\(\int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\) [243]

Optimal result

Integrand size = 32, antiderivative size = 153 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {18 A \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {10 A \cot (c+d x)}{a^3 d}-\frac {A \cot ^3(c+d x)}{3 a^3 d}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {93 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \]

[Out]

18*A*arctanh(cos(d*x+c))/a^3/d-10*A*cot(d*x+c)/a^3/d-1/3*A*cot(d*x+c)^3/a^3/d+2*A*cot(d*x+c)*csc(d*x+c)/a^3/d-
2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3-13/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2-93/5*A*cos(d*x+c)/a^3/d/(1+si
n(d*x+c))

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3045, 3855, 3852, 8, 3853, 2729, 2727} \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {18 A \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {A \cot ^3(c+d x)}{3 a^3 d}-\frac {10 A \cot (c+d x)}{a^3 d}-\frac {93 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}-\frac {13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d} \]

[In]

Int[(Csc[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(18*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (10*A*Cot[c + d*x])/(a^3*d) - (A*Cot[c + d*x]^3)/(3*a^3*d) + (2*A*Cot[c
+ d*x]*Csc[c + d*x])/(a^3*d) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) - (13*A*Cos[c + d*x])/(5*a^3
*d*(1 + Sin[c + d*x])^2) - (93*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3045

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {16 A \csc (c+d x)}{a^3}+\frac {9 A \csc ^2(c+d x)}{a^3}-\frac {4 A \csc ^3(c+d x)}{a^3}+\frac {A \csc ^4(c+d x)}{a^3}+\frac {2 A}{a^3 (1+\sin (c+d x))^3}+\frac {7 A}{a^3 (1+\sin (c+d x))^2}+\frac {16 A}{a^3 (1+\sin (c+d x))}\right ) \, dx \\ & = \frac {A \int \csc ^4(c+d x) \, dx}{a^3}+\frac {(2 A) \int \frac {1}{(1+\sin (c+d x))^3} \, dx}{a^3}-\frac {(4 A) \int \csc ^3(c+d x) \, dx}{a^3}+\frac {(7 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{a^3}+\frac {(9 A) \int \csc ^2(c+d x) \, dx}{a^3}-\frac {(16 A) \int \csc (c+d x) \, dx}{a^3}+\frac {(16 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3} \\ & = \frac {16 A \text {arctanh}(\cos (c+d x))}{a^3 d}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {7 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}-\frac {16 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {(4 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}-\frac {(2 A) \int \csc (c+d x) \, dx}{a^3}+\frac {(7 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{3 a^3}-\frac {A \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a^3 d}-\frac {(9 A) \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d} \\ & = \frac {18 A \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {10 A \cot (c+d x)}{a^3 d}-\frac {A \cot ^3(c+d x)}{3 a^3 d}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {55 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}+\frac {(4 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{15 a^3} \\ & = \frac {18 A \text {arctanh}(\cos (c+d x))}{a^3 d}-\frac {10 A \cot (c+d x)}{a^3 d}-\frac {A \cot ^3(c+d x)}{3 a^3 d}+\frac {2 A \cot (c+d x) \csc (c+d x)}{a^3 d}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {93 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(348\) vs. \(2(153)=306\).

Time = 6.60 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.27 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {A \left (-\frac {29 \cot \left (\frac {1}{2} (c+d x)\right )}{6 d}+\frac {\csc ^2\left (\frac {1}{2} (c+d x)\right )}{2 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 d}+\frac {18 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {18 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right )}{2 d}+\frac {4 \sin \left (\frac {1}{2} (c+d x)\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}-\frac {2}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {26 \sin \left (\frac {1}{2} (c+d x)\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {13}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {186 \sin \left (\frac {1}{2} (c+d x)\right )}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {29 \tan \left (\frac {1}{2} (c+d x)\right )}{6 d}+\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{24 d}\right )}{a^3} \]

[In]

Integrate[(Csc[c + d*x]^4*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(A*((-29*Cot[(c + d*x)/2])/(6*d) + Csc[(c + d*x)/2]^2/(2*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*d) + (
18*Log[Cos[(c + d*x)/2]])/d - (18*Log[Sin[(c + d*x)/2]])/d - Sec[(c + d*x)/2]^2/(2*d) + (4*Sin[(c + d*x)/2])/(
5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5) - 2/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4) + (26*Sin[(c +
d*x)/2])/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - 13/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (1
86*Sin[(c + d*x)/2])/(5*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (29*Tan[(c + d*x)/2])/(6*d) + (Sec[(c + d*x
)/2]^2*Tan[(c + d*x)/2])/(24*d)))/a^3

