3.3.41 \(\int \frac {\csc ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\) [241]

Optimal. Leaf size=113 \[ \frac {4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {A \cot (c+d x)}{a^3 d}-\frac {2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac {31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac {104 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))} \]

[Out]

4*A*arctanh(cos(d*x+c))/a^3/d-94/15*A*cot(d*x+c)/a^3/d+2/5*A*cot(d*x+c)/a^3/d/(1+sin(d*x+c))^3+13/15*A*cot(d*x
+c)/a^3/d/(1+sin(d*x+c))^2+4*A*cot(d*x+c)/a^3/d/(1+sin(d*x+c))

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Rubi [A]
time = 0.29, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3029, 2788, 3855, 3852, 8, 3862, 4007, 4004, 3879} \begin {gather*} -\frac {A \cot (c+d x)}{a^3 d}+\frac {4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {104 A \cot (c+d x)}{15 a^3 d (\csc (c+d x)+1)}+\frac {31 A \cot (c+d x)}{15 a^3 d (\csc (c+d x)+1)^2}-\frac {2 A \cot (c+d x)}{5 a^3 d (\csc (c+d x)+1)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(4*A*ArcTanh[Cos[c + d*x]])/(a^3*d) - (A*Cot[c + d*x])/(a^3*d) - (2*A*Cot[c + d*x])/(5*a^3*d*(1 + Csc[c + d*x]
)^3) + (31*A*Cot[c + d*x])/(15*a^3*d*(1 + Csc[c + d*x])^2) - (104*A*Cot[c + d*x])/(15*a^3*d*(1 + Csc[c + d*x])
)

Rule 8

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

Rule 2788

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 3029

Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[a^n*c^n, Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && 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 3855

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

Rule 3862

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

Rule 3879

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[-Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a),
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 4007

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-(b
*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[
e + f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f
}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\csc ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=(a A) \int \frac {\cot ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\\ &=\frac {A \int \left (\frac {9}{a^2}-\frac {4 \csc (c+d x)}{a^2}+\frac {\csc ^2(c+d x)}{a^2}-\frac {2}{a^2 (1+\csc (c+d x))^3}+\frac {9}{a^2 (1+\csc (c+d x))^2}-\frac {16}{a^2 (1+\csc (c+d x))}\right ) \, dx}{a}\\ &=\frac {9 A x}{a^3}+\frac {A \int \csc ^2(c+d x) \, dx}{a^3}-\frac {(2 A) \int \frac {1}{(1+\csc (c+d x))^3} \, dx}{a^3}-\frac {(4 A) \int \csc (c+d x) \, dx}{a^3}+\frac {(9 A) \int \frac {1}{(1+\csc (c+d x))^2} \, dx}{a^3}-\frac {(16 A) \int \frac {1}{1+\csc (c+d x)} \, dx}{a^3}\\ &=\frac {9 A x}{a^3}+\frac {4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac {3 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))^2}-\frac {16 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))}+\frac {(2 A) \int \frac {-5+2 \csc (c+d x)}{(1+\csc (c+d x))^2} \, dx}{5 a^3}-\frac {(3 A) \int \frac {-3+\csc (c+d x)}{1+\csc (c+d x)} \, dx}{a^3}+\frac {(16 A) \int -1 \, dx}{a^3}-\frac {A \text {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac {2 A x}{a^3}+\frac {4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {A \cot (c+d x)}{a^3 d}-\frac {2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac {31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac {16 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))}-\frac {(2 A) \int \frac {15-7 \csc (c+d x)}{1+\csc (c+d x)} \, dx}{15 a^3}-\frac {(12 A) \int \frac {\csc (c+d x)}{1+\csc (c+d x)} \, dx}{a^3}\\ &=\frac {4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {A \cot (c+d x)}{a^3 d}-\frac {2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac {31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac {4 A \cot (c+d x)}{a^3 d (1+\csc (c+d x))}+\frac {(44 A) \int \frac {\csc (c+d x)}{1+\csc (c+d x)} \, dx}{15 a^3}\\ &=\frac {4 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac {A \cot (c+d x)}{a^3 d}-\frac {2 A \cot (c+d x)}{5 a^3 d (1+\csc (c+d x))^3}+\frac {31 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))^2}-\frac {104 A \cot (c+d x)}{15 a^3 d (1+\csc (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 1.97, size = 167, normalized size = 1.48 \begin {gather*} -\frac {A \left (15 \cot \left (\frac {1}{2} (c+d x)\right )-120 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+120 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {12}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {38}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (-287+79 \cos (2 (c+d x))-354 \sin (c+d x))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}-15 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{30 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/30*(A*(15*Cot[(c + d*x)/2] - 120*Log[Cos[(c + d*x)/2]] + 120*Log[Sin[(c + d*x)/2]] + 12/(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^4 + 38/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*Sin[(c + d*x)/2]*(-287 + 79*Cos[2*(c +
d*x)] - 354*Sin[c + d*x]))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 - 15*Tan[(c + d*x)/2]))/(a^3*d)

