3.1269 \(\int \frac{\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=399 \[ \frac{\left (2 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^5 d}-\frac{2 \left (5 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^5 d}+\frac{2 \left (-9 a^4 b^2+10 a^6-b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 d \sqrt{a^2-b^2}}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}-\frac{\left (5 a^2+b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{3 x \left (2 a^2-b^2\right )}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b^3 d}-\frac{x}{2 b^3} \]

[Out]

-x/(2*b^3) - (3*(2*a^2 - b^2)*x)/b^5 + (Sqrt[a^2 - b^2]*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2
- b^2]])/(a*b^5*d) - (2*Sqrt[a^2 - b^2]*(5*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*b^
5*d) + (2*(10*a^6 - 9*a^4*b^2 - b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^5*Sqrt[a^2 - b^2
]*d) - ArcTanh[Cos[c + d*x]]/(a^3*d) - (3*a*Cos[c + d*x])/(b^4*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b^3*d) + ((
a^2 - b^2)^2*Cos[c + d*x])/(2*a*b^4*d*(a + b*Sin[c + d*x])^2) + (3*(a^2 - b^2)*Cos[c + d*x])/(2*b^4*d*(a + b*S
in[c + d*x])) - ((a^2 - b^2)*(5*a^2 + b^2)*Cos[c + d*x])/(a^2*b^4*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.517162, antiderivative size = 399, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 11, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.407, Rules used = {2897, 3770, 2638, 2635, 8, 2664, 2754, 12, 2660, 618, 204} \[ \frac{\left (2 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^5 d}-\frac{2 \left (5 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a b^5 d}+\frac{2 \left (-9 a^4 b^2+10 a^6-b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 d \sqrt{a^2-b^2}}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}-\frac{\left (5 a^2+b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{3 x \left (2 a^2-b^2\right )}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\sin (c+d x) \cos (c+d x)}{2 b^3 d}-\frac{x}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^5*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

-x/(2*b^3) - (3*(2*a^2 - b^2)*x)/b^5 + (Sqrt[a^2 - b^2]*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2
- b^2]])/(a*b^5*d) - (2*Sqrt[a^2 - b^2]*(5*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a*b^
5*d) + (2*(10*a^6 - 9*a^4*b^2 - b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*b^5*Sqrt[a^2 - b^2
]*d) - ArcTanh[Cos[c + d*x]]/(a^3*d) - (3*a*Cos[c + d*x])/(b^4*d) + (Cos[c + d*x]*Sin[c + d*x])/(2*b^3*d) + ((
a^2 - b^2)^2*Cos[c + d*x])/(2*a*b^4*d*(a + b*Sin[c + d*x])^2) + (3*(a^2 - b^2)*Cos[c + d*x])/(2*b^4*d*(a + b*S
in[c + d*x])) - ((a^2 - b^2)*(5*a^2 + b^2)*Cos[c + d*x])/(a^2*b^4*d*(a + b*Sin[c + d*x]))