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {A \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {128}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {64}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {160}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {176}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {400}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {39}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-144 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{8 d \,a^{3}}\)	\(174\)
default	\(\frac {A \left (\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {128}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {64}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {160}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {176}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {400}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {1}{3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-\frac {39}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-144 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{8 d \,a^{3}}\)	\(174\)
parallelrisch	\(-\frac {\left (432 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-67 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2637 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+8484 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-7 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+11384 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+67 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+7297 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {9934}{5}\right ) A}{24 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(174\)
risch	\(-\frac {4 \left (675 i A \,{\mathrm e}^{9 i \left (d x +c \right )}+135 A \,{\mathrm e}^{10 i \left (d x +c \right )}-3150 i A \,{\mathrm e}^{7 i \left (d x +c \right )}-1710 A \,{\mathrm e}^{8 i \left (d x +c \right )}+5180 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+4572 A \,{\mathrm e}^{6 i \left (d x +c \right )}-3590 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-4906 A \,{\mathrm e}^{4 i \left (d x +c \right )}+925 i A \,{\mathrm e}^{i \left (d x +c \right )}+2081 A \,{\mathrm e}^{2 i \left (d x +c \right )}-212 A \right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} a^{3} d}-\frac {18 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {18 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(208\)
norman	\(\frac {-\frac {A}{24 a d}+\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 a d}-\frac {17 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {17 A \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {7 A \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {A \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}-\frac {66469 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{120 a d}-\frac {8407 A \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}-\frac {3479 A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {3919 A \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {2443 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {845 A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {411 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {18 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\)	\(312\)

[In]

int(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/8/d*A/a^3*(1/3*tan(1/2*d*x+1/2*c)^3-4*tan(1/2*d*x+1/2*c)^2+39*tan(1/2*d*x+1/2*c)-128/5/(tan(1/2*d*x+1/2*c)+1
)^5+64/(tan(1/2*d*x+1/2*c)+1)^4-160/(tan(1/2*d*x+1/2*c)+1)^3+176/(tan(1/2*d*x+1/2*c)+1)^2-400/(tan(1/2*d*x+1/2
*c)+1)-1/3/tan(1/2*d*x+1/2*c)^3+4/tan(1/2*d*x+1/2*c)^2-39/tan(1/2*d*x+1/2*c)-144*ln(tan(1/2*d*x+1/2*c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (145) = 290\).

Time = 0.27 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.88 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\frac {424 \, A \cos \left (d x + c\right )^{6} + 1002 \, A \cos \left (d x + c\right )^{5} - 944 \, A \cos \left (d x + c\right )^{4} - 2074 \, A \cos \left (d x + c\right )^{3} + 531 \, A \cos \left (d x + c\right )^{2} + 1077 \, A \cos \left (d x + c\right ) + 135 \, {\left (A \cos \left (d x + c\right )^{6} - 2 \, A \cos \left (d x + c\right )^{5} - 6 \, A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 9 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - {\left (A \cos \left (d x + c\right )^{5} + 3 \, A \cos \left (d x + c\right )^{4} - 3 \, A \cos \left (d x + c\right )^{3} - 7 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sin \left (d x + c\right ) - 4 \, A\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 135 \, {\left (A \cos \left (d x + c\right )^{6} - 2 \, A \cos \left (d x + c\right )^{5} - 6 \, A \cos \left (d x + c\right )^{4} + 4 \, A \cos \left (d x + c\right )^{3} + 9 \, A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) - {\left (A \cos \left (d x + c\right )^{5} + 3 \, A \cos \left (d x + c\right )^{4} - 3 \, A \cos \left (d x + c\right )^{3} - 7 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sin \left (d x + c\right ) - 4 \, A\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + {\left (424 \, A \cos \left (d x + c\right )^{5} - 578 \, A \cos \left (d x + c\right )^{4} - 1522 \, A \cos \left (d x + c\right )^{3} + 552 \, A \cos \left (d x + c\right )^{2} + 1083 \, A \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right ) - 6 \, A}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{6} - 2 \, a^{3} d \cos \left (d x + c\right )^{5} - 6 \, a^{3} d \cos \left (d x + c\right )^{4} + 4 \, a^{3} d \cos \left (d x + c\right )^{3} + 9 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{3} - 7 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(424*A*cos(d*x + c)^6 + 1002*A*cos(d*x + c)^5 - 944*A*cos(d*x + c)^4 - 2074*A*cos(d*x + c)^3 + 531*A*cos(
d*x + c)^2 + 1077*A*cos(d*x + c) + 135*(A*cos(d*x + c)^6 - 2*A*cos(d*x + c)^5 - 6*A*cos(d*x + c)^4 + 4*A*cos(d
*x + c)^3 + 9*A*cos(d*x + c)^2 - 2*A*cos(d*x + c) - (A*cos(d*x + c)^5 + 3*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^
3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + 4*A)*sin(d*x + c) - 4*A)*log(1/2*cos(d*x + c) + 1/2) - 135*(A*cos(
d*x + c)^6 - 2*A*cos(d*x + c)^5 - 6*A*cos(d*x + c)^4 + 4*A*cos(d*x + c)^3 + 9*A*cos(d*x + c)^2 - 2*A*cos(d*x +
c) - (A*cos(d*x + c)^5 + 3*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + 4*
A)*sin(d*x + c) - 4*A)*log(-1/2*cos(d*x + c) + 1/2) + (424*A*cos(d*x + c)^5 - 578*A*cos(d*x + c)^4 - 1522*A*co
s(d*x + c)^3 + 552*A*cos(d*x + c)^2 + 1083*A*cos(d*x + c) + 6*A)*sin(d*x + c) - 6*A)/(a^3*d*cos(d*x + c)^6 - 2
*a^3*d*cos(d*x + c)^5 - 6*a^3*d*cos(d*x + c)^4 + 4*a^3*d*cos(d*x + c)^3 + 9*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos
(d*x + c) - 4*a^3*d - (a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 - 3*a^3*d*cos(d*x + c)^3 - 7*a^3*d*cos(d*
x + c)^2 + 2*a^3*d*cos(d*x + c) + 4*a^3*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=- \frac {A \left (\int \left (- \frac {\csc ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\right )\, dx + \int \frac {\sin {\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx\right )}{a^{3}} \]