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Maple [A]
time = 0.46, size = 120, normalized size = 1.06

method	result	size
derivativedivides	\(\frac {A \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {32}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {16}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {88}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {28}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right )}{2 d \,a^{3}}\)	\(120\)
default	\(\frac {A \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {32}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {16}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {88}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {28}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}\right )}{2 d \,a^{3}}\)	\(120\)
risch	\(-\frac {4 \left (-320 A \,{\mathrm e}^{4 i \left (d x +c \right )}+150 i A \,{\mathrm e}^{5 i \left (d x +c \right )}+367 A \,{\mathrm e}^{2 i \left (d x +c \right )}-385 i A \,{\mathrm e}^{3 i \left (d x +c \right )}-47 A +205 i A \,{\mathrm e}^{i \left (d x +c \right )}+30 A \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} a^{3} d}-\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{3}}+\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{3}}\)	\(158\)
norman	\(\frac {-\frac {A}{2 a d}+\frac {A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {3811 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}-\frac {893 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {413 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {161 A \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}-\frac {805 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {283 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 a d}-\frac {51 A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/2/d*A/a^3*(tan(1/2*d*x+1/2*c)-1/tan(1/2*d*x+1/2*c)-8*ln(tan(1/2*d*x+1/2*c))-32/5/(tan(1/2*d*x+1/2*c)+1)^5+16
/(tan(1/2*d*x+1/2*c)+1)^4-88/3/(tan(1/2*d*x+1/2*c)+1)^3+28/(tan(1/2*d*x+1/2*c)+1)^2-36/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 519 vs. \(2 (107) = 214\).
time = 0.29, size = 519, normalized size = 4.59 \begin {gather*} -\frac {3 \, A {\left (\frac {\frac {121 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {410 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {610 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {425 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {125 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 5}{\frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} - \frac {5 \, \sin \left (d x + c\right )}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, A {\left (\frac {2 \, {\left (\frac {115 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {185 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {135 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {45 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 32\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/30*(3*A*((121*sin(d*x + c)/(cos(d*x + c) + 1) + 410*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 610*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 + 425*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 125*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5
)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(co
s(d*x + c) + 1)^3 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 5*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a
^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 5*sin(d*x + c)/(a^3*(c
os(d*x + c) + 1))) + 2*A*(2*(115*sin(d*x + c)/(cos(d*x + c) + 1) + 185*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1
35*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 45*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 32)/(a^3 + 5*a^3*sin(d*x + c
)/(cos(d*x + c) + 1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*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 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 15*log(sin(d*x + c)/
(cos(d*x + c) + 1))/a^3))/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (107) = 214\).
time = 0.42, size = 406, normalized size = 3.59 \begin {gather*} \frac {94 \, A \cos \left (d x + c\right )^{4} + 222 \, A \cos \left (d x + c\right )^{3} - 115 \, A \cos \left (d x + c\right )^{2} - 237 \, A \cos \left (d x + c\right ) + 30 \, {\left (A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} - 5 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) - {\left (A \cos \left (d x + c\right )^{3} + 3 \, 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 ) - 30 \, {\left (A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} - 5 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) - {\left (A \cos \left (d x + c\right )^{3} + 3 \, 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 (94 \, A \cos \left (d x + c\right )^{3} - 128 \, A \cos \left (d x + c\right )^{2} - 243 \, 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 )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 5 \, 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 )^{3} + 3 \, 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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(94*A*cos(d*x + c)^4 + 222*A*cos(d*x + c)^3 - 115*A*cos(d*x + c)^2 - 237*A*cos(d*x + c) + 30*(A*cos(d*x +
c)^4 - 2*A*cos(d*x + c)^3 - 5*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) - (A*cos(d*x + c)^3 + 3*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) - 30*(A*cos(d*x + c)^4 - 2*A*cos(d*x +
c)^3 - 5*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) - (A*cos(d*x + c)^3 + 3*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) + (94*A*cos(d*x + c)^3 - 128*A*cos(d*x + c)^2 - 243*A*cos(
d*x + c) - 6*A)*sin(d*x + c) + 6*A)/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^3 - 5*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)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3
*d)*sin(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {A \left (\int \left (- \frac {\csc ^{2}{\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 ^{2}{\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.50, size = 146, normalized size = 1.29 \begin {gather*} -\frac {\frac {120 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {15 \, {\left (8 \, 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 )} + \frac {4 \, {\left (135 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 435 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 605 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 385 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 104 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(120*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 15*A*tan(1/2*d*x + 1/2*c)/a^3 - 15*(8*A*tan(1/2*d*x + 1/2*c)
- A)/(a^3*tan(1/2*d*x + 1/2*c)) + 4*(135*A*tan(1/2*d*x + 1/2*c)^4 + 435*A*tan(1/2*d*x + 1/2*c)^3 + 605*A*tan(
1/2*d*x + 1/2*c)^2 + 385*A*tan(1/2*d*x + 1/2*c) + 104*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

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Mupad [B]
time = 15.77, size = 210, normalized size = 1.86 \begin {gather*} \frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^3\,d}-\frac {4\,A\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {37\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+121\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {514\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {338\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {491\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+A}{d\,\left (2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A - A*sin(c + d*x))/(sin(c + d*x)^2*(a + a*sin(c + d*x))^3),x)

[Out]

(A*tan(c/2 + (d*x)/2))/(2*a^3*d) - (4*A*log(tan(c/2 + (d*x)/2)))/(a^3*d) - (A + (491*A*tan(c/2 + (d*x)/2))/15
+ (338*A*tan(c/2 + (d*x)/2)^2)/3 + (514*A*tan(c/2 + (d*x)/2)^3)/3 + 121*A*tan(c/2 + (d*x)/2)^4 + 37*A*tan(c/2
+ (d*x)/2)^5)/(d*(10*a^3*tan(c/2 + (d*x)/2)^2 + 20*a^3*tan(c/2 + (d*x)/2)^3 + 20*a^3*tan(c/2 + (d*x)/2)^4 + 10
*a^3*tan(c/2 + (d*x)/2)^5 + 2*a^3*tan(c/2 + (d*x)/2)^6 + 2*a^3*tan(c/2 + (d*x)/2)))

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