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3770

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

Rule 2638

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

Rule 2635

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

Rule 8

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

Rule 2664

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

Rule 2754

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x) \cot (c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (\frac{3 \left (-2 a^2+b^2\right )}{b^5}+\frac{\csc (c+d x)}{a^3}+\frac{3 a \sin (c+d x)}{b^4}-\frac{\sin ^2(c+d x)}{b^3}+\frac{\left (a^2-b^2\right )^3}{a b^5 (a+b \sin (c+d x))^3}-\frac{\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )}{a^2 b^5 (a+b \sin (c+d x))^2}+\frac{10 a^6-9 a^4 b^2-b^6}{a^3 b^5 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}+\frac{\int \csc (c+d x) \, dx}{a^3}+\frac{(3 a) \int \sin (c+d x) \, dx}{b^4}-\frac{\int \sin ^2(c+d x) \, dx}{b^3}+\frac{\left (a^2-b^2\right )^3 \int \frac{1}{(a+b \sin (c+d x))^3} \, dx}{a b^5}-\frac{\left (\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\right ) \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^2 b^5}+\frac{\left (10 a^6-9 a^4 b^2-b^6\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^3 b^5}\\ &=-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}-\frac{\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}-\frac{\int 1 \, dx}{2 b^3}-\frac{\left (a^2-b^2\right )^2 \int \frac{-2 a+b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 a b^5}+\frac{\left (\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\right ) \int \frac{a}{a+b \sin (c+d x)} \, dx}{a^2 b^5 \left (-a^2+b^2\right )}+\frac{\left (2 \left (10 a^6-9 a^4 b^2-b^6\right )\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^5 d}\\ &=-\frac{x}{2 b^3}-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}+\frac{\left (a^2-b^2\right ) \int \frac{2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 a b^5}+\frac{\left (\left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a b^5 \left (-a^2+b^2\right )}-\frac{\left (4 \left (10 a^6-9 a^4 b^2-b^6\right )\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^5 d}\\ &=-\frac{x}{2 b^3}-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}+\frac{2 \left (10 a^6-9 a^4 b^2-b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}+\frac{\left (\left (a^2-b^2\right ) \left (2 a^2+b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 a b^5}+\frac{\left (2 \left (a^2-b^2\right )^2 \left (5 a^2+b^2\right )\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 b^5 \left (-a^2+b^2\right ) d}\\ &=-\frac{x}{2 b^3}-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}+\frac{2 \left (10 a^6-9 a^4 b^2-b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}+\frac{\left (\left (a^2-b^2\right ) \left (2 a^2+b^2\right )\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 b^5 d}+\frac{\left (4 \left (a^2-b^2\right ) \left (5 a^2+b^2\right )\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 b^5 d}\\ &=-\frac{x}{2 b^3}-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}-\frac{2 \sqrt{a^2-b^2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a b^5 d}+\frac{2 \left (10 a^6-9 a^4 b^2-b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}-\frac{\left (2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right )\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 b^5 d}\\ &=-\frac{x}{2 b^3}-\frac{3 \left (2 a^2-b^2\right ) x}{b^5}+\frac{\sqrt{a^2-b^2} \left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a b^5 d}-\frac{2 \sqrt{a^2-b^2} \left (5 a^2+b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a b^5 d}+\frac{2 \left (10 a^6-9 a^4 b^2-b^6\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 \sqrt{a^2-b^2} d}-\frac{\tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{3 a \cos (c+d x)}{b^4 d}+\frac{\cos (c+d x) \sin (c+d x)}{2 b^3 d}+\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a b^4 d (a+b \sin (c+d x))^2}+\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 b^4 d (a+b \sin (c+d x))}-\frac{\left (a^2-b^2\right ) \left (5 a^2+b^2\right ) \cos (c+d x)}{a^2 b^4 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.9756, size = 243, normalized size = 0.61 \[ \frac{\frac{2 \left (5 b^2-12 a^2\right ) (c+d x)}{b^5}+\frac{4 \left (-11 a^4 b^2+a^2 b^4+12 a^6-2 b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 b^5 \sqrt{a^2-b^2}}+\frac{2 \left (a^2-b^2\right )^2 \cos (c+d x)}{a b^4 (a+b \sin (c+d x))^2}+\frac{2 \left (5 a^2 b^2-7 a^4+2 b^4\right ) \cos (c+d x)}{a^2 b^4 (a+b \sin (c+d x))}+\frac{4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}-\frac{4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}-\frac{12 a \cos (c+d x)}{b^4}+\frac{\sin (2 (c+d x))}{b^3}}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^5*Cot[c + d*x])/(a + b*Sin[c + d*x])^3,x]

[Out]

((2*(-12*a^2 + 5*b^2)*(c + d*x))/b^5 + (4*(12*a^6 - 11*a^4*b^2 + a^2*b^4 - 2*b^6)*ArcTan[(b + a*Tan[(c + d*x)/
2])/Sqrt[a^2 - b^2]])/(a^3*b^5*Sqrt[a^2 - b^2]) - (12*a*Cos[c + d*x])/b^4 - (4*Log[Cos[(c + d*x)/2]])/a^3 + (4
*Log[Sin[(c + d*x)/2]])/a^3 + (2*(a^2 - b^2)^2*Cos[c + d*x])/(a*b^4*(a + b*Sin[c + d*x])^2) + (2*(-7*a^4 + 5*a
^2*b^2 + 2*b^4)*Cos[c + d*x])/(a^2*b^4*(a + b*Sin[c + d*x])) + Sin[2*(c + d*x)]/b^3)/(4*d)

