3.1270 \(\int \frac{\cos ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=314 \[ \frac{3 \left (2 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^4 d}-\frac{6 \left (-a^4 b^2+2 a^6-b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 b^4 d \sqrt{a^2-b^2}}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}+\frac{2 \left (2 a^2+b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{3 a x}{b^4}+\frac{\cos (c+d x)}{b^3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.4928, antiderivative size = 314, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {2897, 3770, 3767, 8, 2638, 2664, 2754, 12, 2660, 618, 204} \[ \frac{3 \left (2 a^2+b^2\right ) \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 b^4 d}-\frac{6 \left (-a^4 b^2+2 a^6-b^6\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 b^4 d \sqrt{a^2-b^2}}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}+\frac{2 \left (2 a^2+b^2\right ) \left (a^2-b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a^3 d}+\frac{3 a x}{b^4}+\frac{\cos (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

(3*a*x)/b^4 + (3*Sqrt[a^2 - b^2]*(2*a^2 + b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*b^4*d) -
(6*(2*a^6 - a^4*b^2 - b^6)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*b^4*Sqrt[a^2 - b^2]*d) + (3
*b*ArcTanh[Cos[c + d*x]])/(a^4*d) + Cos[c + d*x]/(b^3*d) - Cot[c + d*x]/(a^3*d) - ((a^2 - b^2)^2*Cos[c + d*x])
/(2*a^2*b^3*d*(a + b*Sin[c + d*x])^2) - (3*(a^2 - b^2)*Cos[c + d*x])/(2*a*b^3*d*(a + b*Sin[c + d*x])) + (2*(a^
2 - b^2)*(2*a^2 + b^2)*Cos[c + d*x])/(a^3*b^3*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 3767

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 8

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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 ^4(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=\int \left (\frac{3 a}{b^4}-\frac{3 b \csc (c+d x)}{a^4}+\frac{\csc ^2(c+d x)}{a^3}-\frac{\sin (c+d x)}{b^3}-\frac{\left (a^2-b^2\right )^3}{a^2 b^4 (a+b \sin (c+d x))^3}+\frac{2 \left (2 a^6-3 a^4 b^2+b^6\right )}{a^3 b^4 (a+b \sin (c+d x))^2}-\frac{3 \left (2 a^6-a^4 b^2-b^6\right )}{a^4 b^4 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac{3 a x}{b^4}+\frac{\int \csc ^2(c+d x) \, dx}{a^3}-\frac{\int \sin (c+d x) \, dx}{b^3}-\frac{(3 b) \int \csc (c+d x) \, dx}{a^4}-\frac{\left (a^2-b^2\right )^3 \int \frac{1}{(a+b \sin (c+d x))^3} \, dx}{a^2 b^4}-\frac{\left (3 \left (2 a^6-a^4 b^2-b^6\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^4 b^4}+\frac{\left (2 \left (2 a^6-3 a^4 b^2+b^6\right )\right ) \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{a^3 b^4}\\ &=\frac{3 a x}{b^4}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^3 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}+\frac{2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}+\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^2 b^4}-\frac{\left (2 \left (2 a^6-3 a^4 b^2+b^6\right )\right ) \int \frac{a}{a+b \sin (c+d x)} \, dx}{a^3 b^4 \left (-a^2+b^2\right )}-\frac{\operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}-\frac{\left (6 \left (2 a^6-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^4 b^4 d}\\ &=\frac{3 a x}{b^4}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}-\left (2 \left (\frac{1}{a^2}-\frac{2 a^2}{b^4}+\frac{1}{b^2}\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx-\frac{\left (a^2-b^2\right ) \int \frac{2 a^2+b^2}{a+b \sin (c+d x)} \, dx}{2 a^2 b^4}+\frac{\left (12 \left (2 a^6-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^4 b^4 d}\\ &=\frac{3 a x}{b^4}-\frac{6 \left (2 a^6-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^4 b^4 \sqrt{a^2-b^2} d}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 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^2 b^4}-\frac{\left (4 \left (\frac{1}{a^2}-\frac{2 a^2}{b^4}+\frac{1}{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 )}{d}\\ &=\frac{3 a x}{b^4}-\frac{6 \left (2 a^6-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^4 b^4 \sqrt{a^2-b^2} d}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}+\frac{\left (8 \left (\frac{1}{a^2}-\frac{2 a^2}{b^4}+\frac{1}{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 )}{d}-\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^2 b^4 d}\\ &=\frac{3 a x}{b^4}-\frac{4 \left (\frac{1}{a^2}-\frac{2 a^2}{b^4}+\frac{1}{b^2}\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}-\frac{6 \left (2 a^6-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^4 b^4 \sqrt{a^2-b^2} d}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 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^2 b^4 d}\\ &=\frac{3 a x}{b^4}-\frac{4 \left (\frac{1}{a^2}-\frac{2 a^2}{b^4}+\frac{1}{b^2}\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2} d}-\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^2 b^4 d}-\frac{6 \left (2 a^6-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^4 b^4 \sqrt{a^2-b^2} d}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac{\cos (c+d x)}{b^3 d}-\frac{\cot (c+d x)}{a^3 d}-\frac{\left (a^2-b^2\right )^2 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac{3 \left (a^2-b^2\right ) \cos (c+d x)}{2 a b^3 d (a+b \sin (c+d x))}+\frac{2 \left (a^2-b^2\right ) \left (2 a^2+b^2\right ) \cos (c+d x)}{a^3 b^3 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.20395, size = 332, normalized size = 1.06 \[ \frac{-a^2 b^2 \cos (c+d x)+5 a^4 \cos (c+d x)-4 b^4 \cos (c+d x)}{2 a^3 b^3 d (a+b \sin (c+d x))}+\frac{2 a^2 b^2 \cos (c+d x)+a^4 (-\cos (c+d x))-b^4 \cos (c+d x)}{2 a^2 b^3 d (a+b \sin (c+d x))^2}-\frac{3 \left (-a^4 b^2+a^2 b^4+2 a^6-2 b^6\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^4 b^4 d \sqrt{a^2-b^2}}-\frac{3 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}+\frac{3 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{2 a^3 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{2 a^3 d}+\frac{3 a (c+d x)}{b^4 d}+\frac{\cos (c+d x)}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.21, size = 903, normalized size = 2.9 \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)^2/(a+b*sin(d*x+c))^3,x)

