3.1233 \(\int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=188 \[ \frac {2 b \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {\left (a^2-b^2\right )^2}{a^5 d (a+b \sin (c+d x))}-\frac {4 b \left (a^2-b^2\right ) \csc (c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (\sin (c+d x))}{a^6 d}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6 d} \]

[Out]

-4*b*(a^2-b^2)*csc(d*x+c)/a^5/d+1/2*(2*a^2-3*b^2)*csc(d*x+c)^2/a^4/d+2/3*b*csc(d*x+c)^3/a^3/d-1/4*csc(d*x+c)^4
/a^2/d+(a^4-6*a^2*b^2+5*b^4)*ln(sin(d*x+c))/a^6/d-(a^4-6*a^2*b^2+5*b^4)*ln(a+b*sin(d*x+c))/a^6/d+(a^2-b^2)^2/a
^5/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2721, 894} \[ \frac {\left (a^2-b^2\right )^2}{a^5 d (a+b \sin (c+d x))}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4 d}-\frac {4 b \left (a^2-b^2\right ) \csc (c+d x)}{a^5 d}+\frac {\left (-6 a^2 b^2+a^4+5 b^4\right ) \log (\sin (c+d x))}{a^6 d}-\frac {\left (-6 a^2 b^2+a^4+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6 d}+\frac {2 b \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^4(c+d x)}{4 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]

[Out]

(-4*b*(a^2 - b^2)*Csc[c + d*x])/(a^5*d) + ((2*a^2 - 3*b^2)*Csc[c + d*x]^2)/(2*a^4*d) + (2*b*Csc[c + d*x]^3)/(3
*a^3*d) - Csc[c + d*x]^4/(4*a^2*d) + ((a^4 - 6*a^2*b^2 + 5*b^4)*Log[Sin[c + d*x]])/(a^6*d) - ((a^4 - 6*a^2*b^2
+ 5*b^4)*Log[a + b*Sin[c + d*x]])/(a^6*d) + (a^2 - b^2)^2/(a^5*d*(a + b*Sin[c + d*x]))

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x^5 (a+x)^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {b^4}{a^2 x^5}-\frac {2 b^4}{a^3 x^4}+\frac {-2 a^2 b^2+3 b^4}{a^4 x^3}+\frac {4 b^2 \left (a^2-b^2\right )}{a^5 x^2}+\frac {a^4-6 a^2 b^2+5 b^4}{a^6 x}-\frac {\left (a^2-b^2\right )^2}{a^5 (a+x)^2}+\frac {-a^4+6 a^2 b^2-5 b^4}{a^6 (a+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {4 b \left (a^2-b^2\right ) \csc (c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {2 b \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (\sin (c+d x))}{a^6 d}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6 d}+\frac {\left (a^2-b^2\right )^2}{a^5 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 6.14, size = 187, normalized size = 0.99 \[ -\frac {4 b (a-b) (a+b) \csc (c+d x)}{a^5 d}+\frac {2 b \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {\left (a^2-b^2\right )^2}{a^5 d (a+b \sin (c+d x))}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (\sin (c+d x))}{a^6 d}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]

[Out]

(-4*(a - b)*b*(a + b)*Csc[c + d*x])/(a^5*d) + ((2*a^2 - 3*b^2)*Csc[c + d*x]^2)/(2*a^4*d) + (2*b*Csc[c + d*x]^3
)/(3*a^3*d) - Csc[c + d*x]^4/(4*a^2*d) + ((a^4 - 6*a^2*b^2 + 5*b^4)*Log[Sin[c + d*x]])/(a^6*d) - ((a^4 - 6*a^2
*b^2 + 5*b^4)*Log[a + b*Sin[c + d*x]])/(a^6*d) + (a^2 - b^2)^2/(a^5*d*(a + b*Sin[c + d*x]))

