3.1133 \(\int \frac{\cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=238 \[ \frac{2 \left (-5 a^2 b^2+a^4+4 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 d \sqrt{a^2-b^2}}+\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.703252, antiderivative size = 238, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2724, 3055, 3001, 3770, 2660, 618, 204} \[ \frac{2 \left (-5 a^2 b^2+a^4+4 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^5 d \sqrt{a^2-b^2}}+\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2724

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> -Simp[(Cos[e + f*x]*(a
+ b*Sin[e + f*x])^(m + 1))/(3*a*f*Sin[e + f*x]^3), x] + (-Dist[1/(3*a^2*b*(m + 1)), Int[((a + b*Sin[e + f*x])
^(m + 1)*Simp[6*a^2 - b^2*(m - 1)*(m - 2) + a*b*(m + 1)*Sin[e + f*x] - (3*a^2 - b^2*m*(m - 2))*Sin[e + f*x]^2,
x])/Sin[e + f*x]^3, x], x] - Simp[((3*a^2 + b^2*(m - 2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(3*a^2*b*
f*(m + 1)*Sin[e + f*x]^2), x]) /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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{\cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^3(c+d x) \left (6 \left (a^2-2 b^2\right )-a b \sin (c+d x)-\left (3 a^2-8 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{3 a^2 b}\\ &=-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc ^2(c+d x) \left (-2 b \left (7 a^2-12 b^2\right )+4 a b^2 \sin (c+d x)+6 b \left (a^2-2 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)} \, dx}{6 a^3 b}\\ &=\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac{\int \frac{\csc (c+d x) \left (6 b^2 \left (3 a^2-4 b^2\right )+6 a b \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{a+b \sin (c+d x)} \, dx}{6 a^4 b}\\ &=\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac{\left (b \left (3 a^2-4 b^2\right )\right ) \int \csc (c+d x) \, dx}{a^5}+\frac{\left (a^4-5 a^2 b^2+4 b^4\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{a^5}\\ &=-\frac{b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}+\frac{\left (2 \left (a^4-5 a^2 b^2+4 b^4\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^5 d}\\ &=-\frac{b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}-\frac{\left (4 \left (a^4-5 a^2 b^2+4 b^4\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^5 d}\\ &=\frac{2 \left (a^4-5 a^2 b^2+4 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^5 \sqrt{a^2-b^2} d}-\frac{b \left (3 a^2-4 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^5 d}+\frac{\left (7 a^2-12 b^2\right ) \cot (c+d x)}{3 a^4 d}-\frac{\left (a^2-2 b^2\right ) \cot (c+d x) \csc (c+d x)}{a^3 b d}+\frac{\left (3 a^2-4 b^2\right ) \cot (c+d x) \csc (c+d x)}{3 a^2 b d (a+b \sin (c+d x))}-\frac{\cot (c+d x) \csc ^2(c+d x)}{3 a d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 6.36736, size = 403, normalized size = 1.69 \[ \frac{\left (3 a^2 b-4 b^3\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}+\frac{\left (4 b^3-3 a^2 b\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^5 d}+\frac{a^2 b \cos (c+d x)-b^3 \cos (c+d x)}{a^4 d (a+b \sin (c+d x))}+\frac{\csc \left (\frac{1}{2} (c+d x)\right ) \left (4 a^2 \cos \left (\frac{1}{2} (c+d x)\right )-9 b^2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (9 b^2 \sin \left (\frac{1}{2} (c+d x)\right )-4 a^2 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{6 a^4 d}+\frac{2 \left (-5 a^2 b^2+a^4+4 b^4\right ) \tan ^{-1}\left (\frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (a \sin \left (\frac{1}{2} (c+d x)\right )+b \cos \left (\frac{1}{2} (c+d x)\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^5 d \sqrt{a^2-b^2}}+\frac{b \csc ^2\left (\frac{1}{2} (c+d x)\right )}{4 a^3 d}-\frac{b \sec ^2\left (\frac{1}{2} (c+d x)\right )}{4 a^3 d}-\frac{\cot \left (\frac{1}{2} (c+d x)\right ) \csc ^2\left (\frac{1}{2} (c+d x)\right )}{24 a^2 d}+\frac{\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )}{24 a^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]))/Sqrt[a^2 - b^
2]])/(a^5*Sqrt[a^2 - b^2]*d) + ((4*a^2*Cos[(c + d*x)/2] - 9*b^2*Cos[(c + d*x)/2])*Csc[(c + d*x)/2])/(6*a^4*d)
+ (b*Csc[(c + d*x)/2]^2)/(4*a^3*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a^2*d) + ((-3*a^2*b + 4*b^3)*Lo
g[Cos[(c + d*x)/2]])/(a^5*d) + ((3*a^2*b - 4*b^3)*Log[Sin[(c + d*x)/2]])/(a^5*d) - (b*Sec[(c + d*x)/2]^2)/(4*a
^3*d) + (Sec[(c + d*x)/2]*(-4*a^2*Sin[(c + d*x)/2] + 9*b^2*Sin[(c + d*x)/2]))/(6*a^4*d) + (a^2*b*Cos[c + d*x]
- b^3*Cos[c + d*x])/(a^4*d*(a + b*Sin[c + d*x])) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*a^2*d)

