3.1139 \(\int \frac{\cos ^2(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=182 \[ -\frac{3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \sqrt{a^2-b^2}}-\frac{\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a d (a+b \sin (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.471654, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2890, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac{3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 d \sqrt{a^2-b^2}}-\frac{\left (a^2+6 b^2\right ) \cos (c+d x)}{2 a^3 b d (a+b \sin (c+d x))}+\frac{\left (a^2-3 b^2\right ) \cos (c+d x)}{2 a^2 b d (a+b \sin (c+d x))^2}+\frac{3 b \tanh ^{-1}(\cos (c+d x))}{a^4 d}-\frac{\cot (c+d x)}{a d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2890

Int[cos[(e_.) + (f_.)*(x_)]^4*((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*d*(n + 1)*(m + 1)), Int[(d*Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 1)*Simp[a^2*(n + 1)
*(n + 2) - b^2*(m + n + 2)*(m + n + 3) + a*b*(m + 1)*Sin[e + f*x] - (a^2*(n + 1)*(n + 3) - b^2*(m + n + 2)*(m
+ n + 4))*Sin[e + f*x]^2, x], x], x] - Simp[((a^2*(n + 1) - b^2*(m + n + 2))*Cos[e + f*x]*(d*Sin[e + f*x])^(n
+ 2)*(a + b*Sin[e + f*x])^(m + 1))/(a^2*b*d^2*f*(n + 1)*(m + 1)), x]) /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2
- b^2, 0] && IntegersQ[2*m, 2*n] && LtQ[m, -1] && LtQ[n, -1]

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

Mathematica [A]  time = 2.48228, size = 184, normalized size = 1.01 \[ \frac{-\frac{6 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{a^2 \left (a^2-b^2\right ) \cos (c+d x)}{b (a+b \sin (c+d x))^2}-\frac{a \left (a^2+4 b^2\right ) \cos (c+d x)}{b (a+b \sin (c+d x))}+a \tan \left (\frac{1}{2} (c+d x)\right )-a \cot \left (\frac{1}{2} (c+d x)\right )-6 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.197, size = 489, normalized size = 2.7 \begin{align*}{\frac{1}{2\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{-2}}-6\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{2}}{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}}}-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}}}-10\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}{b}^{3}}{d{a}^{4} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}}}-{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{-2}}-14\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ){b}^{2}}{d{a}^{3} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}}}-5\,{\frac{b}{d{a}^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}}}-3\,{\frac{1}{d{a}^{2}\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}^{2}}{d{a}^{4}\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}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-3\,{\frac{b\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*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)+1/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^2-5/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*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^3-1/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)-1
4/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^2-5/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-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)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.92254, size = 2341, normalized size = 12.86 \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)^2/(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

[-1/4*(2*(a^5 + 5*a^3*b^2 - 6*a*b^4)*cos(d*x + c)^3 - 18*(a^4*b - a^2*b^3)*cos(d*x + c)*sin(d*x + c) + 3*(2*a^
3*b - 4*a*b^3 - 2*(a^3*b - 2*a*b^3)*cos(d*x + c)^2 + (a^4 - a^2*b^2 - 2*b^4 - (a^2*b^2 - 2*b^4)*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^5 + a^3*b^2 - 2*a*b^4)*cos(d*x + c) + 6*(2*a^3*b^2 - 2*a*b^4 - 2*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 +
(a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c) + 1/2) - 6*(2*a^3*b^2 - 2*a
*b^4 - 2*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 + (a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(-
1/2*cos(d*x + c) + 1/2))/(2*(a^7*b - a^5*b^3)*d*cos(d*x + c)^2 - 2*(a^7*b - a^5*b^3)*d + ((a^6*b^2 - a^4*b^4)*
d*cos(d*x + c)^2 - (a^8 - a^4*b^4)*d)*sin(d*x + c)), -1/2*((a^5 + 5*a^3*b^2 - 6*a*b^4)*cos(d*x + c)^3 - 9*(a^4
*b - a^2*b^3)*cos(d*x + c)*sin(d*x + c) + 3*(2*a^3*b - 4*a*b^3 - 2*(a^3*b - 2*a*b^3)*cos(d*x + c)^2 + (a^4 - a
^2*b^2 - 2*b^4 - (a^2*b^2 - 2*b^4)*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*(a^5 + a^3*b^2 - 2*a*b^4)*cos(d*x + c) + 3*(2*a^3*b^2 - 2*a*b^4 - 2*(a^3*b
^2 - a*b^4)*cos(d*x + c)^2 + (a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(1/2*cos(d*x + c)
+ 1/2) - 3*(2*a^3*b^2 - 2*a*b^4 - 2*(a^3*b^2 - a*b^4)*cos(d*x + c)^2 + (a^4*b - b^5 - (a^2*b^3 - b^5)*cos(d*x
+ c)^2)*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2))/(2*(a^7*b - a^5*b^3)*d*cos(d*x + c)^2 - 2*(a^7*b - a^5*b^
3)*d + ((a^6*b^2 - a^4*b^4)*d*cos(d*x + c)^2 - (a^8 - a^4*b^4)*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)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.37205, size = 369, normalized size = 2.03 \begin{align*} -\frac{\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 (\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 )}{\left (a^{2} - 2 \, b^{2}\right )}}{\sqrt{a^{2} - b^{2}} a^{4}} - \frac{6 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a}{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )} - \frac{2 \,{\left (a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 10 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 14 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, a^{2} b\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}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*csc(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

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