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fricas [B]  time = 0.85, size = 542, normalized size = 2.88 \[ \frac {21 \, a^{5} - 82 \, a^{3} b^{2} + 60 \, a b^{4} + 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (18 \, a^{5} - 77 \, a^{3} b^{2} + 60 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4} + {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4} + {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (31 \, a^{4} b - 30 \, a^{2} b^{3} - 6 \, {\left (6 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{7} d \cos \left (d x + c\right )^{4} - 2 \, a^{7} d \cos \left (d x + c\right )^{2} + a^{7} d + {\left (a^{6} b d \cos \left (d x + c\right )^{4} - 2 \, a^{6} b d \cos \left (d x + c\right )^{2} + a^{6} b d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(21*a^5 - 82*a^3*b^2 + 60*a*b^4 + 12*(a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 - 2*(18*a^5 - 77*a^3*b^2
+ 60*a*b^4)*cos(d*x + c)^2 - 12*(a^5 - 6*a^3*b^2 + 5*a*b^4 + (a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 - 2*(a
^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^2 + (a^4*b - 6*a^2*b^3 + 5*b^5 + (a^4*b - 6*a^2*b^3 + 5*b^5)*cos(d*x +
c)^4 - 2*(a^4*b - 6*a^2*b^3 + 5*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 12*(a^5 - 6*a^3*b
^2 + 5*a*b^4 + (a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 - 2*(a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^2 + (a^
4*b - 6*a^2*b^3 + 5*b^5 + (a^4*b - 6*a^2*b^3 + 5*b^5)*cos(d*x + c)^4 - 2*(a^4*b - 6*a^2*b^3 + 5*b^5)*cos(d*x +
c)^2)*sin(d*x + c))*log(-1/2*sin(d*x + c)) - (31*a^4*b - 30*a^2*b^3 - 6*(6*a^4*b - 5*a^2*b^3)*cos(d*x + c)^2)
*sin(d*x + c))/(a^7*d*cos(d*x + c)^4 - 2*a^7*d*cos(d*x + c)^2 + a^7*d + (a^6*b*d*cos(d*x + c)^4 - 2*a^6*b*d*co
s(d*x + c)^2 + a^6*b*d)*sin(d*x + c))

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giac [A]  time = 0.26, size = 278, normalized size = 1.48 \[ \frac {\frac {12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{6}} - \frac {12 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b} + \frac {12 \, {\left (a^{4} b \sin \left (d x + c\right ) - 6 \, a^{2} b^{3} \sin \left (d x + c\right ) + 5 \, b^{5} \sin \left (d x + c\right ) + 2 \, a^{5} - 8 \, a^{3} b^{2} + 6 \, a b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{6}} - \frac {25 \, a^{4} \sin \left (d x + c\right )^{4} - 150 \, a^{2} b^{2} \sin \left (d x + c\right )^{4} + 125 \, b^{4} \sin \left (d x + c\right )^{4} + 48 \, a^{3} b \sin \left (d x + c\right )^{3} - 48 \, a b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{4} \sin \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 8 \, a^{3} b \sin \left (d x + c\right ) + 3 \, a^{4}}{a^{6} \sin \left (d x + c\right )^{4}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(12*(a^4 - 6*a^2*b^2 + 5*b^4)*log(abs(sin(d*x + c)))/a^6 - 12*(a^4*b - 6*a^2*b^3 + 5*b^5)*log(abs(b*sin(d
*x + c) + a))/(a^6*b) + 12*(a^4*b*sin(d*x + c) - 6*a^2*b^3*sin(d*x + c) + 5*b^5*sin(d*x + c) + 2*a^5 - 8*a^3*b
^2 + 6*a*b^4)/((b*sin(d*x + c) + a)*a^6) - (25*a^4*sin(d*x + c)^4 - 150*a^2*b^2*sin(d*x + c)^4 + 125*b^4*sin(d
*x + c)^4 + 48*a^3*b*sin(d*x + c)^3 - 48*a*b^3*sin(d*x + c)^3 - 12*a^4*sin(d*x + c)^2 + 18*a^2*b^2*sin(d*x + c
)^2 - 8*a^3*b*sin(d*x + c) + 3*a^4)/(a^6*sin(d*x + c)^4))/d

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maple [A]  time = 0.88, size = 282, normalized size = 1.50 \[ -\frac {\ln \left (a +b \sin \left (d x +c \right )\right )}{a^{2} d}+\frac {6 b^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,a^{4}}-\frac {5 \ln \left (a +b \sin \left (d x +c \right )\right ) b^{4}}{d \,a^{6}}+\frac {1}{a d \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 b^{2}}{d \,a^{3} \left (a +b \sin \left (d x +c \right )\right )}+\frac {b^{4}}{d \,a^{5} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{4 d \,a^{2} \sin \left (d x +c \right )^{4}}+\frac {1}{d \,a^{2} \sin \left (d x +c \right )^{2}}-\frac {3 b^{2}}{2 d \,a^{4} \sin \left (d x +c \right )^{2}}+\frac {\ln \left (\sin \left (d x +c \right )\right )}{a^{2} d}-\frac {6 \ln \left (\sin \left (d x +c \right )\right ) b^{2}}{d \,a^{4}}+\frac {5 \ln \left (\sin \left (d x +c \right )\right ) b^{4}}{d \,a^{6}}+\frac {2 b}{3 d \,a^{3} \sin \left (d x +c \right )^{3}}-\frac {4 b}{d \,a^{3} \sin \left (d x +c \right )}+\frac {4 b^{3}}{d \,a^{5} \sin \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x)