[Out]

1/2/d/a^3*tan(1/2*d*x+1/2*c)+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)+6/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a+3/d/b^2/(ta
n(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*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*tan(1/2*d*x+1/2*c)^3-6/d/a^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*b^2+4/d/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)^2*a^2+9/d/b/(ta
n(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)^2-3/d/a^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*b-10/d/a^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*b^3+13/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*tan(1/2*d*x+1/2*c)*a+1/d/a/(ta
n(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)-14/d/a^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)*b^2+4/d/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2+1/d/
b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2-5/d/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a
)^2*b-6/d/b^4/(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))*a^2+3/d/b^2/(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))-3/d/a^2/(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^4/(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))*b^2-1/2/d/a^3/tan(1/2*d*x+1/2*c)-3/d/a^4*b*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)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.01052, size = 2634, normalized size = 8.39 \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)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(24*a^6*b*d*x*cos(d*x + c)^2 - 24*a^6*b*d*x + 2*(9*a^5*b^2 - a^3*b^4 - 6*a*b^6)*cos(d*x + c)^3 - 3*(4*a^5
*b + 2*a^3*b^3 + 4*a*b^5 - 2*(2*a^5*b + a^3*b^3 + 2*a*b^5)*cos(d*x + c)^2 + (2*a^6 + 3*a^4*b^2 + 3*a^2*b^4 + 2
*b^6 - (2*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c)^2)*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)) - 6*(3*a^5*b^2 - a^3*b^4 - 2*a*b^6)*cos(d*x + c) + 6*(2
*a*b^6*cos(d*x + c)^2 - 2*a*b^6 + (b^7*cos(d*x + c)^2 - a^2*b^5 - b^7)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/
2) - 6*(2*a*b^6*cos(d*x + c)^2 - 2*a*b^6 + (b^7*cos(d*x + c)^2 - a^2*b^5 - b^7)*sin(d*x + c))*log(-1/2*cos(d*x
+ c) + 1/2) + 2*(6*a^5*b^2*d*x*cos(d*x + c)^2 + 2*a^4*b^3*cos(d*x + c)^3 - 6*(a^7 + a^5*b^2)*d*x - 3*(2*a^6*b
+ a^4*b^3 - 3*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/(2*a^5*b^5*d*cos(d*x + c)^2 - 2*a^5*b^5*d + (a^4*b^6*d*cos
(d*x + c)^2 - (a^6*b^4 + a^4*b^6)*d)*sin(d*x + c)), 1/2*(12*a^6*b*d*x*cos(d*x + c)^2 - 12*a^6*b*d*x + (9*a^5*b
^2 - a^3*b^4 - 6*a*b^6)*cos(d*x + c)^3 - 3*(4*a^5*b + 2*a^3*b^3 + 4*a*b^5 - 2*(2*a^5*b + a^3*b^3 + 2*a*b^5)*co
s(d*x + c)^2 + (2*a^6 + 3*a^4*b^2 + 3*a^2*b^4 + 2*b^6 - (2*a^4*b^2 + a^2*b^4 + 2*b^6)*cos(d*x + c)^2)*sin(d*x
+ c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 3*(3*a^5*b^2 - a^3*b^4 -
2*a*b^6)*cos(d*x + c) + 3*(2*a*b^6*cos(d*x + c)^2 - 2*a*b^6 + (b^7*cos(d*x + c)^2 - a^2*b^5 - b^7)*sin(d*x + c
))*log(1/2*cos(d*x + c) + 1/2) - 3*(2*a*b^6*cos(d*x + c)^2 - 2*a*b^6 + (b^7*cos(d*x + c)^2 - a^2*b^5 - b^7)*si
n(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + (6*a^5*b^2*d*x*cos(d*x + c)^2 + 2*a^4*b^3*cos(d*x + c)^3 - 6*(a^7 +
a^5*b^2)*d*x - 3*(2*a^6*b + a^4*b^3 - 3*a^2*b^5)*cos(d*x + c))*sin(d*x + c))/(2*a^5*b^5*d*cos(d*x + c)^2 - 2*
a^5*b^5*d + (a^4*b^6*d*cos(d*x + c)^2 - (a^6*b^4 + a^4*b^6)*d)*sin(d*x + c))]

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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)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.32316, size = 622, normalized size = 1.98 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )} a}{b^{4}} - \frac{6 \, b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{4}} + \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{3}} - \frac{6 \,{\left (2 \, a^{6} - a^{4} b^{2} + a^{2} b^{4} - 2 \, 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^{4} b^{4}} + \frac{2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a b^{3}}{{\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 )} a^{4} b^{3}} + \frac{2 \,{\left (3 \, a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, a^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 9 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 10 \, b^{6} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 13 \, a^{5} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 14 \, a b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, a^{6} + a^{4} b^{2} - 5 \, a^{2} b^{4}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}^{2} a^{4} b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(6*(d*x + c)*a/b^4 - 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^4 + tan(1/2*d*x + 1/2*c)/a^3 - 6*(2*a^6 - 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)))/(sqrt(a^2 - b^2)*a^4*b^4) + (2*b^4*tan(1/2*d*x + 1/2*c)^3 - a*b^3*tan(1/2*d*x + 1/2*c)^2 + 4*a^4*tan(1
/2*d*x + 1/2*c) + 2*b^4*tan(1/2*d*x + 1/2*c) - a*b^3)/((tan(1/2*d*x + 1/2*c)^3 + tan(1/2*d*x + 1/2*c))*a^4*b^3
) + 2*(3*a^5*b*tan(1/2*d*x + 1/2*c)^3 + 3*a^3*b^3*tan(1/2*d*x + 1/2*c)^3 - 6*a*b^5*tan(1/2*d*x + 1/2*c)^3 + 4*
a^6*tan(1/2*d*x + 1/2*c)^2 + 9*a^4*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*b^4*tan(1/2*d*x + 1/2*c)^2 - 10*b^6*tan(
1/2*d*x + 1/2*c)^2 + 13*a^5*b*tan(1/2*d*x + 1/2*c) + a^3*b^3*tan(1/2*d*x + 1/2*c) - 14*a*b^5*tan(1/2*d*x + 1/2
*c) + 4*a^6 + a^4*b^2 - 5*a^2*b^4)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^4*b^3))/d