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Maple [B]  time = 0.204, size = 988, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)
^2*a+1/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)-6/d/b^4/(1+tan(1/2*d*x+1/2*c)^2)^2*a-12/d/b^5*arcta
n(tan(1/2*d*x+1/2*c))*a^2+5/d/b^3*arctan(tan(1/2*d*x+1/2*c))-5/d*a^2/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x
+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3+1/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*
c)^3+4/d/a^2*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^3-6/d*a^3/b^4/(tan(1/2*d
*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2-9/d*a/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x
+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2+9/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*d*x+1/2*
c)^2+6/d/a^3*b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^2-19/d*a^2/b^3/(tan(1/
2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+11/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1
/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)+8/d/a^2*b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*
c)-6/d*a^3/b^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2+3/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d
*x+1/2*c)*b+a)^2*a+3/d/a/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2+12/d*a^3/b^5/(a^2-b^2)^(1/2)*arct
an(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-11/d*a/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/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/(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/d/a^3*ln(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)/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.84042, size = 2257, normalized size = 5.66 \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)/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(8*a^4*b^3*cos(d*x + c)^3 + 2*(12*a^5*b^2 - 5*a^3*b^4)*d*x*cos(d*x + c)^2 - 2*(12*a^7 + 7*a^5*b^2 - 5*a^
3*b^4)*d*x + (12*a^6 + 13*a^4*b^2 + 3*a^2*b^4 + 2*b^6 - (12*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(12*
a^5*b + a^3*b^3 + 2*a*b^5)*sin(d*x + c))*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*cos(d*x + c)^2 - 2*a
*b*sin(d*x + c) - a^2 - b^2)) - 2*(12*a^6*b + a^4*b^3 - 3*a^2*b^5)*cos(d*x + c) + 2*(b^7*cos(d*x + c)^2 - 2*a*
b^6*sin(d*x + c) - a^2*b^5 - b^7)*log(1/2*cos(d*x + c) + 1/2) - 2*(b^7*cos(d*x + c)^2 - 2*a*b^6*sin(d*x + c) -
a^2*b^5 - b^7)*log(-1/2*cos(d*x + c) + 1/2) - 2*(a^3*b^4*cos(d*x + c)^3 + 2*(12*a^6*b - 5*a^4*b^3)*d*x + 2*(9
*a^5*b^2 - 3*a^3*b^4 - a*b^6)*cos(d*x + c))*sin(d*x + c))/(a^3*b^7*d*cos(d*x + c)^2 - 2*a^4*b^6*d*sin(d*x + c)
- (a^5*b^5 + a^3*b^7)*d), -1/2*(4*a^4*b^3*cos(d*x + c)^3 + (12*a^5*b^2 - 5*a^3*b^4)*d*x*cos(d*x + c)^2 - (12*
a^7 + 7*a^5*b^2 - 5*a^3*b^4)*d*x - (12*a^6 + 13*a^4*b^2 + 3*a^2*b^4 + 2*b^6 - (12*a^4*b^2 + a^2*b^4 + 2*b^6)*c
os(d*x + c)^2 + 2*(12*a^5*b + a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(s
qrt(a^2 - b^2)*cos(d*x + c))) - (12*a^6*b + a^4*b^3 - 3*a^2*b^5)*cos(d*x + c) + (b^7*cos(d*x + c)^2 - 2*a*b^6*
sin(d*x + c) - a^2*b^5 - b^7)*log(1/2*cos(d*x + c) + 1/2) - (b^7*cos(d*x + c)^2 - 2*a*b^6*sin(d*x + c) - a^2*b
^5 - b^7)*log(-1/2*cos(d*x + c) + 1/2) - (a^3*b^4*cos(d*x + c)^3 + 2*(12*a^6*b - 5*a^4*b^3)*d*x + 2*(9*a^5*b^2
- 3*a^3*b^4 - a*b^6)*cos(d*x + c))*sin(d*x + c))/(a^3*b^7*d*cos(d*x + c)^2 - 2*a^4*b^6*d*sin(d*x + c) - (a^5*
b^5 + a^3*b^7)*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)/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.28736, size = 857, normalized size = 2.15 \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)/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - (12*a^2 - 5*b^2)*(d*x + c)/b^5 + 2*(12*a^6 - 11*a^4*b^2 + a^2*b^4
- 2*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)))/(sqr
t(a^2 - b^2)*a^3*b^5) - 2*(6*a^5*b*tan(1/2*d*x + 1/2*c)^7 - a^3*b^3*tan(1/2*d*x + 1/2*c)^7 - 4*a*b^5*tan(1/2*d
*x + 1/2*c)^7 + 12*a^6*tan(1/2*d*x + 1/2*c)^6 + 13*a^4*b^2*tan(1/2*d*x + 1/2*c)^6 - 9*a^2*b^4*tan(1/2*d*x + 1/
2*c)^6 - 6*b^6*tan(1/2*d*x + 1/2*c)^6 + 54*a^5*b*tan(1/2*d*x + 1/2*c)^5 - 9*a^3*b^3*tan(1/2*d*x + 1/2*c)^5 - 1
6*a*b^5*tan(1/2*d*x + 1/2*c)^5 + 36*a^6*tan(1/2*d*x + 1/2*c)^4 + 39*a^4*b^2*tan(1/2*d*x + 1/2*c)^4 - 21*a^2*b^
4*tan(1/2*d*x + 1/2*c)^4 - 12*b^6*tan(1/2*d*x + 1/2*c)^4 + 90*a^5*b*tan(1/2*d*x + 1/2*c)^3 - 27*a^3*b^3*tan(1/
2*d*x + 1/2*c)^3 - 20*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 36*a^6*tan(1/2*d*x + 1/2*c)^2 + 23*a^4*b^2*tan(1/2*d*x +
1/2*c)^2 - 15*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 - 6*b^6*tan(1/2*d*x + 1/2*c)^2 + 42*a^5*b*tan(1/2*d*x + 1/2*c) -
11*a^3*b^3*tan(1/2*d*x + 1/2*c) - 8*a*b^5*tan(1/2*d*x + 1/2*c) + 12*a^6 - 3*a^4*b^2 - 3*a^2*b^4)/((a*tan(1/2*d
*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^3*
b^4))/d