} \left (\frac {1}{4 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}-\frac {7}{12 \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2}}+\frac {35}{24 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}-\frac {35}{8 \sin \left (d x +c \right )}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(219\)
parallelrisch	\(-\frac {53 \left (\frac {588 \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{53}+\frac {12 \left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{53}+\frac {288 \left (3-\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{53}+\frac {4 \left (-\left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-28 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{53}+\left (\left (\cos \left (\frac {5 d x}{2}+\frac {5 c}{2}\right )-\frac {55 \cos \left (\frac {7 d x}{2}+\frac {7 c}{2}\right )}{212}-\frac {55 \cos \left (\frac {9 d x}{2}+\frac {9 c}{2}\right )}{212}-\frac {173 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{106}+\cos \left (\frac {3 d x}{2}+\frac {3 c}{2}\right )\right ) \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2432 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{53}\right ) \left (\csc ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {1160 \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\cos \left (d x +c \right )-3\right )}{53}\right ) a^{2}}{48 d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\)	\(279\)

[In]

int(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/6*I*a^2/(exp(2*I*(d*x+c))-1)^3/(exp(I*(d*x+c))-I)^4/d*(-228*I*exp(8*I*(d*x+c))+75*exp(9*I*(d*x+c))+652*I*ex
p(6*I*(d*x+c))-412*exp(7*I*(d*x+c))-652*I*exp(4*I*(d*x+c))+738*exp(5*I*(d*x+c))+228*I*exp(2*I*(d*x+c))-412*exp
(3*I*(d*x+c))+75*exp(I*(d*x+c)))-49/4*a^2/d*ln(exp(I*(d*x+c))-I)+1/4*a^2/d*ln(exp(I*(d*x+c))+I)+6*a^2/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. 370 vs. \(2 (142) = 284\).

Time = 0.30 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.47 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {150 \, a^{2} \cos \left (d x + c\right )^{4} - 356 \, a^{2} \cos \left (d x + c\right )^{2} + 214 \, a^{2} + 144 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 3 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 147 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{4} - 4 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} - {\left (a^{2} \cos \left (d x + c\right )^{4} - 3 \, a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, {\left (57 \, a^{2} \cos \left (d x + c\right )^{2} - 55 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, {\left (2 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - 3 \, d \cos \left (d x + c\right )^{2} + 2 \, d\right )} \sin \left (d x + c\right ) + 2 \, d\right )}} \]

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(150*a^2*cos(d*x + c)^4 - 356*a^2*cos(d*x + c)^2 + 214*a^2 + 144*(2*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x +
c)^2 + 2*a^2 - (a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 + 2*a^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) + 3*(2*
a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 + 2*a^2 - (a^2*cos(d*x + c)^4 - 3*a^2*cos(d*x + c)^2 + 2*a^2)*sin(d*
x + c))*log(sin(d*x + c) + 1) - 147*(2*a^2*cos(d*x + c)^4 - 4*a^2*cos(d*x + c)^2 + 2*a^2 - (a^2*cos(d*x + c)^4
- 3*a^2*cos(d*x + c)^2 + 2*a^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 4*(57*a^2*cos(d*x + c)^2 - 55*a^2)*sin
(d*x + c))/(2*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 - (d*cos(d*x + c)^4 - 3*d*cos(d*x + c)^2 + 2*d)*sin(d*x +
c) + 2*d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

[In]

integrate(csc(d*x+c)**4*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 = 133, normalized size of antiderivative = 0.89 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {3 \, a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 147 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 144 \, a^{2} \log \left (\sin \left (d x + c\right )\right ) - \frac {2 \, {\left (75 \, a^{2} \sin \left (d x + c\right )^{4} - 114 \, a^{2} \sin \left (d x + c\right )^{3} + 28 \, a^{2} \sin \left (d x + c\right )^{2} + 4 \, a^{2} \sin \left (d x + c\right ) + 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{5} - 2 \, \sin \left (d x + c\right )^{4} + \sin \left (d x + c\right )^{3}}}{24 \, d} \]

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/24*(3*a^2*log(sin(d*x + c) + 1) - 147*a^2*log(sin(d*x + c) - 1) + 144*a^2*log(sin(d*x + c)) - 2*(75*a^2*sin(
d*x + c)^4 - 114*a^2*sin(d*x + c)^3 + 28*a^2*sin(d*x + c)^2 + 4*a^2*sin(d*x + c) + 4*a^2)/(sin(d*x + c)^5 - 2*
sin(d*x + c)^4 + sin(d*x + c)^3))/d

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 294 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 288 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + \frac {3 \, {\left (147 \, a^{2} \sin \left (d x + c\right )^{2} - 330 \, a^{2} \sin \left (d x + c\right ) + 187 \, a^{2}\right )}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {16 \, {\left (33 \, a^{2} \sin \left (d x + c\right )^{3} + 12 \, a^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{2} \sin \left (d x + c\right ) + a^{2}\right )}}{\sin \left (d x + c\right )^{3}}}{48 \, d} \]

[In]

integrate(csc(d*x+c)^4*sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/48*(6*a^2*log(abs(sin(d*x + c) + 1)) - 294*a^2*log(abs(sin(d*x + c) - 1)) + 288*a^2*log(abs(sin(d*x + c))) +
3*(147*a^2*sin(d*x + c)^2 - 330*a^2*sin(d*x + c) + 187*a^2)/(sin(d*x + c) - 1)^2 - 16*(33*a^2*sin(d*x + c)^3
+ 12*a^2*sin(d*x + c)^2 + 3*a^2*sin(d*x + c) + a^2)/sin(d*x + c)^3)/d

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.93 \[ \int \csc ^4(c+d x) \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{8\,d}-\frac {49\,a^2\,\ln \left (\sin \left (c+d\,x\right )-1\right )}{8\,d}-\frac {\frac {25\,a^2\,{\sin \left (c+d\,x\right )}^4}{4}-\frac {19\,a^2\,{\sin \left (c+d\,x\right )}^3}{2}+\frac {7\,a^2\,{\sin \left (c+d\,x\right )}^2}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{3}+\frac {a^2}{3}}{d\,\left ({\sin \left (c+d\,x\right )}^5-2\,{\sin \left (c+d\,x\right )}^4+{\sin \left (c+d\,x\right )}^3\right )}+\frac {6\,a^2\,\ln \left (\sin \left (c+d\,x\right )\right )}{d} \]

[In]

int((a + a*sin(c + d*x))^2/(cos(c + d*x)^5*sin(c + d*x)^4),x)

[Out]

(a^2*log(sin(c + d*x) + 1))/(8*d) - (49*a^2*log(sin(c + d*x) - 1))/(8*d) - ((a^2*sin(c + d*x))/3 + a^2/3 + (7*
a^2*sin(c + d*x)^2)/3 - (19*a^2*sin(c + d*x)^3)/2 + (25*a^2*sin(c + d*x)^4)/4)/(d*(sin(c + d*x)^3 - 2*sin(c +
d*x)^4 + sin(c + d*x)^5)) + (6*a^2*log(sin(c + d*x)))/d