[Out]

-ln(a+b*sin(d*x+c))/a^2/d+6/d/a^4*b^2*ln(a+b*sin(d*x+c))-5/d/a^6*ln(a+b*sin(d*x+c))*b^4+1/a/d/(a+b*sin(d*x+c))
-2/d/a^3*b^2/(a+b*sin(d*x+c))+1/d/a^5/(a+b*sin(d*x+c))*b^4-1/4/d/a^2/sin(d*x+c)^4+1/d/a^2/sin(d*x+c)^2-3/2/d/a
^4/sin(d*x+c)^2*b^2+ln(sin(d*x+c))/a^2/d-6/d/a^4*ln(sin(d*x+c))*b^2+5/d/a^6*ln(sin(d*x+c))*b^4+2/3/d/a^3*b/sin
(d*x+c)^3-4/d/a^3*b/sin(d*x+c)+4/d*b^3/a^5/sin(d*x+c)

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maxima [A]  time = 0.37, size = 189, normalized size = 1.01 \[ \frac {\frac {5 \, a^{3} b \sin \left (d x + c\right ) + 12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \sin \left (d x + c\right )^{4} - 3 \, a^{4} - 6 \, {\left (6 \, a^{3} b - 5 \, a b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, {\left (6 \, a^{4} - 5 \, a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{5} b \sin \left (d x + c\right )^{5} + a^{6} \sin \left (d x + c\right )^{4}} - \frac {12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6}} + \frac {12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{6}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*csc(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 11.81, size = 439, normalized size = 2.34 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^4-62\,a^2\,b^2+64\,b^4\right )-\frac {a^4}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {11\,a^4}{4}-\frac {10\,a^2\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (20\,a\,b^3-\frac {62\,a^3\,b}{3}\right )-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (60\,a^4\,b-96\,a^2\,b^3+32\,b^5\right )}{a}+\frac {5\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6}}{d\,\left (16\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+32\,b\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {\frac {a^2}{16}+\frac {b^2}{8}}{a^4}+\frac {1}{8\,a^2}-\frac {b^2}{2\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b\,\left (32\,a^2+64\,b^2\right )}{64\,a^5}-\frac {b}{4\,a^3}+\frac {4\,b\,\left (\frac {\frac {a^2}{8}+\frac {b^2}{4}}{a^4}+\frac {1}{4\,a^2}-\frac {b^2}{a^4}\right )}{a}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-6\,a^2\,b^2+5\,b^4\right )}{a^6\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^3\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-6\,a^2\,b^2+5\,b^4\right )}{a^6\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(sin(c + d*x)^5*(a + b*sin(c + d*x))^2),x)

[Out]

(tan(c/2 + (d*x)/2)^4*(3*a^4 + 64*b^4 - 62*a^2*b^2) - a^4/4 + tan(c/2 + (d*x)/2)^2*((11*a^4)/4 - (10*a^2*b^2)/
3) + tan(c/2 + (d*x)/2)^3*(20*a*b^3 - (62*a^3*b)/3) - (tan(c/2 + (d*x)/2)^5*(60*a^4*b + 32*b^5 - 96*a^2*b^3))/
a + (5*a^3*b*tan(c/2 + (d*x)/2))/6)/(d*(16*a^6*tan(c/2 + (d*x)/2)^4 + 16*a^6*tan(c/2 + (d*x)/2)^6 + 32*a^5*b*t
an(c/2 + (d*x)/2)^5)) - tan(c/2 + (d*x)/2)^4/(64*a^2*d) + (tan(c/2 + (d*x)/2)^2*((a^2/16 + b^2/8)/a^4 + 1/(8*a
^2) - b^2/(2*a^4)))/d - (tan(c/2 + (d*x)/2)*((b*(32*a^2 + 64*b^2))/(64*a^5) - b/(4*a^3) + (4*b*((a^2/8 + b^2/4
)/a^4 + 1/(4*a^2) - b^2/a^4))/a))/d + (log(tan(c/2 + (d*x)/2))*(a^4 + 5*b^4 - 6*a^2*b^2))/(a^6*d) + (b*tan(c/2
+ (d*x)/2)^3)/(12*a^3*d) - (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^4 + 5*b^4 - 6*a^2*b^2
))/(a^6*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*csc(d*x+c)**5/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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