[In]

integrate(csc(d*x+c)**4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

-A*(Integral(-csc(c + d*x)**4/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x) + Integral(sin(c
+ d*x)*csc(c + d*x)**4/(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1), x))/a**3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 706 vs. \(2 (145) = 290\).

Time = 0.29 (sec) , antiderivative size = 706, normalized size of antiderivative = 4.61 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/120*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 2782*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 9410*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 13645*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9285*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 2580*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 15)/(a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*a^3*sin(d*x + c
)^3/(cos(d*x + c) + 1)^3 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^3*sin(d*x + c)^5/(cos(d*x + c) +
1)^5 + 5*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - 15*(12*sin(d*x +
c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 780*log(sin(d*x + c)/(cos(d*x + c) + 1))/a
^3) - A*((20*sin(d*x + c)/(cos(d*x + c) + 1) - 230*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4777*sin(d*x + c)^3/(
cos(d*x + c) + 1)^3 - 15785*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 22390*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
14940*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 4005*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 5)/(a^3*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 10*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a^3*sin(d*x + c)^8/
(cos(d*x + c) + 1)^8) + 5*(81*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*
x + c)^3/(cos(d*x + c) + 1)^3)/a^3 - 1380*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.39 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2160 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {5 \, {\left (792 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 117 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A\right )}}{a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {48 \, {\left (125 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 445 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 635 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 415 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 108 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}} - \frac {5 \, {\left (A a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 117 \, A a^{6} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{9}}}{120 \, d} \]

[In]

integrate(csc(d*x+c)^4*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/120*(2160*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 5*(792*A*tan(1/2*d*x + 1/2*c)^3 - 117*A*tan(1/2*d*x + 1/2*
c)^2 + 12*A*tan(1/2*d*x + 1/2*c) - A)/(a^3*tan(1/2*d*x + 1/2*c)^3) + 48*(125*A*tan(1/2*d*x + 1/2*c)^4 + 445*A*
tan(1/2*d*x + 1/2*c)^3 + 635*A*tan(1/2*d*x + 1/2*c)^2 + 415*A*tan(1/2*d*x + 1/2*c) + 108*A)/(a^3*(tan(1/2*d*x
+ 1/2*c) + 1)^5) - 5*(A*a^6*tan(1/2*d*x + 1/2*c)^3 - 12*A*a^6*tan(1/2*d*x + 1/2*c)^2 + 117*A*a^6*tan(1/2*d*x +
1/2*c))/a^9)/d

Mupad [B] (verification not implemented)

Time = 15.98 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.05 \[ \int \frac {\csc ^4(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx=-\frac {A\,\left (335\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-35\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+7559\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+24610\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+33170\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+18670\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+1310\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-2375\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-335\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+2160\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10800\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+21600\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+21600\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10800\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2160\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+5\right )}{120\,a^3\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]

[In]

int((A - A*sin(c + d*x))/(sin(c + d*x)^4*(a + a*sin(c + d*x))^3),x)

[Out]

-(A*(335*tan(c/2 + (d*x)/2)^2 - 35*tan(c/2 + (d*x)/2) + 7559*tan(c/2 + (d*x)/2)^3 + 24610*tan(c/2 + (d*x)/2)^4
+ 33170*tan(c/2 + (d*x)/2)^5 + 18670*tan(c/2 + (d*x)/2)^6 + 1310*tan(c/2 + (d*x)/2)^7 - 2375*tan(c/2 + (d*x)/
2)^8 - 335*tan(c/2 + (d*x)/2)^9 + 35*tan(c/2 + (d*x)/2)^10 - 5*tan(c/2 + (d*x)/2)^11 + 2160*log(tan(c/2 + (d*x
)/2))*tan(c/2 + (d*x)/2)^3 + 10800*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^4 + 21600*log(tan(c/2 + (d*x)/2)
)*tan(c/2 + (d*x)/2)^5 + 21600*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^6 + 10800*log(tan(c/2 + (d*x)/2))*ta
n(c/2 + (d*x)/2)^7 + 2160*log(tan(c/2 + (d*x)/2))*tan(c/2 + (d*x)/2)^8 + 5))/(120*a^3*d*tan(c/2 + (d*x)/2)^3*(
tan(c/2 + (d*x)/2) + 1)^5)