3.1325 \(\int \frac{\cos ^3(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=174 \[ \frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{x \left (2 a^2-5 b^2\right )}{2 b^3}+\frac{b \cot (c+d x)}{a^2 d}-\frac{a \cos (c+d x)}{b^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]

[Out]

-((2*a^2 - 5*b^2)*x)/(2*b^3) + (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3
*d) + ((5*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (a*Cos[c + d*x])/(b^2*d) + (b*Cot[c + d*x])/(a^2*d)
- (Cot[c + d*x]*Csc[c + d*x])/(2*a*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b*d)

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Rubi [A]  time = 0.384695, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2896, 3057, 2660, 618, 204, 3770} \[ \frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{x \left (2 a^2-5 b^2\right )}{2 b^3}+\frac{b \cot (c+d x)}{a^2 d}-\frac{a \cos (c+d x)}{b^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

-((2*a^2 - 5*b^2)*x)/(2*b^3) + (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^3
*d) + ((5*a^2 - 2*b^2)*ArcTanh[Cos[c + d*x]])/(2*a^3*d) - (a*Cos[c + d*x])/(b^2*d) + (b*Cot[c + d*x])/(a^2*d)
- (Cot[c + d*x]*Csc[c + d*x])/(2*a*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b*d)

Rule 2896

Int[cos[(e_.) + (f_.)*(x_)]^6*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1))/(a*d*f*(n + 1)), x] +
(Dist[1/(a^2*b^2*d^2*(n + 1)*(n + 2)*(m + n + 5)*(m + n + 6)), Int[(d*Sin[e + f*x])^(n + 2)*(a + b*Sin[e + f*
x])^m*Simp[a^4*(n + 1)*(n + 2)*(n + 3)*(n + 5) - a^2*b^2*(n + 2)*(2*n + 1)*(m + n + 5)*(m + n + 6) + b^4*(m +
n + 2)*(m + n + 3)*(m + n + 5)*(m + n + 6) + a*b*m*(a^2*(n + 1)*(n + 2) - b^2*(m + n + 5)*(m + n + 6))*Sin[e +
f*x] - (a^4*(n + 1)*(n + 2)*(4 + n)*(n + 5) + b^4*(m + n + 2)*(m + n + 4)*(m + n + 5)*(m + n + 6) - a^2*b^2*(
n + 1)*(n + 2)*(m + n + 5)*(2*n + 2*m + 13))*Sin[e + f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(d*
Sin[e + f*x])^(n + 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*d^2*f*(n + 1)*(n + 2)), x] - Simp[(a*(n + 5)*Cos[e +
f*x]*(d*Sin[e + f*x])^(n + 3)*(a + b*Sin[e + f*x])^(m + 1))/(b^2*d^3*f*(m + n + 5)*(m + n + 6)), x] + Simp[(Co
s[e + f*x]*(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^(m + 1))/(b*d^4*f*(m + n + 6)), x]) /; FreeQ[{a, b, d
, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*m, 2*n] && NeQ[n, -1] && NeQ[n, -2] && NeQ[m + n + 5, 0]
&& NeQ[m + n + 6, 0] &&  !IGtQ[m, 0]

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 3770

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac{a \cos (c+d x)}{b^2 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\int \frac{\csc (c+d x) \left (-2 b^2 \left (5 a^2-2 b^2\right )-2 a b \left (a^2-b^2\right ) \sin (c+d x)-2 a^2 \left (2 a^2-5 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{4 a^2 b^2}\\ &=-\frac{\left (2 a^2-5 b^2\right ) x}{2 b^3}-\frac{a \cos (c+d x)}{b^2 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (5 a^2-2 b^2\right ) \int \csc (c+d x) \, dx}{2 a^3}+\frac{\left (a^2-b^2\right )^3 \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^3 b^3}\\ &=-\frac{\left (2 a^2-5 b^2\right ) x}{2 b^3}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{a \cos (c+d x)}{b^2 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}+\frac{\left (2 \left (a^2-b^2\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 b^3 d}\\ &=-\frac{\left (2 a^2-5 b^2\right ) x}{2 b^3}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{a \cos (c+d x)}{b^2 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}-\frac{\left (4 \left (a^2-b^2\right )^3\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 b^3 d}\\ &=-\frac{\left (2 a^2-5 b^2\right ) x}{2 b^3}+\frac{2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^3 d}+\frac{\left (5 a^2-2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{a \cos (c+d x)}{b^2 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\cot (c+d x) \csc (c+d x)}{2 a d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 5.44156, size = 259, normalized size = 1.49 \[ \frac{2 a^3 b^2 \sin (2 (c+d x))+16 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )-a^2 b^3 \csc ^2\left (\frac{1}{2} (c+d x)\right )+a^2 b^3 \sec ^2\left (\frac{1}{2} (c+d x)\right )-20 a^2 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+20 a^2 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )+20 a^3 b^2 c+20 a^3 b^2 d x-8 a^4 b \cos (c+d x)-8 a^5 c-8 a^5 d x-4 a b^4 \tan \left (\frac{1}{2} (c+d x)\right )+4 a b^4 \cot \left (\frac{1}{2} (c+d x)\right )+8 b^5 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-8 b^5 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{8 a^3 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