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Maple [B]  time = 0.178, size = 527, normalized size = 2.2 \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)^4*csc(d*x+c)^4/(a+b*sin(d*x+c))^2,x)

[Out]

1/24/d/a^2*tan(1/2*d*x+1/2*c)^3-1/4/d/a^3*tan(1/2*d*x+1/2*c)^2*b-5/8/d/a^2*tan(1/2*d*x+1/2*c)+3/2/d/a^4*b^2*ta
n(1/2*d*x+1/2*c)+2/d/a^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*b^2*tan(1/2*d*x+1/2*c)-2/d*b^4/a^5/
(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)+2/d/a^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2
*d*x+1/2*c)*b+a)*b-2/d*b^3/a^4/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+2/d/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))-10/d/a^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))*b^2+8/d*b^4/a^5/(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))
-1/24/d/a^2/tan(1/2*d*x+1/2*c)^3+5/8/d/a^2/tan(1/2*d*x+1/2*c)-3/2/d/a^4/tan(1/2*d*x+1/2*c)*b^2+1/4/d/a^3*b/tan
(1/2*d*x+1/2*c)^2+3/d/a^3*b*ln(tan(1/2*d*x+1/2*c))-4/d/a^5*b^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)^4*csc(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.98326, size = 2639, normalized size = 11.09 \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)^4*csc(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/6*(4*(2*a^4 - 3*a^2*b^2)*cos(d*x + c)^3 + 3*((a^2*b - 4*b^3)*cos(d*x + c)^4 + a^2*b - 4*b^3 - 2*(a^2*b - 4
*b^3)*cos(d*x + c)^2 + (a^3 - 4*a*b^2 - (a^3 - 4*a*b^2)*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*(a^4 - 2*a^2*b^2)*cos(d*x + c) +
3*((3*a^2*b^2 - 4*b^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^4)*cos(d*x + c)^2 + (3*a^3*b -
4*a*b^3 - (3*a^3*b - 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 3*((3*a^2*b^2 - 4*b
^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^4)*cos(d*x + c)^2 + (3*a^3*b - 4*a*b^3 - (3*a^3*b
- 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*((7*a^3*b - 12*a*b^3)*cos(d*x + c)^3
- 3*(3*a^3*b - 4*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^5*b*d*cos(d*x + c)^4 - 2*a^5*b*d*cos(d*x + c)^2 + a^5*
b*d - (a^6*d*cos(d*x + c)^2 - a^6*d)*sin(d*x + c)), -1/6*(4*(2*a^4 - 3*a^2*b^2)*cos(d*x + c)^3 + 6*((a^2*b - 4
*b^3)*cos(d*x + c)^4 + a^2*b - 4*b^3 - 2*(a^2*b - 4*b^3)*cos(d*x + c)^2 + (a^3 - 4*a*b^2 - (a^3 - 4*a*b^2)*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))) - 6*(a
^4 - 2*a^2*b^2)*cos(d*x + c) + 3*((3*a^2*b^2 - 4*b^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^
4)*cos(d*x + c)^2 + (3*a^3*b - 4*a*b^3 - (3*a^3*b - 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c
) + 1/2) - 3*((3*a^2*b^2 - 4*b^4)*cos(d*x + c)^4 + 3*a^2*b^2 - 4*b^4 - 2*(3*a^2*b^2 - 4*b^4)*cos(d*x + c)^2 +
(3*a^3*b - 4*a*b^3 - (3*a^3*b - 4*a*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2) + 2*((7*a^
3*b - 12*a*b^3)*cos(d*x + c)^3 - 3*(3*a^3*b - 4*a*b^3)*cos(d*x + c))*sin(d*x + c))/(a^5*b*d*cos(d*x + c)^4 - 2
*a^5*b*d*cos(d*x + c)^2 + a^5*b*d - (a^6*d*cos(d*x + c)^2 - a^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)**4*csc(d*x+c)**4/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.31562, size = 481, normalized size = 2.02 \begin{align*} \frac{\frac{24 \,{\left (3 \, a^{2} b - 4 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{5}} + \frac{48 \,{\left (a^{4} - 5 \, a^{2} b^{2} + 4 \, b^{4}\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^{5}} + \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 36 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}} + \frac{48 \,{\left (a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3} b - a b^{3}\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 )} a^{5}} - \frac{132 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 176 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{3}}{a^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(24*(3*a^2*b - 4*b^3)*log(abs(tan(1/2*d*x + 1/2*c)))/a^5 + 48*(a^4 - 5*a^2*b^2 + 4*b^4)*(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^5) + (a^4*t
an(1/2*d*x + 1/2*c)^3 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^2 - 15*a^4*tan(1/2*d*x + 1/2*c) + 36*a^2*b^2*tan(1/2*d*x
+ 1/2*c))/a^6 + 48*(a^2*b^2*tan(1/2*d*x + 1/2*c) - b^4*tan(1/2*d*x + 1/2*c) + a^3*b - a*b^3)/((a*tan(1/2*d*x +
1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a^5) - (132*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 176*b^3*tan(1/2*d*x + 1/2
*c)^3 - 15*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*a^2*b*tan(1/2*d*x + 1/2*c) + a^3)/
(a^5*tan(1/2*d*x + 1/2*c)^3))/d