(-8*a^5*c + 20*a^3*b^2*c - 8*a^5*d*x + 20*a^3*b^2*d*x + 16*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/S
qrt[a^2 - b^2]] - 8*a^4*b*Cos[c + d*x] + 4*a*b^4*Cot[(c + d*x)/2] - a^2*b^3*Csc[(c + d*x)/2]^2 + 20*a^2*b^3*Lo
g[Cos[(c + d*x)/2]] - 8*b^5*Log[Cos[(c + d*x)/2]] - 20*a^2*b^3*Log[Sin[(c + d*x)/2]] + 8*b^5*Log[Sin[(c + d*x)
/2]] + a^2*b^3*Sec[(c + d*x)/2]^2 + 2*a^3*b^2*Sin[2*(c + d*x)] - 4*a*b^4*Tan[(c + d*x)/2])/(8*a^3*b^3*d)

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Maple [B]  time = 0.124, size = 483, normalized size = 2.8 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{bd} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{1}{bd}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-2\,{\frac{a}{d{b}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{d{b}^{3}}}+5\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{bd}}+2\,{\frac{{a}^{3}}{d{b}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-6\,{\frac{a}{bd\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+6\,{\frac{b}{da\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-2\,{\frac{{b}^{3}}{d{a}^{3}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}-{\frac{5}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

1/8/d/a*tan(1/2*d*x+1/2*c)^2-1/2/d/a^2*tan(1/2*d*x+1/2*c)*b-1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c
)^3-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^2*a+1/d/b/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2
*c)-2/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*a-2/d/b^3*arctan(tan(1/2*d*x+1/2*c))*a^2+5/d/b*arctan(tan(1/2*d*x+1/2*c
))+2/d*a^3/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-6/d/b*a/(a^2-b^2)^(1/2
)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+6/d/a*b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x
+1/2*c)+2*b)/(a^2-b^2)^(1/2))-2/d/a^3*b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1
/2))-1/8/d/a/tan(1/2*d*x+1/2*c)^2-5/2/d/a*ln(tan(1/2*d*x+1/2*c))+1/d/a^3*ln(tan(1/2*d*x+1/2*c))*b^2+1/2/d/a^2*
b/tan(1/2*d*x+1/2*c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.59221, size = 1739, normalized size = 9.99 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[-1/4*(4*a^4*b*cos(d*x + c)^3 + 2*(2*a^5 - 5*a^3*b^2)*d*x*cos(d*x + c)^2 - 2*(2*a^5 - 5*a^3*b^2)*d*x + 2*(a^4
- 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^
2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*c
os(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(2*a^4*b + a^2*b^3)*cos(d*x + c) + (5*a^2*b^3 - 2*b^5 - (
5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - (5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(
d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*b^2*cos(d*x + c)^3 - (a^3*b^2 + 2*a*b^4)*cos(d*x + c))*sin(d
*x + c))/(a^3*b^3*d*cos(d*x + c)^2 - a^3*b^3*d), -1/4*(4*a^4*b*cos(d*x + c)^3 + 2*(2*a^5 - 5*a^3*b^2)*d*x*cos(
d*x + c)^2 - 2*(2*a^5 - 5*a^3*b^2)*d*x - 4*(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sq
rt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 2*(2*a^4*b + a^2*b^3)*cos(d*x + c
) + (5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2) - (5*a^2*b^3 - 2*b^5
- (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*b^2*cos(d*x + c)^3 - (a^3*b^2 + 2*
a*b^4)*cos(d*x + c))*sin(d*x + c))/(a^3*b^3*d*cos(d*x + c)^2 - a^3*b^3*d)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.24983, size = 582, normalized size = 3.34 \begin{align*} \frac{\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} - \frac{4 \,{\left (2 \, a^{2} - 5 \, b^{2}\right )}{\left (d x + c\right )}}{b^{3}} - \frac{4 \,{\left (5 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} + \frac{16 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{\sqrt{a^{2} - b^{2}} a^{3} b^{3}} + \frac{10 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 8 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 16 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 19 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 8 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 8 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 8 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 16 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2} b^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}^{2} a^{3} b^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*((a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 - 4*(2*a^2 - 5*b^2)*(d*x + c)/b^3 - 4*(5*a^2 -
2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 + 16*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor(1/2*(d*x + c)/pi
+ 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^3*b^3) + (10*a^2*b^2*
tan(1/2*d*x + 1/2*c)^6 - 4*b^4*tan(1/2*d*x + 1/2*c)^6 - 8*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 4*a*b^3*tan(1/2*d*x +
1/2*c)^5 - 16*a^4*tan(1/2*d*x + 1/2*c)^4 + 19*a^2*b^2*tan(1/2*d*x + 1/2*c)^4 - 8*b^4*tan(1/2*d*x + 1/2*c)^4 +
8*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 8*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 16*a^4*tan(1/2*d*x + 1/2*c)^2 + 8*a^2*b^2*t
an(1/2*d*x + 1/2*c)^2 - 4*b^4*tan(1/2*d*x + 1/2*c)^2 + 4*a*b^3*tan(1/2*d*x + 1/2*c) - a^2*b^2)/((tan(1/2*d*x +
1/2*c)^3 + tan(1/2*d*x + 1/2*c))^2*a^3*b^2))/d